A `2^@` alcohol can be obtained by the reaction of di-t-butyl ketone and

Options

A. Isopropyl magnesium bromide
B. `t-Butyl` magnesium bromide
C. `EtMgBr`
D. `MeMgBr`

Answer

A, B, C

Explanation

(a,b,c) Due to the steric hindrance of `t-butyl` group in ketone and `G.R` containing bulkyl group of `G.R` with `beta-H`, the reaction fails (i.e., `3^@` alcohol is not obtained). But a hydride ion `(H^(Ө))` transfer from the `beta-position` of `RMgX` to `(C=O)` group takes place through a cyclic transition state to give `2^@` alcohol and alkene.
Therefore, `MeMgBr` or `PhMgBr` without a `beta-H` cannot act as a reducing agent and fail to react with `di-t-butyl` ketone.
`(a),(b)`and `( c)` react with `di-t-butyl` ketone to give `2^@` alcohol and alkene.
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