Question

Which of the following compounds would not give tert-butyl alcohol when treated with excess methyl magnesium bromide followed by acid?

• A. acetyl chloride
• B. acetaldehyde
• C. methyl acetate
• D. acetic anhydride

Answer 1

Answer: (B)

acetaldehyde

Answer 2

The reaction of Grignard reagents with carbonyl compounds and carboxylic acid derivatives leads to the formation of alcohols after hydrolysis. The type of alcohol formed depends on the starting material.

1. Reaction with Aldehydes (except formaldehyde): Aldehydes react with Grignard reagents to form secondary alcohols.

   $RCHO + R'MgX \rightarrow RCH(OMgX)R'$

   $RCH(OMgX)R' + H_3O^+ \rightarrow RCH(OH)R'$ (Secondary alcohol)

   For acetaldehyde ($CH_3CHO$) and methyl magnesium bromide ($CH_3MgBr$):

   $CH_3CHO + CH_3MgBr \rightarrow CH_3CH(OMgBr)CH_3$

   $CH_3CH(OMgBr)CH_3 + H_3O^+ \rightarrow CH_3CH(OH)CH_3$ (Isopropanol, a secondary alcohol)

2. Reaction with Ketones: Ketones react with Grignard reagents to form tertiary alcohols.

   $R_2CO + R'MgX \rightarrow R_2C(OMgX)R'$

   $R_2C(OMgX)R' + H_3O^+ \rightarrow R_2C(OH)R'$ (Tertiary alcohol)

   Tert-butyl alcohol is $(CH_3)_3COH$. To form tert-butyl alcohol using $CH_3MgBr$, the ketone intermediate must be acetone ($CH_3COCH_3$).

   $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$

   $(CH_3)_3COMgBr + H_3O^+ \rightarrow (CH_3)_3COH$ (Tert-butyl alcohol)

3. Reaction with Carboxylic Acid Derivatives (Acid Halides, Esters, Anhydrides): These compounds react with two equivalents of Grignard reagent (if the R group attached to the carbonyl is not H) to form tertiary alcohols via a ketone intermediate.

   • Acetyl chloride ($CH_3COCl$):

     $CH_3COCl + CH_3MgBr \rightarrow CH_3COCH_3 + MgClBr$ (Acetone)

     $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$

     $(CH_3)_3COMgBr + H_3O^+ \rightarrow (CH_3)_3COH$ (Tert-butyl alcohol)

   • Methyl acetate ($CH_3COOCH_3$):

     $CH_3COOCH_3 + CH_3MgBr \rightarrow CH_3COCH_3 + Mg(OCH_3)Br$ (Acetone)

     $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$

     $(CH_3)_3COMgBr + H_3O^+ \rightarrow (CH_3)_3COH$ (Tert-butyl alcohol)

   • Acetic anhydride ($(CH_3CO)_2O$):

     $(CH_3CO)_2O + CH_3MgBr \rightarrow CH_3COCH_3 + CH_3COOMgBr$ (Acetone)

     $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$

     $(CH_3)_3COMgBr + H_3O^+ \rightarrow (CH_3)_3COH$ (Tert-butyl alcohol)

From the analysis, acetyl chloride, methyl acetate, and acetic anhydride react with excess methyl magnesium bromide to form acetone as an intermediate, which then reacts further to give tert-butyl alcohol. Acetaldehyde, being an aldehyde, reacts with methyl magnesium bromide to form isopropanol, which is a secondary alcohol.

Therefore, acetaldehyde is the compound that would not give tert-butyl alcohol.