The compound that will react most readily with NaOH to form methanol is

Question

The compound that will react most readily with NaOH to form methanol is (A) (CH3)4N+I- (B) CH3OCH3 (C) (CH3)3S+I- (D) (CH3)3CCl

Tardigrade Admin · Accepted Answer

Compound (CH3)4N+I- is most reactive due to (i) better leaving group, I- and (ii) due to the fact that the methyl group, with positive N, is not more electron deficient Hence this group is more reactive towards nucleophile, OH- <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/b06e5fea0d6b4dc3e85616a9fa22d949-.png" />

Q. The compound that will react most readily with NaOH to form methanol is

$(C H_{3})_{4} N^{+} I^{-}$

$C H_{3} OC H_{3}$

$(C H_{3})_{3} S^{+} I^{-}$

$(C H_{3})_{3} CCl$

Compound $(C H_{3})_{4} N^{+} I^{-}$ is most reactive due to (i) better leaving group, $I^{-}$ and (ii) due to the fact that the methyl group, with positive N, is not more electron deficient
Hence this group is more reactive towards nucleophile, $O H^{-}$

Q. The compound that will react most readily with NaOH to form methanol is

(CH3​)4​N+I−

CH3​OCH3​

(CH3​)3​S+I−

(CH3​)3​CCl

Compound (CH3​)4​N+I− is most reactive due to (i) better leaving group, I− and (ii) due to the fact that the methyl group, with positive N, is not more electron deficient Hence this group is more reactive towards nucleophile, OH−

$(C H_{3})_{4} N^{+} I^{-}$

$C H_{3} OC H_{3}$

$(C H_{3})_{3} S^{+} I^{-}$

$(C H_{3})_{3} CCl$

Compound $(C H_{3})_{4} N^{+} I^{-}$ is most reactive due to (i) better leaving group, $I^{-}$ and (ii) due to the fact that the methyl group, with positive N, is not more electron deficient
Hence this group is more reactive towards nucleophile, $O H^{-}$