Question

A $2^o$ alcohol as shown can be obtained by the reaction of di-t-butyl ketone and :

• A. Isopropyl magnesium bromide.
• B. t-butyl magnesium bromide.
• C. $EtMgBr$.
• D. $MeMgBr$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Isopropyl magnesium bromide.
B t-butyl magnesium bromide.
C $EtMgBr$.

Due to the steric hindrance of t-butyl group in ketone and G.R containing bulky group or G.R with $\beta -H$, the reaction fails (i.e., $3^o$ alcohol is not obtained). But a hydride ion $\left(H^{\ominus }\right)$ transfer from the $\beta -$ position of $RMgX$ to $(C=O)$ group takes place through a cyclic transition state to give $2^o$ alcohol and alkene.

Therefore, $MeMgBr$ or $PhMgBr$ wihtout a $\beta$-H cannot act as a reducing agent and fail to react with di-t-butyl ketone.

(a) (b) (c) react with di-t-butyl ketone to give $2^o$ alcohol and alkene.