Previous year questions (2013–2025) grouped by marks — 2, 3, 4 & 5 mark questions with answers. 248 questions total.
Chapter-wise Question Distribution
Chapter
2M
3M
4M
5M
Total
Biomolecules
23
49
6
–
78
Aldehydes Ketones Carboxylic Acids
16
14
2
16
48
Amines
7
24
1
6
38
Haloalkanes
12
19
–
1
32
Alcohols Phenols Ethers
8
13
1
6
28
Polymers
1
7
1
–
9
Chemistry in Everyday Life
–
8
1
–
9
Haloalkanes and Haloarenes
3
1
2
–
6
Total
70
135
14
29
248
3 Mark Questions ★
Amines Basicity / PropertiesAsked 16 times
2014201620172019202020232025
2014 How will you convert the following? (i) Nitrobenzene into aniline. (ii) Ethanoic acid into methanamine. (iii) Aniline into N-phenylethanamide. (Write the chemical equations involve...
2014 Account for the following: (i) Primary amines (R NH 2 ) have higher boiling point than tertiary amines (R 3 N). (ii) Aniline does not undergo Friedel-Crafts reaction. (iii) (CH 3 )...
2016 · 3 marks
Give reasons for the following:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(iii) Primary amines have higher boiling point than tertiary amines.
Answer(i) Aniline is a Lewis base while AlCl3 is a Lewis acid. They combine to form a salt, deactivating the benzene ring for electrophilic substitution.
(ii) Due to combined +I effect and solvation effects; (CH3)2NH cation is better solvated than bulky (CH3)3NH+.
(iii) Primary amines have N-H bonds allowing intermolecular hydrogen bonding; tertiary amines have no N-H for H-bonding.
IUPAC Naming of Coordination CompoundsAsked 13 times
201320152016201920202024
2013 · 3 marks
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Neoprene
(iii) Nylon-6,6
Answer(i) Buna-S: Monomers are 1,3-butadiene (CH2=CH–CH=CH2) and Styrene (C6H5CH=CH2)
(ii) Neoprene: Monomer is Chloroprene (2-chloro-1,3-butadiene, CH2=C(Cl)–CH=CH2)
(iii) Nylon-6,6: Monomers are Hexamethylenediamine (H2N–(CH2)6–NH2) and Adipic acid (HOOC–(CH2)4–COOH)
2015 · 3 marks
(i) Which one of the following is a disaccharide: Starch, Maltose, Fructose, Glucose?
(ii) What is the difference between fibrous protein and globular protein?
(iii) Write the name of vitamin whose deficiency causes bone deformities in children.
Answer(i) Maltose is a disaccharide.
(ii) Fibrous proteins: Parallel polypeptide chains, insoluble in water. E.g., keratin, collagen. Globular proteins: Spherical shape, soluble in water. E.g., haemoglobin, insulin.
(iii) Vitamin D deficiency causes bone deformities (rickets) in children.
2016 · 3 marks
(i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
(ii) Why vitamin C cannot be stored in our body?
(iii) What is the difference between a nucleoside and nucleotide?
Answer(i) beta-D-glucose and beta-D-galactose.
(ii) Vitamin C is water soluble, hence excreted through urine and cannot be stored.
(iii) Nucleoside = base + sugar (pentose sugar combines with nitrogen base). Nucleotide = base + sugar + phosphate group (nucleoside combines with phosphate group).
Amino Acids / Proteins / EnzymesAsked 13 times
20142019202020242025
2014 Define the following terms related to proteins: (i) Peptide linkage (ii) Primary structure (iii) Denaturation
2014 Define the following terms: (i) Nucleotide (ii) Anomers (iii) Essential amino acids
2019 · 3 marks
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and glycosidic linkage
(iii) Fibrous protein and globular protein
Answer(i) Amylose is water soluble, straight chain polymer; amylopectin is water insoluble, branched chain polymer.
(ii) Peptide linkage (-CONH-) is formed between two amino acids; glycosidic linkage (-C-O-C-) is an oxide linkage between two monosaccharides.
(iii) In fibrous protein, polypeptide chains run parallel (insoluble in water); in globular protein, chains coil around to give spherical shape (soluble in water).
Carbohydrates (Glucose, Sucrose, Starch)Asked 11 times
201420162018201920202025
2014 · 3 marks
(i) Deficiency of which vitamin causes night-blindness?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose?
Answer(i) Vitamin A deficiency causes night-blindness.
(ii) Uracil is found in RNA only (replaced by thymine in DNA).
(iii) Glucose reaction with HI produces n-hexane which suggests that all the six carbon atoms are linked together in a straight chain.
2014 (i) Deficiency of which vitamin causes rickets? (ii) Give an example for each of fibrous protein and globular protein. (iii) Write the product formed on reaction of D-Glucose with ...
2016 · 3 marks
(i) Write the structural difference between starch and cellulose.
(ii) What type of linkage is present in nucleic acids?
(iii) Give one example each for fibrous protein and globular protein.
Answer(i) Starch is a polymer of alpha-D-glucose units (amylose + amylopectin). Cellulose is a polymer of beta-D-glucose units (unbranched).
(ii) Phosphodiester linkage between the 5' and 3' carbon atoms.
After the ban on plastic bags, students of a school decided to make the people aware of the harmful effects of plastic bags on the environment and Yamuna River. They organized a rally and distributed paper bags.
(i) What values are shown by the students?
(ii) What are bio-degradable polymers? Give one example.
(iii) Is polythene a condensation or the addition polymer?
Answer(i) Environmental awareness, social responsibility, leadership, and civic duty.
(ii) Bio-degradable polymers are polymers that can be decomposed by bacteria/microorganisms. Example: PHBV (poly β-hydroxybutyrate-co-β-hydroxy valerate) or Nylon-2-nylon-6.
(iii) Polythene is an addition polymer (formed by addition polymerization of ethene).
2016 · 3 marks
(i) What is the role of t-butyl peroxide in the polymerization of ethene?
(ii) Identify the monomers in the following polymer: -(-NH-(CH2)6-NH-CO-(CH2)4-CO-)n-
(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Polystyrene, Terylene and Buna-S.
OR Write the mechanism of free radical polymerization of ethene.
Answer(i) t-Butyl peroxide acts as a free radical initiator in the polymerization of ethene.
(ii) Monomers: Hexamethylenediamine (H2N(CH2)6NH2) and Adipic acid (HOOC(CH2)4COOH). This is Nylon-6,6.
OR Mechanism: Step 1 (Initiation): Peroxide decomposes to free radicals which attack ethene.
Step 2 (Propagation): Chain grows by successive addition of monomers.
Step 3 (Termination): Two growing chains combine.
2016 · 3 marks
(i) What is the role of sulphur in the vulcanization of rubber?
(ii) Identify the monomers in the following polymer: -(-O-CH2-CH2-O-CO-C6H4-CO-)n-
(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Terylene, polythene and neoprene.
Answer(i) Sulphur cross-links polymer chains through disulphide bonds (-S-S-), making rubber harder, more elastic, and resistant to temperature changes.
(ii) Monomers: Ethylene glycol (HOCH2CH2OH) and Terephthalic acid (HOOC-C6H4-COOH). This is Terylene (PET).
(a) Which one of the following is a food preservative? Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
Answer(a) Sodium benzoate is a food preservative.
(b) Bithional is added to soap as an antiseptic to reduce odour produced by bacterial decomposition of organic matter on the skin.
(c) Tranquilizers (barbiturates) are the class of drugs used in sleeping pills.
2014 · 3 marks
(i) Give two examples of macromolecules that are chosen as drug targets.
(ii) What are antiseptics? Give an example.
(iii) Why is use of aspartame limited to cold foods and soft drinks?
Answer(i) Proteins and nucleic acids (DNA) are macromolecules chosen as drug targets.
(ii) Antiseptics are chemical substances that prevent the growth of micro-organisms and are safe to apply on living tissues. Example: Dettol (mixture of chloroxylenol and terpineol).
(iii) Aspartame is unstable at high temperatures; it decomposes on heating, so its use is limited to cold foods and soft drinks.
2014 · 3 marks
(i) Name the sweetening agents used in the preparation of sweets for a diabetic patient.
(ii) What are antibodies? Give an example.
(iii) Give two examples of macromolecules that are chosen as drug targets.
Answer(i) Aspartame and saccharin are used as sweetening agents for diabetic patients.
(ii) Antibodies are proteins produced by the immune system in response to foreign substances (antigens). Example: Immunoglobulin (IgG).
(iii) Proteins and nucleic acids are macromolecules chosen as drug targets.
Phenol Reactions / AcidityAsked 8 times
2015202020232024
2015 How do you convert the following: (i) Phenol to anisole (ii) Propan-2-ol to 2-methylpropan-2-ol (iii) Aniline to phenol
2020 · 3 marks
Give the structures of final products expected from the following reactions:
(i) Hydroboration of propene followed by oxidation with H2O2 in alkaline medium.
(ii) Dehydration of (CH3)3C-OH by heating it with 20% H3PO4 at 358K.
(ii) HCHO + CH3MgBr -> CH3CH2OH (Grignard reaction followed by hydrolysis).
(iii) C6H5OH + (CH3CO)2O -> CH3COOC6H5 + CH3COOH.
2023 · 3 marks
(a)
(i) Why is the C-O bond length in phenols less than that in methanol?
(ii) Arrange the following in order of increasing boiling point: Ethoxyethane, Butanal, Butanol, n-butane
(iii) How can phenol be prepared from anisole? Give reaction.
Answer(a)
(i) C-O bond length in phenol is less than methanol because of presence of benzene ring which is aromatic and consisting of double bond. The lone pair present in oxygen is shared with partial conjugation effect while in methanol the lone pair of oxygen shared with normal carbon atom.
(iii) Anisole (C6H5OCH3) + HI -> Phenol (C6H5OH) + CH3I
Other (Biomolecules)Asked 7 times
201420192025
2014 Define the following terms: (i) Glycosidic linkage (ii) Invert sugar (iii) Oligo saccharides
2019 · 3 marks
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents are used in toothpastes?
(iii) Why the use of alitame as artificial sweetener is not recommended?
Answer(i) Tranquilizers.
(ii) Anionic detergents like sodium lauryl sulphate.
(iii) Alitame is high potency sweetener and difficult to control the sweetness of food.
2019 · 3 marks
Define the following terms with suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents
Answer(i) Broad-spectrum antibiotics: Antibiotics effective against a wide range of Gram-positive and Gram-negative bacteria. Example: Chloramphenicol.
(ii) Disinfectants: Chemical substances that kill micro-organisms but cannot be used on living tissues. Example: 1% phenol solution.
(iii) Cationic detergents: Detergents in which the cationic part possesses a long hydrocarbon chain and a positive charge. Example: Cetyltrimethylammonium bromide.
Other (Amines)Asked 6 times
2013201420162017
2013 · 3 marks
Give the structures of A, B and C in the following reactions:
(i) C6H5N2+Cl− ⟶[CuCN] A ⟶[H2O/H+, Δ] B ⟶[H2O/H+, Δ] C
(ii) C6H5NO2 ⟶[Sn+HCl] A ⟶[NaNO2+HCl] B ⟶[H2O/H+, Δ] C
Write the structures of A, B and C in the following:
(a) C6H5-CONH2 --(Br2/aq.KOH) → A --(NaNO2+HCl, 0-5°C) → B --(KI) → C.
(b) CH3-Cl --(KCN) → A --(LiAlH4) → B --(CHCl3+alc.KOH) → C.
Answer(a) A = C6H5NH2 (aniline, Hoffmann bromamide degradation), B = C6H5N2+Cl- (benzenediazonium chloride), C = C6H5I (iodobenzene).
(b) A = CH3CN (methyl cyanide), B = CH3CH2NH2 (ethylamine), C = CH3CH2NC (ethyl isocyanide, carbylamine reaction).
2017 Give reasons: (i) Acetylation of aniline reduces its activation effect. (ii) CH 3 NH 2 is more basic than C 6 H 5 NH 2 . (iii) Although -NH 2 is o/p directing group, yet aniline on...
SN1 vs SN2 MechanismsAsked 6 times
20152016202020242025
2015 Give reasons: (a) n -Butyl bromide has higher boiling point than t -butyl bromide. (b) Racemic mixture is optically inactive. (c) The presence of nitro group ( NO 2 ) at o/p positi...
2016 · 3 marks
Give reasons:
(i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH3-Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) SN1 reactions are accompanied by racemization in optically active alkyl halides.
Answer(i) In chlorobenzene, each carbon atom is sp2 hybridized and due to resonance (+R effect), there is partial double bond character in C-Cl bond making it shorter. In CH3Cl, carbon is sp3 hybridized.
(ii) The dipole moment of chlorobenzene is lower because +R effect of Cl and -I effect oppose each other while in cyclohexyl chloride only -I effect is present. Also sp2 vs sp3 hybridization difference.
(iii) In SN1 reaction, a planar carbocation is formed (sp2 hybridized). The nucleophile can attack from either side equally, leading to formation of racemic mixture.
2020 · 3 marks
Write the product(s) of the following reactions:
(i) Cyclohex-2-enol + PCC ->
(ii) Salicylic acid + (CH3CO)2O/CH3COOH ->
(iii) Cyclohexanone + CH3MgBr followed by H3O+ -> OR
(a) Write the mechanism of the SN1 reaction: (CH3)3C-Br + Aq.NaOH -> (CH3)3C-OH + NaBr
(b) Write the equation for the preparation of 2-methyl-2-methoxypropane by Williamson synthesis.
Answer(i) Cyclohex-2-enone (oxidation of allylic alcohol).
(ii) 2-Acetoxybenzoic acid (Aspirin).
(iii) 1-methylcyclohexanol and 1-methylcyclohexene (via Grignard addition then dehydration).
Step 2: Nucleophilic attack by second alcohol molecule: CH3CH2OH2+ + CH3CH2OH → CH3CH2–O–CH2CH3 + H2O + H+
(b) Acetylation of Salicylic acid: Phenol (OH group) + (CH3CO)2O → Aspirin (acetylsalicylic acid) + CH3COOH. (Acetyl chloride or acetic anhydride may be used)
2016 · 3 marks
Write the main product(s) in each of the following reactions:
(i) CH3-C(CH3)(CH3)-O-CH3 + HI -> ?
(ii) CH3-CH=CH2 --(i) B2H6,
(ii) H2O2/OH-) → ?
(iii) C6H5-OH --(i) aq.NaOH,
(ii) CO2, H+) → ?
Answer(i) CH3OH + (CH3)3C-I (methanol and tert-butyl iodide). The C-O bond of the bulkier group breaks preferentially.
(ii) CH3CH2CH2OH (propan-1-ol, anti-Markovnikov hydroboration-oxidation).
(iii) 2-Hydroxybenzoic acid (Salicylic acid) via Kolbe's reaction.
Other (Aldehydes Ketones Carboxylic Acids)Asked 4 times
201520192024
2015 Predict the products of the following reactions: (i) CH 3 C(=O) CH 3 [(i) H 2 N NH 2 , (ii) KOH/Glycol, Δ] (ii) C 6 H 5 CO CH 3 [NaOH/I 2 ] ? + ? (iii) CH 3 COONa [NaOH/CaO, ...
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. Isomers (A) and (C) give positive Tollens' test whereas isomer (B) does not give Tollens' test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards addition of HCN?
Answer(a) A = CH3CH2CH2CHO (Butanal) B = CH3COCH2CH3 (Butanone) C = (CH3)2CHCHO (2-Methylpropanal) D = CH3CH2CH2CH3 (Butane)
(b) B (Butanone) - as ketones are less reactive towards addition of HCN than aldehydes due to steric effect and inductive effect.
2019 · 3 marks
(i) Out of (CH3)3C-Br and (CH3)3C-I, which one is more reactive towards SN1 and why?
(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(iii) Why dextro and laevo rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?
Answer(i) (CH3)3C-I, due to large size of iodine/better leaving group/lower electronegativity.
(ii) p-Nitrophenol.
(iii) Because enantiomers have identical physical properties such as boiling points, so they cannot be separated by fractional distillation.
2019 · 3 marks
Among all isomers of C4H9Br, identify:
(a) the optically active isomer.
(b) the isomer highly reactive towards SN2.
(c) two isomers giving same product on dehydrohalogenation with alcoholic KOH.
Answer(a) CH3CH2CH(Br)CH3 (2-bromobutane, has chiral carbon).
(b) CH3CH2CH2CH2Br (1-bromobutane, primary, least steric hindrance).
(c) (CH3)3CBr and (CH3)2CHCH2Br both give 2-methylpropene.
Nucleophilic Addition / Aldol / CannizzaroAsked 4 times
20192025
2019 · 3 marks
(a) Give reasons:
(i) Benzoic acid is a stronger acid than acetic acid.
(ii) Methanal is more reactive towards nucleophilic addition than ethanal.
(b) Give simple chemical test to distinguish between propanal and propanone.
Answer(a)
(i) Due to sp2 hybridised carbon and greater resonance stabilization of benzoate ion.
(ii) Methanal has more electrophilic carbonyl carbon (no +I effect) and less steric hindrance.
(b) Tollens' test: Propanal gives silver mirror; propanone does not.
2022-II · 3 marks
(a) What happens when:
(i) Propanone is treated with CH3MgBr and then hydrolysed?
(ii) Ethanal is treated with excess ethanol and acid?
(iii) Methanal undergoes Cannizzaro reaction?
(b) Write the main product in the following reaction:
(i) 2CH3COCl + (CH3)2Cd ->
(ii) CH3CH2CHO --(Zn(Hg)/Conc HCl) →
(iii) C6H5COONa + NaOH --(CaO/heat) →
Answer(a)
(i) When propanone reacts with ethyl magnesium bromide (Grignard's reagent) and then hydrolysed, it forms alcohol: (CH3)2CO + CH3CH2MgBr -> (CH3)2C(OH)CH2CH3 + Mg(OH)Br (2-methylbutan-2-ol).
(ii) When ethanal reacts with excess ethanol and acid gives acetal: CH3CHO + 2C2H5OH -> CH3CH(OC2H5)2 + H2O.
(iii) Methanal will undergo Cannizzaro reaction as it possesses no alpha-hydrogen to form methanol: 2HCHO + conc.KOH -> CH3OH (Methanol) + HCOOK (Potassium methanoate).
(i) Which acid of the following pair would you expect to be stronger? F-CH2-COOH or CH3-COOH (ii) Arrange the following compounds in increasing order of their boiling points: CH3CH2OH, CH3-CHO, CH3-COOH (iii) Give simple chemical test to distinguish between Benzaldehyde and Acetophenone.
