NEET Solved Paper 2017 Question 10
Question: Consider the reactions:
Options:
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
Show Answer
Answer:
Correct Answer: D
Solution:
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{CH _3-CH _2OH}}\ \xrightarrow[573\ K]{Cu}\underset{(A)}{\mathop{CH _3-CHO}}\ \xrightarrow[\Delta ]{{{[Ag{{(NH _3)}_2}]}^{+}} O{H^{-}}} $ $ CH _3-COOH $
BETA
