Q8.3 (i): 3-Methylbutanal

(i) 3-Methylbutanal The parent chain is butanal, which is a 4-carbon aldehyde. A methyl group is attached to carbon 3. CH_3-CH(CH_3)-CH_2-CHO (ii) p -Nitropropiophenone Propiophenone is a phenyl group attached to a propanoyl group ( C_6H_5-CO-CH_2-CH_3 ). A nitro group ( -NO_2 ) is at the para (p)…

Q8.3 (ii): p -Nitropropiophenone

(i) 3-Methylbutanal The parent chain is butanal, which is a 4-carbon aldehyde. A methyl group is attached to carbon 3. CH_3-CH(CH_3)-CH_2-CHO (ii) p -Nitropropiophenone Propiophenone is a phenyl group attached to a propanoyl group ( C_6H_5-CO-CH_2-CH_3 ). A nitro group ( -NO_2 ) is at the para (p)…

Q8.3 (iii): p-Methylbenzaldehyde

(i) 3-Methylbutanal The parent chain is butanal, which is a 4-carbon aldehyde. A methyl group is attached to carbon 3. CH_3-CH(CH_3)-CH_2-CHO (ii) p -Nitropropiophenone Propiophenone is a phenyl group attached to a propanoyl group ( C_6H_5-CO-CH_2-CH_3 ). A nitro group ( -NO_2 ) is at the para (p)…

Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry NCERT Solutions

Q8.1Exercises

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

Solution

(i)

Cyanohydrin: A compound containing a hydroxyl group (-OH) and a cyano group (-CN) attached to the same carbon atom. They are formed by the nucleophilic addition of hydrogen cyanide (HCN) to the carbonyl group of an aldehyde or a ketone. Example Reaction: Acetaldehyde reacts with hydrogen cyanide to form acetaldehyde cyanohydrin.

\mathrm{CH}_{3} \mathrm{CHO} + \mathrm{HCN} \rightarrow \mathrm{CH}_{3} \mathrm{CH(OH)CN}

(ii)

Acetal: A compound that has two alkoxy (-OR) groups attached to the same carbon atom. Acetals are formed when an aldehyde reacts with two equivalents of an alcohol in the presence of an acid catalyst (like dry HCl gas). Example Reaction: Acetaldehyde reacts with ethanol to form 1,1-diethoxyethane (acetaldehyde diethyl acetal).

\mathrm{CH}_{3} \mathrm{CHO} + 2\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \xrightarrow{\mathrm{HCl(g)}} \mathrm{CH}_{3} \mathrm{CH(OC}_{2} \mathrm{H}_{5})_{2} + \mathrm{H}_{2} \mathrm{O}

(iii)

Semicarbazone: A derivative of an aldehyde or ketone formed by condensation reaction with semicarbazide (

H_2N-NH-CO-NH_2

). These are crystalline solids useful for characterization. Example Reaction: Acetone reacts with semicarbazide to form acetone semicarbazone.

\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C=O} + \mathrm{H}_{2} \mathrm{NNHCONH}_{2} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C=NNHCONH}_{2} + \mathrm{H}_{2} \mathrm{O}

(iv)

Aldol: A \beta-hydroxy aldehyde or \beta-hydroxy ketone. The name is derived from 'aldehyde' and 'alcohol'. It is the product of an aldol addition reaction, where two molecules of an aldehyde or ketone (with at least one \alpha-hydrogen) react in the presence of a dilute base. Example Reaction: Two molecules of ethanal (acetaldehyde) react to form 3-hydroxybutanal (an aldol).

2\mathrm{CH}_{3} \mathrm{CHO} \xrightarrow{\text{dil. NaOH}} \mathrm{CH}_{3} \mathrm{CH(OH)CH}_{2} \mathrm{CHO}

(v)

Hemiacetal: A compound containing a hydroxyl group (-OH) and an alkoxy group (-OR) attached to the same carbon atom. They are formed as intermediates during acetal formation from an aldehyde and one equivalent of an alcohol. Example Reaction: Acetaldehyde reacts with one molecule of ethanol in a reversible reaction to form a hemiacetal.

\mathrm{CH}_{3} \mathrm{CHO} + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH(OH)OC}_{2} \mathrm{H}_{5}

(vi)

Oxime: A compound containing the functional group >C=N-OH. Oximes are formed when an aldehyde or a ketone reacts with hydroxylamine (

NH_2OH

). Example Reaction: Propanone (acetone) reacts with hydroxylamine to form propanone oxime (acetoxime).

\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C=O} + \mathrm{H}_{2} \mathrm{NOH} \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C=NOH} + \mathrm{H}_{2} \mathrm{O}

(vii)

Ketal: A compound analogous to an acetal, but derived from a ketone instead of an aldehyde. It has two alkoxy (-OR) groups attached to the carbon that was formerly the carbonyl carbon of the ketone. Example Reaction: Propanone (acetone) reacts with ethylene glycol in the presence of an acid catalyst to form a cyclic ketal.

\mathrm{CH}_{3} \mathrm{COCH}_{3} + \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} \xrightarrow{\mathrm{H}^{+}} \text{cyclic ketal} + \mathrm{H}_{2} \mathrm{O}

(viii)

Imine: A compound containing a carbon-nitrogen double bond (>C=N-). They are formed by the reaction of aldehydes or ketones with primary amines. Example Reaction: Acetaldehyde reacts with methylamine to form an imine.

\mathrm{CH}_{3} \mathrm{CHO} + \mathrm{CH}_{3} \mathrm{NH}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{CH=NCH}_{3} + \mathrm{H}_{2} \mathrm{O}

(ix)

2,4-DNP-derivative: The product formed when an aldehyde or ketone reacts with 2,4-dinitrophenylhydrazine (also known as Brady's reagent). These derivatives are typically brightly colored crystalline solids with sharp melting points, used for identifying carbonyl compounds. Example Reaction: Propanal reacts with 2,4-dinitrophenylhydrazine.

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} + \mathrm{H}_{2} \mathrm{N-NH-C}_{6} \mathrm{H}_{3} (\mathrm{NO}_{2})_{2} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH=N-NH-C}_{6} \mathrm{H}_{3} (\mathrm{NO}_{2})_{2} + \mathrm{H}_{2} \mathrm{O}

(x)

Schiff's base: A specific type of imine, having the general structure R'R''C=NR'''. It is formed from the condensation of a primary amine with an aldehyde or a ketone. The term is often used for imines derived from aromatic amines. Example Reaction: Benzaldehyde reacts with aniline to form benzylideneaniline, a Schiff's base.

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH=NC}_{6} \mathrm{H}_{5} + \mathrm{H}_{2} \mathrm{O}

Q8.2Exercises

Name the following compounds according to IUPAC system of nomenclature:

(i)

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$

(ii)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$

(iii)

$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$

(iv)

$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COCH}_{3}$

(v)

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}$

(vi)

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{COOH}$

(vii)

$\mathrm{OHCC}_{6} \mathrm{H}_{4} \mathrm{CHO}-p$

Solution

(i)

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$ The functional group is an aldehyde (-CHO). The longest carbon chain including the aldehyde carbon has 5 carbons, so the parent alkane is pentane. The suffix is '-al'. Numbering starts from the aldehyde carbon as C-1. There is a methyl group at C-4. IUPAC name: 4-Methylpentanal

(ii)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$ The functional group is a ketone (>C=O). The longest carbon chain containing the carbonyl carbon has 8 carbons (octane). To find the longest chain, we expand the ethyl group: CH3-CH2-CO-CH(CH2CH3)-CH2-CH2-Cl. The chain is Cl-CH2-CH2-CH(CH2-CH3)-CO-CH2-CH3. The longest chain goes through the ethyl branch: Cl(C1)-C2-C3-C4(CH(CH3))-C5O-C6-C7. No, that's not right. The main chain is Cl-CH2-CH2-CH(C2H5)-CO-CH2-CH3. This is Cl-C1-C2-C3H(C4H2-C5H3)-C6O-C7H2-C8H3. So it's an 8-carbon chain. Numbering from left to right gives the carbonyl group position 6. Numbering from right to left gives position 3. So, we number from the right: CH3(1)-CH2(2)-CO(3)-CH(C2H5)(4)-CH2(5)-CH2(6)-Cl(7). The main chain is 7 carbons long. Heptan-3-one. Substituents are chloro at C-7 and ethyl at C-4. IUPAC name: 7-Chloro-4-ethylheptan-3-one

(iii)

$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$ This is an aldehyde with a double bond. The chain has 4 carbons (but-). The aldehyde group gets C-1. The double bond is between C-2 and C-3. The name is formed by replacing 'e' of butene with 'al'. IUPAC name: But-2-enal

(iv)

$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COCH}_{3}$ This is a diketone. The chain has 5 carbons (pentane). The two ketone groups are at C-2 and C-4. The suffix is '-dione'. IUPAC name: Pentane-2,4-dione

(v)

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}$ This is a ketone. The longest chain has 6 carbons (hexane). Numbering from the right gives the carbonyl group position 2. There are methyl groups at C-3, C-3, and C-5. IUPAC name: 3,3,5-Trimethylhexan-2-one

(vi)

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{COOH}$ This is a carboxylic acid. The longest chain including the carboxyl group has 4 carbons (butanoic acid). Numbering starts from the carboxyl carbon as C-1. There are two methyl groups at C-3. Wait, the structure is a tert-butyl group attached to -CH2COOH. So it's (CH3)3C-CH2-COOH. The longest chain is HOOC(1)-CH2(2)-C(CH3)2(3)-CH3(4). So it's a 4-carbon chain. Butanoic acid. There are two methyl groups at C-3. IUPAC name: 3,3-Dimethylbutanoic acid

(vii)

$\mathrm{OHCC}_{6} \mathrm{H}_{4} \mathrm{CHO}-p$ This compound has two aldehyde (-CHO) groups attached to a benzene ring at para positions (1 and 4). When an aldehyde group is attached to a ring, the suffix '-carbaldehyde' is used. Since there are two such groups, the name is based on benzene. IUPAC name: Benzene-1,4-dicarbaldehyde

Q8.3Exercises

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)

3-Methylbutanal The parent chain is butanal, which is a 4-carbon aldehyde. A methyl group is attached to carbon 3.

\mathrm{CH_3-CH(CH_3)-CH_2-CHO}

(ii)

$p$ -Nitropropiophenone Propiophenone is a phenyl group attached to a propanoyl group (

\mathrm{C_6H_5-CO-CH_2-CH_3}

). A nitro group (

\mathrm{-NO_2}

) is at the para (p) position of the phenyl ring.

Structure: A benzene ring with a

\mathrm{-CO-CH_2-CH_3}

group and a

\mathrm{-NO_2}

group at positions 1 and 4, respectively.

(iii)

p-Methylbenzaldehyde Benzaldehyde is an aldehyde group (

\mathrm{-CHO}

) attached to a benzene ring. A methyl group (

\mathrm{-CH_3}

) is at the para (p) position.

Structure: A benzene ring with a

\mathrm{-CHO}

group and a

\mathrm{-CH_3}

group at positions 1 and 4, respectively.

(iv)

4-Methylpent-3-en-2-one The parent chain is a 5-carbon chain (pent) with a ketone at position 2 (-2-one) and a double bond at position 3 (-3-en-). A methyl group is at position 4.

\mathrm{CH_3-C(CH_3)=CH-CO-CH_3}

(v)

4-Chloropentan-2-one The parent chain is a 5-carbon chain (pentan) with a ketone at position 2 (-2-one). A chloro group is at position 4.

\mathrm{CH_3-CO-CH_2-CH(Cl)-CH_3}

(vi)

3-Bromo-4-phenylpentanoic acid The parent chain is a 5-carbon carboxylic acid (pentanoic acid). A bromo group is at position 3 and a phenyl group is at position 4. Numbering starts from the carboxyl carbon.