OR (b)
(i) Which will undergo faster nucleophilic addition reaction? Acetaldehyde or Propanone (ii) What is the composition of Fehling's reagent? (iii) Draw structure of the semicarbazone of Ethanal.
Answer(a)
(i) Among FCH2COOH and CH3COOH, FCH2COOH is stronger due to the presence of electron withdrawing group (-F). As Fluorine is most electronegative element, electronegativity makes it stronger than acetic acid.
(ii) The increasing order of boiling point: CH3CHO < CH3CH2OH < CH3COOH. Ethanoic acid and ethanol have comparatively higher boiling point than ethanal as they both are held by strong hydrogen bonds whereas in ethanal there is dipole-dipole interaction.
(iii) With ammoniacal silver nitrate solution (Tollen's reagent), benzaldehyde forms silver mirror but acetophenone does not give this test. Acetophenone gives positive iodoform test whereas benzaldehyde does not give iodoform test.
OR (b)
(i) Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric hindrance. Thus, acetaldehyde is more reactive than propanone.
(ii) Fehling's solution is a mixture of alkaline solution of copper(II) sulphate (CuSO4) containing sodium potassium tartrate (Rochelle salt - KNaC4H4O6.4H2O).
(iii) Structure of semicarbazone from ethanal: CH3CH=NNHCONH2 + H2O.
p-Block Element Properties/ReactionsAsked 3 times
201420182023
2014 · 3 marks
(a) Draw the structure of major monohalo products in each of the following reactions:
(i) C6H5CH2OH ⟶[PCl5]
(ii) C6H5CH2CH=CH2 + HBr →
(b) Which halogen compound in each of the following pairs will react faster in SN2 reaction?
(ii) Ethanol --(PCC) → Acetaldehyde --(CH3MgBr, then H3O+) → Propan-2-ol.
(b) Step 1 (fast): Protonation of ethanol to form oxonium ion.
Step 2 (slow): Loss of water to form carbocation.
Step 3: Loss of H+ to form ethene.
(c) The -OH group in phenol is highly activating. The lone pair on oxygen is delocalized into the benzene ring, making phenol more electron rich than benzene, facilitating electrophilic substitution.
2019 · 5 marks
(a) Give equations:
(i) Phenol + conc. HNO3.
(ii) Propene + B2H6 followed by H2O2/OH-.
(iii) Sodium t-butoxide + CH3Cl.
(b) Distinguish between butan-1-ol and butan-2-ol.
(c) Arrange in increasing order of acidity: Phenol, Ethanol, Water.
(b) Iodoform test: Butan-2-ol gives yellow ppt of CHI3 with NaOH/I2; butan-1-ol does not.
(c) Ethanol < Water < Phenol.
2020 · 5 marks
(a) An organic compound
(A) having molecular formula C4H8O gives orange red precipitate with 2,4-DNP reagent. It does not reduce Tollen's reagent but gives yellow precipitate of iodoform on heating with NaOH and I2. Compound
(A) on reduction with NaBH4 gives compound
(B) which undergoes dehydration on heating with conc. H2SO4 to form compound (C). Compound
(C) on ozonolysis gives two molecules of ethanal. Identify (A),
(B) and
(C) and write their structures. Write the reactions of compound
(A) with
(i) NaOH/I2 and
(ii) NaBH4.
(b) Give reasons:
(i) Oxidation of propanal is easier than propanone.
(ii) alpha-hydrogen of aldehydes and ketones is acidic in nature.
OR
(a) Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanal
(b) Write the major product(s):
(i) CH3-CH=CH-CH2-CN with (i)DIBAL-H/(ii)H3O+
(ii) CH3-CH2-OH with CrO3
(c) How can you distinguish between propanal and propanone?
Answer(a) A = Butanone (CH3COCH2CH3), B = Butan-2-ol (CH3CH(OH)CH2CH3), C = But-2-ene (CH3CH=CHCH3).
(ii) CH3COCH2CH3 + NaBH4 -> CH3CH(OH)CH2CH3 (Butan-2-ol). (b)
(i) Oxidation of propanal is easier because aldehydes have H attached to carbonyl, ketones have two alkyl groups making oxidation harder.
(ii) alpha-hydrogen is acidic because the carbanion formed is stabilized by resonance with the carbonyl group.
OR (a)
(i) Cyclobutanone cyanohydrin: cyclobutane ring with C(OH)(CN).
(ii) Hemiacetal of ethanal: CH3CH(OR)(OH). (b)
(i) CH3CH=CH-CH2-CHO (DIBAL-H reduces CN to CHO).
(ii) CH3CHO then CH3COOH (CrO3 oxidation).
(c) By iodoform test: Propanone gives yellow ppt of CHI3 with I2/NaOH; propanal does not.
Phenol Reactions / AcidityAsked 4 times
201320142019
2013 · 5 marks
(a) Although phenoxide ion has more number of resonating structures than carboxylate ion, Carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring about the following conversions?
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal
Answer(a)
(i) In phenoxide resonating structures, only structures I and IV carry the negative charge. It is negligibly contributed in the stability of phenoxide ion. Negative charge is present on one oxygen atom.
(ii) In carboxylate ion, negative charge delocalized on both oxygen atoms which is highly electronegative and contributed more in stability of resonating structure of carboxylate ion.
2013 (a) Complete the following reactions: (i) 2H C(=O) H [Conc. KOH] (ii) CH 3 COOH [Br 2 /P] (iii) C 6 H 5 CHO [HNO 3 /H 2 SO 4 , 273-283K] (b) Give simple chemical tests to distingui...
2014 (a) Write the products formed when CH 3 CHO reacts with the following reagents: (i) HCN (ii) H 2 N OH (iii) CH 3 CHO in the presence of dilute NaOH (b) Give simple chemical tests t...
Nucleophilic Addition / Aldol / CannizzaroAsked 4 times
20142020
2014 · 5 marks
(a) Write the products of the following reactions:
(i) Cyclohexanone + NH2OH →
(ii) 2C6H5CHO + Conc. NaOH →
(iii) CH3COOH ⟶[Cl2/P]
(b) Write the chemical equations to illustrate the following name reactions:
2014 (a) Account for the following: (i) CH 3 CHO is more reactive than CH 3 COCH 3 towards reaction with HCN. (ii) Carboxylic acid is a stronger acid than phenol. (b) Write the chemical...
2014 (a) Account for the following: (i) Cl CH 2 COOH is a stronger acid than CH 3 COOH. (ii) Carboxylic acids do not give reactions of carbonyl group. (b) Write the chemical equations t...
IUPAC Naming of Coordination CompoundsAsked 3 times
201720242025
2017 · 5 marks
(a) Write the product(s) in the following reactions:
(a) An organic compound (A) with molecular formula C9H10O forms 2,4-DNP derivative, reduces Fehling solution and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid.
(i) Identify compound (A) and write its IUPAC name.
(ii) Write the reaction of compound (A) with (1) 2,4-Dinitrophenyl hydrazine and (2) Fehling solution
(iii) Write the equation of compound (A) when it undergoes Cannizzaro reaction.
OR (b)
(i) Account for the following:
(1) The alpha-hydrogens of aldehydes and ketones are acidic in nature.
(2) Oxidation of aldehydes is easier than ketones.
(ii) Arrange the following in:
(1) Decreasing reactivity towards nucleophilic addition reaction: propanal, acetone, benzaldehyde.
(2) Increasing order of boiling point: Propane, Ethanol, Dimethylether, Propanal
(iii) Give simple chemical test to distinguish between Benzoic acid and Benzaldehyde.
Answer(a)
(i) 2-Ethyl benzaldehyde
(ii)(1) 2-Ethylbenzaldehyde reacts with 2,4-DNP to form 2,4-DNP derivative + H2O.
(2) Fehling's reaction does not happen with benzaldehyde (aromatic aldehyde).
(iii) Cannizzaro reaction: C6H4(CHO)(CH2CH3) + NaOH -> C6H4(COO-Na+)(CH2CH3) + C6H4(CH2OH)(CH2CH3). Same products are formed.
OR (b) (i)(1) The hydrogen at alpha position of carbonyl group is acidic as it experiences -I effect of the carbonyl group. The proton can be abstracted by a base and the resulting anion is stabilised by keto-enol tautomerism.
(2) Aldehydes are more reactive than ketones and can be easily oxidized. Aldehydes have a hydrogen atom attached to carbonyl carbon which can easily be abstracted. Ketones lack this proton, hence they oxidise only under extreme conditions.
(ii)(1) Propanal > acetone > benzaldehyde
(2) Propane < dimethyl ether < propanal < ethanol
(iii) Benzoic acid reacts with ethanol in the presence of dilute sulphuric acid to form sweet smelling ester Ethyl benzoate. This reaction is not given by benzaldehyde.
Amines Basicity / PropertiesAsked 3 times
20232024
2023 · 5 marks
(a)
(i) Write the reaction involved in Cannizaro's reaction.
(ii) Why are the boiling point of aldehydes and ketones lower than that of corresponding carboxylic acids?
(iii) An organic compound 'A' with molecular formula C5H8O2 is reduced to n-pentane with hydrazone followed by heating with NaOH and Glycol. 'A' forms a dioxime with hydroxylamine and gives a positive Iodoform and Tollen's test. Identify 'A' and give its reaction for Iodoform and Tollen's test.
Answer(a)
(i) Cannizaro Reaction: 2C6H5CHO + Conc. NaOH -> C6H5COONa + C6H5CH2OH (one molecule oxidized to salt, one reduced to alcohol)
(ii) Boiling points of aldehyde and ketones are less than carboxylic acid because carboxylic acid have hydrogen bonding and associated molecules bonding which increases the boiling point of carboxylic acids.
(iii) Compound A is CH3-CO-CH2-CH2-CHO (4-oxopentanal). It is reduced to n-pentane (CH3-CH2-CH2-CH2-CH3) with Zn(Hg)/NH2-NH2. It forms dioxime with NH2OH. Iodoform test: CH3CO- group gives positive iodoform test with I2/NaOH. Tollen's test: -CHO group gives positive Tollen's test.
2023 (a) (I) Give reasons: (i) Aniline on nitration gives good amount of m-nitroaniline, though -NH2 group is o/p directing in electrophilic substitution reactions. (ii) (CH3)2NH is mor...
2024 · 5 marks
(a) An amide 'A' with molecular formula C7H7ON undergoes Hoffmann Bromamide degradation reaction to give amine 'B'. 'B' on treatment with nitrous acid at 273-278 K form 'C' and on treatment with chloroform and ethanolic potassium hydroxide forms 'D'. 'C' on treatment with ethanol gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write the sequence of chemical equations. OR (b)
(i) (1) What is Hinsberg's reagent?
(2) Arrange the following compounds in the increasing order of their basic strength in gaseous phase: C2H5NH2, (C2H5)3N, (C2H5)2NH
(ii) Give reasons for the following:
(1) Methyl amine is more basic than aniline.
(2) Aniline readily reacts with bromine water to give 2,4,6-tribromoaniline.
(3) Primary amines have higher boiling points than tertiary amines.
Answer(a) A = C6H5CONH2 (Benzamide) B = C6H5NH2 (Aniline) C = Benzenediazonium Chloride D = Phenylisocyanide
E = Benzene
Reactions: C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + 2KBr + K2CO3 + 2H2O C6H5NH2 + CHCl3 + KOH -> C6H5NC + 3KCl + 3H2O C6H5N2Cl + C2H5OH -> C6H6 + N2 + HCl + CH3CHO OR (b) (i)(1) The Hinsberg reagent is a chemical reagent used for identification and differentiation of primary, secondary, and tertiary amines. It is benzenesulfonyl chloride (C6H5SO2Cl).
(2) Increasing order of basic strength in gaseous phase: C2H5NH2 < (C2H5)2NH < (C2H5)3N
(ii)(1) In aniline the lone pair of electron remains delocalised in the pi electron ring and thus is not available on nitrogen. However, in methylamine, the lone pair is on NH2 group of methyl and is the source of basicity.
(2) The benzene ring in aniline is highly activated due to strong mesomeric effect of amino group. This readily reacts with bromine water.
(3) In primary amines there are intermolecular hydrogen bonding, so more energy is required to break the attraction. Whereas in tertiary amines there is no intermolecular hydrogen bonding.
2013 · 2 marks
Explain the mechanism of the following reaction: 2CH3–CH2–OH ⟶[H+, 413K] CH3–CH2–O–CH2–CH3 + H2O
(a) Arrange the following compounds in increasing order of their acid strength: p-cresol, p-nitrophenol, phenol.
(b) Write the mechanism (using curved arrow notation) of the following reaction:
CH2=CH2 --H3O+--→ CH3-CH2+ + H2O OR Write the structures of the products when Butan-2-ol reacts with the following:
(a) CrO3
(b) SOCl2
Answer(a) p-cresol < Phenol < p-nitrophenol
(b) Mechanism: (i) Electrophilic attack of H3O+. (ii) Nucleophilic attack of water on carbocation. (iii) Deprotonation.
OR
(a) CH3-CO-CH2-CH3 (Butan-2-one)
(b) CH3-CHCl-CH2-CH3 (2-Chlorobutane)
2023 · 2 marks
For the pair phenol and cyclohexanol, answer the following:
(a) Why is phenol more acidic than cyclohexanol?
(b) Give one chemical test to distinguish between the two.
Answer(a) Cyclohexanol is less acidic as compared to phenol because phenol is an aromatic compound while cyclohexanol is cyclic ring structure containing hydroxyl ion (-OH group). In phenol benzene ring is having double bond and shows sp2 hybridization. Because of more s-character it is more acidic.
(b) When phenol react with Ferric Chloride, it gives violet colour but cyclohexanol remains colourless when react with FeCl3.
2023 · 2 marks
Write the chemical equation involved in the following reactions:
(a) Reimer-Tiemann reaction
(b) Acetylation of Salicylic acid
Answer(a) Reimer-Tiemann Reaction: It is used for ortho-formylation of phenol. Phenol + CHCl3 + KOH -> Salicylaldehyde (o-hydroxybenzaldehyde).
Answer(a) The increasing order towards nucleophilic addition: Butanone < Propanone < Propanal < Ethanal (Due to +I effect).
(b) The increasing order of acidic strength: 4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid (Due to presence of electron withdrawing group).
2022-II · 2 marks
Explain the following reactions:
(a) Wolff-Kishner reduction
(b) Cannizzaro reaction
Answer(a) Wolff-Kishner Reduction Reaction - In this reaction, the reduction of carbonyl compounds to hydrocarbons takes place by heating them with hydrazine and a base to form hydrazone which is further reduced to form methylene group along with nitrogen gas. R2C=O + H2N-NH2 -> R2C=N-NH2 --(glycol+KOH/180°C) → R2CH2 + N2(g)
(b) Cannizzaro Reaction - The self-oxidation reduction (disproportionation) reaction of aldehydes having no alpha-hydrogens when reacts with concentrated alkali is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes react where one is reduced to alcohol and the other is oxidized to carboxylic acid. 2RCHO + NaOH -> RCH2OH + RCOONa.
2022-II · 2 marks
Explain the following reactions:
(a) Clemmensen reaction
(b) Stephen reaction
Answer(a) Clemmensen's reduction: This reaction is used to reduce carbonyl compounds to form simple hydrocarbons in presence of zinc amalgam and concentrated hydrochloric acid. R2C=O --(Zn-Hg/HCl) → R2CH2 + H2O
(b) Stephen Reaction: This reaction is used to synthesize aldehydes from nitriles or cyanides. Firstly, nitriles or cyanides are reduced in the presence of stannous chloride and hydrochloric acid in ethyl acetate solvent to form imine intermediate. Then hydrolysis of this intermediate with water gives corresponding aldehyde. CH3C≡N + 2[H] --(SnCl2/HCl/Ether) → CH3CH=NH + HCl --(H2O/H+) → CH3CHO + NH4Cl.
2022-II · 2 marks
Predict the reagent for carrying out the following transformations:
(a) Benzoyl chloride to Benzaldehyde
(b) Ethanal to 3-hydroxy butanal
(c) Ethanoic acid to 2-chloroethanoic acid
Answer(i) Rosenmund Reduction in presence of Lindlar's Catalyst (H2/Pd-BaSO4).
(ii) Aldol Condensation in presence of dil. NaOH.
(iii) Hell-Volhard Zelinsky Reaction in presence of Cl2/Red Phosphorous. (Any two)
2023 · 2 marks
Do the following conversions in not more than two steps:
Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Justify your answer.
AnswerBenzaldehyde is less reactive than propanal towards nucleophilic addition reactions. In benzaldehyde, there is a resonance of carbonyl group which withdraws electron cloud towards itself, reducing the electrophilicity of the carbonyl carbon.
2024 · 2 marks
(a) Write the stepwise mechanism of nucleophilic addition the carbonyl compounds reactions. OR
(b) How will you convert the following:
(i) Toluene to benzoic acid.
(ii) Ethanol to 3-Hydroxybutanal
Answer(a) Step 1: Electrophilic carbonyl compound forms a sigma bond with nucleophile. Nu attacks C of C=O.
Step 2: Carbon-oxygen pi bond is broken to form negative charged alkoxide.
Step 3: Protonation on alkoxide forming a derivative of alcohol.
(i) Draw the zwitter ion structure for sulphanilic acid.
(ii) How can the activating effect of -NH2 group in aniline be controlled?
Answer(a)
(i) Sulphanilic acid zwitter ion: The -NH2 group gets protonated to -NH3+ and the -SO3H group loses a proton to become -SO3-.
(ii) The activating effect of -NH2 group can be reduced by Friedel-Craft (Alkylation & Acetylation) process because nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group.
2023 · 2 marks
24 (OR): (b)
(i) Complete the reaction with the main product formed:
(ii) CH3-CH2-Br --Sn/KCN → CH3-CH2-CN --Na/C2H5OH/2[H] → CH3-CH2-CH2-NH2 (Propanamine)
2025 · 2 marks
Identify A and B in each of the following reaction sequence:
(a) CH3CH2Cl --NaCN → A --H2/Ni → B
(b) C6H5NH2 --NaNO2/HCl, 0-5C → A --C6H5NH2/H+ → B
Answer(a) A = CH3CH2CN (Ethyl cyanide), B = CH3CH2CH2NH2 (Propanamine).
(b) A = C6H5N2+Cl- (Benzenediazonium chloride), B = C6H5-N=N-C6H4-NH2 (Yellow dye, p-aminoazobenzene).
