Question

The major product of the reaction between CH$_3$CH$_2$ONa and (CH$_3$)$_3$CCl in ethanol is:

• A. CH$_3$CH$_2$OC(CH$_3$)$_3$
• B. CH$_2$ = C(CH$_3$)$_2$
• C. CH$_3$CH$_2$C(CH$_3$)$_3$
• D. CH$_3$CH = CHCH$_3$

Hint:

Tertiary alkyl halides preferentially undergo elimination (E2) instead of substitution due to steric hindrance.

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction Mechanism 
The given reaction involves sodium ethoxide (CH$_3$CH$_2$ONa) and tert-butyl chloride [(CH$_3$)$_3$CCl] in ethanol. Tertiary alkyl halides undergo elimination (E2) rather than substitution due to steric hindrance. 
Step 2: Identifying the Major Product 
Since sodium ethoxide is a strong base, elimination occurs via the E2 mechanism, leading to the formation of an alkene. The product formed is isobutene (\( CH_2 = C(CH_3)_2 \)). Thus, the correct answer is (B).

Answer 2

Step 1: Identify the reactants and their nature
CH₃CH₂ONa is sodium ethoxide, a strong base and a good nucleophile.
(CH₃)₃CCl is tert-butyl chloride, a tertiary alkyl halide.

Step 2: Consider the type of reaction
Tertiary alkyl halides typically do not undergo nucleophilic substitution easily due to steric hindrance.
Instead, elimination reactions (E2 or E1) are favored in presence of a strong base like sodium ethoxide.

Step 3: Predict the product
Elimination of HCl from tert-butyl chloride will give an alkene.
The product formed is isobutylene (2-methylpropene), which has the formula CH₂=C(CH₃)₂.

Step 4: Conclusion
The major product of the reaction between CH₃CH₂ONa and (CH₃)₃CCl in ethanol is CH₂=C(CH₃)₂.
This is due to the base-induced elimination mechanism favored with tertiary alkyl halides.