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NCERT Solutions

Aldehydes, Ketones and Carboxylic Acids

CBSE · Class 12 · Chemistry

NCERT Solutions for Aldehydes, Ketones and Carboxylic Acids — CBSE Class 12 Chemistry.

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A labeled diagram showing the selective partial reduction of nitriles and esters to aldehydes using DIBAL-H (Diisobutylaluminum hydride) at low temperatures, followed by hydrolysis.
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Exercises

8.1What is meant by the following terms? Give an example of the reaction in each case.Show solution
(i) Cyanohydrin: Addition product formed when HCN adds to an aldehyde or ketone. Example:

RCHO+HCN→RCH(OH)CN\mathrm{RCHO + HCN \rightarrow RCH(OH)CN}

(ii) Acetal: Compound formed when an aldehyde reacts with two molecules of alcohol in presence of dry HCl. Example:

RCHO+2ROH→H+RCH(OR)2+H2O\mathrm{RCHO + 2ROH \xrightarrow{H^+} RCH(OR)_2 + H_2O}

(iii) Semicarbazone: Derivative formed when a carbonyl compound reacts with semicarbazide. Example:

R2C=O+H2N−NHCONH2→R2C=NNHCONH2+H2O\mathrm{R_2C=O + H_2N-NHCONH_2 \rightarrow R_2C=NNHCONH_2 + H_2O}

(iv) Aldol: \beta-hydroxy aldehyde or \beta-hydroxy ketone formed by aldol reaction. Example: two molecules of ethanal give aldol:

2CH3CHO→dil. NaOHCH3CH(OH)CH2CHO2\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}

(v) Hemiacetal: Product formed when one molecule of alcohol adds to an aldehyde. Example:

RCHO+ROH⇌RCH(OH)OR\mathrm{RCHO + ROH \rightleftharpoons RCH(OH)OR}

(vi) Oxime: Derivative formed when a carbonyl compound reacts with hydroxylamine. Example:

R2C=O+NH2OH→R2C=NOH+H2O\mathrm{R_2C=O + NH_2OH \rightarrow R_2C=NOH + H_2O}

(vii) Ketal: Cyclic or acyclic derivative formed when a ketone reacts with alcohols; with ethylene glycol it gives a cyclic ketal. Example:

R2C=O+2ROH→H+R2C(OR)2+H2O\mathrm{R_2C=O + 2ROH \xrightarrow{H^+} R_2C(OR)_2 + H_2O}

(viii) Imine: Compound containing the group **>C=NH>C=NH** formed by reaction of a carbonyl compound with ammonia. Example:

R2C=O+NH3→R2C=NH+H2O\mathrm{R_2C=O + NH_3 \rightarrow R_2C=NH + H_2O}

(ix) 2,4-DNP-derivative: Yellow, orange or red derivative formed when a carbonyl compound reacts with 2,4-dinitrophenylhydrazine. Example:

R2C=O+H2NNHC6H3(NO2)2→R2C=NNHC6H3(NO2)2+H2O\mathrm{R_2C=O + H_2NNHC_6H_3(NO_2)_2 \rightarrow R_2C=NNHC_6H_3(NO_2)_2 + H_2O}

(x) Schiff's base: A substituted imine formed when a carbonyl compound reacts with a primary amine. Example:

R2C=O+R′NH2→R2C=NR′+H2O\mathrm{R_2C=O + R'NH_2 \rightarrow R_2C=NR' + H_2O}

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8.1(i)CyanohydrinShow solution
A cyanohydrin is the addition product formed when HCN adds to a carbonyl compound. Example:

RCHO+HCN→RCH(OH)CN\mathrm{RCHO + HCN \rightarrow RCH(OH)CN}

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8.1(ii)AcetalShow solution
An acetal is formed when an aldehyde reacts with two molecules of alcohol in the presence of an acid catalyst. Example:

RCHO+2ROH→H+RCH(OR)2+H2O\mathrm{RCHO + 2ROH \xrightarrow{H^+} RCH(OR)_2 + H_2O}

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8.1(iii)SemicarbazoneShow solution
A semicarbazone is the derivative formed when a carbonyl compound reacts with semicarbazide. Example:

R2C=O+H2N−NHCONH2→R2C=NNHCONH2+H2O\mathrm{R_2C=O + H_2N-NHCONH_2 \rightarrow R_2C=NNHCONH_2 + H_2O}

