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Question

(B) The reaction involves a secondary alkyl fluoride $(CH_3CH_2CH(F)CH_3)$ reacting with a strong base/nucleophile $(CH_3O^-)$ in a protic solvent $(C

Vedclass · Accepted Answer

$CH_3CH=CH-CH_3$

Vedclass · Answer

(B) The reaction involves a secondary alkyl fluoride $(CH_3CH_2CH(F)CH_3)$ reacting with a strong base/nucleophile $(CH_3O^-)$ in a protic solvent $(CH_3OH)$.Since $CH_3O^-$ is a strong base,the elimination reaction $(E2)$ is favored over substitution $(S_N2)$.According to Zaitsev's rule,the more substituted alkene is the major product.Elimination of $H$ from $C_2$ gives $CH_3CH=CHCH_3$ (but$-2-$ene),which is more stable due to hyperconjugation.Elimination of $H$ from $C_4$ gives $CH_3CH_2CH=CH_2$ (but$-1-$ene),which is less stable.Therefore,the major product is $CH_3CH=CHCH_3$.

The major product obtained from the reaction of $CH_3CH_2CH(F)CH_3$ with $CH_3O^- / CH_3OH$ is ..........

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