2025 · 2 marks
Give reasons for the following:
(a) The melting points of alpha-amino acids are generally higher than that of the corresponding carboxylic acids.
(b) Amino acids show amphoteric behaviour.
Answer(a) Amino acids exist as zwitterions in the solid state, which results in strong electrostatic forces of attraction between the positively charged ammonium ion (-NH3+) and the negatively charged carboxylate ion (-COO-). These strong electrostatic forces require more energy to break, leading to higher melting points compared to carboxylic acids which mainly form weaker hydrogen bonds.
(b) Amino acids contain both an acidic (-COOH) group and a basic (-NH2) group. In acidic medium, the amino group accepts a proton, forming -NH3+. In basic medium, the carboxyl group loses a proton, forming -COO-. Due to this dual nature, amino acids can react with both acids and bases, exhibiting amphoteric behaviour.
2024 · 2 marks
(a) Carry out the following conversions:
(i) Nitrobenzene to Aniline
(ii) Aniline to Phenol
OR (b)
(i) Write a chemical test to distinguish between Dimethyl amine and Ethanamine.
(ii) Write the product formed when benzene diazonium chloride is treated with KI.
Answer(a)
(i) Nitrobenzene is reduced to Aniline using Sn/HCl.
(ii) Aniline is first diazotized with NaNO2/HCl at 273-278K to form benzene diazonium chloride, which on hydrolysis gives Phenol + N2 + HCl.
OR (b)
(i) Ethanamine reacts with alcoholic KOH and CHCl3 to form Carbylamine having a characteristic odour. Di-methyl amine does not respond to this test.
(ii) Benzene diazonium chloride reacts with KI to form iodobenzene: C6H5N2Cl + KI -> C6H5I + KCl + N2
2020 · 2 marks
How do antiseptics differ from disinfectants? Name a substance which can be used as a disinfectant as well as an antiseptic.
AnswerAntiseptics are applied to living tissues to kill or prevent growth of microorganisms. Disinfectants are applied to non-living objects. Phenol (0.2% solution is antiseptic, 1% solution is disinfectant).
2020 · 2 marks
Identify the monomers in the following polymers:
(i) [-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]n
(ii) [-CH2-CH(CN)-]n
Answer(i) Monomers: Ethylene glycol (HOCH2CH2OH) and Terephthalic acid (HOOC-C6H4-COOH).
(ii) Monomer: Acrylonitrile (CH2=CH-CN).
2020 · 2 marks
Identify the monomers in the following polymers:
(i) Bakelite-type polymer with phenol-OH and CH2 linkages
(ii) Nylon-type polymer with NH-(CH2)6-NH-CO-(CH2)4-CO units
Answer(i) Monomers: Phenol and Formaldehyde.
(ii) Monomers: Hexamethylenediamine (H2N(CH2)6NH2) and Adipic acid (HOOC(CH2)4COOH).
2020 · 2 marks
Identify the monomers in the following polymers:
(i) [-OCH2CH2-O-C(=O)-C6H4-C(=O)-]n
(ii) Melamine-formaldehyde type polymer with triazine ring and NH-CH2 linkages
Answer(i) Monomers: Ethylene glycol and Terephthalic acid.
(ii) Monomers: Melamine and Formaldehyde.
2020 · 2 marks
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Antiseptics
Answer(i) Antibiotics: Chemical substances produced by microorganisms that inhibit the growth of or destroy other microorganisms, e.g. Penicillin.
(ii) Antiseptics: Chemical substances used to kill or prevent the growth of microorganisms on living tissues, e.g. Dettol, Boric acid.
2020 · 2 marks
Write the reactions showing the presence of following in the open structure of glucose:
(i) a carbonyl group
(ii) Straight chain with six carbon atoms
Answer(i) Reaction of glucose with Br2/H2O (oxidation) -> gluconic acid (proves -CHO group).
(ii) Glucose with HI (prolonged heating) -> n-hexane (proves 6-carbon straight chain: CH2OH-(CHOH)4-CHO -> CH3-CH2-CH2-CH2-CH2-CH3).
2020 · 2 marks
Define the following terms with a suitable example in each:
(i) Tranquilizers
(ii) Anionic detergent
Answer(i) Tranquilizers: Chemical compounds used for the treatment of stress, mild and severe mental diseases, e.g. Valium, Chlordiazepoxide.
(ii) Anionic detergent: Detergents in which the anionic part (long chain hydrophobic group) of the molecule is involved in cleansing action, e.g. Sodium lauryl sulphate.
2020 · 2 marks
Write the reactions showing the presence of following in the open structure of glucose:
(i) an aldehyde group
(ii) a primary alcohol
Answer(i) An aldehyde group: On reduction with sodium amalgam and water, the aldehyde group is reduced to primary alcohol (glucose -> sorbitol).
(ii) A primary alcohol (with nitric acid): glucose + HNO3 (oxidation) -> saccharic acid (both -CHO and -CH2OH oxidized to -COOH).
2020 · 2 marks
Write the reactions showing the presence of following in the open structure of glucose:
(i) five -OH groups
(ii) a carbonyl group
Answer(i) Presence of five -OH groups: Glucose reacts with 5 molecules of acetic anhydride to form pentaacetyl glucose.
(ii) Presence of a carbonyl group: Glucose reacts with HCN to form glucose cyanohydrin, confirming the presence of -CHO group.
2020 · 2 marks
Define the following terms with a suitable example in each:
(i) Antacids
(ii) Artificial Sweetener
Answer(i) Antacids: Chemical substances used to neutralize excess acid in the stomach, e.g. Ranitidine, Cimetidine (or Aluminium hydroxide, Magnesium hydroxide).
(ii) Artificial Sweetener: Chemical substances that provide sweetness without adding calories, e.g. Saccharin, Aspartame.
2023 · 2 marks
Give the reaction of glucose with hydrogen cyanide. Presence of which group is confirmed by this reaction?
AnswerGlucose (CHO-(CHOH)4-CH2OH) + HCN -> Glucocyanohydrin (CH(OH)(CN)-(CHOH)4-CH2OH). The cyano group is present in this reaction, confirming the presence of the aldehyde (-CHO) group in glucose.
2023 · 2 marks
Give the reaction of heating glucose with hydroxylamine. Presence of which group is confirmed by this reaction?
AnswerD-Glucose reacts with hydroxylamine (NH2OH) on heating to form glucoxime (aldo-oxime). In oxime, hydroxyl group is present which is attached to nitrogen of imine group. This confirms the presence of the aldehyde (-CHO) group in glucose.
2023 · 2 marks
Write two differences between DNA and RNA.
AnswerDifferences between DNA and RNA: 1) DNA: Sugar moiety is deoxy ribose. RNA: Sugar moiety is ribose. 2) DNA: It is polymer of long chain of nucleotide. RNA: It is polymer of nucleoside. 3) DNA: Base pairs are A, T, G, C. RNA: Base pairs are A, U, G, C.
2023 · 2 marks
(a) What is the difference between a nucleoside and nucleotide?
(b) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer(a) Nucleotide is composed of a nitrogenous base, sugar and a phosphate group where as Nucleoside is composed of only a nitrogenous base and a phosphate group.
(b) Thymine beta-D-2-deoxyribose and phosphoric acid are obtained as products.
2023 · 2 marks
What are nucleic acids? Why two strands in DNA are not identical but are complementary?
AnswerNucleic Acids: They are naturally occurring chemical compounds that serve as the primary chemical information-carrying molecules in cells. Two main classes are DNA and RNA. Two main strands are held together by hydrogen bonds between specific pair of bases. Cytosine forms hydrogen bond with guanine while adenine forms hydrogen bonds with thymine. As a result they are complementary to each other.
2025 · 2 marks
Write the reaction involved when D-glucose is treated with the following reagents:
(a) What happens when glucose reacts with bromine water? Write chemical equation.
(b) Two bases are mentioned below, identify which is present in DNA and which one is present in RNA: (i) Thymine, (ii) Uracil
Answer(a) Glucose decolourises bromine water and is oxidised to gluconic acid: Glucose + Br2/Water -> Gluconic acid.
(b) Thymine is a nitrogen base for DNA. Uracil is a nitrogen base for RNA.
2024 · 2 marks
(a) What happens when glucose reacts with HI? Write chemical equation.
(b) Which type of bond holds a DNA double helix together?
Answer(a) The reaction of glucose with hydrogen iodide in the presence of heat involves its complete reduction to yield n-hexane. Hydrogen iodide acts as a reducing agent. Glucose + HI -> (heat) -> CH3-CH2-CH2-CH2-CH2-CH3 (n-Hexane)
(b) A DNA helix is held together by hydrogen bonds between the complementary base pairs.
2024 · 2 marks
(a) What happens when Glucose reacts with nitric acid? Write chemical equation.
(b) Write one structural difference between DNA and RNA.
Answer(a) When glucose reacts with nitric acid, it gets oxidised to saccharic acid. Glucose + HNO3 -> Saccharic acid (both -CHO and -CH2OH groups are oxidised to -COOH)
(b) DNA has double strand helix structure. RNA has single strand helix structure.
2013 · 2 marks
Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.
Answer(a) In chlorobenzene, chlorine is attached with a benzene ring through single bond and it is highly electronegative. Any electronegative/electrophilic agent cannot replace it.
(b) There is delocalization of lone pair of electrons present on chlorine atom. Carbon atom of benzene ring is sp2 hybridized. Its structure is stabilized by resonance, so replacement of chlorine atom with other nucleophilic substance will require lots of energy which is not favourable.
2014 · 2 marks
(i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction?
(a) CH3CH2CH2Br (1-bromopropane) and (b) CH3CH(Br)CH3 (2-bromopropane)
(ii) Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation
Answer(i) SN2 reaction occurs more quickly in primary alkyl halides than in secondary/tertiary alkyl halides. (a) is primary alkyl halide and (b) is secondary alkyl halide. So (a) undergoes SN2 reaction faster than (b). Neither is chiral as written. (ii)(a) Inversion of configuration: SN2, because the incoming nucleophile attacks from the backside resulting in inversion.
(b) Racemisation: SN1, in this first a carbocation is formed. The incoming nucleophile can attack from either side resulting in racemic mixture.
2014 · 2 marks
Draw the structure of major monohalo product in each of the following reactions:
(i) Phenol + SOCl2 →
(ii) Cyclohexene + HBr ⟶[Peroxide]
Answer(i) Phenol + SOCl2 → Chlorobenzene (C6H5Cl) is NOT formed easily. Instead with cyclohexanol: Cyclohexanol + SOCl2 → Chlorocyclohexane
(ii) Cyclohexene + HBr ⟶[Peroxide] 3-bromocyclohexene (anti-Markovnikov addition in presence of peroxide)
2014 · 2 marks
Write the mechanism of the following reaction: CH3CH2OH ⟶[HBr] CH3CH2Br + H2O
AnswerMechanism (SN2):
(i) The oxygen lone pair attacks the H–Br molecule: CH3CH2OH + HBr → CH3CH2OH2+ + Br−
(ii) Carbocation formation: CH3CH2OH2+ → CH3CH2+ + H2O
(iii) Attack of nucleophile Br−: CH3CH2+ + Br− → CH3CH2Br
2020 · 2 marks
Draw the structures of the following:
(i) H2S2O8
(ii) XeF6
Answer(i) H2S2O8 (peroxodisulphuric acid) has two S atoms connected by a -O-O- peroxide linkage, each S also bonded to two =O and one -OH.
(ii) XeF6 has distorted octahedral geometry with one lone pair on Xe.
2020 · 2 marks
Draw the structures of the following:
(i) H2S2O7
(ii) BrF5
Answer(i) H2S2O7 (pyrosulphuric acid/oleum) has two SO4 groups sharing one oxygen.
(ii) BrF5 has square pyramidal geometry.
2020 · 2 marks
Draw the structures of the following:
(i) HClO4
(ii) XeOF4
Answer(i) HClO4 (perchloric acid): Cl is central atom bonded to 4 oxygen atoms (3 double bonds, 1 O-H).
(ii) XeOF4: Square pyramidal with one Xe=O bond and 4 Xe-F bonds and one lone pair.
2025 · 2 marks
(a) In the following pair of halogen compounds, which compound undergoes SN1 reaction faster and why? 2-Chloro-2-methylpropane and 3-Chloropropane.
(b) Arrange the following compounds in increasing order of their reactivity towards SN2 displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.
Answer(a) SN1 reaction proceeds via the formation of carbocation. 2-Chloro-2-methylpropane (I) is 3 degree while 3-Chloropropane (II) is 1 degree. Therefore, (I) forms a 3 degree carbocation, while (II) forms a 1 degree carbocation. Since a 3 degree carbocation is more stable than a 2 degree or 1 degree carbocation, (I) undergoes SN1 reaction faster.
(b) The increasing order of reactivity towards SN2 displacement is: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane. Due to steric reasons, the order of reactivity in SN2 reactions follows: 1 degree > 2 degree > 3 degree.
2025 · 2 marks
Give reasons for the following observations:
(a) p-Chloronitrobenzene reacts with (aq)NaOH at 443 K to give p-nitrophenol whereas chlorobenzene reacts with the same reagent at 623 K and 300 atm.
(b) Main product obtained when chloroethane reacts with KCN is propane nitrile while with AgCN it is ethyl isocyanide.
Answer(a) p-Chloronitrobenzene reacts with aq. NaOH at lower temperature to form p-nitrophenol because the nitro group on the benzene ring is electron withdrawing group making the carbon attached to the chlorine significantly more electrophilic, enhancing nucleophilic substitution. Whereas chlorobenzene reacts at 623 K and 300 atm (high temperature and pressure) to overcome the stability of the C-Cl bond due to resonance stabilisation.
(b) In KCN, the bonds are ionic, so carbon atom is free to make bond that's why propane nitrile is formed. While in AgCN the bonds are covalent, so carbon atom is not free to make bond so nitrogen atom of cyanide part will attack on chloroethane to form ethyl isocyanide.
2024 · 2 marks
(a) Which compound in the given pair would undergo SN2 reaction at a faster rate and why? CH3-CH2-I and CH3-CH2-Br
(b) Arrange the following compounds in the increasing order of their boiling points: Butane, 1-Bromobutane, 1-Iodobutane, 1-Chlorobutane
Answer(a) In SN2 reactions, the stronger the nucleophile, the faster the reaction. CH3CH2I undergoes the SN2 reaction faster than CH3CH2Br because iodide (I-) is a better nucleophile than bromide (Br-) due to its larger size and more electron density.
(b) Increasing order of boiling points: Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane. Due to increasing London dispersion forces with increasing halogen atom size.
2024 · 2 marks
Which of the following compounds will react more rapidly by SN2 reaction and why? (a) CH3C(CH3)2Br or CH3CH2CHCH3Br
(b) Arrange the following compounds in the increasing order of their boiling points: CH3CH2CH2CH2Br, (CH3)2CHCH2Br, (CH3)3CBr
Answer(a) 2-bromo-2-methylpropane undergoes SN2 reaction slower than 2-bromobutane. The steric hindrance in tertiary halides makes it more difficult for the nucleophile to approach. Secondary halides typically undergo SN2 reactions faster than tertiary halides.
(b) Increasing order: (CH3)3C-Br < (CH3)2CH-CH2-Br < CH3CH2CH2CH2Br. For isomeric alkyl halides, boiling point decreases with increase in branching due to less surface area and weak intermolecular forces.
2024 · 2 marks
Which of the following compounds will react more rapidly by SN1 reaction and why? (CH3)3Br or CH3CH2Br
(b) Arrange the following compounds in the increasing order of boiling points: Chloromethane, Bromomethane, Dibromomethane, Bromoform
Answer(1) (CH3)3Br undergoes SN1 reaction more rapidly as compared to CH3CH2. Tertiary halides react more rapidly by SN1 reaction due to high stability of tertiary carbocations.
(b) Increasing order of boiling points: Chloromethane < Bromomethane < Dibromomethane < Bromoform
2023 · 2 marks
Why is boiling point of o-dichlorobenzene higher than p-dichlorobenzene but melting point of para isomer is higher than ortho isomer?
AnswerMelting point of p-Chlorobenzene is higher than o-Chlorobenzene because of its closed packed structure which have high intramolecular forces of attraction. That's why high amount of temperature is required to break down the bonds while boiling point of o-chlorobenzene is high because of high dipole interaction present at ortho-position. There is great dipole-dipole forces of attraction between carbon and chlorine atom which results into high boiling point as compared to p-chlorobenzene which have zero dipole moment.
2023 · 2 marks
Write equations for the following: (2 x 1 = 2)
(a) Oxidation of chloroform by air and light
(b) Reaction of chlorobenzene with CH3Cl/anhyd. AlCl3
Why haloarenes are not reactive towards nucleophilic substitution reaction? Give two reasons.
Answer1. Haloarenes are less reactive towards bonds between C-X (carbon & halogen) is shorter because benzene is sp2 hybridized having double bond between two carbons.
2. Haloarenes shows resonating structure in which lone pair present on halogen rotates around the benzene ring and charges are delocalised which make it a stable structure and do not easily substitute the halogen group.
2014 · 2 marks
Write the name of monomers used for getting the following polymers:
(i) Bakelite
(ii) Neoprene
Answer(i) Bakelite: Monomers are formaldehyde (HCHO) and phenol (C6H5OH).
(ii) Neoprene: Monomer is chloroprene (2-chloro-1,3-butadiene, CH2=C(Cl)–CH=CH2).
2014 · 3 marks
(a) Write the mechanism of the following reaction: CH3CH2OH ⟶[HBr] CH3CH2Br + H2O
(b) Write the equation involved in Reimer-Tiemann reaction.
Answer(a) The reaction follows SN2 mechanism:
Step 1: The oxygen lone pair attacks the H–Br molecule, creating a hydronium ion: CH3CH2OH + HBr → CH3CH2OH2+ + Br−
(i) Why is the C-O bond length in phenols less than that in methanol?
(ii) Arrange the following in order of increasing boiling point: Ethoxyethane, Butanal, Butanol, n-butane
(iii) How can phenol be prepared from anisole? Give reaction.
Answer(a)
(i) C-O bond length in phenol is less than methanol because of presence of benzene ring which is aromatic and consisting of double bond. The lone pair present in oxygen is shared with partial conjugation effect while in methanol the lone pair of oxygen shared with normal carbon atom.
(ii) Para-nitro phenol has higher boiling point than ortho-nitrophenol due to intermolecular hydrogen bonding present in para-nitrophenol, which require more energy to break these bonds during boiling. In o-nitrophenol intramolecular hydrogen bonding is present to greater extent. Thus o-nitrophenol is steam volatile due to low boiling point.
