Question

Identify the compound $\left( X \right),\left( Y \right)$ and $\left( Z \right)$ by making an appropriate choice.
$C{H_3}Br\xrightarrow{{Mg/ether}}X\xrightarrow[{water}]{{C{O_2}}}Y\xrightarrow[\Delta ]{{C{H_3}OH,{H^ + }}}Z$
A: $\left( X \right) = C{H_3}MgBr,\left( Y \right) = C{H_3}COOH,\left( Z \right) = C{H_3}COOC{H_3}$
B: $\left( X \right) = C{H_3}C{H_2}Br,\left( Y \right) = C{H_3}C{H_2}OH,\left( Z \right) = C{H_3}C{H_2}C{H_2}C{H_3}$
C: $\left( X \right) = C{H_3}C{H_2}MgBr,\left( Y \right) = C{H_3}C{H_2}COOH,\left( Z \right) = C{H_3}C{H_2}COC{H_3}$
D: $\left( X \right) = C{H_3}COOH,\left( Y \right) = C{H_3}C{H_2}COC{H_3},\left( Z \right) = C{H_3}COOC{H_3}$

Answer 1

There are many reactions in organic chemistry. Most of these reactions are name reactions that is, those reactions have specific names. Each reaction is different from the other. Their reactants, products and catalysts are different. Each reactant or agent used in a reaction plays an important role in obtaining the desired product.

Complete step by step answer:
In this question we were given a reactant and series of reactions and we have to find the product of each reaction. In the first reaction we have to find $X$. The reaction is:
$C{H_3}Br\xrightarrow{{Mg/ether}}X$
In this reaction $C{H_3}Br$ reacts with $Mg$ in the presence of ether. When an alkyl halide reacts with magnesium in the presence of ether Grignard reagent is formed. Grignard reagent is also called Grignard compound which is a compound of general formula $R - Mg - X$ where, $R$ is alkyl chain and $X$ is halogen. In this reaction alkyl halide reacts with $Mg$ in the presence of ether. This means formation of Grignard reagent will take place and the formula of product will be according to the above mentioned general formula. And the formula of the product will be $C{H_3}MgBr$ (here $R$ is $C{H_3}$ and $X$ is $Br$).
So $X$ is $C{H_3}MgBr$.
Second step of the reaction is:
$C{H_3}MgBr\xrightarrow[{water}]{{C{O_2}}}Y$ (Substituting $X$ in given reaction)
In this reaction $C{H_3}MgBr$ reacts with $C{O_2}$ and water. When Grignard reagent ($C{H_3}MgBr$) reacts with carbon dioxide a carboxylic salt is produced which further when reacts with water becomes carboxylic acid. In this reaction product will have two carbons one from Grignard reagent and one from carbon dioxide. Chemical formula of carboxylic acid with two carbons is $C{H_3}COOH$.
Therefore $Y$ is $C{H_3}COOH$.
Third reaction is:
$C{H_3}COOH\xrightarrow[\Delta ]{{C{H_3}OH,{H^ + }}}Z$ (Substituting $Y$ in given reaction)
In this reaction carboxylic acid $C{H_3}COOH$ reacts with alcohol $C{H_3}OH$. When an alcohol reacts with carboxylic acid in the presence of acid then ester and water is formed. Easters are acid derivatives. Acidic hydrogen of carboxylic acid is replaced by alkyl group in ester. In this reaction carboxylic acid is $C{H_3}COOH$, from this acid hydrogen is replaced by $C{H_3}$ (alkyl group) of $C{H_3}OH$. Therefore, compound $Z$ is $C{H_3}COOC{H_3}$ .
Now, we know all our products. $X$ is $C{H_3}MgBr$, $Y$ is $C{H_3}COOH$ and $Z$ is $C{H_3}COOC{H_3}$.
So, the correct answer is option A.

Note:
Wurtz fittig reaction is the chemical reaction of aryl halides with alkyl halides and sodium metal in the presence of dry ether to give substituted aromatic compounds. Aryl halides are also known as haloarene. Aryl halide is an aromatic compound in which one or more hydrogen atoms bonded to an aromatic ring are replaced by a halide.