Question

The reaction of $\alpha$-bromophenylacetone with $\text{NaBH}_4$ in $\text{CH}_3\text{OH}$ followed by treatment with ethylmagnesium bromide and then $\text{H}_3\text{O}^+$ yields which of the following products?

• A.

        OH
        |
  Ph - CH - CH2 - CH3
  

• B.

        OH
        |
  Ph - CH - CH2 - CH2 - CH3
  

• C.

        OH
        |
  Ph - CH - CH2 - CH2 - CH2 - CH3
  

• D.

        OH
        |
  Ph - CH - CH2 - CH2 - CH2 - CH2 - CH3
  

Answer 1

Answer: (B)

      OH
      |
Ph - CH - CH2 - CH2 - CH3

Answer 2

1. Reduction of ketone: $\alpha$-bromophenylacetone is reduced by $\text{NaBH}_4$ to 2-bromo-1-phenylethanol. $\text{PhCOCH}_2\text{Br} \xrightarrow{\text{NaBH}_4/\text{CH}_3\text{OH}} \text{PhCH}(\text{OH})\text{CH}_2\text{Br}$
2. Reaction with Grignard reagent: The Grignard reagent ($\text{C}_2\text{H}_5\text{MgBr}$) reacts with the acidic proton of the alcohol and then displaces the bromide via an $\text{S}_{\text{N}}2$ reaction. $\text{PhCH}(\text{OH})\text{CH}_2\text{Br} \xrightarrow{\text{1. } \text{C}_2\text{H}_5\text{MgBr} \text{ (excess)} \\ \text{2. } \text{H}_3\text{O}^+} \text{PhCH}(\text{OH})\text{CH}_2\text{C}_2\text{H}_5$ The product is 1-phenylpentan-1-ol.