
In the following reaction:
${C_6}{H_5}MgBr + C{O_2} \to A\xrightarrow{{PC{l_5}}}B$
What is A and B?
A. Acid and acid halide
B. Acid and ester
C. Acid halide and acid anhydride
D. Acid anhydride and acid halide
Answer
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Hint: In the question, the reactants provided are phenyl magnesium bromide and carbon dioxide. Phenyl magnesium bromide is an example of Grignard reagent. Carbon dioxide is used as dry ice in this reaction. This reaction is mainly used in the formation of compounds having \[ - COOH\] group.
Complete step by step answer:
We will proceed with the reaction step wise.
Now, as already mentioned the reactants are phenyl magnesium bromide and carbon dioxide (dry ice). Grignard reagents react in the presence of dry ether with $C{O_2}$ to form the salts of carboxylic acids.
The salts formed of carboxylic acids will give the main carboxylic acid on acidification with the mineral acid. The reaction can be represented as follows:
${C_6}{H_5}MgBr + C{O_2}\xrightarrow{{dryether}}{C_6}{H_5}(CO){O^ - }MgB{r^ + }\xrightarrow{{{H_3}{O^ + }}}{C_6}{H_5}COOH$
So, the product formed is benzoic acid. It means that A is an acid.
Now, A will react further with $PC{l_5}$ . In this reaction, the hydroxyl group of carboxylic acid will be replaced by a chlorine atom. The reaction can be represented as follows:
${C_6}{H_5}MgBr + C{O_2}\xrightarrow{{dryether}}{C_6}{H_5}COOH\xrightarrow{{PC{l_5}}}{C_6}{H_5}COCl + POC{l_3} + HCl$
Here, the product formed is benzoyl chloride which falls into the category of acyl chloride. It means that B is an acid halide.
So, the reaction given in the question is
${C_6}{H_5}MgBr + C{O_2} \to {C_6}{H_5}COOH\xrightarrow{{PC{l_5}}}{C_6}{H_5}COCl$
In the last, we can conclude that A and B are acid and acid halide respectively.
So, the correct answer is Option A.
Note: For the conversion of carboxylic acids into acyl chloride; carboxylic acids can be treated with $PC{l_5}$, $PC{l_3}$ and $SOC{l_2}$ (any of these three). The most preferred among these three is $SOC{l_2}$ as the side products formed will be gaseous and that will be released. Thus, the main product acyl chloride is attained with purification.
Complete step by step answer:
We will proceed with the reaction step wise.
Now, as already mentioned the reactants are phenyl magnesium bromide and carbon dioxide (dry ice). Grignard reagents react in the presence of dry ether with $C{O_2}$ to form the salts of carboxylic acids.
The salts formed of carboxylic acids will give the main carboxylic acid on acidification with the mineral acid. The reaction can be represented as follows:
${C_6}{H_5}MgBr + C{O_2}\xrightarrow{{dryether}}{C_6}{H_5}(CO){O^ - }MgB{r^ + }\xrightarrow{{{H_3}{O^ + }}}{C_6}{H_5}COOH$
So, the product formed is benzoic acid. It means that A is an acid.
Now, A will react further with $PC{l_5}$ . In this reaction, the hydroxyl group of carboxylic acid will be replaced by a chlorine atom. The reaction can be represented as follows:
${C_6}{H_5}MgBr + C{O_2}\xrightarrow{{dryether}}{C_6}{H_5}COOH\xrightarrow{{PC{l_5}}}{C_6}{H_5}COCl + POC{l_3} + HCl$
Here, the product formed is benzoyl chloride which falls into the category of acyl chloride. It means that B is an acid halide.
So, the reaction given in the question is
${C_6}{H_5}MgBr + C{O_2} \to {C_6}{H_5}COOH\xrightarrow{{PC{l_5}}}{C_6}{H_5}COCl$
In the last, we can conclude that A and B are acid and acid halide respectively.
So, the correct answer is Option A.
Note: For the conversion of carboxylic acids into acyl chloride; carboxylic acids can be treated with $PC{l_5}$, $PC{l_3}$ and $SOC{l_2}$ (any of these three). The most preferred among these three is $SOC{l_2}$ as the side products formed will be gaseous and that will be released. Thus, the main product acyl chloride is attained with purification.
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