Exams › JEE Advanced › Chemistry
Correct answer: CH3COCH=C(CH3)2
CaC2 + H2O -> C2H2 (acetylene). C2H2 + H2O (HgSO4, H2SO4) -> CH3CHO (Markovnikov addition, Wacker-type). CH3CHO undergoes aldol condensation: 2 CH3CHO --OH^(-)--> CH3CH(OH)CH2CHO --(-H2O, heat)--> CH3CH=CHCHO (but-2-enal/crotonaldehyde). However, with further reaction, another aldol of crotonaldehyde or aldol of acetaldehyde could give a different product. The product CH3COCH=C(CH3)2 is more consistent with 3-molecule aldol of acetaldehyde sequence in some conditions. Let me reconsider: standard base-catalyzed aldol of acetaldehyde gives crotonaldehyde (CH3CH=CHCHO). The answer listed as (A) is mesityl oxide type. This question may be from a specific source where the intended answer is (B) crotonaldehyde.