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8.1(iv)AldolShow solution
An aldol is a \beta-hydroxy aldehyde or \beta-hydroxy ketone formed in aldol reaction. Example:

2CH3CHO→dil. NaOHCH3CH(OH)CH2CHO2\mathrm{CH_3CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH_3CH(OH)CH_2CHO}

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8.1(v)HemiacetalShow solution
A hemiacetal is formed when one molecule of alcohol adds to an aldehyde. Example:

RCHO+ROH⇌RCH(OH)OR\mathrm{RCHO + ROH \rightleftharpoons RCH(OH)OR}

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8.1(vi)OximeShow solution
An oxime is the derivative formed when a carbonyl compound reacts with hydroxylamine. Example:

R2C=O+NH2OH→R2C=NOH+H2O\mathrm{R_2C=O + NH_2OH \rightarrow R_2C=NOH + H_2O}

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8.1(vii)KetalShow solution
A ketal is the product formed when a ketone reacts with alcohols; with ethylene glycol it forms a cyclic ketal. Example:

R2C=O+2ROH→H+R2C(OR)2+H2O\mathrm{R_2C=O + 2ROH \xrightarrow{H^+} R_2C(OR)_2 + H_2O}

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8.1(vii)ImineShow solution
An imine has the group **>C=NH>C=NH and is formed when a carbonyl compound reacts with ammonia**. Example:

R2C=O+NH3→R2C=NH+H2O\mathrm{R_2C=O + NH_3 \rightarrow R_2C=NH + H_2O}

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8.1(ix)2,4-DNP-derivativeShow solution
A 2,4-DNP-derivative is a yellow, orange or red derivative formed when a carbonyl compound reacts with 2,4-dinitrophenylhydrazine. Example:

R2C=O+H2NNHC6H3(NO2)2→R2C=NNHC6H3(NO2)2+H2O\mathrm{R_2C=O + H_2NNHC_6H_3(NO_2)_2 \rightarrow R_2C=NNHC_6H_3(NO_2)_2 + H_2O}

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8.1(x)Schiff's baseShow solution
A Schiff's base is a substituted imine formed when a carbonyl compound reacts with a primary amine. Example:

R2C=O+R′NH2→R2C=NR′+H2O\mathrm{R_2C=O + R'NH_2 \rightarrow R_2C=NR' + H_2O}

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8.2Name the following compounds according to IUPAC system of nomenclature:Show solution
The IUPAC names are:

(i) 3-Methylpentanal

(ii) 4-Chloro-5-ethylhexan-3-one

(iii) But-2-enal

(iv) Pentan-2,4-dione

(v) 2,4,4-Trimethylhexan-2-one

(vi) 2,2-Dimethylbutanoic acid

(vii) Benzene-1,4-dicarbaldehyde

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8.2(i)CH3CH(CH3)CH2CH2CHO\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}Show solution
Choose the longest chain containing the aldehyde carbon. The chain has 5 carbons, so the parent name is pentanal. Number from the aldehyde carbon as 11, and the methyl substituent is on carbon 33.

So the IUPAC name is 3-methylpentanal.

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8.2(ii)CH3CH2COCH(C2H5)CH2CH2Cl\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}(\mathrm{C}_2\mathrm{H}_5)\mathrm{CH}_2\mathrm{CH}_2\mathrm{Cl}Show solution
The longest chain containing the carbonyl group has 6 carbons, so the parent is hexan-3-one. Number from the end nearer the carbonyl group, giving the carbonyl at 33. The substituents are ethyl at 55 and chloro at 44. Alphabetical order gives chloro before ethyl.

So the IUPAC name is 4-chloro-5-ethylhexan-3-one.

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8.2(iii)CH3CH=CHCHO\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCHO}Show solution
The chain has 4 carbons with an aldehyde group at carbon 11 and a double bond between carbons 22 and 33.

So the IUPAC name is but-2-enal.

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8.2(iv)CH3COCH2COCH3\mathrm{CH}_3\mathrm{COCH}_2\mathrm{COCH}_3Show solution
The chain has 5 carbons and two ketone groups at positions 22 and 44.

So the IUPAC name is pentan-2,4-dione.

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8.2(v)CH3CH(CH3)CH2C(CH3)2COCH3\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{C}(\mathrm{CH}_3)_2\mathrm{COCH}_3Show solution
The longest chain containing the carbonyl group has 6 carbons, so the parent is hexan-2-one. Number from the end nearer the carbonyl group so the carbonyl is at 22. There are three methyl substituents: one at 22, one at 44, and one more at 44.