\mathrm{CH_3-CH(C_6H_5)-CH(Br)-CH_2-COOH}

(vii)

$p, p'$ -Dihydroxybenzophenone Benzophenone has a carbonyl group connecting two phenyl rings (

\mathrm{C_6H_5-CO-C_6H_5}

). Dihydroxy means two hydroxyl (

\mathrm{-OH}

) groups are present.

p,p'

indicates that one

\mathrm{-OH}

group is on the para position of each phenyl ring.

Structure: Two benzene rings connected by a

\mathrm{C=O}

group. Each ring has an

\mathrm{-OH}

group para to the point of attachment to the carbonyl carbon.

(viii)

Hex-2-en-4-ynoic acid The parent chain is a 6-carbon carboxylic acid (hex...oic acid). There is a double bond at position 2 (-2-en-) and a triple bond at position 4 (-4-yn-).

\mathrm{CH_3-C \equiv C-CH=CH-COOH}

Q8.4Exercises

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

(i)

$\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}$

(ii)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}$

(iii)

$\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}$

(iv)

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

(v)

O=CC1CCCC1

(vi)

PhCOPh

Solution

(i)

\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}

IUPAC name: The longest carbon chain contains 7 carbons. The carbonyl group is at C-2. Hence, the name is Heptan-2-one.
Common name: The two alkyl groups attached to the carbonyl group are methyl and n-pentyl. Hence, the common name is Methyl n-pentyl ketone.

(ii)

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}

IUPAC name: The parent chain is an aldehyde with 6 carbons (hexanal). Numbering starts from the aldehyde carbon. There is a methyl group at C-2 and a bromo group at C-4. Listing substituents alphabetically, the name is 4-Bromo-2-methylhexanal.
Common name: Not commonly used for such a complex structure.

(iii)

\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}

IUPAC name: This is a straight-chain aldehyde with 7 carbons. Hence, the name is Heptanal.
Common name: The common name for the 7-carbon aldehyde is Enanthaldehyde.

(iv)

\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}

IUPAC name: The parent chain is a 3-carbon aldehyde with a double bond at C-2 (prop-2-enal). A phenyl group (Ph) is attached to C-3. Hence, the name is 3-Phenylprop-2-enal.
Common name: This compound is commonly known as Cinnamaldehyde.

(v)

Cyclopentane ring with a -CHO group attached.

IUPAC name: When an aldehyde group is attached to a ring, the suffix '-carbaldehyde' is used. Hence, the name is Cyclopentanecarbaldehyde.
Common name: Not commonly used.

(vi)

PhCOPh

IUPAC name: The carbonyl group is attached to two phenyl groups. The ketone is considered a derivative of methane. Hence, the name is Diphenylmethanone.
Common name: This compound is very commonly known as Benzophenone.

Q8.5Exercises

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)

The 2,4-dinitrophenylhydrazone of benzaldehyde This is formed by the condensation reaction between benzaldehyde (

C_6H_5CHO

) and 2,4-dinitrophenylhydrazine.

\mathrm{C_6H_5-CH=N-NH-C_6H_4(NO_2)_2}

The two nitro groups are at positions 2 and 4 of the phenyl ring attached to the nitrogen.

(ii)

Cyclopropanone oxime This is formed by the condensation reaction between cyclopropanone and hydroxylamine (

NH_2OH

).

Structure: A cyclopropane ring with a

=N-OH

group attached to one of the ring carbons.

(iii)

Acetaldehydedimethylacetal This acetal is formed by the reaction of acetaldehyde (

CH_3CHO

) with two molecules of methanol (

CH_3OH

).

\mathrm{CH_3-CH(OCH_3)_2}

(iv)

The semicarbazone of cyclobutanone This is formed by the condensation reaction between cyclobutanone and semicarbazide (

NH_2NHCONH_2

).

Structure: A cyclobutane ring with a

=N-NH-CO-NH_2

group attached to one of the ring carbons.

(v)

The ethylene ketal of hexan-3-one This cyclic ketal is formed by the reaction of hexan-3-one (

CH_3CH_2CO(CH_2)_2CH_3

) with ethylene glycol (

HOCH_2CH_2OH

).

Structure: The carbonyl carbon (C-3) of the hexan-3-one chain is bonded to two oxygen atoms, which are part of a five-membered ring:

-O-CH_2-CH_2-O-

. The ethyl and propyl groups remain attached to this carbon.

(vi)

The methyl hemiacetal of formaldehyde This hemiacetal is formed by the reaction of formaldehyde (HCHO) with one molecule of methanol (

CH_3OH

).

\mathrm{HO-CH_2-OCH_3}

Q8.6Exercises

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

(i)

PhMgBr and then $\mathrm{H}_{3} \mathrm{O}^{+}$

(ii)

Tollens' reagent

(iii)

Semicarbazide and weak acid

(iv)

Excess ethanol and acid

(v)

Zinc amalgam and dilute hydrochloric acid

Solution

Cyclohexanecarbaldehyde is an aldehyde with the formula

\mathrm{C_6H_{11}CHO}

.

(i)

Reaction with PhMgBr and then $\mathrm{H_3O^+}$ This is a Grignard reaction. The nucleophilic phenyl group of phenylmagnesium bromide attacks the carbonyl carbon of the aldehyde. Subsequent acidic hydrolysis protonates the resulting alkoxide to form a secondary alcohol.

\mathrm{C_6H_{11}CHO + PhMgBr \rightarrow C_6H_{11}CH(OMgBr)Ph \xrightarrow{H_3O^+} C_6H_{11}CH(OH)Ph}

Product: Cyclohexyl(phenyl)methanol

(ii)

Reaction with Tollens' reagent Tollens' reagent (

[Ag(NH_3)_2]^+OH^−

) is a mild oxidizing agent that oxidizes aldehydes to carboxylate ions. The aldehyde itself is oxidized, and the silver ions are reduced to metallic silver, forming a silver mirror.

\mathrm{C_6H_{11}CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_6H_{11}COO^- + 2Ag(s) + 4NH_3 + 2H_2O}

Products: Cyclohexanecarboxylate ion and silver metal.

(iii)

Reaction with Semicarbazide and weak acid Aldehydes react with semicarbazide (

H_2N-NH-CO-NH_2

) to form semicarbazones. This is a condensation reaction involving the elimination of a water molecule.

\mathrm{C_6H_{11}CHO + H_2N-NH-CO-NH_2 \xrightarrow{H^+} C_6H_{11}CH=N-NH-CO-NH_2 + H_2O}

Product: Cyclohexanecarbaldehyde semicarbazone

(iv)

Reaction with Excess ethanol and acid Aldehydes react with excess alcohol in the presence of an acid catalyst to form acetals. Two molecules of ethanol react with one molecule of the aldehyde.

\mathrm{C_6H_{11}CHO + 2CH_3CH_2OH \xrightarrow{H^+} C_6H_{11}CH(OCH_2CH_3)_2 + H_2O}

Product: Cyclohexanecarbaldehyde diethyl acetal

(v)

Reaction with Zinc amalgam and dilute hydrochloric acid This is the Clemmensen reduction, which reduces the carbonyl group of an aldehyde or ketone to a methylene (

-CH_2-

) or methyl (

-CH_3

) group.

\mathrm{C_6H_{11}CHO \xrightarrow{Zn-Hg, HCl} C_6H_{11}CH_3}

Product: Methylcyclohexane

Q8.7Exercises

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

The conditions for the reactions are:

Aldol condensation: Requires an aldehyde or ketone with at least one $\alpha$ -hydrogen atom, in the presence of a dilute base.
Cannizzaro reaction: Requires an aldehyde with no $\alpha$ -hydrogen atoms, in the presence of a concentrated base.

(i)

Methanal (HCHO): Has no

\alpha

-carbon, hence no

\alpha

-hydrogen. It undergoes the Cannizzaro reaction. Products: Methanol (

CH_3OH

) and Sodium formate (

HCOONa

).

2HCHO + \text{conc. NaOH} \rightarrow CH_3OH + HCOONa

(ii)

2-Methylpentanal: Has one

\alpha

-hydrogen. It undergoes Aldol condensation. The product is a

\beta

-hydroxy aldehyde.

2 \, CH_3CH_2CH_2CH(CH_3)CHO \xrightarrow{\text{dil. } NaOH} CH_3CH_2CH_2CH(CH_3)CH(OH)C(CH_3)(CHO)CH_2CH_2CH_3

Product: 3-Hydroxy-2,4-dimethyl-2-propylheptanal

(iii)

Benzaldehyde ( $C_6H_5CHO$ ): Has no

\alpha

-hydrogen. It undergoes the Cannizzaro reaction. Products: Benzyl alcohol (

C_6H_5CH_2OH

) and Sodium benzoate (

C_6H_5COONa

).

2C_6H_5CHO + \text{conc. NaOH} \rightarrow C_6H_5CH_2OH + C_6H_5COONa

(iv)

Benzophenone ( $C_6H_5COC_6H_5$ ): A ketone with no

\alpha

-hydrogens. It undergoes neither Aldol condensation nor Cannizzaro reaction. (Cannizzaro reaction is characteristic of aldehydes only).

(v)

Cyclohexanone: Has four

\alpha

-hydrogens (two on each adjacent carbon). It undergoes Aldol condensation. The product is 2-(1-hydroxycyclohexyl)cyclohexanone.

(vi)

1-Phenylpropanone ( $C_6H_5COCH_2CH_3$ ): A ketone with two

\alpha

-hydrogens (on the

-CH_2-

group). It undergoes Aldol condensation. The product is 3-hydroxy-2-methyl-1,3-diphenylpentan-1-one.

(vii)

Phenylacetaldehyde ( $C_6H_5CH_2CHO$ ): An aldehyde with two

\alpha

-hydrogens. It undergoes Aldol condensation. The product is 3-hydroxy-2,4-diphenylbutanal.

2 \, C_6H_5CH_2CHO \xrightarrow{\text{dil. } NaOH} C_6H_5CH_2CH(OH)CH(C_6H_5)CHO

(viii)

Butan-1-ol ( $CH_3CH_2CH_2CH_2OH$ ): This is an alcohol, not an aldehyde or ketone. It undergoes neither Aldol condensation nor Cannizzaro reaction.

(ix)

2,2-Dimethylbutanal: An aldehyde with no

\alpha

-hydrogens. It undergoes the Cannizzaro reaction. Products: 2,2-Dimethylbutan-1-ol and Sodium 2,2-dimethylbutanoate.

2 \, CH_3CH_2C(CH_3)_2CHO + \text{conc. NaOH} \rightarrow CH_3CH_2C(CH_3)_2CH_2OH + CH_3CH_2C(CH_3)_2COONa

Q8.8Exercises

How will you convert ethanal into the following compounds?

(i)

Butane-1,3-diol

(ii)

But-2-enal

(iii)

But-2-enoic acid

Solution

The conversions of ethanal into the desired compounds are carried out as follows:

(i) Ethanal to Butane-1,3-diol

This conversion involves two steps: an aldol addition reaction followed by reduction.

Step 1: Aldol Addition Two molecules of ethanal undergo aldol addition in the presence of a dilute base (like dil. NaOH) to form 3-hydroxybutanal.

\mathrm{CH_3CHO + CH_3CHO} \xrightarrow{\text{dil. NaOH}} \underset{\text{3-Hydroxybutanal}}{\mathrm{CH_3CH(OH)CH_2CHO}}

Step 2: Reduction The aldehyde group of 3-hydroxybutanal is reduced to a primary alcohol using a reducing agent like sodium borohydride (

\mathrm{NaBH_4}

).

\underset{\text{3-Hydroxybutanal}}{\mathrm{CH_3CH(OH)CH_2CHO}} \xrightarrow{\mathrm{NaBH_4}} \underset{\text{Butane-1,3-diol}}{\mathrm{CH_3CH(OH)CH_2CH_2OH}}

(ii) Ethanal to But-2-enal

This conversion is an aldol condensation, which is an aldol addition followed by dehydration.

Step 1: Aldol Addition Two molecules of ethanal react in the presence of a dilute base to form 3-hydroxybutanal.

