Home › NCERT Solutions › Class 12 Chemistry › Chapter 8

NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 8

Chapter 8: Aldehydes, Ketones and Carboxylic Acids (Organic Chemistry)

Q: Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

(i) butanone < propanone < propanal < ethanal. (ii) acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.

Q: Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene).

(i), (iv) Oxidation with alkaline KMnO₄ and acidification. (ii) Iodoform reaction, then acid. (iii) Grignard reagent + CO₂, then acid.

Q: 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 (viii) Imine (ix) 2,4-DNP-derivative (x) Schiff's base.

As defined, with the examples above.

Q: 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.

Aldol: (ii), (v), (vi), (vii). Cannizzaro: (i), (iii), (ix). Neither: (iv), (viii).

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

(i) Aldol then NaBH₄. (ii) Aldol then heat. (iii) But-2-enal oxidised by Tollens' reagent.

Q: 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.

The four aldols above (two self and two cross products).

Q: 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.

(i) Tollens'; (ii) iodoform; (iii) NaHCO₃ or FeCl₃; (iv) NaHCO₃; (v) iodoform; (vi) Tollens' or iodoform; (vii) iodoform.

Answers to all 8 intext questions and 20 exercises of Chapter 8, Aldehydes, Ketones and Carboxylic Acids (NCERT Class 12 Chemistry Part II, 2026-27): IUPAC names and structures, preparation, boiling points, reactivity in nucleophilic addition, derivatives (oxime, semicarbazone, 2,4-DNP, acetal), aldol and Cannizzaro reactions, distinguishing tests, acid strength, conversions and structure-finding problems. All 28 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 8 (PDF)

Exam revision? Chapter 8 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 10 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-chemistry/chapter-8-aldehydes-ketones-and-carboxylic-acids

Reactivity towards nucleophilic addition: HCHO > other aldehydes > ketones (steric and +I effects lower it; electron-withdrawing groups raise it). Aldol condensation needs an α-hydrogen; aldehydes with no α-hydrogen give the Cannizzaro reaction with conc. alkali. Tests: Tollens'/Fehling's for aldehydes; iodoform (I₂/NaOH) for CH₃CO– and CH₃CH(OH)–; NaHCO₃ effervescence for carboxylic acids. Electron-withdrawing groups (–F, –Cl, –NO₂), especially near –COOH, increase acid strength.

Intext questions

8.1

Write the structures of the following compounds. (i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec-butyl ketone (vi) 4-Fluoroacetophenone.

Solution

(i) $CH X_{3} CH (OCH X_{3}) CHO$
(ii) $CH X_{3} CH (OH) CH X_{2} CHO$
(iii) a cyclopentane ring with –CHO on C-1 and –OH on C-2
(iv) $CH X_{3} COCH X_{2} CH X_{2} CHO$
(v) $CH X_{3} CH X_{2} CH (CH X_{3}) - CO - CH (CH X_{3}) CH X_{2} CH X_{3}$
(vi) $F - C X_{6} H X_{4} - COCH X_{3}$ (F para to the acetyl group)

As written above.

8.2

Write the structures of products of the following reactions: (i) benzene +

C X_{2} H X_{5} COCl

, anhyd.

AlCl X_{3}

CS X_{2}

(ii)

(C X_{6} H X_{5} CH X_{2}) X_{2} Cd + 2 CH X_{3} COCl

(iii)

H X_{3} C - C \equiv C - H Hg X^{2 +}, H X_{2} SO X_{4}

(iv) 4-nitrotoluene, 1.

CrO X_{2} Cl X_{2}

H X_{3} O X^{+}

Solution

(i) $C X_{6} H X_{5} COCH X_{2} CH X_{3}$ , propiophenone (Friedel–Crafts acylation)
(ii) $2 C X_{6} H X_{5} CH X_{2} COCH X_{3} + CdCl X_{2}$ : 1-phenylpropan-2-one
(iii) $CH X_{3} COCH X_{3}$ , propanone (hydration of the alkyne, via the enol)
(iv) 4-nitrobenzaldehyde, $O X_{2} N - C X_{6} H X_{4} - CHO$ (Etard reaction)

(i) propiophenone; (ii) 1-phenylpropan-2-one; (iii) acetone; (iv) 4-nitrobenzaldehyde.

8.3

Arrange the following compounds in increasing order of their boiling points.

CH X_{3} CHO

CH X_{3} CH X_{2} OH

CH X_{3} OCH X_{3}

CH X_{3} CH X_{2} CH X_{3}

Solution

$CH X_{3} CH X_{2} CH X_{3} < CH X_{3} OCH X_{3} < CH X_{3} CHO < CH X_{3} CH X_{2} OH$

Propane has only weak van der Waals forces; the ether is weakly polar; the aldehyde has stronger dipole–dipole forces; ethanol forms hydrogen bonds.

CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH.

8.4

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Solution

(i) Butanone < propanone < propanal < ethanal. Ketones are less reactive than aldehydes (two alkyl groups give more crowding and more electron release); larger alkyl groups lower reactivity further.
(ii) Acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde. The ketone is least reactive; the electron-releasing –CH₃ lowers the electrophilicity of the carbonyl carbon, while the electron-withdrawing –NO₂ raises it.

(i) butanone < propanone < propanal < ethanal. (ii) acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.

8.5

Predict the products of the following reactions: (i) cyclopentanone +

HO - NH X_{2}

(

H X^{+}

) (ii) cyclohexanone + 2,4-dinitrophenylhydrazine (iii)

R - CH = CH - CHO + NH X_{2} - CO - NH - NH X_{2}

(

H X^{+}

) (iv) acetophenone +

CH X_{3} CH X_{2} NH X_{2}

(

H X^{+}

Solution

(i) Cyclopentanone oxime: the ring carbon becomes $C = N - OH$
(ii) Cyclohexanone 2,4-dinitrophenylhydrazone: ring $C = N - NH - C X_{6} H X_{3} (NO X_{2}) X_{2}$
(iii) The semicarbazone $R - CH = CH - CH = N - NH - CO - NH X_{2}$
(iv) The imine (Schiff base) $C X_{6} H X_{5} C (CH X_{3}) = N - CH X_{2} CH X_{3}$

(i) oxime; (ii) 2,4-DNP derivative; (iii) semicarbazone; (iv) N-ethyl imine of acetophenone.

8.6

Give the IUPAC names of the following compounds: (i)

PhCH X_{2} CH X_{2} COOH

(ii)

(CH X_{3}) X_{2} C = CHCOOH

(iii) cyclopentane ring with –COOH on C-1 and

CH X_{3}

on C-2 (iv) benzoic acid with

NO X_{2}

groups at C-2, C-4 and C-6.

Solution

(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentane-1-carboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid

As listed above.

8.7

Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene).

Solution

(i) $C X_{6} H X_{5} CH X_{2} CH X_{3} KMnO X_{4} / KOH heat C X_{6} H X_{5} COOK H X_{3} O X^{+} C X_{6} H X_{5} COOH$ (the whole side chain is oxidised)
(ii) $C X_{6} H X_{5} COCH X_{3} I X_{2} / NaOH C X_{6} H X_{5} COONa + CHI X_{3}$ ; then $H X_{3} O X^{+} C X_{6} H X_{5} COOH$ (haloform reaction; or oxidise with $KMnO X_{4}$ )
(iii) $C X_{6} H X_{5} Br Mg dry ether C X_{6} H X_{5} MgBr CO X_{2} (dry ice) C X_{6} H X_{5} COOMgBr H X_{3} O X^{+} C X_{6} H X_{5} COOH$
(iv) $C X_{6} H X_{5} CH = CH X_{2} KMnO X_{4} / KOH heat C X_{6} H X_{5} COOK H X_{3} O X^{+} C X_{6} H X_{5} COOH$

(i), (iv) Oxidation with alkaline KMnO₄ and acidification. (ii) Iodoform reaction, then acid. (iii) Grignard reagent + CO₂, then acid.

8.8

Which acid of each pair shown here would you expect to be stronger? (i)

CH X_{3} CO X_{2} H

CH X_{2} FCO X_{2} H

(ii)

CH X_{2} FCO X_{2} H

CH X_{2} ClCO X_{2} H

(iii)

CH X_{2} FCH X_{2} CH X_{2} CO X_{2} H

CH X_{3} CHFCH X_{2} CO X_{2} H

(iv)

F X_{3} C - C X_{6} H X_{4} - COOH

H X_{3} C - C X_{6} H X_{4} - COOH

(both para).

