Question 1

1 What is meant by the following terms ? Give an example of the reaction in each case.
(i) Cyanohydrin (ii) Acetal (iii) Semicarbazone
(iv) Aldol (v) Hemiacetal (vi) Oxime
(vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative
(x) Schiff’s base

Accepted Answer

(i) Cyanohydrin: A compound formed by the addition of hydrogen cyanide (HCN) to an aldehyde or ketone. Example: Acetaldehyde reacts with HCN to form 2-hydroxypropanenitrile (acetaldehyde cyanohydrin).

(ii) Acetal: A compound formed when an aldehyde or ketone reacts with two equivalents of an alcohol under acidic conditions. Example: Ethanal reacts with methanol to form an acetal.

(iii) Semicarbazone: A derivative formed when an aldehyde or ketone reacts with semicarbazide. Example: Acetone reacts with semicarbazide to form acetone semicarbazone.

(iv) Aldol: A β-hydroxy aldehyde or ketone formed by aldol condensation. Example: Two molecules of ethanal undergo aldol condensation to form 3-hydroxybutanal.

(v) Hemiacetal: An intermediate formed when an aldehyde or ketone reacts with one equivalent of an alcohol. Example: Reaction of ethanal with methanol forms a hemiacetal.

(vi) Oxime: A compound formed by the reaction of aldehydes or ketones with hydroxylamine. Example: Acetone reacts with hydroxylamine to form acetone oxime.

(vii) Ketal: A compound formed when a ketone reacts with two equivalents of an alcohol under acidic conditions. Example: Acetone reacts with ethylene glycol to form a ketal.

(viii) Imine: A compound formed by the reaction of aldehydes or ketones with primary amines. Example: Benzaldehyde reacts with methylamine to form an imine.

(ix) 2,4-DNP-derivative: A hydrazone formed by the reaction of aldehydes or ketones with 2,4-dinitrophenylhydrazine. Example: Acetone reacts with 2,4-DNP to form acetone 2,4-dinitrophenylhydrazone.

(x) Schiff’s base: An imine formed by the condensation of primary amines with aldehydes or ketones. Example: Benzaldehyde reacts with aniline to form a Schiff’s base. — Each term is defined with its chemical nature and typical reaction example illustrating its formation.

Question 2

2 Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH(CH3)CH2CH2CH2CHO (ii) CH3CH2COCH(CH3)CH2CH2Cl
(iii) CH3CH=CHCHO (iv) CH3COCH2COCH3
(v) CH3CH(CH3)CHC(CH3)2COCH3 (vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p

Accepted Answer

(i) 4-Methylpentanal
(ii) 6-Chlorohexan-3-one
(iii) Crotonaldehyde (But-2-enal)
(iv) 2,4-Pentanedione
(v) 4,4-Dimethylpentan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) 4-Hydroxybenzaldehyde (p-Hydroxybenzaldehyde) — Each compound is named by identifying the longest carbon chain containing the functional group and numbering it to give the lowest possible numbers to substituents and functional groups according to IUPAC rules.

Question 3

3 Draw the structures of the following compounds.
(i) 3-Methylbutanal (ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

Accepted Answer

Structures:
(i) 3-Methylbutanal: CH3-CH(CH3)-CH2-CHO
(ii) p-Nitropropiophenone: Nitro group at para position on phenyl ring attached to propiophenone
(iii) p-Methylbenzaldehyde: Benzaldehyde with methyl group at para position
(iv) 4-Methylpent-3-en-2-one: Pent-3-en-2-one with methyl at C4
(v) 4-Chloropentan-2-one: Pentan-2-one with chlorine at C4
(vi) 3-Bromo-4-phenylpentanoic acid: Pentanoic acid with bromo at C3 and phenyl at C4
(vii) p,p’-Dihydroxybenzophenone: Benzophenone with hydroxy groups at para positions on both phenyl rings
(viii) Hex-2-en-4-ynoic acid: Hexanoic acid with double bond at C2 and triple bond at C4 — Structures are drawn by placing substituents and functional groups at specified positions on the carbon chain or aromatic ring as per the names.

Question 4

4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO
(v) (vi) PhCOPh

Accepted Answer

(i) Hexan-2-one (Common name: Caproin)
(ii) 5-Bromohexanal
(iii) Heptanal
(iv) Cinnamaldehyde (3-Phenylpropenal)
(v) [Incomplete in text]
(vi) Benzophenone — IUPAC names are derived by identifying the longest chain containing the carbonyl group and numbering it to give the carbonyl carbon the lowest number. Common names are given where well-known.

Note: (v) is incomplete in the source text.

Question 5

5 Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde

Accepted Answer

Structures:
(i) Benzaldehyde 2,4-DNP derivative: Benzaldehyde attached to 2,4-dinitrophenylhydrazine via hydrazone linkage (C=N-NH-aryl)
(ii) Cyclopropanone oxime: Cyclopropanone ring with oxime group (C=NOH)
(iii) Acetaldehydedimethylacetal: Acetaldehyde reacted with two equivalents of methanol forming acetal (CH3CH(OCH3)2)
(iv) Semicarbazone of cyclobutanone: Cyclobutanone reacted with semicarbazide forming C=N-NH-C(=O)-NH2 derivative
(v) Ethylene ketal of hexan-3-one: Hexan-3-one reacted with ethylene glycol forming cyclic ketal at C3
(vi) Methyl hemiacetal of formaldehyde: Formaldehyde reacted with methanol forming hemiacetal (HOCH2OCH3) — Derivatives are formed by characteristic reactions of aldehydes and ketones with specific reagents, structures drawn accordingly.

Question 6

6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(i) PhMgBr and then H3O+
(ii) Tollens’ reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid

Accepted Answer

(i) Reaction with phenylmagnesium bromide (PhMgBr) followed by acid workup gives cyclohexylcarbinol with phenyl group attached to former aldehyde carbon.
(ii) Tollens’ reagent oxidizes the aldehyde group to carboxylic acid, giving cyclohexanecarboxylic acid.
(iii) Reaction with semicarbazide in weak acid forms semicarbazone derivative of cyclohexanecarbaldehyde.
(iv) Excess ethanol and acid catalyze formation of acetal derivative of cyclohexanecarbaldehyde.
(v) Zinc amalgam and dilute HCl reduce the aldehyde to corresponding hydrocarbon, cyclohexane. — Each reagent reacts with the aldehyde functional group in characteristic ways: Grignard adds to carbonyl, Tollens oxidizes aldehyde to acid, semicarbazide forms semicarbazone, ethanol forms acetal, and zinc amalgam reduces aldehyde to alkane.

Question 7

7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde
(iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal

Accepted Answer

Aldol condensation occurs with aldehydes and ketones having α-hydrogens.
Cannizzaro reaction occurs with aldehydes lacking α-hydrogens.

(i) Methanal: No α-hydrogen → Cannizzaro reaction → products: methanol + formic acid
(ii) 2-Methylpentanal: Has α-hydrogen → Aldol condensation → β-hydroxy aldehyde product
(iii) Benzaldehyde: No α-hydrogen → Cannizzaro reaction → benzyl alcohol + benzoic acid
(iv) Benzophenone: Ketone, has α-hydrogen → Aldol condensation possible but less reactive
(v) Cyclohexanone: Ketone with α-hydrogen → Aldol condensation
(vi) 1-Phenylpropanone: Ketone with α-hydrogen → Aldol condensation
(vii) Phenylacetaldehyde: Has α-hydrogen → Aldol condensation
(viii) Butan-1-ol: Alcohol, neither aldol nor Cannizzaro
(ix) 2,2-Dimethylbutanal: No α-hydrogen (due to substitution) → Cannizzaro reaction

Structures of aldol products: β-hydroxy aldehydes or ketones formed by nucleophilic addition of enolate to carbonyl.
Structures of Cannizzaro products: One molecule reduced to alcohol, other oxidized to acid. — The presence or absence of α-hydrogens determines the reaction pathway. Aldol condensation requires α-hydrogens; Cannizzaro occurs when none are present.

Question 8

8 How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid

Accepted Answer

(i) Ethanal to Butane-1,3-diol:
Step 1: Two molecules of ethanal undergo aldol condensation to form 3-hydroxybutanal.
Step 2: Reduction of 3-hydroxybutanal with NaBH4 yields butane-1,3-diol.

