Q.An organic compound 'A' on treatment with ethanoic acid in the presence of hydrochloric acid gas as a catalyst produces an ester 'B'. 'A' on oxidation with $\text{CrO}_3$ in an anhydrous medium gives 'C'. 'C' is heated with concentrated KOH followed by acidification with dilute HCl generates 'A' and 'D'. Three moles of 'D' reacts with $\text{PCl}_3$ gives three moles of compound with molecular formula HCOCl and 'E'. 'D' is reduced to 'A' by lithium aluminium hydride followed by hydrolysis. Write the molecular formulas of the compounds 'A', 'B', 'C', 'D' and 'E'.

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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