Solution

Electron-withdrawing groups stabilise the carboxylate ion and increase acid strength; the effect is larger for more electronegative groups and for groups closer to –COOH.

(i) $CH X_{2} FCO X_{2} H$ (–F withdraws electrons)
(ii) $CH X_{2} FCO X_{2} H$ (F is more electronegative than Cl)
(iii) $CH X_{3} CHFCH X_{2} CO X_{2} H$ (the F is nearer to –COOH)
(iv) $F X_{3} C - C X_{6} H X_{4} - COOH$ (–CF₃ withdraws electrons; –CH₃ releases them)

(i) CH₂FCO₂H; (ii) CH₂FCO₂H; (iii) CH₃CHFCH₂CO₂H; (iv) 4-(trifluoromethyl)benzoic acid.

Exercises

8.1

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 (viii) Imine (ix) 2,4-DNP-derivative (x) Schiff's base.

Solution

(i) Cyanohydrin: the product of adding HCN to a carbonyl group: $CH X_{3} CHO + HCN CH X_{3} CH (OH) CN$
(ii) Acetal: a gem-dialkoxy compound formed from an aldehyde and two molecules of alcohol (dry HCl): $CH X_{3} CHO + 2 C X_{2} H X_{5} OH CH X_{3} CH (OC X_{2} H X_{5}) X_{2} + H X_{2} O$
(iii) Semicarbazone: from a carbonyl compound and semicarbazide: $(CH X_{3}) X_{2} C = O + H X_{2} NNHCONH X_{2} (CH X_{3}) X_{2} C = NNHCONH X_{2} + H X_{2} O$
(iv) Aldol: a β-hydroxy aldehyde (or ketone) from aldol condensation: $2 CH X_{3} CHO dil \cdot NaOH CH X_{3} CH (OH) CH X_{2} CHO$
(v) Hemiacetal: formed from an aldehyde and one molecule of alcohol: $CH X_{3} CHO + C X_{2} H X_{5} OH CH X_{3} CH (OH) OC X_{2} H X_{5}$
(vi) Oxime: from a carbonyl compound and hydroxylamine: $CH X_{3} CHO + NH X_{2} OH CH X_{3} CH = NOH + H X_{2} O$
(vii) Ketal: a cyclic or open gem-dialkoxy compound from a ketone, e.g. with ethylene glycol: $(CH X_{3}) X_{2} C = O + HOCH X_{2} CH X_{2} OH (CH X_{3}) X_{2} C (OCH X_{2} CH X_{2} O) + H X_{2} O$
(viii) Imine: a compound with C=N formed from a carbonyl compound and ammonia or an amine: $(CH X_{3}) X_{2} C = O + RNH X_{2} (CH X_{3}) X_{2} C = NR + H X_{2} O$
(ix) 2,4-DNP derivative: the orange-yellow hydrazone formed with 2,4-dinitrophenylhydrazine: $CH X_{3} CHO + H X_{2} NNHC X_{6} H X_{3} (NO X_{2}) X_{2} CH X_{3} CH = NNHC X_{6} H X_{3} (NO X_{2}) X_{2} + H X_{2} O$
(x) Schiff's base: a substituted imine from an aldehyde and a primary amine: $C X_{6} H X_{5} CHO + C X_{6} H X_{5} NH X_{2} C X_{6} H X_{5} CH = NC X_{6} H X_{5} + H X_{2} O$

As defined, with the examples above.

8.2

Name the following compounds according to IUPAC system of nomenclature: (i)

CH X_{3} CH (CH X_{3}) CH X_{2} CH X_{2} CHO

(ii)

CH X_{3} CH X_{2} COCH (C X_{2} H X_{5}) CH X_{2} CH X_{2} Cl

(iii)

CH X_{3} CH = CHCHO

(iv)

CH X_{3} COCH X_{2} COCH X_{3}

(v)

CH X_{3} CH (CH X_{3}) CH X_{2} C (CH X_{3}) X_{2} COCH X_{3}

(vi)

(CH X_{3}) X_{3} CCH X_{2} COOH

(vii)

OHCC X_{6} H X_{4} CHO

-p.

Solution

(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-enal
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde

As listed above.

8.3

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.

