Identify the major products A and B formed in the following...

The given starting material is 2-methylpentan-3-ol, with the structure $\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH(CH}_3)_2$.

Step 1: Reaction with $\text{SOCl}_2$
Thionyl chloride ($\text{SOCl}_2$) converts alcohols into alkyl chlorides. This reaction proceeds without the formation of a free carbocation, meaning no rearrangement takes place. The $-\text{OH}$ group is directly replaced by a $-\text{Cl}$ atom.
Product A is $\text{CH}_3-\text{CH}_2-\text{CH(Cl)}-\text{CH(CH}_3)_2$, which is named 3-chloro-2-methylpentane.

Step 2: Reaction with alc. $\text{KOH}, \Delta$
Alcoholic $\text{KOH}$ with heat brings about dehydrohalogenation (an E2 elimination reaction). The removal of $\text{HCl}$ can occur in two possible directions from 3-chloro-2-methylpentane:
1. Elimination between C2 and C3 yields $\text{CH}_3-\text{CH}_2-\text{CH}=\text{C(CH}_3)_2$ (2-methylpent-2-ene). This is a trisubstituted alkene with 8 $\alpha$-hydrogens.
2. Elimination between C3 and C4 yields $\text{CH}_3-\text{CH}=\text{CH}-\text{CH(CH}_3)_2$ (4-methylpent-2-ene). This is a disubstituted alkene with 4 $\alpha$-hydrogens.

According to Zaitsev's rule, the more substituted and more stable alkene is formed as the major product. Thus, the major product B is 2-methylpent-2-ene.

Therefore, A = 3-Chloro-2-methylpentane and B = 2-Methylpent-2-ene.

Answer: A = 3-Chloro-2-methylpentane ; B = 2-Methylpent-2-ene