(ii) Ethanal to But-2-enal:
Step 1: Aldol condensation of ethanal gives 3-hydroxybutanal.
Step 2: Dehydration of 3-hydroxybutanal yields But-2-enal (crotonaldehyde).

(iii) Ethanal to But-2-enoic acid:
Step 1: Aldol condensation of ethanal to 3-hydroxybutanal.
Step 2: Oxidation of 3-hydroxybutanal or But-2-enal with mild oxidizing agent (e.g., KMnO4) gives But-2-enoic acid. — Ethanal undergoes aldol condensation to form β-hydroxy aldehyde, which upon dehydration forms α,β-unsaturated aldehyde, and oxidation converts aldehyde to acid.

Question 9

9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Accepted Answer

Possible aldol products:
1) Propanal (nucleophile) + Propanal (electrophile) → 2-Methylpent-2-enal
2) Butanal (nucleophile) + Butanal (electrophile) → 2-Ethylpent-2-enal
3) Propanal (nucleophile) + Butanal (electrophile) → 3-Methylpent-2-enal
4) Butanal (nucleophile) + Propanal (electrophile) → 3-Ethylpent-2-enal

In each case, the aldehyde with α-hydrogen forms enolate ion (nucleophile), attacking the carbonyl carbon of the other aldehyde (electrophile). — Aldol condensation involves enolate ion formation from aldehyde with α-hydrogen acting as nucleophile attacking another aldehyde molecule acting as electrophile. Cross aldol products arise when two different aldehydes are involved.

Question 10

10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Accepted Answer

The compound is o-Hydroxybenzaldehyde (Salicylaldehyde).

Reasoning:
- Molecular formula C9H10O fits salicylaldehyde.
- Forms 2,4-DNP derivative indicating presence of aldehyde or ketone.
- Reduces Tollens’ reagent indicating aldehyde group.
- Undergoes Cannizzaro reaction indicating no α-hydrogen.
- Oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid), consistent with ortho-substituted benzaldehyde.

Hence, the compound is o-Hydroxybenzaldehyde. — The properties and oxidation product identify the compound as salicylaldehyde.

Question 11

11 An organic compound (A) (molecular formula C8H16O) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Accepted Answer

Given:
(A) C8H16O → hydrolysis → (B) carboxylic acid + (C) alcohol
(C) oxidized to (B)
(C) dehydrates to but-1-ene

Interpretation:
(A) is an ester of butanoic acid and butanol.

Reactions:
1) Hydrolysis:
CH3CH2CH2COOCH2CH2CH2CH3 + H2O (acid) → CH3CH2CH2COOH (B) + CH3CH2CH2CH2OH (C)

2) Oxidation:
CH3CH2CH2CH2OH + [O] (CrO3/H2SO4) → CH3CH2CH2COOH

3) Dehydration:
CH3CH2CH2CH2OH → CH2=CHCH2CH3 + H2O

Thus, (A) is butyl butanoate, (B) butanoic acid, (C) butanol. — Hydrolysis of ester yields acid and alcohol; oxidation of alcohol yields acid; dehydration of alcohol yields alkene.

Question 12

12 Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)3CHCOOH, CH3CH2CH2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Accepted Answer

(i) Reactivity towards HCN (nucleophilic addition):
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
Reason: Steric hindrance decreases reactivity; aldehydes more reactive than ketones.

(ii) Acid strength:
(CH3)3CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH
Reason: Electron withdrawing Br increases acidity; position matters.

(iii) Acid strength:
4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid
Reason: Electron donating methoxy decreases acidity; nitro groups increase acidity, more nitro groups stronger acid. — Reactivity and acidity depend on electronic effects and steric hindrance; electron withdrawing groups increase acidity and reactivity towards nucleophiles.