Solution

(i) $(CH X_{3}) X_{2} CHCH X_{2} CHO$
(ii) $O X_{2} N - C X_{6} H X_{4} - COCH X_{2} CH X_{3}$ (para)
(iii) $CH X_{3} - C X_{6} H X_{4} - CHO$ (para)
(iv) $CH X_{3} COCH = C (CH X_{3}) X_{2}$
(v) $CH X_{3} COCH X_{2} CH (Cl) CH X_{3}$
(vi) $HOOC - CH X_{2} - CH (Br) - CH (C X_{6} H X_{5}) - CH X_{3}$
(vii) $HO - C X_{6} H X_{4} - CO - C X_{6} H X_{4} - OH$ (both OH para)
(viii) $HOOC - CH = CH - C \equiv C - CH X_{3}$

As written above.

8.4

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i)

CH X_{3} CO (CH X_{2}) X_{4} CH X_{3}

(ii)

CH X_{3} CH X_{2} CHBrCH X_{2} CH (CH X_{3}) CHO

(iii)

CH X_{3} (CH X_{2}) X_{5} CHO

(iv)

Ph - CH = CH - CHO

(v) cyclopentane ring with –CHO (vi)

PhCOPh

Solution

(i) Heptan-2-one (methyl n-pentyl ketone)
(ii) 4-Bromo-2-methylhexanal (γ-bromo-α-methylcaproaldehyde)
(iii) Heptanal (n-heptaldehyde)
(iv) 3-Phenylprop-2-enal (cinnamaldehyde)
(v) Cyclopentanecarbaldehyde
(vi) Diphenylmethanone (benzophenone)

As listed above.

8.5

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.

Solution

(i) $C X_{6} H X_{5} CH = N - NH - C X_{6} H X_{3} (NO X_{2}) X_{2}$ (NO₂ at 2 and 4 of the second ring)
(ii) a cyclopropane ring in which one carbon is $C = N - OH$
(iii) $CH X_{3} CH (OCH X_{3}) X_{2}$
(iv) a cyclobutane ring in which one carbon is $C = N - NH - CO - NH X_{2}$
(v) $CH X_{3} CH X_{2} - C (- OCH X_{2} CH X_{2} O -) - CH X_{2} CH X_{2} CH X_{3}$ (a five-membered 1,3-dioxolane ring on C-3)
(vi) $HOCH X_{2} OCH X_{3}$

As drawn above.

8.6

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then

H X_{3} O X^{+}

(ii) Tollens' reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid.

Solution

(i) $C X_{6} H X_{11} CH (OH) C X_{6} H X_{5}$ , cyclohexylphenylmethanol
(ii) $C X_{6} H X_{11} COO X^{-}$ (cyclohexanecarboxylate) and a silver mirror
(iii) $C X_{6} H X_{11} CH = NNHCONH X_{2}$ , the semicarbazone
(iv) $C X_{6} H X_{11} CH (OC X_{2} H X_{5}) X_{2}$ , the diethyl acetal
(v) $C X_{6} H X_{11} CH X_{3}$ , methylcyclohexane (Clemmensen reduction)

As listed above.

8.7

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.

Solution

Aldol (have α-hydrogen): (ii), (v), (vi), (vii)

(ii) $CH X_{3} CH X_{2} CH X_{2} CH (CH X_{3}) CH (OH) C (CH X_{3}) (CH X_{2} CH X_{2} CH X_{3}) CHO$ (3-hydroxy-2,4-dimethyl-2-propyloctanal)
(v) 2-(1-hydroxycyclohexyl)cyclohexan-1-one
(vi) $C X_{6} H X_{5} C (OH) (C X_{2} H X_{5}) CH (CH X_{3}) COC X_{6} H X_{5}$
(vii) $C X_{6} H X_{5} CH X_{2} CH (OH) CH (C X_{6} H X_{5}) CHO$ (3-hydroxy-2,4-diphenylbutanal)

Cannizzaro (no α-hydrogen): (i), (iii), (ix)

(i) $2 HCHO + NaOH HCOONa + CH X_{3} OH$
(iii) $2 C X_{6} H X_{5} CHO + NaOH C X_{6} H X_{5} COONa + C X_{6} H X_{5} CH X_{2} OH$
(ix) $2 CH X_{3} CH X_{2} C (CH X_{3}) X_{2} CHO + NaOH CH X_{3} CH X_{2} C (CH X_{3}) X_{2} COONa + CH X_{3} CH X_{2} C (CH X_{3}) X_{2} CH X_{2} OH$

Neither: (iv) benzophenone (a ketone with no α-hydrogen) and (viii) butan-1-ol (not a carbonyl compound).

