question_answer

Consider the reactions: [NEET-2017]

A) A-Ethanol, X-Acetaldehyde, Y-Butanone, Z-Hydrazone

B) A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-hydrazine

C) A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide

D) A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone

Correct Answer: D

Solution :

[d] Since 'A' gives positive silver mirror test therefore, it must be an aldehyde or \[\alpha -\]Hydroxyketone.

Reaction with semicarbazide indicates that A can be an aldehyde or ketone.

Reaction with \[O{{H}^{-}}\]i.e., aldol condensation (by assuming alkali to be dilute) indicates that A is aldehyde as aldol reaction of ketones is reversible and carried out in special apparatus.

These indicates option (4).

\[\underset{(x)}{\mathop{C{{H}_{3}}-C{{H}_{2}}OH}}\,\xrightarrow[573\,K]{Cu}\underset{(A)}{\mathop{C{{H}_{3}}-CHO}}\,\xrightarrow[\Delta ]{{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}},O{{H}^{-}}}\]

\[C{{H}_{3}}-COOH\]