Question

Give the expected organic product when phenylacetic acid, \(\mathrm{PhCH}_{2} \mathrm{COOH}\), is treated with each reagent. (a) \(\mathrm{SOCl}_{2}\) (b) \(\mathrm{NaHCO}_{3}, \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{NaOH}_{2}, \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{MgBr}^{3}\) (one equivalent) (e) \(\mathrm{LiAlH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) (f) \(\mathrm{CH}_{2} \mathrm{~N}_{2}\) (g) \(\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (catalyst)

Short answer

Question: Predict the expected organic product(s) when phenylacetic acid is treated with the following reagents: (a) SOCl2, (b) NaHCO3, H2O, (c) NaOH2, (d) CH3MgBr (one equivalent), (e) LiAlH4 followed by H2O, (f) CH2N2, and (g) CH3OH + H2SO4 (catalyst). Answer: (a) Phenylacetyl chloride: PhCH2COCl (b) Sodium phenylacetate: PhCH2COO-Na+ (c) Error in exercise: NaOH2 doesn't exist (d) Methylated phenylacetic acid: PhCH(CH3)COOH (e) Phenylethanol: PhCH2CH2OH (f) Methyl ester of phenylacetic acid: PhCH2COOCH3 (g) Methyl ester of phenylacetic acid: PhCH2COOCH3

Step-by-step solution

(a) Reaction with SOCl2

In the first reaction, the phenylacetic acid is treated with thionyl chloride (SOCl2). Thionyl chloride is a common reagent used for converting carboxylic acids into acyl chlorides. This reaction occurs via nucleophilic acyl substitution. The oxygen atom in the carboxylic acid attacks the sulfur atom of thionyl chloride, which then leaves as a \(\mathrm{Cl}^{-}\) and replaces the \(\mathrm{OH}\) group with a \(\mathrm{Cl}\) atom. The product will be phenylacetyl chloride, \(\mathrm{PhCH}_{2}\mathrm{COCl}\).

(b) Reaction with NaHCO3, H2O

The second reaction involves treating phenylacetic acid with sodium bicarbonate (NaHCO3) followed by water (H2O). Sodium bicarbonate is a weak base that reacts with the acidic carboxylic acid group to form the sodium salt of carboxylate (phenylacetate, \(\mathrm{PhCH}_{2}\mathrm{COO}^{-}\mathrm{Na}^{+}\)).

(c) Error in Exercise: No clear reaction

There is likely an error in exercise for this part as NaOH2 doesn't exist. You may check the exercise again.

(d) Reaction with CH3MgBr (one equivalent)

With a Grignard reagent (CH3MgBr), phenylacetic acid will react via a nucleophilic acyl substitution. The carbon in the methyl group of the Grignard reagent will attack the carbonyl group of the phenylacetic acid, breaking the \(\mathrm{C=O}\) bond and leading to a methylated phenylacetic acid, \(\mathrm{PhCH}(\mathrm{CH}_{3})\mathrm{COOH}\).

(e) Reaction with LiAlH4 followed by H2O

Lithium aluminum hydride (LiAlH4) is a common reagent for reducing carboxylic acids to primary alcohols. LiAlH4 will reduce the carbonyl group to an alcohol, and the added water quenches the reaction. The final product will be a phenylethanol, \(\mathrm{PhCH}_{2}\mathrm{CH}_{2}\mathrm{OH}\).

(f) Reaction with CH2N2

The reaction of phenylacetic acid with diazomethane (CH2N2) results in the methyl ester of phenylacetic acid. Diazomethane replaces the carboxylic acid group with a methoxy group, forming \(\mathrm{PhCH}_{2}\mathrm{COOCH}_{3}\).

(g) Reaction with CH3OH + H2SO4 (catalyst)

The last reaction involves using methanol (CH3OH) and a catalytic amount of sulfuric acid (H2SO4). This reaction is an esterification of the carboxylic acid group. Methanol will react with the carbonyl group of phenylacetic acid in the presence of the acid catalyst, leading to the methyl ester, \(\mathrm{PhCH}_{2}\mathrm{COOCH}_{3}\).