Aldol: (ii), (v), (vi), (vii). Cannizzaro: (i), (iii), (ix). Neither: (iv), (viii).

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.

Solution

(i) $2 CH X_{3} CHO dil \cdot NaOH CH X_{3} CH (OH) CH X_{2} CHO NaBH X_{4} CH X_{3} CH (OH) CH X_{2} CH X_{2} OH$
(ii) $2 CH X_{3} CHO dil \cdot NaOH CH X_{3} CH (OH) CH X_{2} CHO heat CH X_{3} CH = CHCHO$ (dehydration of the aldol)
(iii) But-2-enal from (ii), then mild oxidation (Tollens' reagent does not touch C=C): $CH X_{3} CH = CHCHO Tollens X^{'} CH X_{3} CH = CHCOOH$

(i) Aldol then NaBH₄. (ii) Aldol then heat. (iii) But-2-enal oxidised by Tollens' reagent.

8.9

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.

Solution

Propanal + propanal (propanal is both): $CH X_{3} CH X_{2} CH (OH) CH (CH X_{3}) CHO$ , 3-hydroxy-2-methylpentanal
Butanal + butanal (butanal is both): $CH X_{3} CH X_{2} CH X_{2} CH (OH) CH (C X_{2} H X_{5}) CHO$ , 2-ethyl-3-hydroxyhexanal
Propanal nucleophile (its enolate), butanal electrophile: $CH X_{3} CH X_{2} CH X_{2} CH (OH) CH (CH X_{3}) CHO$ , 3-hydroxy-2-methylhexanal
Butanal nucleophile, propanal electrophile: $CH X_{3} CH X_{2} CH (OH) CH (C X_{2} H X_{5}) CHO$ , 2-ethyl-3-hydroxypentanal

The four aldols above (two self and two cross products).

8.10

An organic compound with the molecular formula

C X_{9} H X_{10} O

forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Solution

Forms a 2,4-DNP derivative: a carbonyl compound.
Reduces Tollens' reagent: an aldehyde.
Cannizzaro reaction: the –CHO has no α-hydrogen (attached directly to the ring).
Vigorous oxidation gives phthalic acid (1,2-dicarboxylic): two side chains on adjacent ring carbons.

$C X_{9} H X_{10} O = C X_{6} H X_{4} (CHO) (C X_{2} H X_{5})$ : the compound is 2-ethylbenzaldehyde.

2-Ethylbenzaldehyde.

8.11

An organic compound (A) (molecular formula

C X_{8} H X_{16} O X_{2}

) 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.

Solution

A is an ester. Since oxidation of C gives B, both have the same carbon skeleton (4 C each), and C gives but-1-ene, so C is butan-1-ol and B is butanoic acid. A is butyl butanoate.

$CH X_{3} CH X_{2} CH X_{2} COOCH X_{2} CH X_{2} CH X_{2} CH X_{3} + H X_{2} O H X^{+} CH X_{3} CH X_{2} CH X_{2} COOH + CH X_{3} CH X_{2} CH X_{2} CH X_{2} OH$
$CH X_{3} CH X_{2} CH X_{2} CH X_{2} OH H X_{2} CrO X_{4} CH X_{3} CH X_{2} CH X_{2} COOH$
$CH X_{3} CH X_{2} CH X_{2} CH X_{2} OH conc \cdot H X_{2} SO X_{4} heat CH X_{3} CH X_{2} CH = CH X_{2} + H X_{2} O$

A = butyl butanoate, B = butanoic acid, C = butan-1-ol.

8.12

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)

CH X_{3} CH X_{2} CH (Br) COOH

CH X_{3} CH (Br) CH X_{2} COOH

(CH X_{3}) X_{2} CHCOOH

CH X_{3} CH X_{2} CH X_{2} COOH

(acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength).

Solution

(i) Di-tert-butyl ketone < methyl tert-butyl ketone < acetone < acetaldehyde (more crowding lowers reactivity)
(ii) $(CH X_{3}) X_{2} CHCOOH < CH X_{3} CH X_{2} CH X_{2} COOH < CH X_{3} CH (Br) CH X_{2} COOH < CH X_{3} CH X_{2} CH (Br) COOH$ (branching adds +I; Br nearer –COOH strengthens most)
(iii) 4-Methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid

As ordered above.

8.13

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.

Solution

(i) Tollens' test: propanal gives a silver mirror (also red ppt with Fehling's); propanone does not.
(ii) Iodoform test ( $I X_{2} / NaOH$ ): acetophenone ( $C X_{6} H X_{5} COCH X_{3}$ ) gives a yellow precipitate of $CHI X_{3}$ ; benzophenone does not.
(iii) $NaHCO X_{3}$ : benzoic acid gives brisk effervescence of $CO X_{2}$ ; phenol does not. (Or neutral $FeCl X_{3}$ : phenol gives a violet colour.)
(iv) $NaHCO X_{3}$ : benzoic acid effervesces; ethyl benzoate does not.
(v) Iodoform test: pentan-2-one ( $CH X_{3} CO X^{-}$ ) gives yellow $CHI X_{3}$ ; pentan-3-one does not.
(vi) Tollens' test: benzaldehyde gives a silver mirror; acetophenone does not. (Iodoform: acetophenone positive.)
(vii) Iodoform test: ethanal gives yellow $CHI X_{3}$ ; propanal does not.

(i) Tollens'; (ii) iodoform; (iii) NaHCO₃ or FeCl₃; (iv) NaHCO₃; (v) iodoform; (vi) Tollens' or iodoform; (vii) iodoform.

8.14

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.

Solution

First make toluene: $C X_{6} H X_{6} + CH X_{3} Cl anhyd \cdot AlCl X_{3} C X_{6} H X_{5} CH X_{3}$ .

(i) $C X_{6} H X_{5} CH X_{3} KMnO X_{4} / KOH then H X^{+} C X_{6} H X_{5} COOH CH X_{3} OH, H X^{+} C X_{6} H X_{5} COOCH X_{3}$
(ii) $C X_{6} H X_{5} COOH conc \cdot HNO X_{3} / H X_{2} SO X_{4} m - O X_{2} NC X_{6} H X_{4} COOH$ (–COOH is meta-directing)
(iii) $C X_{6} H X_{5} CH X_{3} conc \cdot HNO X_{3} / H X_{2} SO X_{4} p - O X_{2} NC X_{6} H X_{4} CH X_{3}$ (separate from ortho) $KMnO X_{4} then H X^{+} p - O X_{2} NC X_{6} H X_{4} COOH$
(iv) $C X_{6} H X_{5} CH X_{3} Cl X_{2} h ν C X_{6} H X_{5} CH X_{2} Cl KCN C X_{6} H X_{5} CH X_{2} CN H X_{3} O X^{+} C X_{6} H X_{5} CH X_{2} COOH$
(v) $p - O X_{2} NC X_{6} H X_{4} CH X_{3} CrO X_{2} Cl X_{2} then H X_{3} O X^{+} p - O X_{2} NC X_{6} H X_{4} CHO$ (Etard reaction)

Make toluene (CH₃Cl/AlCl₃), then use the sequences above.

8.15

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) Benzaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m-Nitrobenzyl alcohol.

