Question

Identify major product formed in the following reaction. 3-Bromo-2-methylpentane $\xrightarrow{Alc.KOH}$ Major product}

• A. 2-Methylpentan-3-ol
• B. 2-Methylpent-2-ene
• C. 4-Methylpent-3-ene
• D. 4-Methylpentan-3-ol

Hint:

In E2 elimination reactions with alcoholic $KOH$, the major product is generally the most substituted alkene according to Zaitsev's rule. This means the hydrogen is removed from the beta-carbon with the fewest hydrogen atoms, leading to a more stable, more substituted double bond.

Answer 1

The Correct Option is B

Step 1: Identify the reaction type The given reaction: \[ \text{3-Bromo-2-methylpentane} \xrightarrow{\text{alc.\ KOH \] involves an alkyl halide treated with alcoholic KOH. Alcoholic KOH acts as a strong base and favors E2 elimination (dehydrohalogenation), in which a hydrogen atom and a halogen atom are removed from adjacent carbons to form an alkene.
Step 2: Structure of the reactant and identification of $\alpha$ and $\beta$ carbons Structure of 3-Bromo-2-methylpentane: \[ \mathrm{CH_3 - CH(CH_3) - CH(Br) - CH_2 - CH_3} \]

• $\alpha$-carbon: Carbon bearing Br (C3)
• $\beta$-carbons: Adjacent carbons (C2 and C4)

Step 3: Possible elimination pathways (i) Elimination from C2 ($\beta_1$) and C3 ($\alpha$): Removal of H from C2 and Br from C3 gives: \[ \mathrm{CH_3 - C(CH_3)=CH - CH_2 - CH_3} \] This is 2-Methylpent-2-ene. (ii) Elimination from C4 ($\beta_2$) and C3 ($\alpha$): Removal of H from C4 and Br from C3 gives: \[ \mathrm{CH_3 - CH(CH_3) - CH=CH - CH_3} \] This is 4-Methylpent-2-ene.
Step 4: Apply Zaitsev's rule According to Zaitsev's rule, the major product is the more substituted alkene.

• 2-Methylpent-2-ene $\rightarrow$ more substituted (more stable)
• 4-Methylpent-2-ene $\rightarrow$ less substituted

Hence, the major product is: \[ \mathrm{2\text{-}Methylpent\text{-}2\text{-}ene} \]
Step 5: Conclusion \[ \boxed{\text{2-Methylpent-2-ene is the major product \]