\mathrm{2CH_3CHO} \xrightarrow{\text{dil. NaOH}} \underset{\text{3-Hydroxybutanal}}{\mathrm{CH_3CH(OH)CH_2CHO}}

Step 2: Dehydration Upon heating, 3-hydroxybutanal loses a molecule of water to form the

\alpha,\beta

-unsaturated aldehyde, but-2-enal (crotonaldehyde).

\underset{\text{3-Hydroxybutanal}}{\mathrm{CH_3CH(OH)CH_2CHO}} \xrightarrow{\Delta} \underset{\text{But-2-enal}}{\mathrm{CH_3CH=CHCHO}} + \mathrm{H_2O}

(iii) Ethanal to But-2-enoic acid

This conversion involves aldol condensation to form but-2-enal, followed by oxidation.

Step 1: Aldol Condensation First, but-2-enal is prepared from ethanal as shown in part (ii).

\mathrm{2CH_3CHO} \xrightarrow{\text{dil. NaOH, }\Delta} \underset{\text{But-2-enal}}{\mathrm{CH_3CH=CHCHO}} + \mathrm{H_2O}

Step 2: Oxidation The aldehyde group of but-2-enal is oxidized to a carboxylic acid group using a mild oxidizing agent like Tollens' reagent (

[\mathrm{Ag(NH_3)_2}]^+

), which does not affect the carbon-carbon double bond. The resulting carboxylate is then acidified.

\underset{\text{But-2-enal}}{\mathrm{CH_3CH=CHCHO}} \xrightarrow{\text{1. Tollens' Reagent}} \xrightarrow{\text{2. } \mathrm{H_3O^+}} \underset{\text{But-2-enoic acid}}{\mathrm{CH_3CH=CHCOOH}}

Q8.9Exercises

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

The aldol condensation between propanal (

CH_3CH_2CHO

) and butanal (

CH_3CH_2CH_2CHO

) can result in four possible products, as both aldehydes possess

\alpha

-hydrogens and can act as either a nucleophile (after deprotonation by base) or an electrophile.

The four products are formed from two self-condensations and two cross-condensations.

1. Self-condensation of Propanal One molecule of propanal acts as the electrophile, and the enolate of another propanal molecule acts as the nucleophile.

Electrophile: Propanal ( $CH_3CH_2CHO$ )
Nucleophile: Propanal enolate ( $CH_3\overline{C}HCHO$ )

Reaction:

CH_3CH_2CHO + CH_3CH_2CHO \xrightarrow{\text{dil. } OH^-, \Delta} \underset{\text{2-Methylpent-2-enal}}{CH_3CH_2CH=C(CH_3)CHO} + H_2O

Structure and Name:

Structure: $CH_3CH_2CH=C(CH_3)CHO$
Name: 2-Methylpent-2-enal

2. Self-condensation of Butanal One molecule of butanal acts as the electrophile, and the enolate of another butanal molecule acts as the nucleophile.

Electrophile: Butanal ( $CH_3CH_2CH_2CHO$ )
Nucleophile: Butanal enolate ( $CH_3CH_2\overline{C}HCHO$ )

Reaction:

CH_3CH_2CH_2CHO + CH_3CH_2CH_2CHO \xrightarrow{\text{dil. } OH^-, \Delta} \underset{\text{2-Ethylhex-2-enal}}{CH_3CH_2CH_2CH=C(CH_2CH_3)CHO} + H_2O

Structure and Name:

Structure: $CH_3CH_2CH_2CH=C(CH_2CH_3)CHO$
Name: 2-Ethylhex-2-enal

3. Cross-condensation: Butanal (Electrophile) + Propanal (Nucleophile) Butanal acts as the electrophile, and the enolate of propanal acts as the nucleophile.

Electrophile: Butanal ( $CH_3CH_2CH_2CHO$ )
Nucleophile: Propanal enolate ( $CH_3\overline{C}HCHO$ )

Reaction:

CH_3CH_2CH_2CHO + CH_3CH_2CHO \xrightarrow{\text{dil. } OH^-, \Delta} \underset{\text{2-Methylhex-2-enal}}{CH_3CH_2CH_2CH=C(CH_3)CHO} + H_2O

Structure and Name:

Structure: $CH_3CH_2CH_2CH=C(CH_3)CHO$
Name: 2-Methylhex-2-enal

4. Cross-condensation: Propanal (Electrophile) + Butanal (Nucleophile) Propanal acts as the electrophile, and the enolate of butanal acts as the nucleophile.

Electrophile: Propanal ( $CH_3CH_2CHO$ )
Nucleophile: Butanal enolate ( $CH_3CH_2\overline{C}HCHO$ )

Reaction:

CH_3CH_2CHO + CH_3CH_2CH_2CHO \xrightarrow{\text{dil. } OH^-, \Delta} \underset{\text{2-Ethylpent-2-enal}}{CH_3CH_2CH=C(CH_2CH_3)CHO} + H_2O

Structure and Name:

Structure: $CH_3CH_2CH=C(CH_2CH_3)CHO$
Name: 2-Ethylpent-2-enal

Q8.10Exercises

An organic compound with the molecular formula $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$ 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

Let's deduce the structure of the organic compound with the molecular formula

\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}

based on the given reactions.

Molecular Formula and Degree of Unsaturation: The formula is $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$ . The degree of unsaturation (DBE) is calculated as: DBE = C + 1 - H/2 = 9 + 1 - 10/2 = 5. A DBE of 4 or more suggests the presence of a benzene ring. A benzene ring accounts for 4 units of unsaturation (3 double bonds + 1 ring). The remaining one unit of unsaturation corresponds to a carbonyl group (C=O).
Reaction with 2,4-DNP: The compound forms a 2,4-DNP derivative. This is a characteristic test for carbonyl compounds, confirming the presence of an aldehyde or a ketone group.
Reduces Tollens' reagent: The compound reduces Tollens' reagent to form a silver mirror. This is a positive test for an aldehyde group (-CHO). Ketones do not give this test. Therefore, the compound is an aldehyde.
Undergoes Cannizzaro reaction: The compound undergoes the Cannizzaro reaction. This reaction is characteristic of aldehydes that do not have an $\alpha$ -hydrogen atom. This means the -CHO group is attached to a carbon atom that has no hydrogen atoms bonded to it (e.g., a tertiary carbon or a carbon in a benzene ring that is also bonded to another substituent).
Vigorous Oxidation: On vigorous oxidation (e.g., with hot conc. $\mathrm{KMnO_4}$ ), it gives 1,2-benzenedicarboxylic acid (phthalic acid). This indicates that the compound is a disubstituted benzene with two substituents at adjacent positions (ortho positions). During vigorous oxidation, any alkyl side chain on a benzene ring is oxidized to a carboxylic acid group (-COOH), and an aldehyde group is also oxidized to a -COOH group.

Structure Determination:

The compound is an ortho-disubstituted benzene. One substituent is the aldehyde group (-CHO).
Let the other substituent be R. The structure is $\mathrm{C_6H_4(CHO)(R)}$ .
The molecular formula is $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$ . The $\mathrm{C_6H_4CHO}$ part accounts for $\mathrm{C_7H_5O}$ .
Subtracting this from the molecular formula gives the formula for R: $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O} - \mathrm{C_7H_5O} = \mathrm{C_2H_5}$ .
So, the substituent R is an ethyl group ( $\mathrm{-CH_2CH_3}$ ).
The compound is 2-ethylbenzaldehyde.

Verification:

Structure: 2-Ethylbenzaldehyde. It is an ortho-disubstituted benzene derivative.
Formula: $\mathrm{C_9H_{10}O}$ . Correct.
Aldehyde tests: It has a -CHO group, so it gives positive 2,4-DNP and Tollens' tests.
Cannizzaro reaction: The -CHO group is attached to a carbon of the benzene ring which is also attached to the ethyl group. This carbon has no hydrogen atom. Thus, it lacks an $\alpha$ -hydrogen and will undergo the Cannizzaro reaction.
Oxidation: Vigorous oxidation of 2-ethylbenzaldehyde oxidizes both the -CHO group and the $\mathrm{-CH_2CH_3}$ group to -COOH groups, yielding 1,2-benzenedicarboxylic acid.

Reactions:

Cannizzaro Reaction: $2 \times (\text{2-Ethylbenzaldehyde}) \xrightarrow{\text{Conc. KOH}} (\text{2-Ethylbenzyl alcohol}) + (\text{Potassium 2-ethylbenzoate})$
Vigorous Oxidation: $\underset{\text{2-Ethylbenzaldehyde}}{\mathrm{C_6H_4(CHO)(C_2H_5)}} \xrightarrow{\mathrm{KMnO_4, H^+, \Delta}} \underset{\text{1,2-Benzenedicarboxylic acid}}{\mathrm{C_6H_4(COOH)_2}}$

Conclusion: The identified organic compound is 2-Ethylbenzaldehyde.

Q8.11Exercises

An organic compound (A) (molecular formula $\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}_{2}$ ) 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

Let's identify the compounds (A), (B), and (C) by analyzing the given reaction sequence.

Compound (A) has the molecular formula $\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}_{2}$ . It is hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). This reaction is characteristic of an ester. Reaction (i): $\underset{\text{(A)} \mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}_{2}}}{\text{Ester}} + \mathrm{H_2O} \xrightarrow{\mathrm{H^+}} \underset{\text{(B)}}{\text{Carboxylic acid}} + \underset{\text{(C)}}{\text{Alcohol}}$
Oxidation of (C) gives (B). This implies that the alcohol (C) and the carboxylic acid (B) have the same number of carbon atoms. Since the ester (A) is formed from acid (B) and alcohol (C), the total number of carbon atoms in (A) is the sum of carbon atoms in (B) and (C). Let $n$ be the number of carbon atoms in (B) and (C). Total carbons in (A) = $n + n = 2n = 8$ . Therefore, $n = 4$ . So, (B) is a carboxylic acid with 4 carbon atoms ( $\mathrm{C_4}$ -acid), and (C) is an alcohol with 4 carbon atoms ( $\mathrm{C_4}$ -alcohol).
Dehydration of (C) gives but-1-ene. The alcohol (C) is a butanol. The dehydration of an alcohol to form an alkene follows specific rules. The formation of but-1-ene as the product points towards the starting alcohol being butan-1-ol. Dehydration of butan-2-ol would primarily yield but-2-ene (Saytzeff's rule). Reaction (ii): $\underset{\text{(C)}}{\mathrm{CH_3CH_2CH_2CH_2OH}} \xrightarrow{\text{conc. } \mathrm{H_2SO_4}, \Delta} \underset{\text{But-1-ene}}{\mathrm{CH_3CH_2CH=CH_2}} + \mathrm{H_2O}$ Thus, compound (C) is Butan-1-ol.
Identification of (B). Compound (B) is obtained by the oxidation of (C) (Butan-1-ol) with chromic acid. Oxidation of a primary alcohol gives a carboxylic acid with the same number of carbon atoms. Reaction (iii): $\underset{\text{(C) Butan-1-ol}}{\mathrm{CH_3CH_2CH_2CH_2OH}} \xrightarrow{\mathrm{H_2CrO_4}} \underset{\text{(B) Butanoic acid}}{\mathrm{CH_3CH_2CH_2COOH}}$ Thus, compound (B) is Butanoic acid.
Identification of (A). Compound (A) is an ester formed from carboxylic acid (B) (Butanoic acid) and alcohol (C) (Butan-1-ol). The ester is butyl butanoate. Esterification Reaction: $\underset{\text{(B) Butanoic acid}}{\mathrm{CH_3CH_2CH_2COOH}} + \underset{\text{(C) Butan-1-ol}}{\mathrm{HOCH_2CH_2CH_2CH_3}} \rightleftharpoons \underset{\text{(A) Butyl butanoate}}{\mathrm{CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3}} + \mathrm{H_2O}$ The molecular formula of butyl butanoate is $\mathrm{C_8H_{16}O_2}$ , which matches that of compound (A). Thus, compound (A) is Butyl butanoate.

