Q.(a) An organic Compound (A) of molecular formula $C_7H_6O$ undergoes Cannizaro reaction. Compound (A) also reacts with Chlorine in the presence of Conc. $FeCl_3$ to give Compound (B). Compound (A) reacts with Chlorine in the absence of catalyst to give Compound (C). Identify A, B and C with suitable reactions. OR

The Cannizzaro reaction occurs only with aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking an alpha-hydrogen undergo self oxidation-reduction (disproportionation) with concentrated alkali to give an alcohol and a salt of a carboxylic acid:

2 HCHO + NaOH --> CH3OH + HCOONa

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The Cannizzaro reaction (base-induced disproportionation into an alcohol and a carboxylate) only happens for aldehydes with NO alpha-hydrogen — because an alpha-H aldehyde instead undergoes the much faster aldol condensation.

Check each option for an alpha-hydrogen (an H on the carbon directly attached to -CHO):

1. cyclohexanecarbaldehyde — the ring carbon attached to CHO carries an alpha-H → undergoes aldol, not Cannizzaro.
2. CH3CHO (acetaldehyde) — the CH3 carbon has alpha-H's → aldol.
3. 2-methylpropanal, (CH3)2CH-CHO — the CH carbon has an alpha-H → aldol. …

The Cannizzaro reaction is given only by aldehydes with NO $\alpha$-hydrogen; CH3CHO has $\alpha$-hydrogens (on its methyl group), so it is oxidised/reduced by the aldol route instead and does not undergo Cannizzaro.

The Cannizzaro reaction is a base-mediated disproportionation (self oxidation–reduction) of an aldehyde lacking $\alpha$-hydrogens: two molecules of the aldehyde react with concentrated $NaOH$, one being oxidised to the carboxylate and the other reduced to the alcohol. If $\alpha$-hydrogens ARE present, the base instead deprotonates the $\alpha$-carbon and the aldehyde undergoes base-catalysed aldol condensation, which is much faster.

• (B) benzaldehyde ($C_6H_5CHO$): the carbon bonded to CHO is an aromatic ring carbon with no removable $\alpha$-H for enolisation — undergoes Cannizzaro (classic example, gives benzyl alcohol + sodium benzoate).
• (C) $HCHO$ (formaldehyde): has no $\alpha$-carbon at all — undergoes Cannizzaro (gives methanol + sodium formate). …

The Cannizzaro reaction is a self oxidation-reduction of aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking alpha-hydrogen react in concentrated alkali — one is oxidised to the acid (salt) and the other reduced to the alcohol.

• HCHO has no alpha-H → gives Cannizzaro reaction: …

Cannizzaro reaction (base-catalysed disproportionation of an aldehyde to an alcohol and a carboxylate) only happens for aldehydes that have NO alpha-hydrogen.

Cannizzaro's reaction requires an aldehyde lacking alpha-hydrogens, because if alpha-H were present, concentrated base would instead deprotonate it and trigger aldol condensation, which is much faster.

• (a) o-methylbenzaldehyde: the carbonyl carbon is attached directly to the aromatic ring — no alpha-carbon with H exists on that side, so it undergoes Cannizzaro.
• (c) HCHO (formaldehyde): the carbonyl carbon has only H atoms on it, no alpha-carbon at all — undergoes Cannizzaro (indeed reacts fastest of all). …

Formaldehyde has no α-H → Cannizzaro disproportionation with NaOH giving methanol + sodium formate.

Formaldehyde (HCHO) has no alpha-hydrogen atom, so it cannot undergo aldol condensation. Instead, with concentrated NaOH it undergoes the Cannizzaro reaction — a self oxidation-reduction (disproportionation):

2 HCHO + NaOH → CH3OH + HCOONa

One molecule of formaldehyde is reduced to methyl alcohol (CH3OH) and the other is oxidised to sodium formate (the sodium salt of formic acid).

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