Question

Explain reason:
(a) Grignard reagent should be prepared under anhydrous conditions.
(b) ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CHClC}{\text{H}}_{\text{3}}$ is hydrolysed more easily with KOH than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{C}{\text{H}}_2\text{Cl}$.

Answer 1

Hint – Here you need to know that Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes. In ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CHClC}{\text{H}}_{\text{3}}$ carbocation is formed then it stabilizes the structure. Knowing this will solve your problem.

Complete answer:
(a) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes. Therefore, Grignard reagents should be prepared under anhydrous conditions.
(b) When $H_{2}O$ attacks the $2^0$ alkyl halide ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CHClC}{\text{H}}_{\text{3}}$ ​, the carbocation thus formed has resonance and hyper-conjugation to stabilize it.
In ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{C}{\text{H}}_2\text{Cl}$, only resonance is present.
Hence, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{CHClC}{\text{H}}_{\text{3}}$ becomes more stable than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{C}{\text{H}}_2\text{Cl}$ so it is hydrolysed more easily.

Note – Whenever you face such problems of chemistry you need to draw the structures and observe the arrangements of electrons in lone pairs and shared pairs. According to that, electronegativity of elements decides what could happen to make the structure more stable. When Grignard reagents (RMgX) are treated with the hydroxyl group it forms alkanes.