Summary of Reactions Involved:

Hydrolysis of (A): $\underset{\text{(A) Butyl butanoate}}{\mathrm{CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3}} + \mathrm{H_2O} \xrightarrow{\mathrm{H^+}} \underset{\text{(B) Butanoic acid}}{\mathrm{CH_3CH_2CH_2COOH}} + \underset{\text{(C) Butan-1-ol}}{\mathrm{CH_3CH_2CH_2CH_2OH}}$
Oxidation of (C): $\underset{\text{(C) Butan-1-ol}}{\mathrm{CH_3CH_2CH_2CH_2OH}} \xrightarrow{\mathrm{H_2CrO_4}} \underset{\text{(B) Butanoic acid}}{\mathrm{CH_3CH_2CH_2COOH}}$
Dehydration of (C): $\underset{\text{(C) Butan-1-ol}}{\mathrm{CH_3CH_2CH_2CH_2OH}} \xrightarrow{\text{conc. } \mathrm{H_2SO_4}, \Delta} \underset{\text{But-1-ene}}{\mathrm{CH_3CH_2CH=CH_2}} + \mathrm{H_2O}$

Q8.12Exercises

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)

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{COOH}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{COOH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}$ , $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (acid strength)

(iii)

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

Solution

(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)

The addition of HCN to carbonyl compounds is a nucleophilic addition reaction. The reactivity of the carbonyl group towards nucleophiles is influenced by two main factors:

Steric hindrance: Bulky groups attached to the carbonyl carbon hinder the approach of the nucleophile.
Electronic effects: Electron-donating alkyl groups decrease the positive charge (electrophilicity) on the carbonyl carbon, making it less reactive.

Let's analyze the given compounds:

Acetaldehyde ( $CH_3CHO$ ): One methyl group and one hydrogen atom. Minimal steric hindrance and electronic effect.
Acetone ( $CH_3COCH_3$ ): Two methyl groups. More steric hindrance and electron-donating effect than acetaldehyde.
Methyl tert-butyl ketone ( $(CH_3)_3CCOCH_3$ ): One methyl and one very bulky tert-butyl group. Significant steric hindrance.
Di-tert-butyl ketone ( $(CH_3)_3CCOC(CH_3)_3$ ): Two extremely bulky tert-butyl groups. Maximum steric hindrance, which makes the carbonyl carbon almost inaccessible.

Based on increasing steric hindrance and electron-donating effect of alkyl groups, the reactivity towards HCN decreases.

Reactivity order: Acetaldehyde > Acetone > Methyl tert-butyl ketone > Di-tert-butyl ketone.

Therefore, the increasing order of reactivity towards HCN is:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) $CH_3CH_2CH(Br)COOH$ , $CH_3CH(Br)CH_2COOH$ , $(CH_3)_2CHCOOH$ , $CH_3CH_2CH_2COOH$ (acid strength)

The acid strength of carboxylic acids depends on the stability of the carboxylate anion formed after the loss of a proton.

Electron-withdrawing groups (EWGs) like halogens (-Br) increase the acidity by stabilizing the carboxylate anion through the -I (negative inductive) effect.
Electron-donating groups (EDGs) like alkyl groups decrease the acidity by destabilizing the carboxylate anion through the +I (positive inductive) effect.
The inductive effect decreases with distance.

Let's analyze the given compounds:

$CH_3CH_2CH_2COOH$ (Butanoic acid): Has a propyl group (+I effect).
$(CH_3)_2CHCOOH$ (2-Methylpropanoic acid): Has an isopropyl group which has a stronger +I effect than a propyl group.
$CH_3CH(Br)CH_2COOH$ (3-Bromobutanoic acid): Has a Br atom (EWG, -I effect) at the $\gamma$ -position.
$CH_3CH_2CH(Br)COOH$ (2-Bromobutanoic acid): Has a Br atom at the $\alpha$ -position. The -I effect is strongest when the EWG is closest to the -COOH group.

The order of acid strength will be: (Compound with strongest EWG at

\alpha

-position) > (Compound with EWG at a farther position) > (Compound with weakest EDG) > (Compound with stronger EDG).

Acid strength order:

CH_3CH_2CH(Br)COOH > CH_3CH(Br)CH_2COOH > CH_3CH_2CH_2COOH > (CH_3)_2CHCOOH

.

Therefore, the increasing order of acid strength is:

$(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$

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

The acid strength of substituted benzoic acids depends on the nature of the substituent on the benzene ring.

Electron-withdrawing groups (EWGs) like -NO $_2$ increase the acid strength by stabilizing the benzoate anion. The -NO $_2$ group has a strong -I and -R effect.
Electron-donating groups (EDGs) like -OCH $_3$ decrease the acid strength by destabilizing the benzoate anion. The -OCH $_3$ group has a -I effect but a much stronger +R (resonance) effect.

Let's analyze the given compounds:

Benzoic acid: The reference compound.
4-Methoxybenzoic acid: The -OCH $_3$ group at the para position is a strong electron-donating group (+R effect), making it a weaker acid than benzoic acid.
4-Nitrobenzoic acid: The -NO $_2$ group at the para position is a strong electron-withdrawing group (-R and -I effect), making it a stronger acid than benzoic acid.
3,4-Dinitrobenzoic acid: Two -NO $_2$ groups provide a very strong electron-withdrawing effect, making it the strongest acid among the given compounds.

The order of acid strength will be: (Two EWGs) > (One EWG) > (No substituent) > (One EDG).

Acid strength order: 3,4-Dinitrobenzoic acid > 4-Nitrobenzoic acid > Benzoic acid > 4-Methoxybenzoic acid.

Therefore, the increasing order of acid strength is:

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid

Q8.13Exercises

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)

Propanal and Propanone

Test: Tollen's Test
Reagent: Tollen's reagent (ammoniacal silver nitrate solution).
Observation: Propanal, being an aldehyde, gives a silver mirror on the inner side of the test tube. Propanone, a ketone, does not react.
Reaction: $CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$

(ii)

Acetophenone and Benzophenone

Test: Iodoform Test
Reagent: Iodine solution and sodium hydroxide ( $I_2/NaOH$ ).
Observation: Acetophenone ( $C_6H_5COCH_3$ ) has a methyl group attached to the carbonyl carbon, so it gives a yellow precipitate of iodoform ( $CHI_3$ ). Benzophenone ( $C_6H_5COC_6H_5$ ) does not have this structural feature and does not react.
Reaction: $C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow + 3NaI + 3H_2O$

(iii)

Phenol and Benzoic acid

Test: Sodium Bicarbonate Test
Reagent: Aqueous sodium bicarbonate solution ( $NaHCO_3$ ).
Observation: Benzoic acid is a carboxylic acid and is strong enough to react with sodium bicarbonate, producing brisk effervescence due to the evolution of carbon dioxide ( $CO_2$ ) gas. Phenol is a weaker acid and does not react.
Reaction: $C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$

(iv)

Benzoic acid and Ethyl benzoate

Test: Sodium Bicarbonate Test
Reagent: Aqueous sodium bicarbonate solution ( $NaHCO_3$ ).
Observation: Benzoic acid reacts to produce brisk effervescence of $CO_2$ gas. Ethyl benzoate is an ester and does not react.
Reaction: $C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$

(v)

Pentan-2-one and Pentan-3-one

Test: Iodoform Test
Reagent: Iodine solution and sodium hydroxide ( $I_2/NaOH$ ).
Observation: Pentan-2-one ( $CH_3COCH_2CH_2CH_3$ ) is a methyl ketone and gives a yellow precipitate of iodoform. Pentan-3-one ( $CH_3CH_2COCH_2CH_3$ ) is not a methyl ketone and does not give this test.
Reaction: $CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \rightarrow CH_3CH_2CH_2COONa + CHI_3 \downarrow + 3NaI + 3H_2O$

(vi)

Benzaldehyde and Acetophenone

Test: Tollen's Test
Reagent: Tollen's reagent.
Observation: Benzaldehyde, an aldehyde, forms a silver mirror. Acetophenone, a ketone, does not react.
Reaction: $C_6H_5CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_6H_5COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$ (Alternatively, the Iodoform test can be used, where acetophenone gives a positive test and benzaldehyde does not.)

(vii)

Ethanal and Propanal

Test: Iodoform Test
Reagent: Iodine solution and sodium hydroxide ( $I_2/NaOH$ ).
Observation: Ethanal ( $CH_3CHO$ ) contains the $CH_3CO-$ group and gives a yellow precipitate of iodoform. Propanal ( $CH_3CH_2CHO$ ) does not have this group and will not give a positive test.
Reaction: $CH_3CHO + 3I_2 + 4NaOH \rightarrow HCOONa + CHI_3 \downarrow + 3NaI + 3H_2O$

Q8.14Exercises

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

(i)

Methyl benzoate from Benzene

Friedel-Crafts Alkylation: Benzene is treated with chloromethane in the presence of anhydrous aluminium chloride to form toluene. $C_6H_6 + CH_3Cl \xrightarrow{Anhyd. AlCl_3} C_6H_5CH_3 + HCl$
Oxidation: Toluene is oxidized using alkaline potassium permanganate, followed by acidification, to form benzoic acid. $C_6H_5CH_3 \xrightarrow{(i) KMnO_4/KOH, \Delta (ii) H_3O^+} C_6H_5COOH$
Esterification: Benzoic acid is heated with methanol in the presence of a small amount of concentrated sulphuric acid to form methyl benzoate. $C_6H_5COOH + CH_3OH \xrightarrow{H^+, \Delta} C_6H_5COOCH_3 + H_2O$

(ii)

m-Nitrobenzoic acid from Benzene

Friedel-Crafts Acylation: Benzene is treated with acetyl chloride in the presence of anhydrous aluminium chloride to form acetophenone. The acetyl group is a meta-directing group. $C_6H_6 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} C_6H_5COCH_3 + HCl$
Nitration: Acetophenone is nitrated with a mixture of concentrated nitric acid and concentrated sulphuric acid to give m-nitroacetophenone. $C_6H_5COCH_3 \xrightarrow{Conc. HNO_3, Conc. H_2SO_4} m-NO_2C_6H_4COCH_3$
Oxidation: The m-nitroacetophenone is oxidized using a strong oxidizing agent like sodium hypochlorite (or alkaline $KMnO_4$ ) followed by acidification to form m-nitrobenzoic acid. This is the haloform reaction. $m-NO_2C_6H_4COCH_3 \xrightarrow{(i) NaOCl (ii) H_3O^+} m-NO_2C_6H_4COOH + CHCl_3$

(iii)

p-Nitrobenzoic acid from Benzene

Friedel-Crafts Alkylation: Benzene is converted to toluene by reacting with chloromethane in the presence of anhydrous aluminium chloride. $C_6H_6 + CH_3Cl \xrightarrow{Anhyd. AlCl_3} C_6H_5CH_3 + HCl$
Nitration: Toluene is nitrated with a mixture of concentrated nitric acid and concentrated sulphuric acid. The methyl group is ortho, para-directing. The p-nitrotoluene is the major product and can be separated from the ortho-isomer. $C_6H_5CH_3 \xrightarrow{Conc. HNO_3, Conc. H_2SO_4} p-NO_2C_6H_4CH_3 + o-NO_2C_6H_4CH_3$
Oxidation: The separated p-nitrotoluene is oxidized using alkaline potassium permanganate followed by acidification to yield p-nitrobenzoic acid. $p-NO_2C_6H_4CH_3 \xrightarrow{(i) KMnO_4/KOH, \Delta (ii) H_3O^+} p-NO_2C_6H_4COOH$

(iv)

Phenylacetic acid from Benzene

Friedel-Crafts Alkylation: Benzene is converted to toluene. $C_6H_6 + CH_3Cl \xrightarrow{Anhyd. AlCl_3} C_6H_5CH_3 + HCl$
Side-chain Chlorination: Toluene is treated with chlorine in the presence of sunlight (UV light) to form benzyl chloride. $C_6H_5CH_3 + Cl_2 \xrightarrow{h\nu} C_6H_5CH_2Cl + HCl$
Cyanide Substitution: Benzyl chloride is treated with alcoholic potassium cyanide to form phenylacetonitrile (benzyl cyanide). $C_6H_5CH_2Cl + KCN(alc.) \rightarrow C_6H_5CH_2CN + KCl$
Hydrolysis: Phenylacetonitrile is hydrolyzed with an acid or alkali to form phenylacetic acid. $C_6H_5CH_2CN \xrightarrow{H_3O^+, \Delta} C_6H_5CH_2COOH + NH_4^+ $