Solution

(i) $CH X_{3} COCH X_{3} NaBH X_{4} CH X_{3} CH (OH) CH X_{3} conc \cdot H X_{2} SO X_{4} heat CH X_{3} CH = CH X_{2}$
(ii) $C X_{6} H X_{5} COOH SOCl X_{2} C X_{6} H X_{5} COCl H X_{2}, Pd - BaSO X_{4} C X_{6} H X_{5} CHO$ (Rosenmund reduction)
(iii) $C X_{2} H X_{5} OH PCC CH X_{3} CHO dil \cdot NaOH CH X_{3} CH (OH) CH X_{2} CHO$
(iv) $C X_{6} H X_{6} CH X_{3} COCl anhyd \cdot AlCl X_{3} C X_{6} H X_{5} COCH X_{3} conc \cdot HNO X_{3} / H X_{2} SO X_{4} m - O X_{2} NC X_{6} H X_{4} COCH X_{3}$
(v) $C X_{6} H X_{5} CHO C X_{6} H X_{5} MgBr then H X_{3} O X^{+} (C X_{6} H X_{5}) X_{2} CHOH PCC (or CrO X_{3}) (C X_{6} H X_{5}) X_{2} CO$
(vi) $C X_{6} H X_{5} Br Mg dry ether C X_{6} H X_{5} MgBr CH X_{3} CHO then H X_{3} O X^{+} C X_{6} H X_{5} CH (OH) CH X_{3}$
(vii) $C X_{6} H X_{5} CHO + CH X_{3} CHO dil \cdot NaOH heat C X_{6} H X_{5} CH = CHCHO H X_{2}, Ni C X_{6} H X_{5} CH X_{2} CH X_{2} CH X_{2} OH$
(viii) $C X_{6} H X_{5} CHO HCN C X_{6} H X_{5} CH (OH) CN H X_{3} O X^{+} C X_{6} H X_{5} CH (OH) COOH$
(ix) $C X_{6} H X_{5} COOH conc \cdot HNO X_{3} / H X_{2} SO X_{4} m - O X_{2} NC X_{6} H X_{4} COOH B X_{2} H X_{6} m - O X_{2} NC X_{6} H X_{4} CH X_{2} OH$ (diborane reduces –COOH but not –NO₂)

As in the two-step sequences above.

8.16

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation.

Solution

(i) Acetylation: introduction of an acetyl group ( $CH X_{3} CO X^{-}$ ) in place of an active hydrogen of –OH, –NH₂ etc., using acetyl chloride or acetic anhydride (often with pyridine): $C X_{6} H X_{5} NH X_{2} + (CH X_{3} CO) X_{2} O C X_{6} H X_{5} NHCOCH X_{3} + CH X_{3} COOH$
(ii) Cannizzaro reaction: aldehydes with no α-hydrogen, heated with conc. alkali, disproportionate: one molecule is reduced to an alcohol and another oxidised to the salt of an acid: $2 HCHO + NaOH (conc \cdot) CH X_{3} OH + HCOONa$
(iii) Cross aldol condensation: aldol condensation between two different aldehydes or ketones. If both have α-hydrogens, four products form. It is useful when one partner has no α-hydrogen, e.g. $C X_{6} H X_{5} CHO + CH X_{3} COCH X_{3} NaOH C X_{6} H X_{5} CH = CHCOCH X_{3}$ (Claisen–Schmidt)
(iv) Decarboxylation: loss of $CO X_{2}$ from a carboxylic acid; the sodium salt heated with soda lime (NaOH + CaO) gives a hydrocarbon with one carbon less: $CH X_{3} COONa + NaOH CaO heat CH X_{4} + Na X_{2} CO X_{3}$

As described, with the examples above.

8.17

Complete each synthesis by giving missing starting material, reagent or products: (i) ethylbenzene,

KMnO X_{4}

, KOH, heat (ii) phthalic acid,

SOCl X_{2}

, heat (iii)

C X_{6} H X_{5} CHO

H X_{2} NCONHNH X_{2}

(iv) benzene → benzophenone (v) 4-oxocyclohexane-1-carbaldehyde,

[Ag (NH X_{3}) X_{2}] X^{+}

(vi) 2-formylbenzoic acid, NaCN/HCl (vii)

C X_{6} H X_{5} CHO + CH X_{3} CH X_{2} CHO

, dil. NaOH, heat (viii)

CH X_{3} COCH X_{2} COOC X_{2} H X_{5}

, (i)

NaBH X_{4}

(ii)

H X^{+}

(ix) cyclohexanol,

CrO X_{3}

(x) methylenecyclohexane → cyclohexanecarbaldehyde (xi) ? → (i)

O X_{3}

(ii) Zn–

H X_{2} O

→ 2 cyclohexanone.