Question 13

13 Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone (ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal

Accepted Answer

(i) Propanal and Propanone:
Tollens’ test: Propanal gives positive silver mirror; Propanone does not.
(ii) Acetophenone and Benzophenone:
2,4-DNP test: Both positive; however, benzophenone does not give iodoform test, acetophenone does.
(iii) Phenol and Benzoic acid:
Ferric chloride test: Phenol gives violet color; benzoic acid does not.
(iv) Benzoic acid and Ethyl benzoate:
Sodium bicarbonate test: Benzoic acid reacts to give CO2; ester does not.
(v) Pentan-2-one and Pentan-3-one:
Iodoform test: Pentan-2-one positive; Pentan-3-one negative.
(vi) Benzaldehyde and Acetophenone:
Tollens’ test: Benzaldehyde positive; acetophenone negative.
(vii) Ethanal and Propanal:
Iodoform test: Both positive; however, propanal gives slower reaction. — Chemical tests exploit functional group differences such as aldehyde vs ketone, acid vs ester, phenol vs acid, and position of methyl groups for iodoform test.

Question 14

14 How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom
(i) Methyl benzoate (ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.

Accepted Answer

(i) Methyl benzoate:
Step 1: Benzene → Benzoic acid (by side chain oxidation of toluene)
Step 2: Benzoic acid + Methanol (acid catalyst) → Methyl benzoate

(ii) m-Nitrobenzoic acid:
Step 1: Benzene → Benzoic acid
Step 2: Nitration of benzoic acid → m-Nitrobenzoic acid (due to -COOH meta directing)

(iii) p-Nitrobenzoic acid:
Step 1: Benzene → p-Nitrotoluene (nitration)
Step 2: Oxidation of methyl group → p-Nitrobenzoic acid

(iv) Phenylacetic acid:
Step 1: Benzene → Benzyl chloride (chlorination)
Step 2: Benzyl chloride + NaCN → Benzyl cyanide
Step 3: Hydrolysis of benzyl cyanide → Phenylacetic acid

(v) p-Nitrobenzaldehyde:
Step 1: Benzene → p-Nitrotoluene (nitration)
Step 2: Side chain oxidation of methyl group → p-Nitrobenzaldehyde (partial oxidation) — Preparation involves functional group transformations starting from benzene using nitration, oxidation, substitution, and esterification reactions.

Question 15

15 How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m-Nitrobenzyl alcohol

Accepted Answer

(i) Propanone to Propene:
Step 1: Propanone to 2-propanol (reduction)
Step 2: Dehydration of 2-propanol to propene

(ii) Benzoic acid to Benzaldehyde:
Step 1: Reduction of benzoic acid to benzyl alcohol
Step 2: Oxidation of benzyl alcohol to benzaldehyde

(iii) Ethanol to 3-Hydroxybutanal:
Step 1: Ethanol to acetaldehyde (oxidation)
Step 2: Aldol condensation of acetaldehyde to 3-hydroxybutanal

(iv) Benzene to m-Nitroacetophenone:
Step 1: Friedel-Crafts acylation of benzene with acetyl chloride
Step 2: Nitration to give m-nitroacetophenone

(v) Benzaldehyde to Benzophenone:
Step 1: Benzaldehyde to benzoin (benzoin condensation)
Step 2: Oxidation of benzoin to benzil
Step 3: Benzil to benzophenone (via benzil rearrangement) [if allowed]

(vi) Bromobenzene to 1-Phenylethanol:
Step 1: Bromobenzene to phenylmagnesium bromide
Step 2: Reaction with acetaldehyde followed by acid workup

(vii) Benzaldehyde to 3-Phenylpropan-1-ol:
Step 1: Benzaldehyde + allyl bromide (Grignard reagent)
Step 2: Acid workup to give 3-phenylpropan-1-ol

(viii) Benzaldehyde to α-Hydroxyphenylacetic acid:
Step 1: Benzaldehyde + hydrogen cyanide to form cyanohydrin
Step 2: Hydrolysis of cyanohydrin to α-hydroxyphenylacetic acid

(ix) Benzoic acid to m-Nitrobenzyl alcohol:
Step 1: Nitration of benzoic acid to m-nitrobenzoic acid
Step 2: Reduction of m-nitrobenzoic acid to m-nitrobenzyl alcohol — Conversions involve functional group interconversions, use of Grignard reagents, Friedel-Crafts reactions, oxidations, reductions, and condensations.

and Carboxylic

and Carboxylic — Study Notes

Introduction

Nomenclature of Carboxylic Acids

Physical Properties

Practice Questions — and Carboxylic

All 5 Chapters in Chemistry-II