So the IUPAC name is 2,4,4-trimethylhexan-2-one.

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8.2(vi)(CH3)3CCH2COOH(\mathrm{CH}_3)_3\mathrm{CCH}_2\mathrm{COOH}Show solution
The carboxyl carbon is numbered 11. The longest chain has 4 carbons, so the parent acid is butanoic acid. Two methyl groups are attached to carbon 22.

So the IUPAC name is 2,2-dimethylbutanoic acid.

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8.2(vii)OHCC6H4CHO−p\mathrm{OHCC}_6\mathrm{H}_4\mathrm{CHO}-pShow solution
The structure has a benzene ring with two aldehyde groups at the 1,4-positions. Therefore the IUPAC name is benzene-1,4-dicarbaldehyde.

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8.3Draw the structures of the following compounds.Show solution
The structures are:

(i) 3-Methylbutanal: CH3CH(CH3)CH2CHO\mathrm{CH_3CH(CH_3)CH_2CHO}

(ii) p-Nitropropiophenone: $\mathrm{p\!-
O_2NC_6H_4COCH_2CH_3}$

(iii) p-Methylbenzaldehyde: $\mathrm{p\!-
CH_3C_6H_4CHO}$

(iv) 4-Methylpent-3-en-2-one: CH3COCH=C(CH3)CH3\mathrm{CH_3COCH=C(CH_3)CH_3}

(v) 4-Chloropentan-2-one: CH3COCH2CH(Cl)CH3\mathrm{CH_3COCH_2CH(Cl)CH_3}

(vi) 3-Bromo-4-phenylpentanoic acid: HOOCCH2CH(Br)CH(C6H5)CH3\mathrm{HOOCCH_2CH(Br)CH(C_6H_5)CH_3}

(vii) p,p'-Dihydroxybenzophenone: $\mathrm{HO\!-
C_6H_4COC_6H_4\!-
OH}withboth with both \mathrm{-OH}$ groups para to the carbonyl group

(viii) Hex-2-en-4-ynoic acid: HOOCCH=CHC≡CCH3\mathrm{HOOCCH=CHC\equiv CCH_3}

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8.3(i)3-MethylbutanalShow solution
The parent chain has 4 carbons including the aldehyde carbon. Numbering starts at the aldehyde carbon, and the methyl substituent is on carbon 33.

Structure: **CH3CH(CH3)CH2CHO\mathrm{CH_3CH(CH_3)CH_2CHO}**

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8.3(ii)pp-NitropropiophenoneShow solution
This is propiophenone with a para-nitro substituent on the benzene ring.

Structure: $\mathrm{p\!-
O_2NC_6H_4COCH_2CH_3}$

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8.3(iii)pp-MethylbenzaldehydeShow solution
This is benzaldehyde with a para-methyl substituent on the ring.

Structure: $\mathrm{p\!-
CH_3C_6H_4CHO}$

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8.3(iv)4-Methylpent-3-en-2-oneShow solution
The structure of 4-methylpent-3-en-2-one is:

CH3COCH=C(CH3)CH3\mathrm{CH_3COCH=C(CH_3)CH_3}

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8.3(v)4-Chloropentan-2-oneShow solution
The structure of 4-chloropentan-2-one is:

CH3COCH2CH(Cl)CH3\mathrm{CH_3COCH_2CH(Cl)CH_3}

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8.3(vi)3-Bromo-4-phenylpentanoic acidShow solution
The longest chain containing the carboxyl group has five carbons, so the parent is pentanoic acid. Numbering starts from the –COOH carbon as carbon 1. The substituents are bromo at C-3 and phenyl at C-4.

So the IUPAC name is 3-bromo-4-phenylpentanoic acid.

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8.3(vii)p,p′p,p'-DihydroxybenzophenoneShow solution
In benzophenone, two benzene rings are attached to a carbonyl group. With hydroxy groups at the 4 and 4′ positions, the IUPAC name is 4,4′-dihydroxybenzophenone.

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8.3(viii)Hex-2-en-4-ynoic acidShow solution
The structure has a six-carbon carboxylic acid chain. Numbering starts from the –COOH carbon as C-1. The double bond is at C-2 and the triple bond at C-4, so the name is hex-2-en-4-ynoic acid.

