Q.Which of the following compound will give disproportionation reaction in the presence of alkali?

The Cannizzaro reaction is a base-mediated disproportionation unique to aldehydes with no alpha-hydrogen, where one molecule is reduced to an alcohol and another is oxidised to a carboxylate salt.

Aldehydes that have no alpha-hydrogen (i.e., no hydrogen atom on the carbon adjacent to the -CHO group, such as benzaldehyde, formaldehyde, or trimethylacetaldehyde) cannot undergo the usual base-catalysed aldol condensation, since that requires removing an alpha-hydrogen to form an enolate.

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The Cannizzaro reaction requires an aldehyde with no alpha-hydrogen atoms (i.e., not enolizable). Among the given options, only $(CH_3)_3C{-}CHO$ (pivalaldehyde) has no $\alpha$-H, so it alone undergoes the reaction. The correct option is (B).

The Cannizzaro (base-catalysed self oxidation-reduction/disproportionation) reaction occurs only with aldehydes that lack an alpha-hydrogen atom.

An aldehyde with alpha-hydrogens instead undergoes base-catalysed aldol condensation (since the alpha-H is acidic enough to be removed by strong base, forming an enolate that attacks another aldehyde molecule), NOT the Cannizzaro reaction.

Checking each option:

• CH3-CHO (acetaldehyde): has alpha-H (on the CH3 carbon) - undergoes aldol, not Cannizzaro.
• CH3-CH2-CHO (propanal): has alpha-H - undergoes aldol.
• (CH3)2-CH-CHO (isobutyraldehyde): has alpha-H (on the CH carbon) - undergoes aldol. …

Cannizzaro's reaction needs an aldehyde with NO alpha-hydrogen; acetaldehyde has alpha-hydrogens and undergoes a different reaction (aldol condensation) instead.

Cannizzaro's reaction (base-mediated disproportionation into one molecule of alcohol and one of the carboxylate salt) occurs only for aldehydes that have no $\alpha$-hydrogen — because if $\alpha$-hydrogens were present, the far faster aldol condensation pathway would dominate instead.

Checking each option for $\alpha$-hydrogens:

• (a) 1-methylcyclohexane-1-carbaldehyde: the carbon bearing $-CHO$ has no H on it (fully substituted, quaternary-type carbon holding both $-CHO$ and $-CH_3$ plus two ring bonds) → no $\alpha$-H → gives Cannizzaro.
• (b) Benzaldehyde: the carbonyl carbon is attached directly to the aromatic ring, with no adjacent sp³ carbon bearing H → no $\alpha$-H → gives Cannizzaro (the classic textbook example). …

Formaldehyde and benzaldehyde undergo Cannizzaro reaction precisely because they LACK an alpha-hydrogen, not because they have one.

Assertion (A) is correct: aldehydes that have no $\alpha$-hydrogen atom (such as HCHO and $C_6H_5CHO$) cannot undergo aldol condensation; instead, when treated with concentrated alkali, two molecules of such an aldehyde undergo a self oxidation-reduction (disproportionation) reaction called the Cannizzaro reaction, giving one molecule of the corresponding alcohol and one molecule of the corresponding carboxylate salt.

$2HCHO \xrightarrow{conc.\,NaOH} CH_3OH + HCOONa$ …

Only aldehydes without any alpha-hydrogen atom (so they cannot undergo aldol condensation) undergo the Cannizzaro reaction on treatment with concentrated alkali.

The Cannizzaro reaction is a self-oxidation-reduction (disproportionation) reaction in which two molecules of an aldehyde lacking alpha-hydrogens, when treated with concentrated (50%) NaOH or KOH, give one molecule of an alcohol (by reduction) and one molecule of a carboxylate salt (by oxidation).

Example: 2HCHO + NaOH -> CH3OH + HCOONa (formaldehyde disproportionates to methanol and sodium formate).

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Benzaldehyde has no $\alpha$-hydrogen, so it cannot undergo aldol condensation; instead, concentrated alkali makes it undergo the Cannizzaro reaction — a self-oxidation-reduction (disproportionation) between two molecules of the aldehyde.

Benzaldehyde ($C_6H_5CHO$) has no hydrogen on the carbon adjacent to the carbonyl (the ring carbon takes that position), so it cannot enolise and cannot undergo aldol condensation. When treated with a concentrated (strong) base such as NaOH, it instead undergoes the Cannizzaro reaction: hydroxide ion adds to the carbonyl of one molecule, and the resulting alkoxide intermediate transfers a hydride ion to the carbonyl carbon of a second benzaldehyde molecule. This is a dispropor …

Formaldehyde has no alpha-hydrogen, so concentrated NaOH disproportionates it via the Cannizzaro reaction into sodium formate and methanol.

Aldehydes lacking an alpha-hydrogen (such as formaldehyde, HCHO, and benzaldehyde) cannot undergo aldol condensation. Instead, when treated with concentrated alkali (typically 50% NaOH), they undergo a self-oxidation-reduction (disproportionation) called the Cannizzaro reaction: one molecule is oxidised to the carboxylate salt while another is reduced to the alcohol.

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Cannizzaro needs no α-H; acetaldehyde has α-H, so it is the exception.

Cannizzaro's reaction: 2 RCHO + conc. NaOH → RCH₂OH + RCOO⁻ (self oxidation–reduction). It occurs only for aldehydes lacking an α-hydrogen. …

Formaldehyde undergoes Cannizzaro's reaction; acetaldehyde does not.

The Cannizzaro reaction is a disproportionation of an aldehyde in concentrated alkali: one molecule is oxidised to a carboxylate ion and another is reduced to an alcohol. This reaction is possible only for aldehydes that lack an alpha-hydrogen (an H on the carbon next to the -CHO), because aldehydes with alpha-hydrogens undergo aldol condensation instead.

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