Discussion
The product of the following reaction is
\(\mathrm{C_{6}H_{5}CHO + C_{6}H_{5}COCH_{3} \xrightarrow[293\ K]{NaOH}}\)
Answer: A
The product is C6H5CH = CHCOC6H5 (benzalacetophenone, chalcone).
This is a Claisen-Schmidt (cross aldol) condensation. Benzaldehyde has no \(\alpha\)-hydrogen, so it acts only as the electrophile. NaOH removes an \(\alpha\)-hydrogen from the CH3 of acetophenone, and the carbanion adds to the aldehyde carbonyl. The \(\beta\)-hydroxy ketone formed loses water, because the resulting double bond is conjugated with both the ring and the C=O:
C6H5CHO + CH3COC6H5 gives C6H5CH=CHCOC6H5 + H2O.
Stilbene (C6H5CH=CHC6H5) has lost the carbonyl, and benzoin (C6H5CH(OH)COC6H5) comes from two benzaldehydes with cyanide. Benzil and deoxybenzoin are not formed here.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development