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Question #43974

Difficulty: 8/10 | 🟢 Free

The Question

Trichloroacetaldehyde, \ce{CCl3CHO (aq)}, is a highly electrophilic carbonyl compound that forms a stable gem-diol hydrate (Khyd≫1K_{hyd} \gg 1) in aqueous media. When 1.0 equivalent of \ce{CCl3CHO (aq)} is reacted with an excess of concentrated \ce{NaOD (aq)} in \ce{D2O (l)} at 298 K, the expected Cannizzaro reaction is completely suppressed due to a kinetically favored alternative pathway. A rapid fragmentation occurs, yielding a haloform derivative (P\mathbf{P}) and a carboxylate salt (Q\mathbf{Q}). Identify the major organic products P\mathbf{P} and Q\mathbf{Q}.

Options

  • A.

    P=\ceCDCl3(l)\mathbf{P} = \ce{CDCl3 (l)}, Q=\ceHCOO−(aq)\mathbf{Q} = \ce{HCOO- (aq)}

  • B.

    P=\ceCHCl3(l)\mathbf{P} = \ce{CHCl3 (l)}, Q=\ceDCOO−(aq)\mathbf{Q} = \ce{DCOO- (aq)}

  • C.

    P=\ceCCl3CH2OD(aq)\mathbf{P} = \ce{CCl3CH2OD (aq)}, Q=\ceCCl3COO−(aq)\mathbf{Q} = \ce{CCl3COO- (aq)}

  • D.

    P=\ceCDCl3(l)\mathbf{P} = \ce{CDCl3 (l)}, Q=\ceDCOO−(aq)\mathbf{Q} = \ce{DCOO- (aq)}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Cannizzaro Mechanism|Hydrate Stability vs. Carbonyl|Reduction Selectivity (Hydrides)

Why This Question Is Tricky

Common Mistakes

This question tests the 'Chemical Exception' to the Cannizzaro reaction.

Students often blindly apply the rule 'no α\alpha-hydrogens = Cannizzaro'.

The trap introduces chloral (\ce{CCl3CHO}), where the tetrahedral intermediate prefers C-C bond cleavage over a hydride shift due to the exceptional leaving group ability of the \ceCCl3−\ce{CCl3-} carbanion (Haloform-type cleavage).

Furthermore, the stable hydrate equilibrium kinetically suppresses the bimolecular Cannizzaro pathway.

The use of \ce{NaOD}/\ce{D2O} tests mechanistic atom-tracing, ensuring students recognize that the non-acidic formyl hydrogen remains on the formate (\ce{HCOO-}), while the carbanion abstracts a deuteron to form deuterochloroform (\ce{CDCl3}).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAldehydes, Ketones, and Carboxylic Acids - Class 12
ConceptCannizzaro Mechanism
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree