JEE Challenger
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ChemMedMain 2026 • 5th Apr - Shift 2

Condensation of Acetaldehyde with Semicarbazide and Hydrolysis of Hemiacetal

Given below are two statements :

Statement I : The condensation reaction between CH3−CH=O\mathrm{CH_3-CH=O} and H2N−N∣H−C∥O−NH2\mathrm{H_2N-\underset{\displaystyle\substack{\textstyle|\\ \textstyle\text{H}}}{N}-\underset{\displaystyle\substack{\textstyle\|\\ \textstyle\text{O}}}{C}-NH_2} under optimum pH will produce CH3−CH=N−C∥O−N∣H−NH2\mathrm{CH_3-CH=N-\underset{\displaystyle\substack{\textstyle\|\\ \textstyle\text{O}}}{C}-\underset{\displaystyle\substack{\textstyle|\\ \textstyle\text{H}}}{N}-NH_2}

Statement II : The molecule, Ph−CH∣OH−O−CH3\mathrm{Ph-\mkern-8mu\underset{\displaystyle\begin{array}{l}|\\\text{OH}\end{array}}{CH}\mkern-8mu-O-CH_3} will generate Ph−CH=O\mathrm{Ph-CH=O} in the presence of dilute acid.

In the light of the above statements, choose the correct answer from the options given below :

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Correct

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and Ketones

Step-by-Step Solution

To evaluate the correctness of the given statements, let us analyze them individually:

Analysis of Statement I:

Semicarbazide has the chemical formula H2N−NH−C(=O)−NH2\text{H}_2\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2 and contains two distinct types of −NH2-\text{NH}_2 groups:

  1. Amide −NH2-\text{NH}_2 group: The lone pair of electrons on this nitrogen atom is involved in resonance with the adjacent carbonyl group: H2N−NH−CO∥−NH2⟷H2N−NH−CO−∣=NH2+\text{H}_2\text{N}-\text{NH}-\overset{\underset{\parallel}{\text{O}}}{\text{C}}-\text{NH}_2 \longleftrightarrow \text{H}_2\text{N}-\text{NH}-\overset{\underset{\vert}{\text{O}^-}}{\text{C}}=\text{NH}_2^+ Due to this delocalization, the nucleophilicity of this nitrogen is significantly reduced.

  2. Hydrazine-type −NH2-\text{NH}_2 group: The lone pair on the terminal nitrogen atom of the H2N−NH−\text{H}_2\text{N}-\text{NH}- moiety is not directly conjugated with the carbonyl group. Thus, it remains localized and far more nucleophilic.

During the nucleophilic addition-elimination reaction with acetaldehyde (CH3−CHO\text{CH}_3-\text{CHO}) under optimum pH (pH≈3.5−4.5\text{pH} \approx 3.5 - 4.5), attack occurs exclusively via the more nucleophilic hydrazine-type −NH2-\text{NH}_2 group to form acetaldehyde semicarbazone: CH3−CHO+H2N−NH−CO∥−NH2→optimum pHCH3−CH=N−NH−CO∥−NH2+H2O\text{CH}_3-\text{CHO} + \text{H}_2\text{N}-\text{NH}-\overset{\underset{\parallel}{\text{O}}}{\text{C}}-\text{NH}_2 \xrightarrow{\text{optimum pH}} \text{CH}_3-\text{CH}=\text{N}-\text{NH}-\overset{\underset{\parallel}{\text{O}}}{\text{C}}-\text{NH}_2 + \text{H}_2\text{O}

Statement I incorrectly depicts the product as CH3−CH=N−C(=O)−NH−NH2\text{CH}_3-\text{CH}=\text{N}-\text{C}(=\text{O})-\text{NH}-\text{NH}_2, which would falsely correspond to condensation involving the less nucleophilic amide nitrogen.

Therefore, Statement I is false.

Analysis of Statement II:

The given molecule, Ph−CH(OH)(OCH3)\text{Ph}-\text{CH}(\text{OH})(\text{OCH}_3), contains both a hydroxyl group (−OH-\text{OH}) and an alkoxy group (−OCH3-\text{OCH}_3) attached to the same carbon atom, making it a hemiacetal (specifically, benzaldehyde methyl hemiacetal).

Hemiacetals are unstable in aqueous acidic media and undergo rapid hydrolysis:

  1. Protonation of the methoxy oxygen gives a good leaving group (CH3OH\text{CH}_3\text{OH}): Ph−CH(OH)(OCH3)+H+⇌Ph−CH(OH)(H+O−CH3)\text{Ph}-\text{CH}(\text{OH})(\text{OCH}_3) + \text{H}^+ \rightleftharpoons \text{Ph}-\text{CH}(\text{OH})(\overset{+}{\text{H}}\text{O}-\text{CH}_3)
  2. Loss of methanol generates a resonance-stabilized oxocarbenium ion: Ph−CH(OH)(H+O−CH3)⇌Ph−CH=O+H+CH3OH\text{Ph}-\text{CH}(\text{OH})(\overset{+}{\text{H}}\text{O}-\text{CH}_3) \rightleftharpoons \text{Ph}-\text{CH}=\overset{+}{\text{O}}\text{H} + \text{CH}_3\text{OH}
  3. Deprotonation yields benzaldehyde: Ph−CH=O+H⇌Ph−CH=O+H+\text{Ph}-\text{CH}=\overset{+}{\text{O}}\text{H} \rightleftharpoons \text{Ph}-\text{CH}=\text{O} + \text{H}^+

Thus, in the presence of dilute acid, the hemiacetal hydrolyzes to regenerate benzaldehyde (Ph−CH=O\text{Ph}-\text{CH}=\text{O}).

Therefore, Statement II is true.

Conclusion:

  • Statement I is false.
  • Statement II is true.

This corresponds to Option D.

Condensation of Acetaldehyde with Semicarbazide and Hydrolysis of Hemiacetal | Chemistry PYQ Solution - JEE Challenger