Question

Predict the % product formation in the given reaction :
$CH_3CH_2CBr(CH_3)_2 \xrightarrow{KOH(alc.)} CH_3CH=C(CH_3)_2 \ (X) + CH_3CH_2C(CH_3)=CH_2 \ (Y)$}

• A. X = 29%, Y = 71%
• B. X = 71%, Y = 29%
• C. X = 50%, Y = 50%
• D. X = 0%, Y = 100%

Hint:

Small bases (e.g., $NaOH$, $KOH$, $NaOEt$, $NaOMe$) favor the Zaitsev product (more substituted alkene). Bulky bases (e.g., potassium tert-butoxide, LDA) suffer from steric hindrance and thus favor the Hofmann product (less substituted alkene, formed by removing the most accessible proton).

Answer 1

The Correct Option is B

Step 1: Concept:
The given reaction is a dehydrohalogenation of a tertiary alkyl halide (2-bromo-2-methylbutane) using an alcoholic base ($KOH$), which proceeds primarily via an E2 elimination mechanism. We need to identify the major and minor alkene products and their relative percentages.

Step 2: Key Formula or Approach:
The regioselectivity of E2 eliminations using small, unhindered bases (like $OH^-$ from $KOH(alc)$ or ethoxide) is governed by Zaitsev's Rule.
Zaitsev's Rule states that the major product will be the most highly substituted, and therefore most thermodynamically stable, alkene.

Step 3: Step-by-step Explanation:

• The substrate, 2-bromo-2-methylbutane, has $\beta$-hydrogens on two different types of adjacent carbon atoms.

• Elimination involving the $\beta$-hydrogen from the $CH_2$ group (C-3) yields 2-methyl-2-butene. This is product (X). It has three alkyl groups attached to the double bond, making it a trisubstituted alkene.

• Elimination involving a $\beta$-hydrogen from one of the $CH_3$ groups (C-1 or the methyl branch) yields 2-methyl-1-butene. This is product (Y). It has only two alkyl groups attached to the double bond, making it a disubstituted alkene.

• Because $KOH(alc.)$ is a relatively small base, it can easily access the more sterically hindered internal protons to form the more stable alkene.

• Therefore, product (X), the Zaitsev product, will be the major product.

• Historically and experimentally, the dehydrohalogenation of 2-bromo-2-methylbutane with small bases yields approximately 71% of the trisubstituted alkene (X) and 29% of the disubstituted alkene (Y).

• This specific 71:29 ratio is a classic textbook example used to illustrate regiochemistry in E2 eliminations.

Step 4: Final Answer:
The major product X forms around 71%, and the minor product Y forms around 29%. This matches option (B).