(v)

p-Nitrobenzaldehyde from Benzene

Friedel-Crafts Alkylation: Benzene is converted to toluene. $C_6H_6 + CH_3Cl \xrightarrow{Anhyd. AlCl_3} C_6H_5CH_3 + HCl$
Nitration: Toluene is nitrated to form p-nitrotoluene as the major product, which is then separated. $C_6H_5CH_3 \xrightarrow{Conc. HNO_3, Conc. H_2SO_4} p-NO_2C_6H_4CH_3$
Etard Reaction: p-Nitrotoluene is treated with chromyl chloride ( $CrO_2Cl_2$ ) in carbon disulphide, followed by hydrolysis, to form p-nitrobenzaldehyde. $p-NO_2C_6H_4CH_3 \xrightarrow{(i) CrO_2Cl_2, CS_2 (ii) H_3O^+} p-NO_2C_6H_4CHO$

Q8.15Exercises

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 $\alpha$ -Hydroxyphenylacetic acid

(ix)

Benzoic acid to m- Nitrobenzyl alcohol

Solution

(i)

Propanone to Propene

Reduction: Propanone is reduced to propan-2-ol using a reducing agent like sodium borohydride ( $NaBH_4$ ). $CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3$
Dehydration: Propan-2-ol is heated with concentrated sulphuric acid to cause dehydration, forming propene. $CH_3CH(OH)CH_3 \xrightarrow{Conc. H_2SO_4, \Delta} CH_2=CH-CH_3 + H_2O$

(ii)

Benzoic acid to Benzaldehyde

Chlorination: Benzoic acid is treated with thionyl chloride ( $SOCl_2$ ) to form benzoyl chloride. $C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl$
Rosenmund Reduction: Benzoyl chloride is catalytically hydrogenated using palladium on barium sulphate ( $Pd/BaSO_4$ ) to give benzaldehyde. $C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$

(iii)

Ethanol to 3-Hydroxybutanal

Oxidation: Ethanol is oxidized to ethanal using pyridinium chlorochromate (PCC), a mild oxidizing agent. $CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$
Aldol Condensation: Ethanal undergoes aldol condensation in the presence of a dilute base (like $dil. NaOH$ ) to form 3-hydroxybutanal. $2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO$

(iv)

Benzene to m-Nitroacetophenone

Friedel-Crafts Acylation: Benzene is treated with acetyl chloride in the presence of anhydrous aluminium chloride to form acetophenone. The acetyl group is a meta-director. $C_6H_6 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} C_6H_5COCH_3 + HCl$
Nitration: Acetophenone is nitrated using a mixture of concentrated nitric acid and concentrated sulphuric acid to yield m-nitroacetophenone. $C_6H_5COCH_3 \xrightarrow{Conc. HNO_3/Conc. H_2SO_4} m-NO_2C_6H_4COCH_3$

(v)

Benzaldehyde to Benzophenone

Oxidation: Benzaldehyde is oxidized to benzoic acid using an oxidizing agent like acidified potassium permanganate ( $KMnO_4$ ). $C_6H_5CHO \xrightarrow{KMnO_4/H^+} C_6H_5COOH$
Reaction with Benzene: Benzoic acid is converted to benzoyl chloride with $PCl_5$ or $SOCl_2$ . Benzoyl chloride then reacts with benzene in a Friedel-Crafts acylation reaction to form benzophenone. $C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{C_6H_6, Anhyd. AlCl_3} C_6H_5COC_6H_5$

(vi)

Bromobenzene to 1-Phenylethanol

Grignard Reagent Formation: Bromobenzene is treated with magnesium metal in dry ether to form phenylmagnesium bromide (a Grignard reagent). $C_6H_5Br + Mg \xrightarrow{Dry~ether} C_6H_5MgBr$
Reaction with Aldehyde: Phenylmagnesium bromide is reacted with ethanal ( $CH_3CHO$ ), followed by acid hydrolysis, to produce 1-phenylethanol. $C_6H_5MgBr + CH_3CHO \xrightarrow{(i) Dry~ether (ii) H_3O^+} C_6H_5CH(OH)CH_3$

(vii)

Benzaldehyde to 3-Phenylpropan-1-ol

Cross Aldol Condensation: Benzaldehyde is treated with ethanal in the presence of a dilute base. The enolate from ethanal attacks benzaldehyde, and subsequent heating causes dehydration to form cinnamaldehyde. $C_6H_5CHO + CH_3CHO \xrightarrow{dil. NaOH, \Delta} C_6H_5CH=CHCHO + H_2O$
Reduction: Cinnamaldehyde is reduced using a strong reducing agent like Lithium Aluminium Hydride ( $LiAlH_4$ ) or catalytic hydrogenation ( $H_2/Ni$ ) which reduces both the double bond and the aldehyde group to give 3-phenylpropan-1-ol. $C_6H_5CH=CHCHO \xrightarrow{H_2/Ni} C_6H_5CH_2CH_2CH_2OH$

(viii)

Benzaldehyde to $\alpha$ -Hydroxyphenylacetic acid

Cyanohydrin Formation: Benzaldehyde reacts with hydrogen cyanide (HCN) to form benzaldehyde cyanohydrin. $C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$
Hydrolysis: The cyanohydrin is subjected to acid hydrolysis to convert the cyano group ( $-CN$ ) into a carboxylic acid group ( $-COOH$ ), yielding $\alpha$ -hydroxyphenylacetic acid (mandelic acid). $C_6H_5CH(OH)CN \xrightarrow{H_3O^+, \Delta} C_6H_5CH(OH)COOH$

(ix)

Benzoic acid to m-Nitrobenzyl alcohol

Nitration: Benzoic acid is nitrated with a mixture of concentrated nitric acid and concentrated sulphuric acid. The $-COOH$ group is meta-directing, so m-nitrobenzoic acid is formed. $C_6H_5COOH \xrightarrow{Conc. HNO_3/Conc. H_2SO_4} m-NO_2C_6H_4COOH$
Reduction: m-Nitrobenzoic acid is treated with a strong reducing agent like Lithium Aluminium Hydride ( $LiAlH_4$ ) followed by hydrolysis. $LiAlH_4$ reduces the carboxylic acid group to a primary alcohol group but does not affect the nitro group. $m-NO_2C_6H_4COOH \xrightarrow{(i) LiAlH_4 (ii) H_2O} m-NO_2C_6H_4CH_2OH$

Q8.16Exercises

Describe the following:

(i)

Acetylation

(ii)

Cannizzaro reaction

(iii)

Cross aldol condensation

(iv)

Decarboxylation

Solution

(i)

Acetylation Acetylation is the process of introducing an acetyl group (

CH_3CO-

) into a molecule, typically containing an active hydrogen atom like in alcohols, phenols, or amines. The reaction is usually carried out using acetylating agents like acetyl chloride (

CH_3COCl

) or acetic anhydride (

(CH_3CO)_2O

). The reaction is often performed in the presence of a base like pyridine, which neutralizes the acid byproduct (HCl or

CH_3COOH

) and catalyzes the reaction. For example, the acetylation of ethanol: $CH_3CH_2OH + CH_3COCl \xrightarrow{Pyridine} CH_3CH_2OCOCH_3 + HCl$ (Ethyl acetate)

(ii)

Cannizzaro Reaction Aldehydes which do not have an

\alpha

-hydrogen atom, such as formaldehyde (

HCHO

) and benzaldehyde (

C_6H_5CHO

), undergo self-oxidation and reduction (disproportionation) on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to the corresponding alcohol, while another molecule is oxidized to the sodium or potassium salt of the corresponding carboxylic acid. For example, the Cannizzaro reaction of benzaldehyde: $2C_6H_5CHO + NaOH(conc.) \rightarrow C_6H_5CH_2OH + C_6H_5COONa$ (Benzyl alcohol) (Sodium benzoate)

(iii)

Cross Aldol Condensation Cross aldol condensation is an aldol condensation reaction that occurs between two different aldehydes or ketones, or one aldehyde and one ketone. If both reactants have

\alpha

-hydrogens, a complex mixture of four products is formed (two from self-condensation and two from cross-condensation). However, if one of the aldehydes does not have an

\alpha

-hydrogen (like benzaldehyde or formaldehyde), the reaction is more useful as it leads to a limited number of products. The enolate ion is formed from the carbonyl compound containing the

\alpha

-hydrogen, which then attacks the carbonyl carbon of the other compound. For example, the reaction between ethanal and propanal: $CH_3CHO + CH_3CH_2CHO \xrightarrow{dil. NaOH} \text{Mixture of 4 products}$ A more directed example is the reaction between benzaldehyde (no

\alpha

-H) and acetophenone: $C_6H_5CHO + C_6H_5COCH_3 \xrightarrow{OH^-, \Delta} C_6H_5CH=CHCOC_6H_5 + H_2O$ (Benzalacetophenone)

(iv)

Decarboxylation Decarboxylation is a chemical reaction that removes a carboxyl group (

-COOH

) from a molecule, releasing carbon dioxide (

CO_2

). This reaction is common for carboxylic acids. The ease of decarboxylation depends on the stability of the carbanion formed as an intermediate. Carboxylic acids are typically decarboxylated by heating with soda lime (a mixture of

NaOH

and

CaO

). The reaction proceeds via the sodium salt of the acid. For example, the decarboxylation of sodium ethanoate: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$

eta

-keto acids decarboxylate readily upon gentle heating. $CH_3COCH_2COOH \xrightarrow{\Delta} CH_3COCH_3 + CO_2$

Q8.17Exercises

Complete each synthesis by giving missing starting material, reagent or products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Solution

(i)

The starting material is hexan-1-ol. The reaction is oxidation of a primary alcohol to a carboxylic acid using a strong oxidizing agent like Jones reagent (

CrO_3-H_2SO_4

) or acidic/alkaline

KMnO_4

. $CH_3(CH_2)_4CH_2OH \xrightarrow{CrO_3-H_2SO_4} CH_3(CH_2)_4COOH$ The missing starting material is Hexan-1-ol,

CH_3(CH_2)_4CH_2OH

.

(ii)

The starting material is an ester which on hydrolysis gives a carboxylate and an alcohol. The products are cyclohexanecarboxylate and ethanol. Therefore, the starting ester is Ethyl cyclohexanecarboxylate. $C_6H_{11}COOC_2H_5 + NaOH \xrightarrow{H_2O} C_6H_{11}COONa + C_2H_5OH \xrightarrow{H^+} C_6H_{11}COOH$

(iii)

This is the Etard reaction. Toluene or its derivative is oxidized to the corresponding benzaldehyde using chromyl chloride (

CrO_2Cl_2

) in a non-polar solvent like

CS_2

or

CCl_4

, followed by hydrolysis. The product is p-nitrobenzaldehyde. Thus, the starting material must be p-nitrotoluene. $p-NO_2C_6H_4CH_3 \xrightarrow{(i) CrO_2Cl_2, CS_2 (ii) H_3O^+} p-NO_2C_6H_4CHO$

(iv)

This is Friedel-Crafts acylation of benzene. Benzene reacts with an acid chloride in the presence of anhydrous

AlCl_3

to form a ketone. The product is acetophenone. Therefore, the reagent must be acetyl chloride ( $CH_3COCl$ ) or acetic anhydride ( $(CH_3CO)_2O$ ). $C_6H_6 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} C_6H_5COCH_3 + HCl$

(v)

This is the haloform reaction followed by acidification. The product is butanoic acid. The haloform reaction works on methyl ketones. The starting material must be a methyl ketone with a total of 5 carbon atoms, which is Pentan-2-one. The other product,

CHCl_3

, confirms the use of a hypochlorite reagent. $CH_3CH_2CH_2COCH_3 \xrightarrow{NaOCl} CH_3CH_2CH_2COONa + CHCl_3 \xrightarrow{H^+} CH_3CH_2CH_2COOH$ The missing starting material is Pentan-2-one,

CH_3CH_2CH_2COCH_3

.

