Question

Complete the reaction:
${C_6}{H_5}COOH + C{H_3}MgI \to$ ?
(A) ${C_6}{H_5}COOMgI$
(B) $C{H_4}$
(C) Both A and B
(D) none

Answer 1

In the above reaction, nucleophile is involved. The Grignard reagent provides the required nucleophile. The alkyl group abstracts the acidic proton and the magnesium iodide will attack the conjugate base.

Complete step by step solution:
Grignard reagent is an organometallic compound in organic chemistry. The Grignard reagent exhibits nucleophilic qualities. The Grignard reagent has the ability to form new carbon-carbon bonds. When Grignard's reagent is present in a protic solvent, it accepts a proton from it. After accepting a proton, it becomes inert and does not react. It is an organomagnesium compound with general formula RMgX, here R is the alkyl group, Mg is magnesium and X is halogen atom. The Grignard reagent was discovered by the French chemist Victor Grignard. The Grignard reagent is prepared by the reaction of magnesium metal with alkyl or aryl halide through radical mechanism.
The Grignard reagent does acid-base reaction, substitution reaction and addition reaction. Of the following reaction, Grignard reagent prefers first acid-base, then substitution reaction and then addition reaction.
Here, the reactant is benzoic acid. Benzoic acid releases hydrogen ion as it is an acid. Grignard releases methanide ions. Methanide ion being a nucleophile, reacts with hydrogen ion and forms methane. The remaining part combines to form ${C_6}{H_5}COOMgI$ .
The reaction that takes place is as follows,
${C_6}{H_5}COOH + C{H_3}MgI \to C{H_4} + {C_6}{H_5}COOMgI$
Hence the product obtained is $C{H_4}$ and ${C_6}{H_5}COOMgI$ .
Therefore, the correct answer is option C.

Note:
Benzoic acid is the simplest carboxylic acid. Benzoic acid and their salts are used as preservatives. Benzoic acid is used in the treatment of fungal diseases. It is used in the industrial synthesis of many organic substances.