2023 · 3 marks
(b) What happens when (OR)
(i) Anisole is treated with CH3Cl anhydrous AlCl3?
(ii) Phenol is oxidised with Na2Cr2O7/H+?
(iii) (CH3)3C-OH is heated with Cu/573 K?
Write chemical equation in support of your answer.
Answer(b)
(i) Friedel craft Alkylation reaction will take place. Anisole + CH3Cl --(Anhy. AlCl3) → o-methyl anisole + p-methyl anisole
(ii) A conjugated diketone is produced - Benzoquinone. Phenol --(Na2Cr2O7/H+) → Benzoquinone
(iii) Dehydration takes place and alkene is formed. (CH3)3C-OH --(Cu/573K, -H2O) → CH3-C(CH3)=CH2 (2-methylpropene)
2024 · 3 marks
Write chemical equations for the following reactions (Do any three):
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. Isomers (A) and (C) give positive Tollens' test whereas isomer (B) does not give Tollens' test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards addition of HCN?
Answer(a) A = CH3CH2CH2CHO (Butanal) B = CH3COCH2CH3 (Butanone) C = (CH3)2CHCHO (2-Methylpropanal) D = CH3CH2CH2CH3 (Butane)
(b) B (Butanone) - as ketones are less reactive towards addition of HCN than aldehydes due to steric effect and inductive effect.
(i) Benzoic acid is a stronger acid than acetic acid.
(ii) Methanal is more reactive towards nucleophilic addition than ethanal.
(b) Give simple chemical test to distinguish between propanal and propanone.
Answer(a)
(i) Due to sp2 hybridised carbon and greater resonance stabilization of benzoate ion.
(ii) Methanal has more electrophilic carbonyl carbon (no +I effect) and less steric hindrance.
(b) Tollens' test: Propanal gives silver mirror; propanone does not.
2022-II · 3 marks
(a) Complete the following: (i) CH3CN --(1. DIBAL-H / 2. H2O) → 'A' --(H2N-OH / H+) → 'B' (ii) Write IUPAC name of the following compound: 3-bromobenzaldehyde (iii) Write a chemical test to distinguish between the following compounds: Phenol and Benzoic acid OR
(b) Convert the following: (i) Benzoic acid to Benzaldehyde (ii) Propan-1-ol to 2-Bromopropanoic acid (iii) Acetaldehyde to But-2-enal
(iii) Phenol and Benzoic Acid can be distinguished by iron chloride (FeCl3) test. Phenol gives violet colouration with neutral FeCl3 solution while benzoic acid gives buff coloured precipitate of ferric benzoate.
(i) Propanone is treated with CH3MgBr and then hydrolysed?
(ii) Ethanal is treated with excess ethanol and acid?
(iii) Methanal undergoes Cannizzaro reaction?
(b) Write the main product in the following reaction:
(i) 2CH3COCl + (CH3)2Cd ->
(ii) CH3CH2CHO --(Zn(Hg)/Conc HCl) →
(iii) C6H5COONa + NaOH --(CaO/heat) →
Answer(a)
(i) When propanone reacts with ethyl magnesium bromide (Grignard's reagent) and then hydrolysed, it forms alcohol: (CH3)2CO + CH3CH2MgBr -> (CH3)2C(OH)CH2CH3 + Mg(OH)Br (2-methylbutan-2-ol).
(ii) When ethanal reacts with excess ethanol and acid gives acetal: CH3CHO + 2C2H5OH -> CH3CH(OC2H5)2 + H2O.
(iii) Methanal will undergo Cannizzaro reaction as it possesses no alpha-hydrogen to form methanol: 2HCHO + conc.KOH -> CH3OH (Methanol) + HCOOK (Potassium methanoate).
(i) Which acid of the following pair would you expect to be stronger? F-CH2-COOH or CH3-COOH (ii) Arrange the following compounds in increasing order of their boiling points: CH3CH2OH, CH3-CHO, CH3-COOH (iii) Give simple chemical test to distinguish between Benzaldehyde and Acetophenone.
OR (b)
(i) Which will undergo faster nucleophilic addition reaction? Acetaldehyde or Propanone (ii) What is the composition of Fehling's reagent? (iii) Draw structure of the semicarbazone of Ethanal.
Answer(a)
(i) Among FCH2COOH and CH3COOH, FCH2COOH is stronger due to the presence of electron withdrawing group (-F). As Fluorine is most electronegative element, electronegativity makes it stronger than acetic acid.
(ii) The increasing order of boiling point: CH3CHO < CH3CH2OH < CH3COOH. Ethanoic acid and ethanol have comparatively higher boiling point than ethanal as they both are held by strong hydrogen bonds whereas in ethanal there is dipole-dipole interaction.
(iii) With ammoniacal silver nitrate solution (Tollen's reagent), benzaldehyde forms silver mirror but acetophenone does not give this test. Acetophenone gives positive iodoform test whereas benzaldehyde does not give iodoform test.
OR (b)
(i) Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric hindrance. Thus, acetaldehyde is more reactive than propanone.
(ii) Fehling's solution is a mixture of alkaline solution of copper(II) sulphate (CuSO4) containing sodium potassium tartrate (Rochelle salt - KNaC4H4O6.4H2O).
(iii) Structure of semicarbazone from ethanal: CH3CH=NNHCONH2 + H2O.
2025 · 3 marks
(A) Explain the following reactions and write chemical equation involved:
(a) Wolff-Kishner reduction
(b) Etard reaction
(c) Cannizzaro reaction.
OR
(B) Write the structures of A, B and C in the following sequence of reactions:
(a) CH3COOH --SOCl2 → A --H2,Pd-BaSO4 → B --H2N-NH2 → C.
(b) CH3CN --1.DIBAL-H, 2.H2O → A --Dil.NaOH → B --Delta → C.
Answer(A)(a) Wolff-Kishner reduction: Treating the carbonyl compound with hydrazine (N2H4) and a strong base (KOH) in a high-temperature reaction converts ketone/aldehyde to alkane.
(b) Etard reaction: Oxidises toluene to benzaldehyde using CrO2Cl2.
(c) Cannizzaro reaction: A base-catalysed reaction where non-enolisable aldehydes undergo disproportionation. 2HCHO + NaOH -> CH3OH + HCOONa.
OR (B)(a) A = CH3COCl, B = CH3CHO, C = CH3CH=N-NH2. CH3COOH + SOCl2 -> CH3COCl + HCl + SO2. CH3COCl + H2 (Pd-BaSO4) -> CH3CHO + HCl. CH3CHO + NH2NH2 -> CH3CH=NNH2 + H2O.
(b) A = CH3CHO, B = CH3CHOHCH2CHO, C = CH3CH=CHCHO. CH3CN + DIBAL-H/H2O -> CH3CHO. CH3CHO + dil.NaOH -> CH3CHOHCH2CHO (aldol). CH3CHOHCH2CHO --Delta → CH3CH=CHCHO + H2O.
2024 · 3 marks
Give the structure of the major product expected from the following reactions:
(a) Reaction of ethanal with methyl-magnesium bromide followed by hydrolysis.
(b) Hydration of But-1-ene in the presence of dilute sulphuric acid.
(c) Only the aldehyde group reacts with NaCN/HCl to form cyanohydrin. The -CHO gets converted to -CH(OH)(CN).
2024 · 3 marks
Give the structure of the major product expected from the following reactions:
(a) Reaction of propanal with methyl magnesium bromide followed by hydrolysis.
(b) Reaction of phenol with Br2 in CS2.
(c) Reaction of propene with diborane followed by oxidation.
Answer(a) The nucleophile MeMgBr attacks the carbonyl carbon of propanal to form the final product followed by hydrolysis to form the secondary alcohol: CH3CH2CH(OH)CH3 (Butan-2-ol).
(b) Reaction of phenol with Br2 in CS2 at 273K gives 2-bromophenol (minor product) + 4-bromophenol (major product).
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Phenol and Benzoic acid
(ii) Propanal and Propanone
(b) Which one of the given compounds is a stronger acid and why? CH2FCH2CH2COOH or CH3CHFCH2COOH
Answer(a)
(i) Phenol with neutral ferric chloride: colour change (green, blue purple or red). No change with benzoic acid.
(ii) Propanal with Tollen's Reagent: On heating, silver mirror is formed on inner walls of test tube. Aldehyde group reduces Tollen's reagent. No change with propanone.
(b) 3-fluorobutanoic acid (CH3CHFCH2COOH) is stronger than 4-fluorobutanoic acid (CH2FCH2CH2COOH) due to the distance. In inductive effect, distance increases the strength decreases. The fluorine is closer to COOH in CH3CHFCH2COOH.
2013 · 3 marks
Give the structures of A, B and C in the following reactions:
(i) C6H5N2+Cl− ⟶[CuCN] A ⟶[H2O/H+, Δ] B ⟶[H2O/H+, Δ] C
(ii) C6H5NO2 ⟶[Sn+HCl] A ⟶[NaNO2+HCl] B ⟶[H2O/H+, Δ] C
(i) Primary amines (R–NH2) have higher boiling point than tertiary amines (R3N).
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
Answer(i) Primary amines have N–H bonds available for intermolecular hydrogen bonding. Tertiary amines do not have N–H bonds, so they cannot form H-bonds. Hence primary amines have higher boiling points.
(ii) Aniline does not undergo Friedel-Crafts reaction because the Lewis acid catalyst (AlCl3) reacts with the lone pair on nitrogen of aniline forming a salt, which deactivates the benzene ring.
(iii) In aqueous solution, (CH3)2NH is more basic than (CH3)3N because solvation of the substituted ammonium cation stabilizes it. Greater size of ion means lesser solvation and less stabilization. So the order in water is: primary > secondary > tertiary.
2014 · 3 marks
Give the structures of A, B and C in the following reactions:
Write the structures of A, B and C in the following:
(a) C6H5-CONH2 --(Br2/aq.KOH) → A --(NaNO2+HCl, 0-5°C) → B --(KI) → C.
(b) CH3-Cl --(KCN) → A --(LiAlH4) → B --(CHCl3+alc.KOH) → C.
Answer(a) A = C6H5NH2 (aniline, Hoffmann bromamide degradation), B = C6H5N2+Cl- (benzenediazonium chloride), C = C6H5I (iodobenzene).
(b) A = CH3CN (methyl cyanide), B = CH3CH2NH2 (ethylamine), C = CH3CH2NC (ethyl isocyanide, carbylamine reaction).
2016 · 3 marks
Give reasons for the following:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(iii) Primary amines have higher boiling point than tertiary amines.
Answer(i) Aniline is a Lewis base while AlCl3 is a Lewis acid. They combine to form a salt, deactivating the benzene ring for electrophilic substitution.
(ii) Due to combined +I effect and solvation effects; (CH3)2NH cation is better solvated than bulky (CH3)3NH+.
(iii) Primary amines have N-H bonds allowing intermolecular hydrogen bonding; tertiary amines have no N-H for H-bonding.
2017 · 3 marks
Give reasons:
(i) Acetylation of aniline reduces its activation effect.
(ii) CH3NH2 is more basic than C6H5NH2.
(iii) Although -NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.
Answer(i) Due to resonance, the electron pair of nitrogen atom gets delocalised towards carbonyl group/resonating structures, reducing activation effect.
(ii) Because of +I effect in methylamine, electron density at nitrogen increases whereas in aniline, resonance takes place and electron density on nitrogen decreases.
(iii) Due to protonation of aniline in acidic medium forming anilinium ion. Nitration of anilinium ion gives m-nitroaniline.
2017 · 3 marks
Write the structures of compounds A, B and C in the following reactions:
(a) CH3-COOH --NH3/Δ--→ A --Br2/KOH(aq)--→ B --CHCl3 + alc. KOH--→ C
(b) C6H5N2+BF4- --NaNO2/Cu, Δ--→ A --Fe/HCl--→ B --CH3COCl/pyridine--→ C
An aromatic compound 'A' on heating with Br2 and KOH forms a compound 'B' of molecular formula C6H7N which on reacting with CHCl3 and alcoholic KOH produces a foul smelling compound 'C'. Write the structure and IUPAC names of compound A, B and C.
AnswerA = Benzamide (C6H5CONH2). B = Aniline (C6H5NH2) via Hofmann bromamide degradation. C = Phenylisocyanide/Benzeneisonitrile (C6H5NC) via carbylamine reaction.
2019 · 3 marks
Write the structures of main products when benzene diazonium chloride reacts with:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) Aromatic primary amines cannot be prepared by Gabriel's phthalimide synthesis.
(iii) Aliphatic amines are stronger bases than ammonia.
Answer(i) Aniline being a Lewis base forms a complex with AlCl3 (Lewis acid). The amino group is not in a position to activate the benzene ring towards electrophilic substitution.
(ii) Aromatic primary amines cannot be prepared by Gabriel's phthalimide synthesis because haloarenes react with potassium phthalimide and they are little reactive, so bond cleavage does not take place.
(iii) Aliphatic amines are stronger bases than ammonia because the alkyl group has +I effect which increases the electron density on nitrogen atom.
2020 · 3 marks
Arrange the following compounds as directed:
(i) In increasing order of solubility in water: (CH3)2NH, CH3NH2, C6H5NH2
(ii) In decreasing order of basic strength in aqueous solution: (CH3)3N, (CH3)2NH, CH3NH2
(iii) In increasing order of boiling point: (C2H5)2NH, (C2H5)3N, C2H5NH2
Answer(i) C6H5NH2 < (CH3)2NH < CH3NH2.
(ii) (CH3)2NH > (CH3)3N > CH3NH2.
(iii) (C2H5)3N < (C2H5)2NH < C2H5NH2.
2022-II · 3 marks
(a) Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Aniline does not undergo Friedel-Crafts reaction. (iii) Primary amines have higher boiling points than tertiary amines. OR (b)
(i) Arrange the following compounds in the increasing order of their basic strength in aqueous solution: CH3NH2, (CH3)3N, (CH3)2NH (ii) What is Hinsberg's reagent? (iii) What is the role of pyridine in the acylation reaction of amines?
Answer(a)
(i) Aniline has higher pKb than methylamine because aniline undergoes resonance and the electrons on N-atom are delocalized in benzene ring. Thus, electrons are less available to donate whereas in methylamine electron density on nitrogen atom is greater than aniline.
(ii) Aniline does not undergo Friedel-Crafts reaction because aniline acts as a strong base which reacts with AlCl3 to form salt. Thus, due to the presence of positive charge on N-atom, electrophilic substitution is deactivated in benzene ring.
(iii) Boiling point of primary amines is higher than that of tertiary amines because tertiary amines do not form hydrogen bond due to absence of H-atoms. Primary amines have two hydrogen atoms attached to electronegative N-atom leading to greater magnitude of H-bonding.
OR (b)
(i) The increasing order of basic strength in aqueous solution: (CH3)3N < CH3NH2 < (CH3)2NH (Due to steric hindrance and +I effect of alkyl groups).
(ii) The benzene sulphonyl chloride (C6H5SO2Cl) is called Hinsberg reagent. It is an organo-sulphur compound used for detection and distinction of primary, secondary and tertiary amine in a given sample.
(iii) Pyridine is used in acylation of amines because it acts as a strong base which helps in removing the side product HCl from the reaction mixture. It acts as an acceptor for the acid by-product formed during the reaction.
2022-II · 3 marks
A compound 'A' on reduction with iron scrap and hydrochloric acid gives compound 'B' with molecular formula C6H7N. Compound 'B' on reaction with CHCl3 and alcoholic KOH produces an obnoxious smell of carbylamine due to the formation of 'C'. Identify 'A', 'B' and 'C' and write the chemical reactions involved.
AnswerA is Nitrobenzene, B is Aniline, C is Phenyl isocyanide. C6H5NO2 + Fe/HCl -> C6H5NH2 (Aniline, B) C6H5NH2 + CHCl3 + 3KOH --(warm) → C6H5NC (Phenyl isocyanide, C) + 3KCl + 3H2O
2022-II · 3 marks
A primary amine 'A' C2H5N reacts with alkyl halide (C2H5I) to give secondary amine 'B'. 'B' reacts with C6H5SO2Cl to give a solid 'C' which is insoluble in alkali. Identify 'A', 'B', 'C' and write all the chemical reactions involved.
AnswerA = Ethyl amine (CH3CH2NH2), B = Diethyl amine (CH3CH2NHCH3CH2), C = N,N-diethyl benzene sulphonamide. CH3CH2NH2 + CH3CH2I -> CH3CH2NHCH2CH3 (A -> B, secondary amine) CH3CH2NHCH2CH3 + C6H5SO2Cl -> C6H5SO2N(C2H5)2 (B -> C, sulphonamide insoluble in alkali)
2022-II · 3 marks
Give reasons:
(a) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines.
(b) Aniline does not give Friedel-Crafts reaction.
(c) Although -NH2 group is o/p directing in electrophilic substitution reaction, yet aniline on nitration gives good yield of m-nitroaniline.
Answer(a) Ammonolysis of alkyl halides leads to the formation of mixture of primary, secondary, tertiary amines and quaternary salts. It is because every time nucleophilic substitution reaction takes place in which amine acts as a nucleophile and form primary amine which further react and form secondary amine, which again react with the alkyl halide to form the tertiary amine, and further leads to the formation of quaternary salt. Thus, it is difficult to get the pure amine.
(b) Aniline does not undergo Friedel-Crafts reaction because aniline acts as a strong base which reacts with AlCl3 to form salt. Thus, due to the presence of positive charge on N-atom electrophilic substitution is deactivated in benzene ring.
(c) Nitration is carried out in an acidic medium. In a strongly acidic medium, aniline is protonated to give anilinium ion (which is meta-directing). For this reason, aniline on nitration gives a substantial amount of m-nitroaniline (47%).
2023 · 3 marks
(a) Illustrate Sandmeyer's reaction with an equation.
(b) Explain, why (CH3)2NH is more basic than (CH3)3N in aqueous solution.
(b) The order of basicity is (CH3)2-NH > (CH3)3-N because alkyl group is small, there is no steric hindrance to H-bonding. So, nature of alkyl group is responsible for basicity of 2° amine. Secondly, there is inductive effect which is important for salvation effect.
2025 · 3 marks
(a) Arrange the following compounds in increasing order of their boiling point: (CH3)2NH, CH3CH2NH2, CH3CH2OH.
(b) Give plausible explanation for each of the following:
(i) Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
(ii) Amides are less basic than amines.
