Question

$(25)$HO$\xrightarrow{\substack{CH_3MgBr \\ then \ H_2O}}$

Answer 1

Tetrahydrofuran

Answer 2

The reaction proceeds in two main steps:

1. Acid-base reaction: The acidic proton of the alcohol is abstracted by the strong base, $CH_3MgBr$, forming a magnesium alkoxide and methane gas. $$\text{HO-(CH}_2\text{)}_4\text{-Cl} + \text{CH}_3\text{MgBr} \longrightarrow \text{CH}_4 + \text{MgBr}^+\text{ }^-\text{O-(CH}_2\text{)}_4\text{-Cl}$$
2. Intramolecular SN2 reaction: The alkoxide acts as an internal nucleophile, attacking the carbon bonded to the chlorine atom, leading to the formation of a 5-membered cyclic ether and displacement of the chloride ion. $$\text{MgBr}^+\text{ }^-\text{O-(CH}_2\text{)}_4\text{-Cl} \longrightarrow \text{Tetrahydrofuran} + \text{MgClBr}$$

The product is Tetrahydrofuran.

4-chlorobutan-1-ol reacts with $CH_3MgBr$. First, the alcohol's acidic proton is removed by $CH_3MgBr$, forming an alkoxide. This alkoxide then undergoes an intramolecular SN2 reaction, with the oxygen attacking the carbon bearing the chlorine, displacing it to form a stable 5-membered cyclic ether, tetrahydrofuran. Workup with $H_2O$ quenches the reaction.