Question

When o-bromoanisole is treated with NaNH2 in NH3 (I), a reaction proceeding via a benzyne intermediate, what two isomeric products are possible? (Check two products). o-methoxyaniline m-methoxyaniline p-methoxyaniline In practice, only one isomer is formed. Which one is it? o-methoxyaniline m-methoxyaniline p-methoxyaniline Why is this product favored? (Select the major factor). The methoxy group is too bulky and causes a steric hindrance to amide ion attack at the ortho- position. The reaction actually proceeds by nucleophilic aromatic substitution (addition-elimination) mechanism. The inductively electron-withdrawing methoxy group better stabilizes the anion in the ortho- position than in the meta- position. The incoming amide anion nucleophile is repelled by the electrons of the methoxy group.

Answer 1

o-bromoanisole reacts with sodamide in presence of liqiud ammonia to form two isomeric products Thev are m-methoxy aniline and o-methoxvaniline OCH3 OCH3 OCH3 NH2 Br NaNH2 NH3 (1) NH m-methoxyaniline o-methoxyaniline o-bromoanisole (Major) (Minor) Thus, m-methoxy aniline is the major isomer mechanism: This is an addition-elimination mechanism, occured via benzyne intermediate o-bromoanisole reacts with NaNH2 to form benzyne intermediate OCH3 OCH o-bromoanisole The nucleophile (NH2 attacks the benzyne intermediate at either C2 carbon or C3 carbon atom to give the two carbanions OCH3 OCH3 OCH Attacks at C3 NH NH2 stable carbanion OCH3 OCH OCH3 NH NH Attacks at C2 unstable carbanion Therefore, the major product is m-methoxy aniline. Because methoxy group is an electron withdrawing group (-I group, inductive effect), which stabilizes carbanion formed at ortho position Thus, option 3 is correct.

Answer 2

When o-bromoanisole is treated with NaNH2 in NH3 (1), a reaction proceeding via a benzyne intermediat what two isomeric products are possible? (Check two products). o-methoxyaniline m-methoxyaniline p-methoxyaniline In practice, only one isomer is formed. Which one is it? O o-methoxyaniline m-methoxyaniline O p-methoxyaniline Why is this product favored? (Select the major factor) The reaction actually proceeds by nucleophilic aromatic substitution (addition-elimination) mechanism. The methoxy group is too bulky and causes a steric hindrance to amide ion attack at the ortho- position. The incoming amide anion nucleophile is repelled by the electrons Oof the methoxy group. The inductively electron-withdrawing methoxy group better stabilizes the anion in the ortho-position than in the meta- position.

Answer 3

1st: ortho and meta

2nd: meta

3rd: third answer choice