Question

2-Phenylbutan-2-ol can not be prepared by

• A. PhMgBr + $\xrightarrow[ether]{} H^{\oplus}$
• B. $CH_3MgBr + Ph - \overset{O}{\underset{||}{C}} - C_2H_5$ $\xrightarrow[ether]{} H^{\oplus}$
• C. $C_2H_5MgBr + Ph - \overset{O}{\underset{||}{C}} - CH_3$ $\xrightarrow[ether]{} H^{\oplus}$
• D. $CH_3CH_2CH_2MgBr + PhCHO$ $\xrightarrow[ether]{} H^{\oplus}$

Answer 1

Answer: (D)

4

Answer 2

To determine which reaction cannot prepare 2-phenylbutan-2-ol, we first need to understand the structure of 2-phenylbutan-2-ol and the general reaction of Grignard reagents with carbonyl compounds.

The structure of 2-phenylbutan-2-ol is:

$\mathrm{CH_3CH_2-C(OH)(Ph)-CH_3}$

This is a tertiary alcohol. Tertiary alcohols are typically formed by the reaction of a Grignard reagent with a ketone. Aldehydes (except formaldehyde) yield secondary alcohols, and formaldehyde yields primary alcohols.

Let's analyze each option:

1. PhMgBr + Butan-2-one ($\mathrm{CH_3COCH_2CH_3}$)

   • Reactants: Phenylmagnesium bromide (PhMgBr) and Butan-2-one (a ketone).

   • Reaction: The phenyl group (Ph-) from the Grignard reagent attacks the carbonyl carbon of butan-2-one.

     $\mathrm{CH_3COCH_2CH_3 + PhMgBr \xrightarrow{ether} CH_3C(OMgBr)(Ph)CH_2CH_3 \xrightarrow{H_3O^+} CH_3C(OH)(Ph)CH_2CH_3}$

   • Product: 2-phenylbutan-2-ol.

   • This reaction can prepare 2-phenylbutan-2-ol.

2. $\mathrm{CH_3MgBr + Ph-CO-C_2H_5}$ (Propiophenone)

   • Reactants: Methylmagnesium bromide ($\mathrm{CH_3MgBr}$) and Propiophenone (a ketone).

   • Reaction: The methyl group ($\mathrm{CH_3-}$) from the Grignard reagent attacks the carbonyl carbon of propiophenone.

     $\mathrm{Ph-CO-C_2H_5 + CH_3MgBr \xrightarrow{ether} Ph-C(OMgBr)(CH_3)-C_2H_5 \xrightarrow{H_3O^+} Ph-C(OH)(CH_3)-C_2H_5}$

   • Product: 2-phenylbutan-2-ol.

   • This reaction can prepare 2-phenylbutan-2-ol.

3. $\mathrm{C_2H_5MgBr + Ph-CO-CH_3}$ (Acetophenone)

   • Reactants: Ethylmagnesium bromide ($\mathrm{C_2H_5MgBr}$) and Acetophenone (a ketone).

   • Reaction: The ethyl group ($\mathrm{C_2H_5-}$) from the Grignard reagent attacks the carbonyl carbon of acetophenone.

     $\mathrm{Ph-CO-CH_3 + C_2H_5MgBr \xrightarrow{ether} Ph-C(OMgBr)(C_2H_5)-CH_3 \xrightarrow{H_3O^+} Ph-C(OH)(C_2H_5)-CH_3}$

   • Product: 2-phenylbutan-2-ol.

   • This reaction can prepare 2-phenylbutan-2-ol.

4. $\mathrm{CH_3CH_2CH_2MgBr + PhCHO}$ (Benzaldehyde)

   • Reactants: n-Propylmagnesium bromide ($\mathrm{CH_3CH_2CH_2MgBr}$) and Benzaldehyde (an aldehyde).

   • Reaction: The n-propyl group ($\mathrm{CH_3CH_2CH_2-}$) from the Grignard reagent attacks the carbonyl carbon of benzaldehyde.

     $\mathrm{PhCHO + CH_3CH_2CH_2MgBr \xrightarrow{ether} Ph-CH(OMgBr)-CH_2CH_2CH_3 \xrightarrow{H_3O^+} Ph-CH(OH)-CH_2CH_2CH_3}$

   • Product: 1-phenylbutan-1-ol. This is a secondary alcohol, not 2-phenylbutan-2-ol.

   • This reaction cannot prepare 2-phenylbutan-2-ol.

Therefore, the reaction that cannot prepare 2-phenylbutan-2-ol is option (4).