Nucleophiles such as ammonia and its derivatives, \(H_2N-Z\), add to the carbonyl group of aldehydes and ketones. The reaction is reversible and catalysed by acid. The equilibrium favours product formation due to rapid dehydration of the intermediate to form \(-C=N-Z\), where Z = alkyl, aryl, OH, NH2, C6H5NH, NHCONH2, etc.
What is formed when acetaldehyde reacts with methylamine in the presence of a trace of acid?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → CH3CH=N–CH3
The passage states that a reagent \( \ce{H2N-Z} \) adds to the carbonyl group of an aldehyde or ketone in the presence of acid, and the intermediate is rapidly dehydrated to give >C=N–Z, where Z may be an alkyl group. Methylamine is \( \ce{H2N-CH3} \), so Z = –CH3:
\( \displaystyle \ce{CH3CHO + H2N-CH3} \xrightarrow{\ce{H^+}} \ce{CH3CH=N-CH3 + H2O} \)
The C=O oxygen leaves as water and a C=N double bond forms: the product is a substituted imine (Schiff’s base), N-methylethanimine.
Why the other options are incorrect
Option (2) CH3CH2–NH–CH3 — Incorrect. This is a secondary amine with a C–N single bond; it would need a further reduction of the C=N bond, and no reducing agent is present.
Option (3) CH3CO–NH–CH3 — Incorrect. This is an amide; it keeps a C=O group, while in this reaction the carbonyl oxygen is lost as water.
Option (4) CH3CH(OH)–NH–CH3 — Incorrect. This is the addition intermediate; it loses water rapidly, so it is not the product.