Question

Consider the reactions :- Identify $A, X, Y$ and $Z$

• A. A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide
• B. A - Ethanal , X - Ethanol , Y - But - 2 - enal , Z - Semicarbazone
• C. A- Ethanol, X- Acet aldehyde, Y- Butanone, Z- Hydrazone
• D. A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-hydrazine

Answer 1

The Correct Option is B

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 $OH ^{-}$ 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 (2).

$\underset{(X)}{CH _{3}- CH _{2} OH}{ ->[{Cu}][{573\,K}]} {\underset{(A)}{CH3 -CHO} ->[{(Ag(NH_3)_2)^+,OH^-}][{\Delta}] CH3 -COOH}$