Solution

(i) $C X_{6} H X_{5} COOK$ (potassium benzoate; benzoic acid after acidification)
(ii) $C X_{6} H X_{4} (COCl) X_{2}$ (phthaloyl chloride)
(iii) $C X_{6} H X_{5} CH = NNHCONH X_{2}$ (benzaldehyde semicarbazone)
(iv) Reagent: $C X_{6} H X_{5} COCl$ with anhydrous $AlCl X_{3}$ (Friedel–Crafts acylation)
(v) 4-Oxocyclohexane-1-carboxylate ion (only –CHO is oxidised; the ketone is untouched) and a silver mirror
(vi) 2-[Cyano(hydroxy)methyl]benzoic acid (HCN adds to the –CHO to form a cyanohydrin)
(vii) $C X_{6} H X_{5} CH = C (CH X_{3}) CHO$ (2-methyl-3-phenylprop-2-enal; cross aldol followed by dehydration)
(viii) $CH X_{3} CH (OH) CH X_{2} COOC X_{2} H X_{5}$ (NaBH₄ reduces the ketone, not the ester)
(ix) Cyclohexanone
(x) Reagents: (1) $B X_{2} H X_{6}$ , then $H X_{2} O X_{2} / OH X^{-}$ (gives cyclohexylmethanol), (2) PCC
(xi) Starting material: cyclohexylidenecyclohexane (two cyclohexane rings joined by a C=C)

As listed above.

8.18

Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. (ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Solution

(i) In 2,2,6-trimethylcyclohexanone, the three methyl groups next to the C=O crowd the carbonyl carbon (steric hindrance), so the cyanide ion cannot easily attack it. Cyclohexanone has no such hindrance.

(ii) Semicarbazide is $H X_{2} N - NH - CO - NH X_{2}$ . The –NH₂ attached directly to C=O is part of an amide: its lone pair is delocalised onto the carbonyl oxygen by resonance, so it is not nucleophilic. Only the –NH₂ on the hydrazine end (attached to NH) has a free lone pair and attacks the carbonyl compound.

(iii) Esterification is reversible: $RCOOH + R X^{'} OH RCOOR X^{'} + H X_{2} O$ . Removing water or the ester as it forms shifts the equilibrium to the right (Le Chatelier's principle) and increases the yield of ester.

(i) Steric hindrance by the methyl groups. (ii) One –NH₂ is deactivated by resonance with C=O. (iii) The reaction is reversible; removing a product drives it forward.

8.19

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Solution

Empirical formula: C = $\frac{69.77}{12} = 5.81$ , H = $\frac{11.63}{1} = 11.63$ , O = $\frac{18.60}{16} = 1.16$ . Dividing by 1.16 gives C : H : O = 5 : 10 : 1, so $C X_{5} H X_{10} O$ (mass 86, equal to the molecular mass).

No Tollens' reaction: a ketone.
Bisulphite addition and positive iodoform test: a methyl ketone, $CH X_{3} CO X^{-}$ .
Oxidation to ethanoic and propanoic acids: the chain breaks next to C=O into a 2-carbon and a 3-carbon part.

The compound is pentan-2-one, $CH X_{3} COCH X_{2} CH X_{2} CH X_{3}$ .

Pentan-2-one, CH₃COCH₂CH₂CH₃.

8.20

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Solution

In the carboxylate ion, the negative charge is shared equally by two oxygen atoms (highly electronegative); its two resonance structures are equivalent, which gives large stabilisation. In the phenoxide ion, most resonance structures put the negative charge on carbon atoms of the ring, which are less electronegative; these structures contribute less, and the charge mostly stays on one oxygen. So the carboxylate ion is much more stable, and carboxylic acids are stronger acids than phenol.

In carboxylate, the charge is spread over two equivalent, electronegative O atoms; in phenoxide, it is spread onto less electronegative carbons, which stabilises it less.

Cover of Complete Chemistry Notes for Class 12 (NCERT 2026-27, CBSE Board 2027)

Done with the answers?Now revise the whole chapter: Class 12 Chemistry notes

For NCERT Chemistry Class 12 Part I and Part II (2026-27 edition) - all 10 chapters, CBSE board 2027
Typed concept notes with every reaction, mechanism and named reaction
Every NCERT intext and exercise question solved

₹199₹24920% off · all 10 chapters · instant PDF

Read free sample

What is inside the Chapter 8 notes →

Board exam? Add the Class 12 Chemistry PYQ book (503 solved board questions) · ₹99

← Chapter 7: Alcohols, Phenols and Ethers Chapter 9: Amines →

More free NCERT solutions for Class 12Class 12 Maths · 10 chapters Class 12 Physics · 14 chapters Class 12 Biology · 13 chapters All classes →

Found a mistake or need help with a question? Message us on WhatsApp. Free study tips and new solutions: join our WhatsApp Channel.