Answer(a) Stronger intermolecular H-bonding results in associated molecules that have higher boiling points. Alcohol > Amine, as O-H bond is more polar than N-H bond. Thus, the increasing order of boiling point is: (CH3)2NH < CH3CH2NH2 < CH3CH2OH. (b)
(i) Aromatic primary amines cannot be formed by this reaction as the aryl halides do not undergo nucleophilic substitution with potassium phthalimide which involves the cleavage of C-X bond. Aryl halides have double bond character in C-X bond.
(ii) In amides, the lone pair of nitrogen is involved in resonance with the carbonyl group and is not available for coordinate bond formation with a proton. But in amine the lone pair of nitrogen available for protonation that's why amides are less basic than amines.
2025 · 3 marks
(a) Arrange the following in decreasing order of pKb: Aniline, p-nitroaniline, p-methylaniline.
(b) Account for the following:
(i) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(ii) Methylamine in water reacts with FeCl3 to precipitate hydrated ferric oxide.
Answer(a) p-Nitroaniline > Aniline > p-Methylaniline (decreasing order of pKb, i.e. increasing basicity). (b)
(i) Diazonium salts of aromatic amines are stabilised by resonance with the aromatic ring, where the positive charge on the nitrogen is delocalised into the ring. Aliphatic amines do not have this resonance stabilisation, and their diazonium salts are less stable.
(ii) When methylamine is added to water, it acts as a weak base, producing hydroxide ions (OH-). These hydroxide ions then react with the ferric ions (Fe3+) from ferric chloride (FeCl3), resulting in the precipitation of hydrated ferric oxide. CH3NH2 + H2O -> CH3NH3+ + OH-. FeCl3 -> Fe3+ + 3Cl-. 2Fe3+ + 6OH- -> 2Fe(OH)3 or Fe2O3.3H2O.
2024 · 3 marks
A compound 'X' with molecular formula C3H9N reacts with C6H5SO2Cl to give a solid, insoluble in alkali. Identify 'X' and give the IUPAC name of the product. Write the reaction involved.
AnswerSince the compound formed on reaction with C6H5SO2Cl is insoluble in alkali, the amine is a secondary amine. The structure of compound X is CH3NHCH2CH3. IUPAC name is N-Methyl ethanamine.
2013 · 3 marks
After watching a programme on TV about the adverse effects of junk food and soft drinks on the health of school children, Sonali, a student of Class XII, discussed the issue with the school principal. Principal immediately instructed the canteen contractor to replace the fast food with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision was welcomed by the parents and the students.
(a) What values are expressed by Sonali and the Principal of the school?
(b) Give two examples of water-soluble vitamins.
Answer(a) Sonali and principal discussed about the harmful effects of fast food on the health of students. They expressed the value of healthy food because healthy food is low in carbohydrates and fat. Healthy food contains high amount of minerals and vitamins which are important for growth and overall development of the students.
(b) Vitamin B and C are water soluble. E.g., legumes, leafy vegetable and citrus.
2014 · 3 marks
(i) Deficiency of which vitamin causes night-blindness?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose?
Answer(i) Vitamin A deficiency causes night-blindness.
(ii) Uracil is found in RNA only (replaced by thymine in DNA).
(iii) Glucose reaction with HI produces n-hexane which suggests that all the six carbon atoms are linked together in a straight chain.
2014 · 3 marks
(i) Deficiency of which vitamin causes rickets?
(ii) Give an example for each of fibrous protein and globular protein.
(iii) Write the product formed on reaction of D-Glucose with Br2 water.
Answer(i) Deficiency of Vitamin D and lack of calcium causes rickets.
(ii) Fibrous protein: α-keratin (found in hairs, nails, collagen). Globular protein: Haemoglobin (carries oxygen in blood).
(iii) Gluconic acid (C6H12O7) is formed. On reacting glucose with bromine water, glucose undergoes oxidation reaction to form gluconic acid.
2014 · 3 marks
Define the following terms related to proteins:
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Answer(i) A peptide bond is created when two amino acids come together between the amino group of one molecule and the carboxylic acid group of another molecule, creating a covalent link with loss of water (–CO–NH– linkage).
(ii) The linear arrangement of a protein's amino acid structural building blocks constitutes the primary structure of a peptide. It describes the sequence from N-terminal to C-terminal end.
(iii) Denaturation is the process by which proteins or nucleic acids lose their quaternary, tertiary, and secondary structures through application of external stress such as heat, acid/base, or organic solvents.
2014 · 3 marks
Define the following terms:
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Oligo saccharides
Answer(i) Glycosidic bond or glycosidic linkage connects a sugar molecule to another molecule by losing one water molecule. The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage.
(ii) Invert sugar: A product of the hydrolysis of sucrose that contains glucose and fructose in a fixed ratio (equimolar). Found naturally in fruits and honey.
(iii) Oligo saccharides: A saccharide polymer known as an oligosaccharide contains a few (usually three to ten) simple sugars (monosaccharides).
2014 · 3 marks
Define the following terms:
(i) Nucleotide
(ii) Anomers
(iii) Essential amino acids
Answer(i) Nucleotide: A unit formed by the attachment of a base to the 1' position of sugar is known as nucleoside. When nucleoside is linked to phosphoric acid at 5'-position, it forms a nucleotide.
(ii) Anomers: These are the cyclic monosaccharides which differ at C-1 in aldose and C-2 in ketose, also known as epimers or stereoisomers.
(iii) Essential amino acids: An amino acid that cannot be produced by the organism and must thus be given in the food is known as an essential amino acid or indispensable amino acid.
2015 · 3 marks
(i) Which one of the following is a disaccharide: Starch, Maltose, Fructose, Glucose?
(ii) What is the difference between fibrous protein and globular protein?
(iii) Write the name of vitamin whose deficiency causes bone deformities in children.
Answer(i) Maltose is a disaccharide.
(ii) Fibrous proteins: Parallel polypeptide chains, insoluble in water. E.g., keratin, collagen. Globular proteins: Spherical shape, soluble in water. E.g., haemoglobin, insulin.
(iii) Vitamin D deficiency causes bone deformities (rickets) in children.
2016 · 3 marks
(i) What is the role of t-butyl peroxide in the polymerization of ethene?
(ii) Identify the monomers in the following polymer: -(-NH-(CH2)6-NH-CO-(CH2)4-CO-)n-
(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Polystyrene, Terylene and Buna-S.
OR Write the mechanism of free radical polymerization of ethene.
Answer(i) t-Butyl peroxide acts as a free radical initiator in the polymerization of ethene.
(ii) Monomers: Hexamethylenediamine (H2N(CH2)6NH2) and Adipic acid (HOOC(CH2)4COOH). This is Nylon-6,6.
OR Mechanism: Step 1 (Initiation): Peroxide decomposes to free radicals which attack ethene.
Step 2 (Propagation): Chain grows by successive addition of monomers.
Step 3 (Termination): Two growing chains combine.
2016 · 3 marks
(i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
(ii) Why vitamin C cannot be stored in our body?
(iii) What is the difference between a nucleoside and nucleotide?
Answer(i) beta-D-glucose and beta-D-galactose.
(ii) Vitamin C is water soluble, hence excreted through urine and cannot be stored.
(iii) Nucleoside = base + sugar (pentose sugar combines with nitrogen base). Nucleotide = base + sugar + phosphate group (nucleoside combines with phosphate group).
2016 · 3 marks
(i) What is the role of sulphur in the vulcanization of rubber?
(ii) Identify the monomers in the following polymer: -(-O-CH2-CH2-O-CO-C6H4-CO-)n-
(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Terylene, polythene and neoprene.
Answer(i) Sulphur cross-links polymer chains through disulphide bonds (-S-S-), making rubber harder, more elastic, and resistant to temperature changes.
(ii) Monomers: Ethylene glycol (HOCH2CH2OH) and Terephthalic acid (HOOC-C6H4-COOH). This is Terylene (PET).
(a) Write the product when D-glucose reacts with conc. HNO3.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins.
Answer(a) Polysaccharides: Carbohydrates that give large number of monosaccharide units on hydrolysis. E.g., starch, cellulose, glycogen.
(b) Denatured protein: Proteins that lose their biological activity when secondary and tertiary structures are destroyed. E.g., curdling of milk, coagulation of egg.
(c) Essential amino acids: Amino acids which cannot be synthesised in the body and must be obtained from diet. E.g., Valine, Leucine.
(b) Due to presence of both carboxyl (-COOH, acidic) and amino (-NH2, basic) groups. In aqueous solution, zwitterion is formed.
(c) α-helix has intramolecular hydrogen bonding with regular coiling; β-pleated has intermolecular H-bonding with chains stretched and arranged side by side.
2019 · 3 marks
Write the structure of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
Answer(i) Nylon-6,6: Hexamethylenediamine (H2N-(CH2)6-NH2) and Adipic acid (HOOC-(CH2)4-COOH).
(ii) Glyptal: Ethylene glycol and Phthalic acid.
(iii) Buna-S: 1,3-Butadiene (CH2=CH-CH=CH2) and Styrene (C6H5CH=CH2).
2019 · 3 marks
(i) Is polypropylene {CH2-CH(CH3)}n a homopolymer or copolymer? Give reason.
(ii) Write the monomers of melamine-formaldehyde polymer.
(iii) What is the role of sulphur in vulcanization of rubber?
Answer(i) Homopolymer because it is formed from only one type of monomer (propylene).
(ii) Melamine and formaldehyde.
(iii) Sulphur forms cross-links between polymer chains making rubber harder and more elastic.
2019 · 3 marks
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents are used in toothpastes?
(iii) Why the use of alitame as artificial sweetener is not recommended?
Answer(i) Tranquilizers.
(ii) Anionic detergents like sodium lauryl sulphate.
(iii) Alitame is high potency sweetener and difficult to control the sweetness of food.
2019 · 3 marks
Define the following terms with suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents
Answer(i) Broad-spectrum antibiotics: Antibiotics effective against a wide range of Gram-positive and Gram-negative bacteria. Example: Chloramphenicol.
(ii) Disinfectants: Chemical substances that kill micro-organisms but cannot be used on living tissues. Example: 1% phenol solution.
(iii) Cationic detergents: Detergents in which the cationic part possesses a long hydrocarbon chain and a positive charge. Example: Cetyltrimethylammonium bromide.
2019 · 3 marks
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and glycosidic linkage
(iii) Fibrous protein and globular protein
Answer(i) Amylose is water soluble, straight chain polymer; amylopectin is water insoluble, branched chain polymer.
(ii) Peptide linkage (-CONH-) is formed between two amino acids; glycosidic linkage (-C-O-C-) is an oxide linkage between two monosaccharides.
(iii) In fibrous protein, polypeptide chains run parallel (insoluble in water); in globular protein, chains coil around to give spherical shape (soluble in water).
2019 · 3 marks
Write chemical reactions to show that open structure of D-glucose contains the following:
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(iii) Why soaps are biodegradable whereas detergents are non-biodegradable?
Answer(i) Bithional is added as antiseptic to reduce odour from bacterial decomposition of organic matter on skin.
(ii) Mg(OH)2 is insoluble and does not increase pH above neutrality; NaHCO3 being soluble can make stomach alkaline causing more acid production.
(iii) Soaps have straight chain hydrocarbons easily decomposed by microorganisms; detergents have branched chains resistant to bacterial decomposition.
2019 · 3 marks
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics
Answer(i) Antibiotics: Chemical substances that inhibit growth of or destroy micro-organisms. Example: Penicillin.
(ii) Artificial sweeteners: Compounds giving sweetening effect without adding calories. Example: Saccharin, Aspartame.
(iii) Analgesics: Drugs that relieve pain without causing loss of consciousness. Example: Aspirin (non-narcotic), Morphine (narcotic).
2019 · 3 marks
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins
Answer(i) Oligosaccharides: Carbohydrates yielding two to ten monosaccharide units on hydrolysis. Example: Sucrose.
(ii) Denaturation: When protein loses its biological activity due to change in temperature or pH, H-bonds are disturbed, globules unfold. Example: Boiling of egg.
(iii) Vitamins: Organic compounds required in small amounts for specific biological functions. Example: Vitamin A.
2019 · 3 marks
Write the reactions when D-glucose is treated with:
(i) Br2 water
(ii) H2N-OH
(iii) (CH3CO)2O
Answer(i) D-Glucose + Br2 water -> Gluconic acid (COOH(CHOH)4CH2OH).
(ii) D-Glucose + NH2OH -> Glucose oxime (CH=N-OH(CHOH)4CH2OH).
(b) Arrange in increasing order of intermolecular forces: Terylene, Buna-N, Polystyrene.
(c) Define biodegradable polymers with example.
Answer(a)
(i) Bakelite.
(ii) Natural rubber/Cellulose.
(b) Buna-N < Polystyrene < Terylene.
(c) Polymers degraded by micro-organisms in reasonable time. Example: PHBV, Nylon-2-nylon-6.
2019 · 3 marks
(a) Write product when D-glucose reacts with Br2(aq).
(b) What bonding provides stability to alpha-helix of protein?
(c) Name the vitamin whose deficiency causes pernicious anaemia.
Answer(a) Gluconic acid.
(b) Intramolecular hydrogen bonding.
(c) Vitamin B12.
2019 · 3 marks
Define:
(a) Invert sugar
(b) Native protein
(c) Nucleotide
Answer(a) Equimolar mixture of glucose and fructose from sucrose hydrolysis; rotation changes from dextro to laevo.
(b) Protein in biological system with unique 3D structure and biological activity.
(c) Unit of nitrogenous base + pentose sugar + phosphate.
2020 · 3 marks
Differentiate between following:
(i) Amylose and Amylopectin
(ii) Globular protein and Fibrous protein
(iii) Nucleotide and Nucleoside
Answer(i) Amylose is a straight chain polymer of D-glucose linked by 1,4-glycosidic linkage. Amylopectin is a branched chain polymer linked by alpha-1,4 and alpha-1,6 glycosidic linkage.
(ii) Globular protein: polypeptide chains arranged as coils, spherical shape, water soluble. Fibrous protein: chains run parallel, thread-like structure, insoluble in water.
(iii) Nucleotide consists of a nitrogenous base, sugar and phosphate groups. Nucleoside consists of a nitrogenous base covalently bonded to a sugar without phosphate group.
2020 · 3 marks
Define the following terms with a suitable example in each:
(a) Polysaccharides
(b) Denatured protein
(c) Fibrous protein
Answer(a) Polysaccharides: Complex long chains of monosaccharides linked by glycosidic bonds, e.g. starch, cellulose.
(b) Denatured protein: When native protein is subjected to physical change like change in temperature or pH, the hydrogen bonds are disturbed, helix gets uncoiled and protein loses its biological activity, e.g. coagulation of egg.
(c) Fibrous protein: When polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, e.g. Keratin, myosin.
2020 · 3 marks
(i) What are the hydrolysis products of DNA?
(ii) What happens when D-glucose is treated with Bromine water?
(iii) What is the effect of denaturation on the structure of proteins?
Answer(i) On hydrolysis of DNA, the products are pentose sugar (deoxyribose), phosphoric acid and bases (adenine, guanine, cytosine, thymine).
(ii) When D-glucose is treated with bromine water, D-gluconic acid is formed (the -CHO group is oxidized to -COOH).
(iii) On denaturation, secondary and tertiary proteins get converted into primary proteins. Denaturation disrupts the normal alpha-helix and beta sheets and uncoils it into a random shape.
2020 · 3 marks
Write the name and structures of monomer(s) in the following polymers:
(i) Nylon-6
(ii) PVC
(iii) Neoprene
Answer(i) Nylon-6: Monomer is Caprolactam.
(ii) PVC: Monomer is Vinyl chloride (CH2=CHCl).
(iii) Neoprene: Monomer is Chloroprene (CH2=C(Cl)-CH=CH2).
2020 · 3 marks
Write the name and structures of monomers in the following polymers:
(i) Nylon-6,6
(ii) Terylene
(iii) PHBV
Answer(i) Nylon-6,6: Hexamethylenediamine (H2N(CH2)6NH2) and Adipic acid (HOOC(CH2)4COOH).
(ii) Terylene: Ethylene glycol (HOCH2CH2OH) and Terephthalic acid (HOOC-C6H4-COOH).
(iii) PHBV: 3-hydroxybutanoic acid and 3-hydroxypentanoic acid.
2023 · 3 marks
Give reasons for any 3 of the following observations:
(a) Penta-acetate of glucose does not react with hydroxylamine.
(b) Amino acids behave like salts.
(c) Water soluble vitamins must be taken regularly in diet.
(d) The two strands in DNA are complimentary to each other.
Answer(a) Because hydroxyl amine reacts with the aldehyde group to form oxime. In penta-acetate of glucose, the aldehyde group is acetylated (protected), so it does not react with hydroxylamine.
(b) In a solution, amino acids exist as Zwitter ions: H3N+-CH(R)-COO-. Because of (+) and (-) charge they exist as solids and form salts.
(c) Water soluble vitamins can not retain in body for longer time. If someone takes high diet of vitamin B and C they would not be as harmful as they dissolve in water and excrete outside of body.
(d) Two DNA strands are complementary because they are connected with each other through base pair C,G,T by hydrogen bonding and run parallel to each other (5'->3' and 3'->5').
2025 · 3 marks
(a) What is the difference between native protein and denatured protein?
(b) Which one of the following is a disaccharide? Glucose, Lactose, Amylose, Fructose.
(c) Which vitamin is responsible for the coagulation of blood?
Answer(a) Protein found in a biological system with unique three-dimensional structure and biological activity is called native protein. When a protein in its native form is subjected to changes in temperature, etc, its structure is destroyed and it loses its biological activity. The protein thus formed is called denatured protein.
(b) Lactose.
(c) Vitamin K.
2025 · 3 marks
(a) Write the reaction when D-glucose reacts with the following:
(ii) D-glucose + 5(CH3CO)2O -> Glucose penta acetate + 5CH3COOH.
(b) Vitamin C cannot be stored in large amounts in our body because it is water-soluble, meaning it dissolves in water. As a result, the body excretes excess vitamin C through urine.
2025 · 3 marks
Define the following terms:
(a) Native protein
(b) Nucleotide
(c) Essential amino acid
Answer(a) A native protein is a protein that has folded into its unique three-dimensional structure under physiological conditions and possesses its specific biological activity.
(b) A nucleotide is the basic building block of nucleic acids (DNA and RNA). It consists of three components: a phosphoric acid group, a pentose sugar and a nitrogenous base.