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8.4Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.Show solution
The IUPAC names are:

1. 3-Bromo-4-phenylpentanoic acid
2. 4,4′-Dihydroxybenzophenone
3. Hex-2-en-4-ynoic acid

Method used: name the parent chain containing the principal functional group, number from the carboxyl carbon, then assign positions to the substituents and multiple bonds.

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8.4(i)CH3CO(CH2)4CH3\mathrm{CH}_3\mathrm{CO}(\mathrm{CH}_2)_4\mathrm{CH}_3Show solution
For $
\mathrm{CH_3CO(CH_2)_4CH_3}$, the longest chain containing the carbonyl group has 7 carbons:

CH3−CO−(CH2)4−CH3\mathrm{CH_3-CO-(CH_2)_4-CH_3}

Number from the end nearer the carbonyl group, so the carbonyl is at C-2. Therefore the IUPAC name is heptan-2-one.

Wait — the given formula is exactly:
CH3CO(CH2)4CH3\mathrm{CH_3CO(CH_2)_4CH_3} = CH3−CO−CH2−CH2−CH2−CH2−CH3\mathrm{CH_3-CO-CH_2-CH_2-CH_2-CH_2-CH_3}, which is heptan-2-one.

The common name is methyl n-hexyl ketone.

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8.4(ii)CH3CH2CHBrCH2CH(CH3)CHO\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHBrCH}_2\mathrm{CH}(\mathrm{CH}_3)\mathrm{CHO}Show solution
Take the longest chain containing the aldehyde group. The aldehyde carbon is C-1.

Structure: CH3CH2CHBrCH2CH(CH3)CHO\mathrm{CH_3CH_2CHBrCH_2CH(CH_3)CHO}

Numbering from the −CHO-CHO carbon:
- C-2 has methyl
- C-4 has bromo

So the IUPAC name is 4-bromo-2-methylhexanal.

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8.4(iii)CH3(CH2)5CHO\mathrm{CH}_3(\mathrm{CH}_2)_5\mathrm{CHO}Show solution
The structure CH3(CH2)5CHO\mathrm{CH_3(CH_2)_5CHO} has 7 carbons in the chain including the aldehyde carbon. Therefore the IUPAC name is heptanal. The common name is oenanthaldehyde.

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8.4(iv)Ph−CH=CH−CHO\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}Show solution
For Ph−CH=CH−CHO\mathrm{Ph-CH=CH-CHO}, the parent chain is prop-2-enal and the phenyl group is at C-3 when numbering starts from the aldehyde carbon.

So the IUPAC name is 3-phenylprop-2-enal. The common name is cinnamaldehyde.

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8.4(vi)PhCOPh\mathrm{PhCOPh}Show solution
PhCOPh\mathrm{PhCOPh} has two phenyl groups attached to a carbonyl carbon, so it is benzophenone. Its IUPAC name is also accepted as diphenylmethanone.

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8.5Draw structures of the following derivatives.Show solution
The structures are:

1. 2,4-Dinitrophenylhydrazone of benzaldehyde: benzaldehyde condensed with 2,4-DNP to give the hydrazone, i.e. C6H5CH=N−NH−C6H3(NO2)2\mathrm{C_6H_5CH=N-NH-C_6H_3(NO_2)_2}.
2. Cyclopropanone oxime: cyclopropanone carbonyl converted to C=NOH\mathrm{C=NOH}.
3. Acetaldehyde dimethyl acetal: CH3CH(OCH3)2\mathrm{CH_3CH(OCH_3)_2}.
4. Semicarbazone of cyclobutanone: cyclobutanone carbonyl converted to C=NNHCONH2\mathrm{C=NNHCONH_2}.
5. Ethylene ketal of hexan-3-one: cyclic ketal formed with ethylene glycol at the carbonyl carbon of hexan-3-one.
6. Methyl hemiacetal of formaldehyde: HO−CH2−OCH3\mathrm{HO-CH_2-OCH_3}.

These are drawn by replacing the carbonyl oxygen appropriately with the corresponding derivative group.

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8.5(i)The 2,4-dinitrophenylhydrazone of benzaldehydeShow solution
Benzaldehyde reacts with 2,4-dinitrophenylhydrazine to form the corresponding 2,4-DNP hydrazone. The product structure is:

C6H5CH=N−NH−C6H3(NO2)2\mathrm{C_6H_5CH=N-NH-C_6H_3(NO_2)_2}

This is the condensation product of the aldehyde with 2,4-DNP.

