Question

What would be the major product formed in the reaction?

\[ CH_3-CH(CH_3)-CH_2Cl \xrightarrow[(CH_3)_3CO^-]{\text{heat}} X \]

• A. \( CH_3-CH(CH_3)-CH_2-O-C(CH_3)_3 \)
• B. \( CH_2-C(CH_3)_2-O-C(CH_3)_3 \)
• C. \( CH_2=C(CH_3)-CH_3 \)
• D. \( CH_3-CH(CH_3)-O-C(CH_3)_3 \)

Hint:

Bulky bases like tert-butoxide favor elimination (E2) reactions and often give less substituted (Hofmann) alkenes.

Answer 1

The Correct Option is C

Step 1: Identify the reaction conditions.
The reagent \( (CH_3)_3CO^- \) is tert-butoxide ion, which is a strong and bulky base.
Heat is also provided, which favors elimination over substitution.

Step 2: Determine reaction type.
Due to the bulky nature of tert-butoxide, it prefers elimination (E2 mechanism) rather than substitution.
Thus, dehydrohalogenation occurs, removing \( H \) and \( Cl \) to form an alkene.

Step 3: Identify possible elimination product.
The substrate is \( CH_3-CH(CH_3)-CH_2Cl \).
Elimination of \( H \) from the adjacent carbon and \( Cl \) gives:
\[ CH_2=C(CH_3)-CH_3 \]

Step 4: Reason for product formation.
Although bulky bases often give Hofmann product (less substituted alkene), here only one elimination pathway is feasible leading to the given alkene.

Step 5: Conclusion.
Thus, the major product formed is:
\[ \boxed{CH_2=C(CH_3)-CH_3} \]