(c) Essential amino acids are amino acids that the body can't make, so they must be supplied by diet. The nine essential amino acids are: Histidine, Isoleucine, Leucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Valine.
2025 · 3 marks
Define the following terms:
(a) Glycosidic linkage
(b) Invert sugar
(c) Oligosaccharides
Answer(a) A glycosidic linkage (or glycosidic bond) is a type of ethereal linkage that joins a sugar molecule (monosaccharide) to another molecule. This bond is crucial in the formation of complex carbohydrates like disaccharides, oligosaccharides and polysaccharides.
(b) Invert sugar is a mixture of glucose and fructose produced by the hydrolysis of sucrose (table sugar) into its component monosaccharides. This process is called inversion because the optical rotation of the sugar changes direction.
(c) Oligosaccharides are a class of carbohydrates that consist of a few monosaccharide units (typically 2 to 10 units) joined together by glycosidic bonds. They are intermediate in size between disaccharides and polysaccharides.
2025 · 3 marks
(a) Write the product obtained when D-glucose reacts with H2N-OH.
(b) As amino acids contain both basic -NH2 group and acidic -COOH group, they are said to be amphoteric.
(c) Vitamin C is water-soluble so it cannot be stored in the body. Excess amount of vitamin C is excreted through urine.
2025 · 3 marks
Define the following terms:
(a) Native protein
(b) Nucleotide
(c) Essential amino acid
Answer(a) The sequence in which amino acids are linked together with the help of peptide bonds determines the native state of a protein. A native protein is a protein that has folded into its unique three-dimensional structure under physiological conditions and possesses its specific biological activity.
(b) A nucleotide is the basic building block of nucleic acids (DNA and RNA). It consists of three components: a phosphoric acid group, a pentose sugar and a nitrogenous base.
(c) Essential amino acids are amino acids that the body can't make, so they must be supplied by diet. The nine essential amino acids are: Histidine, Isoleucine, Leucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Valine.
2024 · 3 marks
Explain the following terms:
(a) Essential amino acids
(b) Peptide bond
(c) Denaturation
Answer(a) Essential amino acids are a group of amino acids that cannot be synthesized by the human body in sufficient amounts or at all. There are nine essential amino acids: Histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan and valine.
(b) A peptide bond is a type of covalent bond that forms between the amino group (-NH2) of one amino acid and the carboxyl group (-COOH) of another amino acid, resulting in the formation of a peptide linkage.
(c) Denaturation refers to the process by which the native structure of a biological molecule, such as a protein or nucleic acid, is disrupted, leading to the loss of its biological activity. It can occur due to heat, pH extremes, organic solvents, detergents, and mechanical agitation.
2024 · 3 marks
Define the following terms:
(a) Non-essential amino acids
(b) Monosaccharides
(c) Anomers
Answer(a) Non-essential amino acids: The amino acids which can be synthesized in the body are known as non-essential amino acids. For e.g., glycine, alanine, glutamine, etc.
(b) Monosaccharides: Monosaccharides include non-hydrolysable carbohydrates. They are soluble in water. Those containing an aldehydic group are called aldoses while others containing a ketonic group are called ketoses.
(c) Anomers: The isomers which differ only in the configuration of the hydroxyl group at C-1 are called anomers and are referred to as alpha and beta-forms.
2024 · 3 marks
Define the following terms:
(a) Glycosidic linkage
(b) Primary structure of protein
(c) Disaccharides
Answer(a) Glycosidic linkage: Linkage between two monosaccharides units through oxygen atom is called glycosidic linkage.
(b) Primary structure of protein: The specific sequence of linkage of amino acids in each polypeptide chain of the protein is called primary structure of protein.
(c) Disaccharides: The sugar formed when two monosaccharides units joined together through a glycosidic linkage is called disaccharide.
2013 · 3 marks
(a) Which one of the following is a food preservative? Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
Answer(a) Sodium benzoate is a food preservative.
(b) Bithional is added to soap as an antiseptic to reduce odour produced by bacterial decomposition of organic matter on the skin.
(c) Tranquilizers (barbiturates) are the class of drugs used in sleeping pills.
2014 · 3 marks
(i) Give two examples of macromolecules that are chosen as drug targets.
(ii) What are antiseptics? Give an example.
(iii) Why is use of aspartame limited to cold foods and soft drinks?
Answer(i) Proteins and nucleic acids (DNA) are macromolecules chosen as drug targets.
(ii) Antiseptics are chemical substances that prevent the growth of micro-organisms and are safe to apply on living tissues. Example: Dettol (mixture of chloroxylenol and terpineol).
(iii) Aspartame is unstable at high temperatures; it decomposes on heating, so its use is limited to cold foods and soft drinks.
2014 · 3 marks
(i) Name the sweetening agents used in the preparation of sweets for a diabetic patient.
(ii) What are antibodies? Give an example.
(iii) Give two examples of macromolecules that are chosen as drug targets.
Answer(i) Aspartame and saccharin are used as sweetening agents for diabetic patients.
(ii) Antibodies are proteins produced by the immune system in response to foreign substances (antigens). Example: Immunoglobulin (IgG).
(iii) Proteins and nucleic acids are macromolecules chosen as drug targets.
2017 · 3 marks
Define the following:
(i) Anionic detergents
(ii) Broad spectrum antibiotics
(iii) Antiseptic
Answer(i) Anionic detergents are sodium salts of sulfonated long chain alcohols or hydrocarbons. The anionic part (long chain) is involved in cleansing action.
(ii) Broad spectrum antibiotics are those antibiotics which are effective against a wide range of gram-positive and gram-negative bacteria. E.g., Chloramphenicol.
(iii) Antiseptics are chemical substances which prevent the growth of micro-organisms and may even kill them. They are applied to living tissues. E.g., Dettol.
2017 · 3 marks
Define the following:
(i) Cationic detergents
(ii) Narrow spectrum antibiotics
(iii) Disinfectants
Answer(i) Cationic detergents are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions. The cationic part has a long hydrocarbon chain and a positive charge.
(ii) Narrow spectrum antibiotics are effective against a limited range of bacteria (either gram-positive or gram-negative).
(iii) Disinfectants kill micro-organisms but are not safe for living tissues. They are applied to inanimate objects. E.g., 1% phenol solution.
2017 · 3 marks
Define the following:
(i) Anionic detergents
(ii) Limited spectrum antibiotics
(iii) Tranquilizers
Answer(i) Anionic detergents are sodium salts of sulfonated long chain alcohols or hydrocarbons. The anionic part is involved in cleansing.
(ii) Limited spectrum antibiotics are effective only against specific types of bacteria (gram-positive or gram-negative, not both).
(iii) Tranquilizers are chemical compounds used for treatment of stress and mild or severe mental diseases. They reduce anxiety. E.g., Equanil.
2017 · 3 marks
Define the following:
(a) Anionic detergents
(b) Narrow spectrum antibiotics
(c) Antacids
Answer(a) Anionic detergents: Sodium salts of sulfonated long chain alcohols/hydrocarbons.
(b) Narrow spectrum antibiotics: Effective against limited range of bacteria.
(c) Antacids: Substances that neutralize excess acid in the stomach. E.g., Ranitidine, Cimetidine.
2018 · 3 marks
(a) Why is bithional added to soap?
(b) What is tincture of iodine? Write its one use.
(c) Among the following, which one acts as a food preservative? Aspartame, Aspirin, Sodium benzoate, Paracetamol
Answer(a) Bithional is added to soap as an antiseptic to impart antiseptic properties.
(b) Tincture of iodine is 2-3% solution of iodine in alcohol-water mixture. Used as an antiseptic for wounds.
(c) Sodium benzoate acts as a food preservative.
2014 · 3 marks
(a) Draw the structure of major monohalo products in each of the following reactions:
(i) C6H5CH2OH ⟶[PCl5]
(ii) C6H5CH2CH=CH2 + HBr →
(b) Which halogen compound in each of the following pairs will react faster in SN2 reaction?
(ii) C6H5CH2CH=CH2 + HBr → C6H5CH2CHBrCH3 (Markovnikov addition)
(b)
(i) CH3I reacts faster because I− is a better leaving group than Br− (weaker C–I bond).
(ii) CH3Cl reacts faster in SN2 as it is a primary halide with less steric hindrance than (CH3)3CCl (tertiary).
2015 · 3 marks
Give reasons:
(a) n-Butyl bromide has higher boiling point than t-butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (–NO2) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer(a) n-Butyl bromide has a larger surface area, leading to stronger van der Waals' forces and higher boiling point.
(b) Racemic mixture contains equal amounts of two enantiomers. Rotation due to one enantiomer is cancelled by the other, making it optically inactive.
(c) –NO2 acts as electron withdrawing group (−I effect). At o/p positions, it stabilizes the negative charge in the transition state during nucleophilic attack, increasing reactivity.
2016 · 3 marks
Give reasons:
(i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH3-Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) SN1 reactions are accompanied by racemization in optically active alkyl halides.
Answer(i) In chlorobenzene, each carbon atom is sp2 hybridized and due to resonance (+R effect), there is partial double bond character in C-Cl bond making it shorter. In CH3Cl, carbon is sp3 hybridized.
(ii) The dipole moment of chlorobenzene is lower because +R effect of Cl and -I effect oppose each other while in cyclohexyl chloride only -I effect is present. Also sp2 vs sp3 hybridization difference.
(iii) In SN1 reaction, a planar carbocation is formed (sp2 hybridized). The nucleophile can attack from either side equally, leading to formation of racemic mixture.
(c) Methylenecyclohexane or 1-methylcyclohexene (Saytzeff product, more substituted alkene is major).
2019 · 3 marks
(i) Out of (CH3)3C-Br and (CH3)3C-I, which one is more reactive towards SN1 and why?
(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(iii) Why dextro and laevo rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?
Answer(i) (CH3)3C-I, due to large size of iodine/better leaving group/lower electronegativity.
(ii) p-Nitrophenol.
(iii) Because enantiomers have identical physical properties such as boiling points, so they cannot be separated by fractional distillation.
2019 · 3 marks
Among all isomers of C4H9Br, identify:
(a) the optically active isomer.
(b) the isomer highly reactive towards SN2.
(c) two isomers giving same product on dehydrohalogenation with alcoholic KOH.
Answer(a) CH3CH2CH(Br)CH3 (2-bromobutane, has chiral carbon).
(b) CH3CH2CH2CH2Br (1-bromobutane, primary, least steric hindrance).
(c) (CH3)3CBr and (CH3)2CHCH2Br both give 2-methylpropene.
2020 · 3 marks
Identify A, B, C, D, E and F in the following: CH3CH(CH3)CH2Br --alc.KOH → A --HBr → B --Na/dry ether → C; CH3CH(CH3)CH2Br --Mg/dry ether → D --H2O → E; CH3CH(CH3)CH2Br --NaOC2H5 → F
AnswerA is CH3-C(=CH2)-CH3 (2-methylpropene), B is CH3-C(Br)(CH3)-CH3 (2-bromo-2-methylpropane), C is CH3-C(CH3)2-C(CH3)2-CH3 (2,2,3,3-tetramethylbutane via Wurtz), D is CH3CH(CH3)CH2MgBr (Grignard reagent), E is CH3CH(CH3)CH3 (2-methylpropane), F is CH3CH(CH3)CH2OC2H5 (ethyl isobutyl ether).
2020 · 3 marks
(i) Write the structure of major alkene formed by beta-elimination of 2,2,3-trimethyl-3-bromopentane with sodium ethoxide in ethanol.
(ii) Which one of the compounds in the following pairs is chiral? (Br on secondary carbon vs Br on primary carbon)
(b) Why is chlorobenzene resistant to nucleophilic substitution reaction?
Answer(a)
(i) Enantiomers are molecules that are non-superimposable mirror images of each other. They arise from chirality and have the unique property of rotating polarised light in opposite directions.
(ii) A racemic mixture is a mixture that contains equal amounts of two enantiomers of a chiral molecule. It is a 1:1 mixture of two mirror-image molecules that are non-superimposable (enantiomers).
(b) In chlorobenzene, the halogen atom is bonded to the highly electronegative sp2 hybridised carbon atom and lone pair of halogen taking part in resonance with benzene ring. So, nucleophilic substitution reaction is not possible.
2024 · 3 marks
Account for the following:
(a) Haloalkanes react with AgCN to form isocyanide as main product.
(b) Allyl chloride shows high reactivity towards SN1 reaction.
(c) Haloarenes are extremely less reactive towards nucleophilic substitution reactions.
Answer(a) Cyanide ion is an ambident ion and it reacts with alkyl halides to form isocyanides. AgCN is covalent and replaces the halogen via SN2 mechanism, when the lone pair of nitrogen in AgCN is attached to the carbon bearing the halogen yielding isocyanide.
(b) Allyl chloride undergoes heterolysis to form a carbocation. This carbocation is resonance stabilised and is stable. Hence SN1 mechanism that involves the formation of a carbocation in the transition state is favoured.
(c) Halogens are highly electronegative and draw the delocalized pi cloud of the benzene ring towards itself, which makes the ring electron deficient and do not undergo electrophilic attack.
2024 · 3 marks
Account for the following:
(a) Haloalkanes react with NaCN to form both cyanides and isocyanides.
(b) Haloarenes do not undergo nucleophilic substitution reaction easily.
(c) Benzyl chloride gives SN1 reaction.
Answer(a) On reacting with NaCN, haloalkane form cyanide and isocyanide because of ionic character of NaCN.
(b) The partial double bond nature of C-X bond in a haloarene is resonance stabilised due to presence of aryl group than a haloalkane. Therefore, haloarenes can't be cleaved by a nucleophile easily and hence they are less reactive towards nucleophilic substitution reactions.
(c) Benzyl chloride gives SN1 reaction because the intermediate carbocation formed in slowest step is stabilised through resonance.
2024 · 3 marks
(a) Write the IUPAC name of the given compound (cyclohexane with Cl and Br substituents).
(b) The presence of -NO2 group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions. Give reason to explain the above statement.
(c) What happens when ethyl chloride is treated with alcoholic potassium hydroxide?
Answer(a) 3-Bromo-1-chlorocyclohexene
(b) The nitro group present on ortho or para position, increases the reactivity due to increased delocalization of negative charge involving nitro group. The nitro group present at ortho or para position withdraws the electron density from the benzene ring, facilitating the nucleophile attack. The carbanion formed is stabilised through resonance and NO2 group.
(c) When ethyl chloride (C2H5Cl) reacts with alcoholic potassium hydroxide (alc. KOH), it undergoes an elimination reaction known as the E2 (bimolecular elimination) reaction. C2H5Cl + alc. KOH -> C2H4 + KCl + H2O. Ethene is formed.
2024 · 3 marks
(a) Write the IUPAC name of the given compound (cyclohexane with Cl and Br substituents).
(b) The presence of -NO2 group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions. Give reason.
(c) What happens when ethyl chloride is treated with alcoholic potassium hydroxide?
Answer(a) 3-Bromo-1-chlorocyclohexene
(b) The nitro group present at ortho or para position increases the reactivity due to increased delocalization of negative charge involving nitro group, withdrawing electron density from the benzene ring, facilitating the nucleophile attack.
(c) C2H5Cl + alc. KOH -> C2H4 + KCl + H2O. Ethene is formed through E2 elimination.
2024 · 3 marks
(a) Write the IUPAC name of the given compound (benzene with Br and Cl substituents).
(b) Why are haloalkanes more reactive towards nucleophilic substitution reactions than haloarenes?
(c) What happens when ethyl chloride is treated with aqueous KOH?
Answer(a) 1-bromo-4-chlorobenzene
(b) Haloalkanes are generally more reactive towards nucleophilic substitution reactions as compared to haloarenes because the carbon-halogen bond is generally weaker in haloalkanes as compared to haloarenes. The C-X bond in aryl halide acquires a partial double bond character due to resonance.
(c) When ethyl chloride is treated with aqueous KOH it undergoes an elimination reaction known as dehydrohalogenation: CH3CH2Cl + aq. KOH -> CH3-CH2-OH + KCl
2024 · 3 marks
(a) Write the IUPAC name of the given compound: CH2=CH-CH2-CH2-Cl
(b) Why is thionyl chloride preferred for preparing alkyl halides from alcohols?
(c) What happens when methyl bromide reacts with KCN?
Answer(a) 1-Chloro-But-3-ene
(b) Thionyl chloride is preferred for the preparation of alkyl halide from alcohol because the byproducts formed in this reaction are gases (SO2 and HCl). These are easily separated from the desired product.
(c) CH3Br + KCN -> CH3CN + KBr. Methyl cyanide or acetonitrile forms which on hydrolysis gives carboxylic acid.
2023 · 3 marks
Answer any three of the following:
(a) Which isomer of C5H10 gives a single monochloro compound C5H9Cl in bright sunlight?
(b) Arrange the following compounds in increasing order of reactivity towards SN2 reaction: 2-Bromopentane, 1-Bromopentane, 2-Bromo-2-methylbutane
(c) Why p-dichlorobenzene has higher melting point than those of ortho and meta-isomers?
(d) Identify A and B in the following: (Br-cyclopentane) --(Mg/Dry ether) → A --(H2O) → B
Answer(a) The hydrocarbon is either alkene or cycloalkane. It does not react with chlorine in dark. Hence it cannot be alkene. Hence, it is cycloalkene. The compound is cyclopentane (C5H10).
(c) It is due to greater symmetry of para-isomer that fits in the crystal better as compared to o- & m-isomers.
(d) A -> Cyclobutyl magnesium bromide (MgBr). B -> 1-cyclobutyl ethanol (CH3-C(OH)(H)-cyclobutyl).
2013 · 3 marks
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Neoprene
(iii) Nylon-6,6
Answer(i) Buna-S: Monomers are 1,3-butadiene (CH2=CH–CH=CH2) and Styrene (C6H5CH=CH2)
(ii) Neoprene: Monomer is Chloroprene (2-chloro-1,3-butadiene, CH2=C(Cl)–CH=CH2)
(iii) Nylon-6,6: Monomers are Hexamethylenediamine (H2N–(CH2)6–NH2) and Adipic acid (HOOC–(CH2)4–COOH)
2014 · 3 marks
After the ban on plastic bags, students of a school decided to make the people aware of the harmful effects of plastic bags on the environment and Yamuna River. They organized a rally and distributed paper bags.
(i) What values are shown by the students?
(ii) What are bio-degradable polymers? Give one example.
(iii) Is polythene a condensation or the addition polymer?
Answer(i) Environmental awareness, social responsibility, leadership, and civic duty.