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8.5(ii)Cyclopropanone oximeShow solution
Cyclopropanone oxime is the oxime formed when the carbonyl group of cyclopropanone reacts with hydroxylamine, giving the structure with the group **C=NOH\mathrm{C=NOH}** on the cyclopropane ring.

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8.5(iii)AcetaldehydedimethylacetalShow solution
Acetaldehyde dimethyl acetal is formed by addition of two methanol molecules to acetaldehyde. The structure is:

CH3CH(OCH3)2\mathrm{CH_3CH(OCH_3)_2}

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8.5(iv)The semicarbazone of cyclobutanoneShow solution
Cyclobutanone reacts with semicarbazide to form the corresponding semicarbazone, having the carbonyl group converted into **C=N−NHCONH2\mathrm{C=N-NHCONH_2}** on the cyclobutane ring.

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8.5(v)The ethylene ketal of hexan-3-oneShow solution
Hexan-3-one reacts with ethylene glycol to form the corresponding cyclic ketal. The product is the ethylene ketal of hexan-3-one.

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8.5(vi)The methyl hemiacetal of formaldehydeShow solution
The methyl hemiacetal of formaldehyde has one hydroxyl group and one methoxy group on the same carbon. Its structure is:

HOCH2OCH3\mathrm{HOCH_2OCH_3}

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8.6Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.Show solution
For cyclohexanecarbaldehyde:

- With **PhMgBr followed by H3O+\mathrm{H_3O^+}, the Grignard reagent adds to the aldehyde carbonyl to give a secondary alcohol.
- With Tollens' reagent, the aldehyde is oxidised to the corresponding carboxylic acid salt.
- With semicarbazide and weak acid, it forms the semicarbazone.
- With excess ethanol and acid, it forms the acetal.
- With zinc amalgam and dilute HCl, the aldehyde group is reduced to methyl**, giving the corresponding hydrocarbon.

These follow the standard reactions of aldehydes given in the chapter.

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8.6(i)PhMgBr\mathrm{PhMgBr} and then H3O+\mathrm{H}_3\mathrm{O}^+Show solution
A Grignard reagent, PhMgBr, adds to the aldehyde carbonyl carbon. After acidic hydrolysis, the carbonyl carbon becomes an alcohol carbon bearing Ph, H, and the cyclohexyl group.

So the product is 1-phenylcyclohexan-1-ol.

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8.6(ii)Tollens' reagentShow solution
Tollens' reagent oxidises the aldehyde group to the corresponding carboxylate, which on work-up corresponds to cyclohexanecarboxylic acid.

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8.6(iii)Semicarbazide and weak acidShow solution
An aldehyde reacts with semicarbazide in weak acid to form the corresponding semicarbazone. Therefore the product is cyclohexanecarbaldehyde semicarbazone.

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8.6(iv)Excess ethanol and acidShow solution
With excess ethanol and acid, an aldehyde forms the acetal. So cyclohexanecarbaldehyde gives cyclohexanecarbaldehyde diethyl acetal.

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8.6(v)Zinc amalgam and dilute hydrochloric acidShow solution
Zn(Hg)/dil. HCl causes Clemmensen reduction, converting the aldehyde group (−CHO)(-CHO) into (−CH3)(-CH_3).

So cyclohexanecarbaldehyde becomes methylcyclohexane (also written as cyclohexylmethane in some contexts).

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8.7Which 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.Show solution
Classify each compound by whether it has an **b1b1-hydrogen.

- Methanal**: no b1b1-hydrogen, so it undergoes Cannizzaro reaction.
- 2-Methylpentanal: has b1b1-hydrogen, so it undergoes aldol condensation.
- Benzaldehyde: no b1b1-hydrogen, so it undergoes Cannizzaro reaction.
- Benzophenone: no b1b1-hydrogen and is a ketone, so neither.
- Cyclohexanone: has b1b1-hydrogen, so it undergoes aldol condensation.
- 1-Phenylpropanone: has b1b1-hydrogen, so it undergoes aldol condensation.
- Phenylacetaldehyde: has b1b1-hydrogen, so it undergoes aldol condensation.
- Butan-1-ol: not a carbonyl compound, so neither.
- 2,2-Dimethylbutanal: aldehyde but no b1b1-hydrogen, so Cannizzaro reaction.