(vi)

This is the Cannizzaro reaction. An aldehyde with no

\alpha

-hydrogen atoms undergoes disproportionation in the presence of a concentrated base. Benzaldehyde is treated with concentrated

NaOH

. One molecule is reduced to benzyl alcohol and the other is oxidized to sodium benzoate. The missing products are Benzyl alcohol ( $C_6H_5CH_2OH$ ) and Sodium benzoate ( $C_6H_5COONa$ ). $2C_6H_5CHO + \text{conc. } NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$

(vii)

This is the reduction of a carboxylic acid to a primary alcohol. Butanoic acid is reduced to butan-1-ol. This requires a strong reducing agent like lithium aluminium hydride (

LiAlH_4

) followed by hydrolysis. The missing reagent is (i) $LiAlH_4$ / Ether (ii) $H_2O$ . $CH_3CH_2CH_2COOH \xrightarrow{(i) LiAlH_4 (ii) H_2O} CH_3CH_2CH_2CH_2OH$

(viii)

This is the hydrolysis of a nitrile (cyanide) to a carboxylic acid. The product is 2-hydroxypropanoic acid (lactic acid). The starting material must be the corresponding nitrile, which is 2-hydroxypropanenitrile (acetaldehyde cyanohydrin). $CH_3CH(OH)CN \xrightarrow{H_3O^+} CH_3CH(OH)COOH$ The missing starting material is 2-hydroxypropanenitrile,

CH_3CH(OH)CN

.

(ix)

This is the hydrolysis of an anhydride to form two molecules of carboxylic acid. The product is benzoic acid. The starting material must be benzoic anhydride. $(C_6H_5CO)_2O + H_2O \xrightarrow{H^+} 2C_6H_5COOH$ The missing starting material is Benzoic anhydride,

(C_6H_5CO)_2O

.

(x)

This is an aldol condensation between two molecules of butanal, followed by heating which causes dehydration (loss of water) to form an

\alpha, \beta

-unsaturated aldehyde. The product is 2-ethylhex-2-enal. $2CH_3CH_2CH_2CHO \xrightarrow{dil. NaOH, \Delta} CH_3CH_2CH_2CH=C(C_2H_5)CHO + H_2O$ The missing product is 2-Ethylhex-2-enal,

CH_3CH_2CH_2CH=C(C_2H_5)CHO

.

(xi) This is the catalytic hydrogenation of an unsaturated aldehyde. The reagent is hydrogen gas with a nickel catalyst. This will reduce both the carbon-carbon double bond and the aldehyde group. The starting material is but-2-enal. $CH_3CH=CHCHO \xrightarrow{H_2/Ni} CH_3CH_2CH_2CH_2OH$ The missing product is Butan-1-ol,

CH_3CH_2CH_2CH_2OH

.

Q8.18Exercises

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 $-\mathrm{NH}_{2}$ 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)

Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

Explanation: The formation of cyanohydrin is a nucleophilic addition reaction where a cyanide ion (

\mathrm{CN}^-

) attacks the carbonyl carbon. The reactivity of ketones in nucleophilic addition reactions is influenced by steric hindrance around the carbonyl group.

Cyclohexanone: The carbonyl group in cyclohexanone is relatively unhindered, allowing the nucleophile ( $\mathrm{CN}^-$ ) to attack it easily, leading to a good yield of the corresponding cyanohydrin. $\text{Cyclohexanone} + \mathrm{HCN} \rightarrow \text{Cyclohexanone cyanohydrin}$
2,2,6-trimethylcyclohexanone: This molecule has three bulky methyl groups. Two of these methyl groups are at the C-2 position and one at the C-6 position, which are adjacent to the carbonyl group (at C-1). These methyl groups create significant steric hindrance, shielding the carbonyl carbon from the attack of the nucleophile ( $\mathrm{CN}^-$ ). Therefore, the reaction does not proceed or proceeds in very poor yield.

(ii)

There are two

-\mathrm{NH}_{2}

groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

Explanation: The structure of semicarbazide is

H_2N(1)-C(O)-NH(2)-NH_2(3)

. There are two

-\mathrm{NH}_{2}

groups, one attached to the carbonyl carbon (position 1) and the other attached to another nitrogen atom (position 3).

The lone pair of electrons on the nitrogen atom at position 1 is involved in resonance with the carbonyl group. This delocalization makes the lone pair less available for nucleophilic attack. The resonating structures are: $H_2N-C(O)-NH-NH_2 \leftrightarrow {H_2N}^+=C(O^-)-NH-NH_2$ Due to this resonance, the nitrogen at position 1 is less nucleophilic.
The lone pair of electrons on the nitrogen atom at position 3 is not involved in resonance. It is freely available for nucleophilic attack on the carbonyl group of an aldehyde or a ketone.

Therefore, only the

-\mathrm{NH}_{2}

group at position 3 acts as a nucleophile and 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.

Explanation: The reaction of a carboxylic acid with an alcohol to form an ester, known as Fischer esterification, is a reversible reaction.

The general reaction is:

\mathrm{RCOOH} + \mathrm{R'OH} \xrightarrow{\mathrm{H}^+} \mathrm{RCOOR'} + \mathrm{H_2O}

\text{Carboxylic acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{Water}

According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract the effect of the change.

In this equilibrium, water and ester are the products. If either water or the ester is removed from the reaction mixture as it is formed, the equilibrium will shift in the forward direction to produce more products. This increases the yield of the ester. Water is often removed by distillation or by using a dehydrating agent.

Q8.19Exercises

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

The problem requires us to determine the structure of an organic compound based on its composition and chemical properties.

Step 1: Determine the empirical formula

Given composition:

Carbon (C) = 69.77%
Hydrogen (H) = 11.63%
Oxygen (O) = 100% - (69.77% + 11.63%) = 18.6%

We can determine the ratio of atoms by dividing the percentage by the atomic mass of each element.

Element	Percentage	Atomic Mass	Relative moles	Simplest Ratio
Carbon (C)	69.77	12.01	$\frac{69.77}{12.01} = 5.81$	$\frac{5.81}{1.16} \approx 5$
Hydrogen (H)	11.63	1.008	$\frac{11.63}{1.008} = 11.54$	$\frac{11.54}{1.16} \approx 10$
Oxygen (O)	18.60	16.00	$\frac{18.60}{16.00} = 1.16$	$\frac{1.16}{1.16} = 1$

The simplest whole-number ratio of C:H:O is 5:10:1. Therefore, the empirical formula of the compound is

\mathrm{C_5H_{10}O}

.

Step 2: Determine the molecular formula

The empirical formula mass = (5

\times

12.01) + (10

\times

1.008) + (1

\times

16.00) = 60.05 + 10.08 + 16.00 = 86.13 g/mol. The given molecular mass is 86.

n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{86}{86.13} \approx 1

Thus, the molecular formula is the same as the empirical formula:

\mathrm{C_5H_{10}O}

.

Step 3: Deduce the structure from chemical properties

It does not reduce Tollens' reagent: This indicates that the compound is not an aldehyde. Since the formula $\mathrm{C_5H_{10}O}$ corresponds to a saturated acyclic aldehyde or ketone (general formula $\mathrm{C_nH_{2n}O}$ ), the compound must be a ketone.
It forms an addition compound with sodium hydrogensulphite ( $\mathrm{NaHSO_3}$ ): This confirms the presence of a carbonyl group (C=O). This reaction is typical for most aldehydes and methyl ketones.
It gives a positive iodoform test: This is a specific test for the presence of a methyl keto group ( $\mathrm{CH_3-CO-}$ ) or a group that can be oxidized to it (like $\mathrm{CH_3-CH(OH)-}$ ). Since we know it is a ketone, it must contain the $\mathrm{CH_3-CO-}$ group.
Combining the formula and iodoform test: The molecular formula is $\mathrm{C_5H_{10}O}$ and it has a $\mathrm{CH_3-CO-}$ group. The remaining part of the molecule must be $\mathrm{C_5H_{10}O} - \mathrm{C_2H_3O} = \mathrm{C_3H_7}$ . The structure is $\mathrm{C_3H_7-CO-CH_3}$ . The $\mathrm{C_3H_7}$ group can be n-propyl ( $\mathrm{CH_3CH_2CH_2-}$ ) or isopropyl ( $\mathrm{ (CH_3)_2CH- }$ ). This gives two possible structures:
- Pentan-2-one: $\mathrm{CH_3-CO-CH_2CH_2CH_3}$
- 3-Methylbutan-2-one: $\mathrm{CH_3-CO-CH(CH_3)_2}$
On vigorous oxidation it gives ethanoic and propanoic acid: According to Popoff's rule, during the oxidation of an unsymmetrical ketone, the C-C bond cleavage occurs such that the keto group stays with the smaller alkyl group. However, in practice, cleavage can occur on both sides.
- Let's consider Pentan-2-one: $\mathrm{CH_3-CO-CH_2CH_2CH_3}$ Cleavage of the $\mathrm{C_2-C_3}$ bond gives ethanoic acid and propanoic acid. $\mathrm{CH_3-CO | -CH_2CH_2CH_3} \xrightarrow{\text{[O]}} \mathrm{CH_3COOH} + \mathrm{CH_3CH_2COOH}$ This matches the given products.
- Let's consider 3-Methylbutan-2-one: $\mathrm{CH_3-CO-CH(CH_3)_2}$ Cleavage would give ethanoic acid ( $\.mathrm{CH_3COOH}$ ) and propanone ( $\.mathrm{CH_3COCH_3}$ ). Propanone would further oxidize to give more ethanoic acid and $\.mathrm{CO_2}$ . This does not yield propanoic acid.

Therefore, the compound must be Pentan-2-one.

Final Answer: The possible structure of the compound is Pentan-2-one.

\mathrm{CH_3-CO-CH_2CH_2CH_3}

Q8.20Exercises

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

Solution

Carboxylic acids are stronger acids than phenols because the conjugate base of a carboxylic acid (carboxylate ion) is more stabilized by resonance than the conjugate base of a phenol (phenoxide ion).

Let's compare the stability of the two conjugate bases:

1. Carboxylic Acid and Carboxylate Ion

A carboxylic acid (

\mathrm{RCOOH}

) dissociates in water to form a carboxylate ion (

\mathrm{RCOO}^-

) and a hydronium ion.

\mathrm{RCOOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{RCOO}^- + \mathrm{H_3O}^+

The carboxylate ion is stabilized by resonance. The negative charge is delocalized over two highly electronegative oxygen atoms. The two contributing resonance structures are equivalent.

Resonance in Carboxylate Ion:

\mathrm{R-C(=O)-O^-} \leftrightarrow \mathrm{R-C(-O^-)=O}

Key features of resonance in carboxylate ion:

There are two equivalent resonance structures.
The negative charge is delocalized over two electronegative oxygen atoms.

2. Phenol and Phenoxide Ion

A phenol (

\mathrm{C_6H_5OH}

) dissociates in water to form a phenoxide ion (

\mathrm{C_6H_5O}^-

) and a hydronium ion.

\mathrm{C_6H_5OH} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_6H_5O}^- + \mathrm{H_3O}^+

The phenoxide ion is also stabilized by resonance. The negative charge is delocalized from the oxygen atom into the benzene ring. This results in five resonance structures.

Resonance in Phenoxide Ion: The negative charge is on the oxygen atom in two structures, and on the ortho and para carbon atoms of the benzene ring in the other three structures.

Key features of resonance in phenoxide ion:

There are five resonance structures, but they are not equivalent.
In three of the five structures, the negative charge is on a less electronegative carbon atom.
The structures where the negative charge is on carbon are less stable contributors to the resonance hybrid than the structures where it is on the more electronegative oxygen atom.