(ii) Bio-degradable polymers are polymers that can be decomposed by bacteria/microorganisms. Example: PHBV (poly β-hydroxybutyrate-co-β-hydroxy valerate) or Nylon-2-nylon-6.
(iii) Polythene is an addition polymer (formed by addition polymerization of ethene).
2015 · 3 marks
Write the names and structures of the monomers of the following polymers:
(i) Nylon-6,6
(ii) PHBV
(iii) Neoprene
Answer(i) Nylon-6,6: Hexamethylenediamine (H2N(CH2)6NH2) and Adipic acid (HOOC(CH2)4COOH)
(ii) PHBV: 3-hydroxybutanoic acid and 3-hydroxypentanoic acid
(ii) Melamine-formaldehyde: Melamine and Formaldehyde (HCHO)
(iii) Buna-S: 1,3-Butadiene (CH2=CH-CH=CH2) and Styrene (C6H5CH=CH2)
2017 · 3 marks
Write the structures of the monomers used for getting the following polymers:
(i) Nylon-6
(ii) Melamine-formaldehyde polymer
(iii) Teflon
Answer(i) Nylon-6: Caprolactam
(ii) Melamine-formaldehyde: Melamine and Formaldehyde (HCHO)
(iii) Teflon: Tetrafluoroethylene (CF2=CF2)
2017 · 3 marks
Write the structures of the monomers used for getting the following polymers:
(a) Polyvinyl chloride (PVC)
(b) Melamine-formaldehyde polymer
(c) Buna-N
Answer(a) PVC: Vinyl chloride (CH2=CHCl)
(b) Melamine-formaldehyde: Melamine and Formaldehyde (HCHO)
(c) Buna-N: 1,3-Butadiene and Acrylonitrile (CH2=CHCN)
2025 · 4 marks
Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o- and p-positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols. Answer the following:
(a) What happens when phenol reacts with
(i) Br2/CS2
(ii) Conc. HNO3
(b) Why does phenol not undergo protonation readily?
(c) Which is a stronger acid - phenol or cresol? Give reasons.
OR
(c) Write the IUPAC name of the product formed in the Reimer-Tiemann reaction.
Answer(a)
(i) Phenol + Br2/CS2 at 273K gives 4-Bromophenol (major) and 2-Bromophenol (minor).
(ii) Phenol + conc. HNO3 gives 2,4,6-Trinitrophenol (Picric acid).
(b) One lone pair of oxygen in phenol is involved in resonance with benzene ring, thus generating a positive charge on oxygen. It will not easily undergo protonation.
(c) Phenol is a stronger acid. In cresol, due to +I effect and hyperconjugation effect of methyl group, the density of electron towards oxygen atom increases, destabilising the conjugate base. While in phenol, the phenoxide ion is resonance stabilised and that makes it a stronger acid than cresol.
OR
(c) The product is salicylaldehyde: IUPAC: 2-Hydroxybenzaldehyde.
2023 · 4 marks
The carbon-oxygen double bond is polarised in aldehydes and ketones due to higher electronegativity of oxygen relative to carbon. Therefore they undergo nucleophilic addition reactions with a number of nucleophiles such as HCN, NaHSO3, alcohols, ammonia derivatives and Grignard reagents. Aldehydes are easily oxidised by mild oxidising agents as compared to ketones. The carbonyl group of carboxylic acid does not give reactions of aldehydes and ketones. Carboxylic acids are considerably more acidic than alcohols and most of simple phenols. Answer the following:
(a) Write the name of the product when an aldehyde reacts with excess alcohol in presence of dry HCl.
(b) Why carboxylic acid is a stronger acid than phenol?
(c)
(i) Arrange the following compounds in increasing order of their reactivity towards CH3MgBr: CH3CHO, (CH3)3C-C(=O)-CH3, CH3-C(=O)-CH3
(ii) Write the product: C6H4(CHO)(O) + NH2CONHNH2 -> ?
Answer(a) It yields alkoxyalcohol intermediate, known as hemiacetals which further reacts to give gem-dialkoxy compound known as acetal.
(b) Due to the resonance in carboxylic acids, the negative atom (oxygen atom), whereas, in alcohol or phenols, the negative charge is on the less electronegative atom.
(c)
(i) (CH3)3C-C(=O)-CH3 < CH3-C(=O)-CH3 < CH3CHO
(ii) Semicarbazone is formed: C6H5-C=N-NH-CO-NH2 + H2O
(ii) Write a chemical test to distinguish between propanal and propanone. (OR)
Answer(i) The -CHO group is selectively oxidized by Tollen's reagent to give carboxylate salt.
(ii) Iodoform test: Propanone gives iodoform test (yellow precipitate of CHI3 with I2/NaOH) while propanal does not, as propanone has a CH3CO- group.
2025 · 4 marks
Amines have a lone pair of electrons on nitrogen atom due to which they behave as Lewis base. [Passage about basicity of amines]. Answer the following:
(a) Arrange the following in the increasing order of their basic character. Give reason: p-nitroaniline, aniline, p-methylaniline.
(b) Why pKb of aniline is more than that of methylamine? (c)
(i) Arrange the following in the increasing order of their basic character in an aqueous solution: (CH3)3N, (CH3)2NH, NH3, CH3NH2.
OR (c)
(ii) Why ammonolysis of alkyl halides is not a good method to prepare pure amines?
Answer(a) p-nitroaniline < aniline < p-methylaniline (increasing basicity). Basicity increases with electron donating groups and decreases with electron withdrawing groups.
(b) In aniline, the lone pair of electrons on the N atom is delocalised over the benzene ring. As a result, the electron density on nitrogen decreases. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher. (c)
(ii) The primary amine formed can react with the alkyl halide to form the secondary amine, which in turn will again react with the alkyl halide to form the tertiary amine, which also reacts with the alkyl halide leading to the formation of quaternary salt. So a mixture of products is formed.
2016 · 4 marks
Due to hectic and busy schedule, Mr. Angad made his life full of tensions and anxiety. He started taking sleeping pills without consulting the doctor. Mr. Deepak advised him to stop and suggested yoga, meditation and physical exercises.
(a) What are the values (at least two) displayed by Mr. Deepak?
(b) Why is it not advisable to take sleeping pills without consulting doctor?
(c) What are tranquilisers? Give two examples.
Answer(a) Values: Caring, responsible, health awareness, true friendship.
(b) Sleeping pills can be habit-forming (addictive) and can have serious side effects including drowsiness, dependence, and even overdose.
(c) Tranquilisers are drugs used for the treatment of stress, anxiety, and mental diseases. Examples: Valium (Diazepam), Chlordiazepoxide (Librium).
2016 · 4 marks
Due to hectic schedule, Mr. Singh started taking junk food and eating irregularly. He felt severe chest pain. Mr. Khanna took him to doctor who diagnosed acidity. Mr. Khanna advised home food, yoga and meditation.
(i) What are the values displayed by Mr. Khanna?
(ii) What are antacids? Give one example.
(iii) Would it be advisable to take antacids for a long period of time? Give reason.
Answer(i) Values: Caring, responsibility, awareness about health, true friendship.
(ii) Antacids are substances that neutralize excess acid in the stomach. Example: Ranitidine / Aluminium hydroxide / Milk of magnesia.
(iii) No, long-term use can cause rebound acidity, kidney problems and mineral absorption issues.
2017 · 4 marks
After watching a programme on TV about the presence of carcinogens (cancer causing agents) potassium bromate and potassium iodate in bread and other bakery products, Ritu a class XII student decided to aware others about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct canteen contractor to stop selling sandwiches, pizza, burgers and other bakery products to the students. Principal took an immediate action and instructed the canteen contractor to replace the bakery products with some proteins and vitamins rich food like fruits, salads, sprouts etc. The decision was welcomed by the parents and students. After reading the above passage, answer the following questions:
(i) What are the values (at least two) displayed by Ritu?
(ii) Which polysaccharide component of carbohydrates is commonly present in bread?
(iii) Write the two types of secondary structure of proteins.
(iv) Give two examples of water soluble vitamins.
Answer(i) Concerned, caring, socially alert, leadership (or any other 2 values).
(ii) Starch.
(iii) α-Helix and β-pleated sheets.
(iv) Vitamin B / B1 / B2 / B6 / C (any two).
2023 · 4 marks
Carbohydrates are optically active polyhydroxy aldehydes and ketones. They are also called saccharides. All those carbohydrates which reduce Fehling's solution and Tollen's reagent are referred to as reducing sugars. Glucose, the most important source of energy for mammals, is obtained by the hydrolysis of starch. Vitamins are accessory food factors required in the diet. Proteins are the polymers of alpha-amino acids and perform various structural and dynamic functions in the organisms. Deficiency of vitamins leads to many diseases. Answer the following:
(a) The penta-acetate of glucose does not react with Hydroxylamine. What does it indicate?
(b) Why cannot vitamin C be stored in our body?
(c) Define the following as related to proteins:
(i) Peptide linkage
(ii) Denaturation
Answer(a) It indicates the absence of free -CHO group.
(b) As it is water-soluble. It is hard to be stored in our body. It is usually secreted via urine.
(c)
(i) Peptide linkage is an amide formed between two amino acid known as peptide bond. NH2-CH-CO-NH-C-CO-NH-C
(ii) Denaturation: The unfolding or breaking up of a protein converted into primary structure in which linear chains of amino acids are attached.
2023 · 4 marks
Q32 (c) OR: Define the following as related to carbohydrates:
(i) Anomers
(ii) Glycosidic linkage (OR)
Answer(i) Anomers are cyclic monosaccharides or glycosides that are epimers, differing from each other in the configuration of C-2 if they are ketoses.
(ii) The two monosaccharides joined together by a glycosidic linkage formed by the loss of a water molecule for eg C1 of alpha-D-glucose and C2 of beta-D-fructose are held together by glycosidic linkage.
2024 · 4 marks
Certain organic compounds are required in small amounts in our diet but their deficiency causes specific disease. These compounds are called vitamins. Based on this:
(a) What is the other name of vitamin B6?
(b) Name the vitamin whose deficiency causes increased blood clotting time.
(c) Xerophthalmia is caused by the deficiency of which vitamin? Give two sources of this vitamin.
OR
(c) Why can't vitamin C be stored in our body? Name the disease caused by the deficiency of this vitamin.
Answer(a) Pyridoxine
(b) Vitamin K
(c) Vitamin A deficiency. Sources: Fish, dairy, eggs.
OR
(c) Since vitamin C is soluble in water, so it dissolves in water and can be eliminated out from the body through urine and hence cannot be stored. Deficiency causes scurvy (dry, brittle and coiled hair, gum bleeding).
2015 · 4 marks
Mr. Roy, the principal of one reputed school organized a seminar to discuss the serious issue of diabetes and depression in students. They resolved to ban junk food and introduce healthy snacks. After six months, Mr. Roy conducted a health survey and discovered a tremendous improvement in health of students.
(i) What are the values (at least two) displayed by Mr. Roy?
(ii) As a student, how can you spread awareness about this issue?
(iii) What are tranquilisers? Give an example.
(iv) Why is use of aspartame limited to cold foods and drinks?
Answer(i) Values: Awareness about health, caring for students' wellbeing, proactive leadership, social responsibility.
(ii) By organizing awareness campaigns, distributing pamphlets, and educating peers about the harmful effects of junk food.
(iii) Tranquilisers are chemical compounds used for the treatment of stress, anxiety, and mental diseases. They are neurologically active drugs. Example: Valium (diazepam), chlordiazepoxide.
(iv) Aspartame decomposes at high temperature and becomes unstable, so it is limited to cold foods and drinks.
2023 · 4 marks
Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both SN1 and SN2 mechanisms. SN1 is a two step reaction while SN2 is a single step reaction. For any haloalkane which mechanism is followed depends on factors such as structure of haloalkane, properties of leaving group, nucleophilic reagent and solvent. Influences of solvent polarity: In SN1 reaction, the polarity of the system increases from the reactant to the transition state, because a polar solvent has a greater effect on the transition state than the reactant, thereby reducing activation energy and accelerating the reaction. In SN2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. Answer the following questions:
(a) Why racemisation occurs in SN1?
(b) Why is ethanol less polar than water?
(c) Which one of the following in each pair is more reactive towards SN2 reaction?
(i) CH3-CH2-I or CH3CH2-Cl
(ii) Chlorocyclohexane or (Chloromethyl)cyclohexane
Answer(a) In SN1 mechanism, there is intermediate carbocation formed. Due to which racemization of the product takes place or D and L form formed.
(b) In ethanol alkyl chain is present which is responsible for non-polar nature. Water has high value of dipole moment than ethanol. That's why water is more polar than ethanol.
(c)
(i) CH3-CH2-I will react faster than CH3-CH2-Cl because I is bigger in size and more polarized atom. Its bond dissociation enthalpy is less so it easily reacts with other substances.
(ii) (Chloromethyl)cyclohexane is more reactive compared to chlorocyclohexane because in chlorocyclohexane bond length is less and it is closely attached with cyclohexane while 1-methyl-1-chlorocyclohexane is less stable, one extra methyl group is attached which makes it more reactive towards SN1 mechanism.
2023 · 4 marks
31 (OR) (c): Arrange the following in the increasing order of their reactivity towards SN1 reactions:
(ii) 1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane
2018 · 4 marks
Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags. Answer the following:
(a) Write the values (at least two) shown by Shyam.
(b) Write one structural difference between low-density polythene and high-density polythene.
(c) Why did Shyam refuse to accept the items in polythene bags?
(d) What is a biodegradable polymer? Give an example.
(b) Low-density polythene (LDPE) has highly branched chain structure; High-density polythene (HDPE) has linear chain structure with little branching.
(c) Because polythene bags are non-biodegradable, cause environmental pollution, and their use is banned by the government with heavy penalty.
(d) Biodegradable polymer: A polymer that is degraded by microorganisms in a reasonable time. E.g., PHBV (poly-β-hydroxybutyrate-co-β-hydroxyvalerate), PGA.
2017 · 5 marks
(a) Write the product(s) in the following reactions:
(ii) Ethanol --(PCC) → Acetaldehyde --(CH3MgBr, then H3O+) → Propan-2-ol.
(b) Step 1 (fast): Protonation of ethanol to form oxonium ion.
Step 2 (slow): Loss of water to form carbocation.
Step 3: Loss of H+ to form ethene.
(c) The -OH group in phenol is highly activating. The lone pair on oxygen is delocalized into the benzene ring, making phenol more electron rich than benzene, facilitating electrophilic substitution.
2019 · 5 marks
(a) Account for the following:
(i) o-nitrophenol is more steam volatile than p-nitrophenol.
(ii) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butylmethylether.
(b) Write the reaction involved in:
(i) Reimer-Tiemann reaction
(ii) Friedel-Crafts alkylation of phenol.
(c) Give simple chemical test to distinguish between ethanol and phenol.
Answer(a)
(i) o-Nitrophenol has intramolecular H-bonding; p-nitrophenol has intermolecular H-bonding making it less volatile.
(ii) t-butyl chloride is tertiary so elimination is favored over substitution with strong base CH3O-. (b)
(i) Phenol + CHCl3 + 3KOH -> Salicylaldehyde.
(ii) Phenol + RCOCl --(AlCl3) → C-acylation and O-acylation products.
(c) Add neutral FeCl3: Phenol gives violet coloration, ethanol does not. Or iodoform test: ethanol gives yellow ppt of CHI3 with NaOH/I2, phenol does not.
2019 · 5 marks
(a) Give equations:
(i) Phenol + conc. HNO3.
(ii) Propene + B2H6 followed by H2O2/OH-.
(iii) Sodium t-butoxide + CH3Cl.
(b) Distinguish between butan-1-ol and butan-2-ol.
(c) Arrange in increasing order of acidity: Phenol, Ethanol, Water.
(c) Kolbe's: Phenol + NaOH -> Na-phenoxide --(CO2, high T/P then H+) → Salicylic acid.
2025 · 5 marks
(A) An organic compound 'A', molecular formula C2H6O oxidises with CrO3 to form a compound 'B'. Compound 'B' on warming with iodine and aqueous solution of NaOH gives a yellow precipitate of compound 'C'. When compound 'A' is heated with conc. H2SO4 at 413 K gives a compound 'D', which on reaction with excess HI gives compound 'E'. Identify compounds 'A', 'B', 'C', 'D' and 'E' and write chemical equations involved.
OR (B)(a) Write chemical equations of the following reactions:
(i) Phenol is treated with conc. HNO3
(ii) Propene is treated with B2H6 followed by oxidation by H2O2/OH-.
(iii) Sodium t-butoxide is treated with CH3Cl.
(b) Give a simple chemical test to distinguish between butan-1-ol and butan-2-ol.
(c) Arrange the following in increasing order of acid strength: phenol, ethanol, water.
(b) Lucas test: Butan-1-ol (primary) reacts very slowly, while butan-2-ol (secondary) reacts more quickly with Lucas reagent showing turbidity within a few minutes.
(c) Ethanol (C2H5OH) < Water (H2O) < Phenol (C6H5OH). Phenol is strongest acid because its conjugate base is stabilised by resonance.
2013 · 5 marks
(a) Although phenoxide ion has more number of resonating structures than carboxylate ion, Carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring about the following conversions?
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal
Answer(a)
(i) In phenoxide resonating structures, only structures I and IV carry the negative charge. It is negligibly contributed in the stability of phenoxide ion. Negative charge is present on one oxygen atom.
(ii) In carboxylate ion, negative charge delocalized on both oxygen atoms which is highly electronegative and contributed more in stability of resonating structure of carboxylate ion.
(i) Ethanal and Propanal: Both give positive Tollen's test. To distinguish, use Iodoform test – ethanal gives positive iodoform test (yellow ppt of CHI3) but propanal does not.
(ii) Benzoic acid and Phenol: Ferric chloride test – Phenol gives violet colour with FeCl3 while benzoic acid does not.
2014 · 5 marks
(a) Write the products of the following reactions:
(i) Cyclohexanone + NH2OH →
(ii) 2C6H5CHO + Conc. NaOH →
(iii) CH3COOH ⟶[Cl2/P]
(b) Write the chemical equations to illustrate the following name reactions:
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
(ii) Carboxylic acid is a stronger acid than phenol.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro's reaction
(c) Out of CH3CH2–CO–CH3 and CH3CH2–CH2–CO–CH3, which gives iodoform test?
Answer(a)
(i) In acetone, alkyl chain is present on both sides of carbonyl group causing steric hindrance and −I effect making it less reactive towards nucleophilic attack. Acetaldehyde does not have this, making it more reactive.
