In a set of the given reactions, acetic acid yielded a product C. $CH_3COOH + PCl_5 \rightarrow A \xrightarrow[Anh.\ AlCl_3]{C_6H_6} B \xrightarrow[ether]{C_2H_5MgBr} C$ Product C would be
A
$CH_3CH(OH)C_2H_5$
B
$CH_3COC_6H_5$
C
$CH_3CH(OH)C_6H_5$
D
$CH_3-C(OH)(C_2H_5)-C_6H_5$
The answer is D.
Solution
Step 1: $PCl_5$ converts acetic acid into acetyl chloride (A): $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$ Step 2: Friedel-Crafts acylation of benzene with acetyl chloride gives acetophenone (B): $CH_3COCl + C_6H_6 \xrightarrow{anhyd.\ AlCl_3} CH_3COC_6H_5 + HCl$ Step 3: Ethylmagnesium bromide adds to the ketone; the ethyl group attacks the carbonyl carbon. Hydrolysis then gives a tertiary alcohol (C): $CH_3COC_6H_5 + C_2H_5MgBr \rightarrow CH_3C(OMgBr)(C_2H_5)C_6H_5 \xrightarrow{H_2O} CH_3C(OH)(C_2H_5)C_6H_5$ So C is 2-phenylbutan-2-ol, $CH_3-C(OH)(C_2H_5)-C_6H_5$.
ConceptAcid chloride to ketone to tertiary alcohol
Aldehydes, Ketones and Carboxylic Acids in past papers
91 questions from this chapter have appeared across 27 exam years.