Comparison and Conclusion

Although the phenoxide ion has more resonance structures (five) than the carboxylate ion (two), the resonance in the carboxylate ion is more effective and leads to greater stability for two main reasons:

Equivalence of Structures: The two resonance structures of the carboxylate ion are identical in energy (equivalent). Equivalent resonance structures contribute equally to the resonance hybrid and lead to a very high degree of stabilization. The five resonance structures of the phenoxide ion are not equivalent and contribute differently, leading to less effective stabilization.
Charge Delocalization: In the carboxylate ion, the negative charge is spread over two highly electronegative oxygen atoms. In the phenoxide ion, the charge is spread over one electronegative oxygen atom and three less electronegative carbon atoms. Delocalization of charge onto a more electronegative atom is more stabilizing.

Because the carboxylate ion is significantly more stable than the phenoxide ion, the equilibrium for the dissociation of a carboxylic acid lies further to the right than that for a phenol. This means that carboxylic acids have a greater tendency to donate a proton and are therefore stronger acids than phenols.

Q8.2Intext Question

Write the structures of products of the following reactions;

(i)

(ii)

$\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\right)_{2} \mathrm{Cd}+2 \mathrm{CH}_{3} \mathrm{COCl} \rightarrow$

(iii)

(iv)

Solution

The products of the following reactions are:

(i)

C_6H_6 + CH_3COCl \xrightarrow{\text{Anhydrous } AlCl_3} C_6H_5COCH_3 + HCl

This is the Friedel-Crafts acylation of benzene. The electrophile, acylium ion (

CH_3CO^+

), is generated from acetyl chloride and anhydrous aluminium chloride. It attacks the benzene ring to form acetophenone. Product: Acetophenone (

C_6H_5COCH_3

)

(ii)

(C_6H_5CH_2)_2Cd + 2CH_3COCl \rightarrow 2C_6H_5CH_2COCH_3 + CdCl_2

This reaction involves an organocadmium reagent (dibenzylcadmium) and an acid chloride (acetyl chloride). Organocadmium reagents are specific for the preparation of ketones from acid chlorides. The benzyl group from the organocadmium compound replaces the chlorine atom of the acid chloride. Product: 1-Phenylpropan-2-one (

C_6H_5CH_2COCH_3

)

(iii)

CH_3-C\equiv CH + H_2O \xrightarrow{Hg^{2+}, H_2SO_4} [CH_3-C(OH)=CH_2] \rightleftharpoons CH_3-CO-CH_3

This is the hydration of an alkyne (propyne) in the presence of mercuric sulphate and sulphuric acid. According to Markovnikov's rule, the addition of water places the

-OH

group on the more substituted carbon, forming an unstable enol intermediate. This enol rapidly tautomerises to the more stable keto form, propanone. Product: Propanone (

CH_3COCH_3

)

(iv)

p-NO_2C_6H_4CH_3 \xrightarrow{\text{(i) } CrO_2Cl_2, CS_2 \text{ (ii) } H_3O^+} p-NO_2C_6H_4CHO

This is the Etard reaction. The reagent chromyl chloride (

CrO_2Cl_2

) oxidizes the methyl group of p-nitrotoluene to an aldehyde group. The reaction proceeds via a chromium complex intermediate, which is then hydrolysed to yield the aldehyde. The nitro group (

-NO_2

) is unaffected by this reagent. Product: p-Nitrobenzaldehyde (

p-NO_2C_6H_4CHO

)

Q8.3Intext Question

Arrange the following compounds in increasing order of their boiling points. $\mathrm{CH}_{3} \mathrm{CHO}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$

Solution

The boiling points of compounds depend on the strength of intermolecular forces of attraction. The compounds given are

CH_3CHO

(ethanal),

CH_3CH_2OH

(ethanol),

CH_3OCH_3

(methoxymethane), and

CH_3CH_2CH_3

(propane). All have comparable molar masses (44-46 g/mol).

$CH_3CH_2CH_3$ (Propane): It is a nonpolar alkane. The only intermolecular forces are weak London dispersion forces. Therefore, it has the lowest boiling point.
$CH_3OCH_3$ (Methoxymethane): It is an ether, which is weakly polar. It exhibits dipole-dipole interactions, which are stronger than the dispersion forces in propane. Thus, its boiling point is higher than that of propane.
$CH_3CHO$ (Ethanal): It is an aldehyde with a polar carbonyl group. It has stronger dipole-dipole interactions than methoxymethane due to the higher polarity of the $C=O$ bond compared to the $C-O-C$ bonds. Hence, its boiling point is higher than that of methoxymethane.
$CH_3CH_2OH$ (Ethanol): It is an alcohol. The presence of the $-OH$ group allows ethanol molecules to form strong intermolecular hydrogen bonds with each other. Hydrogen bonds are much stronger than dipole-dipole interactions or London dispersion forces. Consequently, ethanol has the highest boiling point among the given compounds.

The increasing order of boiling points is based on the increasing strength of intermolecular forces: Van der Waals forces < Dipole-dipole interactions < Hydrogen bonding.

Final Answer: The increasing order of boiling points is:

CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH

(Propane < Methoxymethane < Ethanal < Ethanol)

Q8.6Intext Question

Give the IUPAC names of the following compounds:

(i)

$\mathrm{Ph} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$

(ii)

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOOH}$

(iii)

CC1CCCC1C(=O)O

(iv)

O=C(O)c1c(N+[O-])cc(N+[O-])cc1N+[O-]

Solution

(i)

$\mathrm{PhCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ The parent chain is a three-carbon carboxylic acid, which is propanoic acid. The phenyl group (Ph) is a substituent at carbon 3 (numbering starts from the carboxyl carbon). Therefore, the IUPAC name is 3-Phenylpropanoic acid.

(ii)

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOOH}$ The parent chain contains four carbon atoms, including the carboxyl group, so it is a butenoic acid. Numbering starts from the carboxyl carbon. The double bond is between C-2 and C-3, so it is but-2-enoic acid. A methyl group is attached to C-3. Therefore, the IUPAC name is 3-Methylbut-2-enoic acid.

(iii)

2-Methylcyclopentanecarboxylic acid CC1CCCC1C(=O)O The structure shows a carboxylic acid group attached to a cyclopentane ring. The ring is the parent structure. The principal functional group is the carboxylic acid, so the numbering of the ring starts at the carbon atom bearing the -COOH group. A methyl group is attached to the adjacent carbon, which is C-2. Therefore, the IUPAC name is 2-Methylcyclopentanecarboxylic acid.

(iv)

2,4,6-Trinitrobenzoic acid O=C(O)c1c(N+[O-])cc(N+[O-])cc1N+[O-] The parent compound is benzoic acid. There are three nitro (-NO_2) groups attached to the benzene ring at positions 2, 4, and 6 with respect to the carboxylic acid group. Therefore, the IUPAC name is 2,4,6-Trinitrobenzoic acid.

Q8.7Intext Question

Show how each of the following compounds can be converted to benzoic acid.

(i)

Ethylbenzene

(ii)

Acetophenone

(iii)

Bromobenzene

(iv)

Phenylethene (Styrene)

Solution

The following conversions to benzoic acid can be achieved as shown:

(i)

Ethylbenzene to Benzoic acid Ethylbenzene can be oxidized to benzoic acid by using a strong oxidizing agent like alkaline potassium permanganate (

KMnO_4

) or acidic potassium dichromate (

K_2Cr_2O_7

). The entire alkyl side-chain is oxidized to a carboxyl group.

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3} \xrightarrow{\mathrm{KMnO}_{4}-\mathrm{KOH}, \Delta} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK} \xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}

(ii)

Acetophenone to Benzoic acid Acetophenone, being a methyl ketone, undergoes the haloform reaction. It is treated with sodium hypohalite (e.g., sodium hypoiodite, prepared from

I_2

and NaOH) to form sodium benzoate, which on acidification gives benzoic acid.

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3} + 3\mathrm{NaOI} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa} + \mathrm{CHI}_{3} + 2\mathrm{NaOH}

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa} \xrightarrow{\mathrm{H}^{+}/\mathrm{H}_{2} \mathrm{O}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}

(iii)

Bromobenzene to Benzoic acid Bromobenzene is first converted into a Grignard reagent (phenylmagnesium bromide) by reacting it with magnesium in dry ether. The Grignard reagent is then treated with solid carbon dioxide (dry ice), followed by acid hydrolysis to produce benzoic acid.

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br} + \mathrm{Mg} \xrightarrow{\text{Dry ether}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr} + \mathrm{O=C=O} \xrightarrow{\text{Dry ether}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOMgBr} \xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}

(iv)

Phenylethene (Styrene) to Benzoic acid Phenylethene can be vigorously oxidized using hot, alkaline potassium permanganate. The double bond is cleaved, and the carbon attached to the benzene ring is oxidized to a carboxyl group.

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2} \xrightarrow{\mathrm{KMnO}_{4}-\mathrm{KOH}, \Delta} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK} \xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}

Q8.8Intext Question

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

(i)

$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ or $\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$

(ii)

$\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$ or $\mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}$

(iii)

$\mathrm{CH}_{2} \mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ or $\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{CO}_{2} \mathrm{H}$

(iv)

O=C(O)c1ccc(C(F)(F)F)cc1 or Cc1ccc(C(=O)O)cc1

Solution

The strength of a carboxylic acid is determined by the stability of its conjugate base (carboxylate anion) after donating a proton. Electron-withdrawing groups (EWGs) stabilize the anion by dispersing the negative charge, thus increasing acidity. Electron-donating groups (EDGs) destabilize the anion, decreasing acidity.

(i)

$\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$ is stronger than $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ . Fluorine is a highly electronegative atom and acts as an electron-withdrawing group (-I effect). It pulls electron density away from the carboxylate group, stabilizing the resulting fluoroacetate anion. The methyl group in acetic acid is weakly electron-donating (+I effect), which slightly destabilizes the acetate anion. Therefore, fluoroacetic acid is a stronger acid than acetic acid.

(ii)

$\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}$ is stronger than $\mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}$ . Both fluorine and chlorine are electron-withdrawing. However, fluorine is more electronegative than chlorine. Consequently, the -I effect of fluorine is stronger than that of chlorine. This leads to greater stabilization of the fluoroacetate anion compared to the chloroacetate anion. Thus, fluoroacetic acid is a stronger acid.

(iii)

$\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ is stronger than $\mathrm{CH}_{2} \mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ . The inductive effect (-I effect) of the electron-withdrawing fluorine atom decreases with increasing distance from the carboxyl group. In 3-fluorobutanoic acid (

\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{CO}_{2} \mathrm{H}

), the fluorine atom is on carbon-3, which is closer to the -COOH group than the fluorine atom in 4-fluorobutanoic acid (

\mathrm{CH}_{2} \mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}

), which is on carbon-4. The closer proximity results in a stronger acid-strengthening effect. Therefore, 3-fluorobutanoic acid is stronger.

(iv)

4-(Trifluoromethyl)benzoic acid is stronger than 4-Methylbenzoic acid. O=C(O)c1ccc(C(F)(F)F)cc1 or Cc1ccc(C(=O)O)cc1 The trifluoromethyl group (-CF_3) is a very powerful electron-withdrawing group due to the strong -I effect of the three fluorine atoms. It strongly stabilizes the benzoate anion. In contrast, the methyl group (-CH_3) is an electron-donating group (through +I effect and hyperconjugation), which destabilizes the benzoate anion. Therefore, 4-(Trifluoromethyl)benzoic acid is a significantly stronger acid than 4-methylbenzoic acid (p-toluic acid).