(ii) Carboxylic acid is more acidic because of its stabilising resonating structure of carboxylate ions. Carboxylic acid has two electronegative oxygen atoms, while in phenol the acidic character is due to presence of phenoxide ions with one oxygen atom.
(c) CH3CH2–CO–CH3 gives iodoform test because it has the CO–CH3 group. CH3CH2–CH2–CO–CH3 also gives iodoform test. Both give positive iodoform test as they contain CO–CH3 group.
2014 · 5 marks
(a) Write the products formed when CH3CHO reacts with the following reagents:
(i) HCN
(ii) H2N–OH
(iii) CH3CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ferric chloride test: Phenol gives violet colour with FeCl3; benzoic acid does not.
(ii) Iodoform test: Propanone gives positive iodoform test (yellow ppt of CHI3) with NaOH + I2. Propanal does not give iodoform test.
2014 · 5 marks
(a) Account for the following:
(i) Cl–CH2COOH is a stronger acid than CH3COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro's reaction
(c) Out of CH3CH2–CO–CH3 and CH3CH2–CH2–CO–CH3, which gives iodoform test?
Answer(a)
(i) ClCH2COOH is stronger acid than CH3COOH because Cl is an electron withdrawing group (−I effect) which disperses electron density away from the O–H bond, making release of H+ easier.
(ii) Carboxylic acids contain carbonyl group but do not undergo nucleophilic addition because the lone pair of oxygen in –OH is in resonance with C=O, reducing the electrophilic character of carbonyl carbon.
(c) Both give positive iodoform test as they contain CO–CH3 group.
2016 · 5 marks
Write the structures of A, B, C, D and E in the following reactions: C6H6 --(CH3COCl/Anhyd.AlCl3) → A --(Zn-Hg/Conc.HCl) → B --(KMnO4-KOH,delta/H3O+) → C. A --(NaOH) → D + E. OR
(a) Write the chemical equation for Cannizzaro reaction.
(b) Draw the structure of semicarbazone of ethanal.
(c) Why pKa of F-CH2-COOH is lower than that of Cl-CH2-COOH?
(d) Write the product: CH3-CH=CH-CH2CN --(i) DIBAL-H,
(ii) H2O) → ?
(e) How can you distinguish between propanal and propanone?
AnswerA = C6H5COCH3 (Acetophenone), B = C6H5CH2CH3 (Ethylbenzene), C = C6H5COOH (Benzoic acid), D = C6H5COONa (Sodium benzoate), E = CHI3 (Iodoform). OR
(a) 2HCHO --(conc.NaOH) → HCOONa + CH3OH.
(b) CH3CH=N-NHCONH2.
(c) Stronger -I effect of fluorine makes FCH2COOH more acidic (lower pKa).
(d) CH3-CH=CH-CH2-CHO (Pent-3-ene-1-al).
(e) Tollens' test: Propanal gives silver mirror; propanone does not.
2016 · 5 marks
(a) Write structures of A and B:
(i) CH3COCl --(H2/Pd-BaSO4) → A --(H2N-OH) → B.
(ii) CH3MgBr --(1. CO2, 2. H3O+) → A --(PCl5) → B.
(b) Distinguish between:
(i) C6H5-COCH3 and C6H5-CHO,
(ii) CH3COOH and HCOOH.
(c) Arrange in increasing boiling points: CH3CHO, CH3COOH, CH3CH2OH.
OR
(a) Write equation for Wolff-Kishner reduction.
(b) Arrange in increasing reactivity towards nucleophilic addition: C6H5COCH3, CH3-CHO, CH3COCH3.
(c) Why carboxylic acid does not give reactions of carbonyl group?
(i) CH3CH=CH-CO-CH3 gives iodoform test (has COCH3), CH3CH2-CO-CH=CH2 does not.
(ii) Benzaldehyde reacts with Tollen's reagent; benzoic acid gives brisk effervescence with NaHCO3.
OR (a)
(i) Propane (Clemmensen reduction).
(ii) C6H6 + Na2CO3 (decarboxylation).
(iii) CH2=CH-CH2-CHO (aldehyde via DIBAL-H reduction of nitrile).
(b) C6H5COCH3 < CH3COCH3 < CH3CHO < HCHO.
(c) CH3CH=N-NH-C6H3(NO2)2.
2020 · 5 marks
(a) An organic compound
(A) having molecular formula C4H8O gives orange red precipitate with 2,4-DNP reagent. It does not reduce Tollen's reagent but gives yellow precipitate of iodoform on heating with NaOH and I2. Compound
(A) on reduction with NaBH4 gives compound
(B) which undergoes dehydration on heating with conc. H2SO4 to form compound (C). Compound
(C) on ozonolysis gives two molecules of ethanal. Identify (A),
(B) and
(C) and write their structures. Write the reactions of compound
(A) with
(i) NaOH/I2 and
(ii) NaBH4.
(b) Give reasons:
(i) Oxidation of propanal is easier than propanone.
(ii) alpha-hydrogen of aldehydes and ketones is acidic in nature.
OR
(a) Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanal
(b) Write the major product(s):
(i) CH3-CH=CH-CH2-CN with (i)DIBAL-H/(ii)H3O+
(ii) CH3-CH2-OH with CrO3
(c) How can you distinguish between propanal and propanone?
Answer(a) A = Butanone (CH3COCH2CH3), B = Butan-2-ol (CH3CH(OH)CH2CH3), C = But-2-ene (CH3CH=CHCH3).
(ii) CH3COCH2CH3 + NaBH4 -> CH3CH(OH)CH2CH3 (Butan-2-ol). (b)
(i) Oxidation of propanal is easier because aldehydes have H attached to carbonyl, ketones have two alkyl groups making oxidation harder.
(ii) alpha-hydrogen is acidic because the carbanion formed is stabilized by resonance with the carbonyl group.
OR (a)
(i) Cyclobutanone cyanohydrin: cyclobutane ring with C(OH)(CN).
(ii) Hemiacetal of ethanal: CH3CH(OR)(OH). (b)
(i) CH3CH=CH-CH2-CHO (DIBAL-H reduces CN to CHO).
(ii) CH3CHO then CH3COOH (CrO3 oxidation).
(c) By iodoform test: Propanone gives yellow ppt of CHI3 with I2/NaOH; propanal does not.
2023 · 5 marks
(a)
(i) Write the reaction involved in Cannizaro's reaction.
(ii) Why are the boiling point of aldehydes and ketones lower than that of corresponding carboxylic acids?
(iii) An organic compound 'A' with molecular formula C5H8O2 is reduced to n-pentane with hydrazone followed by heating with NaOH and Glycol. 'A' forms a dioxime with hydroxylamine and gives a positive Iodoform and Tollen's test. Identify 'A' and give its reaction for Iodoform and Tollen's test.
Answer(a)
(i) Cannizaro Reaction: 2C6H5CHO + Conc. NaOH -> C6H5COONa + C6H5CH2OH (one molecule oxidized to salt, one reduced to alcohol)
(ii) Boiling points of aldehyde and ketones are less than carboxylic acid because carboxylic acid have hydrogen bonding and associated molecules bonding which increases the boiling point of carboxylic acids.
(iii) Compound A is CH3-CO-CH2-CH2-CHO (4-oxopentanal). It is reduced to n-pentane (CH3-CH2-CH2-CH2-CH3) with Zn(Hg)/NH2-NH2. It forms dioxime with NH2OH. Iodoform test: CH3CO- group gives positive iodoform test with I2/NaOH. Tollen's test: -CHO group gives positive Tollen's test.
2023 · 5 marks
34 (OR): (b)
(i) Give a chemical test to distinguish between ethanol acid and ethanoic acid.
(ii) Why is the alpha-hydrogens of aldehydes and ketones are acidic in nature?
(iii) An organic compound 'A' with molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.
(1) Identify 'A', 'B' and 'C'.
(2) Out of 'B' and 'C', which will have higher boiling point? Give reason.
Answer(b)
(i) Sodium bicarbonate (NaHCO3) is used to distinguish ethanol and ethanoic acid. Ethanoic acid gives brisk Red fumes with sodium bicarbonate. Ethanol do not react with sodium bicarbonate.
(ii) alpha-Hydrogen is always attached with oxygen which is highly electronegative and having lone pair of electrons. Alpha hydrogen atom when donates electron it acts as an acid.
(2) Ethanoic acid will have higher boiling point because it contains associate molecules which participate in hydrogen bonding. H-bonding is strong bonding which requires high temperature to boil.
2025 · 5 marks
(A)(a) Carry out the following conversions:
(i) Ethanal to But-2-enal
(ii) Propanoic acid to ethane.
(b) An alkene A with molecular formula C5H10 on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C.
OR
(B) An organic compound
(A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to get a carboxylic acid
(B) and an alcohol (C). Oxidation of
(C) with chromic acid produced (B).
(C) on dehydration gives But-1-ene. Identify (A),
(B) and
(C) and write chemical equations for the reactions involved.
Answer(A)(a)
(i) Ethanal to But-2-enal by aldol condensation: 2CH3CHO -> CH3CH(OH)CH2CHO (3-Hydroxybutanal) -> CH3CH=CH-CHO (But-2-enal) + H2O.
(a) An organic compound (A) with molecular formula C9H10O forms 2,4-DNP derivative, reduces Fehling solution and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid.
(i) Identify compound (A) and write its IUPAC name.
(ii) Write the reaction of compound (A) with (1) 2,4-Dinitrophenyl hydrazine and (2) Fehling solution
(iii) Write the equation of compound (A) when it undergoes Cannizzaro reaction.
OR (b)
(i) Account for the following:
(1) The alpha-hydrogens of aldehydes and ketones are acidic in nature.
(2) Oxidation of aldehydes is easier than ketones.
(ii) Arrange the following in:
(1) Decreasing reactivity towards nucleophilic addition reaction: propanal, acetone, benzaldehyde.
(2) Increasing order of boiling point: Propane, Ethanol, Dimethylether, Propanal
(iii) Give simple chemical test to distinguish between Benzoic acid and Benzaldehyde.
Answer(a)
(i) 2-Ethyl benzaldehyde
(ii)(1) 2-Ethylbenzaldehyde reacts with 2,4-DNP to form 2,4-DNP derivative + H2O.
(2) Fehling's reaction does not happen with benzaldehyde (aromatic aldehyde).
(iii) Cannizzaro reaction: C6H4(CHO)(CH2CH3) + NaOH -> C6H4(COO-Na+)(CH2CH3) + C6H4(CH2OH)(CH2CH3). Same products are formed.
OR (b) (i)(1) The hydrogen at alpha position of carbonyl group is acidic as it experiences -I effect of the carbonyl group. The proton can be abstracted by a base and the resulting anion is stabilised by keto-enol tautomerism.
(2) Aldehydes are more reactive than ketones and can be easily oxidized. Aldehydes have a hydrogen atom attached to carbonyl carbon which can easily be abstracted. Ketones lack this proton, hence they oxidise only under extreme conditions.
(ii)(1) Propanal > acetone > benzaldehyde
(2) Propane < dimethyl ether < propanal < ethanol
(iii) Benzoic acid reacts with ethanol in the presence of dilute sulphuric acid to form sweet smelling ester Ethyl benzoate. This reaction is not given by benzaldehyde.
2015 · 5 marks
An aromatic compound 'A' of molecular formula C7H7ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions: (C7H7ON) A ⟶[Br2+KOH] C6H5NH2 ⟶[NaNO2+HCl, 273K] B ⟶[CH3CH2OH, KI] C C6H5NH2 ⟶[CHCl3+NaOH] D, and with KI → E
AnswerA = C6H5NHCOCH3 (acetanilide, Hoffmann bromamide degradation gives aniline) C6H5NH2 + NaNO2 + HCl → C6H5N2+Cl− (B, benzenediazonium chloride) B + CH3CH2OH, KI → C6H5I or C6H5OH (C) D = C6H5NC (phenyl isocyanide, carbylamine reaction)
E = C6H5I (iodobenzene from Sandmeyer reaction with KI)
2015 · 5 marks
(a) Write the structures of main products when aniline reacts with the following reagents:
(i) Br2/water
(ii) HCl
(iii) (CH3CO)2O/pyridine
(b) Arrange the following in the increasing order of their boiling point:
C2H5NH2, C2H5OH, (CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds: (CH3)2NH and (CH3)3N
(c) Hinsberg test: (CH3)2NH (secondary amine) with benzenesulphonyl chloride forms a precipitate which is insoluble in KOH. (CH3)3N (tertiary amine) does not react with benzenesulphonyl chloride.
2018 · 5 marks
(a) Write the reactions involved in the following:
(i) Hoffmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
(a) Write the structures of the main products of the following reactions:
(i) Aniline + (CH3CO)2O --Pyridine--→
(ii) Benzenesulphonyl chloride + (CH3)2NH →
(iii) Benzenediazonium chloride + CH3CH2OH →
(b) Give a simple chemical test to distinguish between aniline and N,N-dimethylaniline.
(c) Arrange the following in the increasing order of their pKb values: C6H5NH2, C2H5NH2, C6H5NHCH3
(i) Aniline on nitration gives good amount of m-nitroaniline, though -NH2 group is o/p directing in electrophilic substitution reactions.
(ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(iii) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines.
(II) Write the reaction involved in the following:
(i) Carbyl amine test
(ii) Gabriel phthalimide synthesis
Answer(a) (I) (i) Nitration is carried out in an acidic medium. In a strongly acidic medium, aniline is protonated to give anilinium ion which is meta-directing. Therefore, aniline on nitration gives a substantial amount of m-nitroaniline.
(ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. As the number of methyl groups increases, the extent of hydration decreases due to steric hindrance. Greater is the extent of hydration, greater is the stability of ion and greater is the basic strength of amine.
(iii) Ammonolysis of alkyl halides leads to the formation of a mixture of primary, secondary, tertiary amines and quaternary salts which is very difficult to separate and obtain pure amine.
(b) (I) Write the structures of A, B and C in the following reactions:
(i) C6H5N2+Cl- --(CuCN) → A --(H2O/H+) → B --(NH3/Delta) → C
(ii) C6H5NO2 --(Fe/HCl) → A --(NaNO2+HCl/273K) → B --(C2H5OH) → C
(II) Why aniline does not undergo Friedel-Crafts reaction? (III) Arrange the following in increasing order of their boiling point: C2H5OH, C2H5NH2, (C2H5)3N (OR)
(ii) C6H5NO2 --(Fe/HCl) → C6H5NH2/Aniline (A) --(NaNO2+HCl/273K) → C6H5N2+Cl-/Benzene diazonium chloride (B) --(C2H5OH) → C6H6 + N2 + ... Actually: B --(C2H5OH) → C6H5OH? No. Let me reconsider. B (Benzene diazonium chloride) --(C2H5OH) → C6H6 (Benzene). So C = Benzene. Actually from the PDF: C -> Benzoic acid (COOH on benzene).
(II) It is due to the fact that AlCl3 being electron deficient acts as a Lewis base and attacks on the lone pair of nitrogen present in aniline to form insoluble complex which precipitates out and does not proceed. (III) Increasing order of boiling points: (CH3)3N < C2H5NH2 < C2H5OH
2024 · 5 marks
(a) An amide 'A' with molecular formula C7H7ON undergoes Hoffmann Bromamide degradation reaction to give amine 'B'. 'B' on treatment with nitrous acid at 273-278 K form 'C' and on treatment with chloroform and ethanolic potassium hydroxide forms 'D'. 'C' on treatment with ethanol gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write the sequence of chemical equations. OR (b)
(i) (1) What is Hinsberg's reagent?
(2) Arrange the following compounds in the increasing order of their basic strength in gaseous phase: C2H5NH2, (C2H5)3N, (C2H5)2NH
(ii) Give reasons for the following:
(1) Methyl amine is more basic than aniline.
(2) Aniline readily reacts with bromine water to give 2,4,6-tribromoaniline.
(3) Primary amines have higher boiling points than tertiary amines.
Answer(a) A = C6H5CONH2 (Benzamide) B = C6H5NH2 (Aniline) C = Benzenediazonium Chloride D = Phenylisocyanide
E = Benzene
Reactions: C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + 2KBr + K2CO3 + 2H2O C6H5NH2 + CHCl3 + KOH -> C6H5NC + 3KCl + 3H2O C6H5N2Cl + C2H5OH -> C6H6 + N2 + HCl + CH3CHO OR (b) (i)(1) The Hinsberg reagent is a chemical reagent used for identification and differentiation of primary, secondary, and tertiary amines. It is benzenesulfonyl chloride (C6H5SO2Cl).
(2) Increasing order of basic strength in gaseous phase: C2H5NH2 < (C2H5)2NH < (C2H5)3N
(ii)(1) In aniline the lone pair of electron remains delocalised in the pi electron ring and thus is not available on nitrogen. However, in methylamine, the lone pair is on NH2 group of methyl and is the source of basicity.
(2) The benzene ring in aniline is highly activated due to strong mesomeric effect of amino group. This readily reacts with bromine water.
(3) In primary amines there are intermolecular hydrogen bonding, so more energy is required to break the attraction. Whereas in tertiary amines there is no intermolecular hydrogen bonding.
2020 · 5 marks
(a) Account for the following:
(i) Tendency to show -2 oxidation state decreases from oxygen to tellurium.
(ii) Acidic character increases from HF to HI.
(iii) Moist SO2 gas acts as a reducing agent.
(b) Draw the structure of an oxoacid of sulphur containing S-O-S linkage.
(c) Complete the following equation: XeF2 + H2O -> OR
(a) Among the hydrides of group 16, write the hydride which is
(i) a strong reducing agent
(ii) has maximum bond angle
(iii) most thermally stable. Give suitable reason in each.
(b) Complete the following equations: S + H2SO4(conc.) -> ; Cl2 + NaOH(cold and dilute) ->
Answer(a)
(i) As size increases from O to Te, tendency to gain 2 electrons decreases due to decrease in electronegativity.
(ii) Bond strength decreases from H-F to H-I (bond length increases), so ease of releasing H+ increases, making HI most acidic.
(iii) Moist SO2 acts as reducing agent because it can be oxidized to SO3/H2SO4 by accepting oxygen.
(b) H2S2O7 (pyrosulphuric acid) has S-O-S linkage.
(c) 2XeF2 + 2H2O -> 2Xe + 4HF + O2.
OR (a)
(i) H2Te (largest size, weakest bond).
(ii) H2O (maximum bond angle ~104.5° due to small size and high electronegativity of O).
(iii) H2O (most thermally stable due to strongest O-H bond).