Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) is refluxed with metallic sodium in dry ethers the major product obtained is (a) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)

The given reaction is a reaction between an alcohol and metallic sodium in a dry ether. This is called a sodium metal reduction reaction. It is a form of reduction reaction where the metallic sodium reduces the alcohol.

In a sodium metal reduction reaction, the hydrogen atom of the \(\mathrm{OH}\) group on the alcohol undergoes abstraction by the sodium metal. The sodium metal donates an electron to the hydrogen atom, effectively releasing it as a hydride ion (\(\mathrm{H}^{-}\)). Concurrently, the sodium replaces the hydrogen on the \(\mathrm{OH}\) group, forming a sodium alkoxide intermediate. The hydride ion then attacks the carbon atom bonded to the halogen atom (\(\mathrm{Br}\)), resulting in the formation of a double bond and the expulsion of the halogen ion (\(\mathrm{Br}^-\)).

To find the major product, apply the sodium metal reduction mechanism to the given alcohol: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\). First, the sodium metal abstracts the hydrogen from the \(\mathrm{OH}\) group: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{Na} \rightarrow \mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa} + \mathrm{H}^{-}\) Next, the hydride ion attacks the carbon atom bonded to the bromine atom, causing the double bond formation and the expulsion of the bromide ion: \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa} + \mathrm{H}^{-} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} + \mathrm{NaBr}\)

The major product formed after refluxing the given alcohol with metallic sodium in dry ethers is: (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)