Q8.1Intext Questions

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

The structures of the given compounds are as follows:

(i)

α-Methoxypropionaldehyde The parent aldehyde is propionaldehyde (

CH_3CH_2CHO

). The α-carbon is the carbon atom adjacent to the aldehyde group (C-2). A methoxy group (

-OCH_3

) is attached to this α-carbon. Structure:

CH_3-CH(OCH_3)-CHO

(ii)

3-Hydroxybutanal The parent aldehyde is butanal (

CH_3CH_2CH_2CHO

). A hydroxy group (

-OH

) is present at position 3. Structure:

CH_3-CH(OH)-CH_2-CHO

(iii)

2-Hydroxycyclopentane carbaldehyde 'Carbaldehyde' indicates that a

-CHO

group is attached to a ring system, in this case, a cyclopentane ring. A hydroxy group (

-OH

) is at position 2 of the ring, with the carbon attached to the

-CHO

group being C-1. Structure:

  CHO
 /  \

(CH-OH)-CH_2 | | CH_2----CH_2

Or, represented linearly, a cyclopentane ring with a CHO group and an OH group on adjacent carbons.

(iv)

4-Oxopentanal The parent aldehyde is pentanal (

CH_3CH_2CH_2CH_2CHO

). The term 'oxo' indicates a ketone group (

=O

) is present as a substituent. It is at position 4. Structure:

CH_3-CO-CH_2-CH_2-CHO

(v)

Di-sec-butyl ketone A ketone has a carbonyl group (

C=O

) bonded to two alkyl groups. Here, both groups are sec-butyl. A sec-butyl group is

CH_3CH_2CH(CH_3)-

. Structure:

(CH_3CH_2CH(CH_3))_2C=O

or

CH_3CH_2CH(CH_3)-CO-CH(CH_3)CH_2CH_3

Its IUPAC name is 5-methylheptan-3-one, but the structure is derived from the common name given.

(vi)

4-Fluoroacetophenone Acetophenone is a benzene ring attached to an acetyl group (

-COCH_3

). A fluorine atom (

-F

) is attached at position 4 (the para position) of the benzene ring. Structure: A benzene ring with a

-COCH_3

group at C-1 and a

-F

atom at C-4.

Q8.4Intext Questions

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. Hint: Consider steric effect and electronic effect.

Solution

The reactivity of aldehydes and ketones in nucleophilic addition reactions is governed by two main factors:

Steric Effect: The presence of bulky groups around the carbonyl carbon hinders the approach of the nucleophile, decreasing reactivity.
Electronic Effect: Electron-donating groups (like alkyl groups, $+I$ effect) decrease the positive charge (electrophilicity) on the carbonyl carbon, making it less reactive. Electron-withdrawing groups (like $-NO_2$ , $-I, -R$ effect) increase the electrophilicity of the carbonyl carbon, making it more reactive.

In general, aldehydes are more reactive than ketones because they have only one alkyl group (and a small H atom) compared to two alkyl groups in ketones, resulting in less steric hindrance and less deactivation by electron donation.

(i) Ethanal, Propanal, Propanone, Butanone.

Ketones vs Aldehydes: Propanone and Butanone are ketones and are less reactive than Ethanal and Propanal, which are aldehydes.
Comparing Aldehydes: Ethanal ( $CH_3CHO$ ) has a methyl group, while Propanal ( $CH_3CH_2CHO$ ) has a larger ethyl group. The ethyl group causes more steric hindrance and has a stronger electron-donating ( $+I$ ) effect than the methyl group. Thus, Propanal is less reactive than Ethanal.
Comparing Ketones: Butanone ( $CH_3COCH_2CH_3$ ) has a methyl and an ethyl group attached to the carbonyl carbon, whereas Propanone ( $CH_3COCH_3$ ) has two methyl groups. Butanone is sterically more hindered and has a stronger electron-donating effect from the ethyl group compared to the second methyl group in propanone. Therefore, Butanone is less reactive than Propanone.

The order of reactivity is: Butanone < Propanone < Propanal < Ethanal.

(ii) Benzaldehyde, $p$ -Tolualdehyde, $p$ -Nitrobenzaldehyde, Acetophenone.

Ketone vs Aldehydes: Acetophenone ( $C_6H_5COCH_3$ ) is a ketone, while the other three are aldehydes. Due to greater steric hindrance (phenyl and methyl vs phenyl and H) and the electron-donating methyl group, Acetophenone is the least reactive.
Comparing Aromatic Aldehydes: The reactivity of Benzaldehyde, $p$ $p$ -Tolualdehyde, and $p$ $p$ -Nitrobenzaldehyde depends on the substituent on the para position of the benzene ring.
- $p$ -Tolualdehyde ( $p-CH_3-C_6H_4-CHO$ ): The methyl group ( $-CH_3$ ) is electron-donating ( $+I$ effect, hyperconjugation), which deactivates the carbonyl group towards nucleophilic attack. It is less reactive than benzaldehyde.
- Benzaldehyde ( $C_6H_5CHO$ ): This serves as the reference compound.
- $p$ -Nitrobenzaldehyde ( $p-NO_2-C_6H_4-CHO$ ): The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group ( $-R$ and $-I$ effect). It withdraws electron density from the carbonyl carbon, increasing its electrophilicity and making it the most reactive of the three aldehydes.

The order of reactivity is: Acetophenone <

p

-Tolualdehyde < Benzaldehyde <

p

-Nitrobenzaldehyde.

Final Answer:

(i)

The increasing order of reactivity is: Butanone < Propanone < Propanal < Ethanal.

(ii)

The increasing order of reactivity is: Acetophenone <

p

-Tolualdehyde < Benzaldehyde <

p

-Nitrobenzaldehyde.

Q8.5Intext Questions

Predict the products of the following reactions:

(i)

(ii)

(iii)

(iv)

Solution

The products of the given reactions are predicted as follows:

(i)

This is the reaction of cyclohexanone with semicarbazide (

H_2N-NH-CO-NH_2

). It is a condensation reaction where the carbonyl oxygen of the ketone and two hydrogen atoms from the terminal

-NH_2

group of semicarbazide are eliminated as a water molecule, forming a semicarbazone.

C_6H_{10}O + H_2N-NH-CO-NH_2 \xrightarrow{H^+} C_6H_{10}=N-NH-CO-NH_2 + H_2O

Product: Cyclohexanone semicarbazone.

(ii)

This reaction involves the reduction of a lactone (a cyclic ester) with Diisobutylaluminium hydride (DIBAL-H), followed by hydrolysis. DIBAL-H reduces esters to aldehydes. In the case of a lactone, the ring is cleaved, and the ester group is reduced to an aldehyde, while the ester oxygen becomes a hydroxyl group.

The reactant is 4-hydroxypentanoic acid lactone (γ-valerolactone). The reduction opens the ring to form a hydroxy aldehyde.

\text{Lactone} \xrightarrow{\text{(i) DIBAL-H (ii) } H_2O} \text{Hydroxy aldehyde}

Product: 4-Hydroxy-5-methylhexanal. The structure is

CH_3-CH(OH)-CH(CH_3)-CH_2-CH_2-CHO

.

(iii)

This is the reaction of toluene (

C_6H_5CH_3

) with potassium permanganate (

KMnO_4

) in the presence of potassium hydroxide (

KOH

). Hot, alkaline

KMnO_4

is a strong oxidizing agent that oxidizes the alkyl side chain of a benzene ring to a carboxyl group (

-COOH

). The reaction first produces the potassium salt of the carboxylic acid (potassium benzoate), which upon acidification (workup with

H_3O^+

) gives the carboxylic acid.

C_6H_5CH_3 \xrightarrow{KMnO_4, KOH, \Delta} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH

Product: Benzoic acid (

C_6H_5COOH

).

(iv)

This reaction shows the addition of Hydrogen Cyanide (HCN) to a ketone (cyclohexanone), followed by hydrolysis. The first step is a nucleophilic addition of the cyanide ion (

CN^-

) to the carbonyl carbon, forming a cyanohydrin. The second step is the acid-catalyzed hydrolysis of the nitrile group (

-C\equiv N

) to a carboxylic acid group (

-COOH

).

Step 1: Formation of cyanohydrin.

C_6H_{10}O + HCN \rightarrow C_6H_{10}(OH)CN

Step 2: Hydrolysis of the cyanohydrin.

C_6H_{10}(OH)CN + 2H_2O \xrightarrow{H^+, \Delta} C_6H_{10}(OH)COOH + NH_3

The overall reaction converts the carbonyl group into a hydroxyl group and a carboxyl group on the same carbon atom.

Product: 1-Hydroxycyclohexanecarboxylic acid.

Aldehydes, Ketones and Carboxylic AcidsClass 12 Chemistry NCERT Solutions

What is meant by the following terms ? Give an example of the reaction in each case.

Cyanohydrin

Acetal

Semicarbazone

Aldol

Hemiacetal

Oxime

Ketal

Imine

2,4-DNP-derivative

Schiff's base

Solution

Name the following compounds according to IUPAC system of nomenclature:

CH3CH(CH3)CH2CH2CHO\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}CH3​CH(CH3​)CH2​CH2​CHO

CH3CH2COCH(C2H5)CH2CH2Cl\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}CH3​CH2​COCH(C2​H5​)CH2​CH2​Cl

CH3CH=CHCHO\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}CH3​CH=CHCHO

CH3COCH2COCH3\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COCH}_{3}CH3​COCH2​COCH3​

CH3CH(CH3)CH2C(CH3)2COCH3\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}CH3​CH(CH3​)CH2​C(CH3​)2​COCH3​

(CH3)3CCH2COOH\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{COOH}(CH3​)3​CCH2​COOH

OHCC6H4CHO−p\mathrm{OHCC}_{6} \mathrm{H}_{4} \mathrm{CHO}-pOHCC6​H4​CHO−p

Solution

Draw the structures of the following compounds.

3-Methylbutanal

ppp-Nitropropiophenone

p-Methylbenzaldehyde

4-Methylpent-3-en-2-one

4-Chloropentan-2-one

3-Bromo-4-phenylpentanoic acid

p,pp, pp,p '-Dihydroxybenzophenone

Hex-2-en-4-ynoic acid

Solution

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

CH3CO(CH2)4CH3\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}CH3​CO(CH2​)4​CH3​

CH3CH2CHBrCH2CH(CH3)CHO\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}CH3​CH2​CHBrCH2​CH(CH3​)CHO

CH3(CH2)5CHO\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}CH3​(CH2​)5​CHO

Ph−CH=CH−CHO\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}Ph−CH=CH−CHO

O=CC1CCCC1

PhCOPh

Solution

Draw structures of the following derivatives.

The 2,4-dinitrophenylhydrazone of benzaldehyde

Cyclopropanone oxime

Acetaldehydedimethylacetal

The semicarbazone of cyclobutanone

The ethylene ketal of hexan-3-one

The methyl hemiacetal of formaldehyde

Solution

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

PhMgBr and then H3O+\mathrm{H}_{3} \mathrm{O}^{+}H3​O+

Tollens' reagent

Semicarbazide and weak acid

Excess ethanol and acid

Zinc amalgam and dilute hydrochloric acid

Solution

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.

Methanal

2-Methylpentanal

Benzaldehyde

Benzophenone

Cyclohexanone

1-Phenylpropanone

Phenylacetaldehyde

Butan-1-ol

2,2-Dimethylbutanal

Solution

How will you convert ethanal into the following compounds?

Butane-1,3-diol

But-2-enal

But-2-enoic acid

Solution

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

An organic compound with the molecular formula C9H10O\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}C9​H10​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

Solution

Arrange the following compounds in increasing order of their property as indicated:

Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN )

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

Solution

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$

$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$

$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COCH}_{3}$

$\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}$

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{COOH}$

$\mathrm{OHCC}_{6} \mathrm{H}_{4} \mathrm{CHO}-p$

$p$ -Nitropropiophenone

$p, p$ '-Dihydroxybenzophenone

$\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}$

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}$

$\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}$

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

PhMgBr and then $\mathrm{H}_{3} \mathrm{O}^{+}$

An organic compound with the molecular formula $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$ forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Benazaldehyde to $\alpha$ -Hydroxyphenylacetic acid

There are two $-\mathrm{NH}_{2}$ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

$\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\right)_{2} \mathrm{Cd}+2 \mathrm{CH}_{3} \mathrm{COCl} \rightarrow$

$\mathrm{Ph} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOOH}$