Question

In the following reaction, ${{C}_{6}}{{H}_{5}}C{{H}_{2}}Br\xrightarrow[(ii){{H}_{3}}{{O}^{+}}]{(i)Mg,\,ether}X$, the product ‘X’ is:
(A) ${{C}_{6}}{{H}_{5}}C{{H}_{2}}OC{{H}_{2}}{{C}_{6}}{{H}_{5}}$
(B) ${{C}_{6}}{{H}_{5}}C{{H}_{2}}OH$
(C) ${{C}_{6}}{{H}_{5}}C{{H}_{3}}$
(D) ${{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}{{C}_{6}}{{H}_{5}}$

Answer 1

Hint: A halogen derivative of alkane reacts with magnesium metal to form a very famous reagent. That reagent is an organometallic compound having a central magnesium atom directly bonded to an alkyl group and a halogen atom.


Complete step by step solution:
-Grignard reagent is an organometallic compound having a magnesium atom directly bonded to an alkyl group and a halogen atom.
-An organometallic compound is a compound in which a metal atom is directly bonded to a carbon atom. Some examples of organometallic compounds are methyl lithium, methyl magnesium bromide, trimethylaluminum, etc.
-When an alkyl halide reacts with magnesium metal in the presence of dry conditions which is provided using dry ether, alkyl magnesium halide is formed. This alkyl magnesium halide is known as Grignard’s reagent.
-In this case, benzyl bromide first reacts with magnesium metal in the presence of dry ether to form Benzyl magnesium bromide.

-Then, Benzyl magnesium bromide undergoes hydrolysis to give toluene.

-We can also represent this reaction as follows,
${{C}_{6}}{{H}_{5}}C{{H}_{2}}Br\xrightarrow[(ii){{H}_{3}}{{O}^{+}}]{(i)Mg,\,ether}{{C}_{6}}{{H}_{5}}C{{H}_{3}}$
Therefore, in the reaction, ${{C}_{6}}{{H}_{5}}C{{H}_{2}}Br\xrightarrow[(ii){{H}_{3}}{{O}^{+}}]{(i)Mg,\,ether}X$, the product ‘X’ is Toluene.

So, the correct answer is option (C) ${{C}_{6}}{{H}_{5}}C{{H}_{3}}$


Note: Carefully read these types of questions. The order in which the reactants present on the reaction arrow react with the starting compound is very important. When the numbers are reversed then the reaction might not take place or products obtained will be different.