Expected products:
- Cannizzaro gives one molecule of alcohol and one of carboxylate salt.
- Aldol gives a b2b2-hydroxy aldehyde/ketone, which may dehydrate to an b1,b2b1,b2-unsaturated carbonyl compound.

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8.7(i)MethanalShow solution
Methanal has **no b1b1-hydrogen, so it cannot undergo aldol condensation. In strong alkali it undergoes self oxidation and reduction to give the Cannizzaro reaction**.

For methanal:
2HCHO+NaOH→HCOONa+CH3OH\mathrm{2HCHO + NaOH \rightarrow HCOONa + CH_3OH}

So the correct classification is Cannizzaro reaction.

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8.7(ii)2-MethylpentanalShow solution
2-Methylpentanal has an **b1b1-hydrogen, so it can form an enolate ion and undergo aldol condensation**. It does not undergo Cannizzaro reaction because it has an b1b1-hydrogen.

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8.7(iii)BenzaldehydeShow solution
In **reactions with NaHSO3_3 or HCN, benzaldehyde is more reactive than ketones because it has only one carbon group attached to the carbonyl carbon and less steric hindrance. In contrast, ketones are less reactive due to greater steric and electron-donating effects from two alkyl/aryl groups. So among the listed compounds, benzaldehyde** is the aldehyde and would be more reactive than ketones in nucleophilic addition.

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8.7(iv)Benzophenone
8.7(v)Cyclohexanone
8.7(vi)1-Phenylpropanone
8.7(vii)Phenylacetaldehyde
8.7(viii)Butan-1-ol
8.7(ix)2,2-Dimethylbutanal
8.8How will you convert ethanal into the following compounds?
8.8(i)Butane-1,3-diol
8.8(ii)But-2-enal
8.8(iii)But-2-enoic acid
8.9Write 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.
8.10An organic compound with the molecular formula C9H10O\mathrm{C_9H_{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.
8.11An organic compound (A) (molecular formula C9H10O3\mathrm{C_9H_{10}O_3}) 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.
8.12Arrange the following compounds in increasing order of their property as indicated:
8.12(i)Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
8.12(ii)CH3CH2CH(Br)COOH\mathrm{CH_3CH_2CH(Br)COOH}, CH3CH(Br)CH2COOH\mathrm{CH_3CH(Br)CH_2COOH}, (CH3)2CHCOOH(\mathrm{CH_3})_2\mathrm{CHCOOH}, CH3CH2CH2COOH\mathrm{CH_3CH_2CH_2COOH} (acid strength)
8.12(iii)Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
8.13(i)Propanal and Propanone
8.13(ii)Acetophenone and Benzophenone
8.13(iii)Phenol and Benzoic acid
8.13(iv)Benzoic acid and Ethyl benzoate
8.13(v)Pentan-2-one and Pentan-3-one
8.13(vi)Benzaldehyde and Acetophenone
8.13(vii)Ethanal and Propanal
8.14How 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.
8.14(i)Methyl benzoate
8.14(ii)mm-Nitrobenzoic acid
8.14(iii)pp-Nitrobenzoic acid
8.14(iv)Phenylacetic acid
8.14(v)pp-Nitrobenzaldehyde.
8.15How will you bring about the following conversions in not more than two steps?
8.15(i)Propanone to Propene
8.15(ii)Benzoic acid to Benzaldehyde
8.15(iii)Ethanol to 3-Hydroxybutanal
8.15(iv)Benzene to mm-Nitroacetophenone
8.15(v)Benzaldehyde to Benzophenone
8.15(vi)Bromobenzene to 1-Phenylethanol
8.15(vii)Benzaldehyde to 3-Phenylpropan-1-ol
8.15(viii)Benazaldehyde to α\alpha-Hydroxyphenylacetic acid
8.15(ix)Benzoic acid to mm-Nitrobenzyl alcohol
8.16Describe the following:
8.16(i)Acetylation
8.16(ii)Cannizzaro reaction
8.16(iii)Cross aldol condensation
8.16(iv)Decarboxylation
8.17Complete each synthesis by giving missing starting material, reagent or products
8.18(i)Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
8.18(ii)There are two −NH2-\mathrm{NH}_2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
8.18(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.
8.19An organic compound contains 69.77%69.77\% carbon, 11.63%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.
8.20Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

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