Question

What is the product of the reaction between 1-chloro-2-(trifluoromethyl)benzene and sodamide (\( NaNH_2 \)) in liquid ammonia?

• A. Methoxybenzene
• B. Aniline
• C. 3-(trifluoromethyl) aniline
• D. 2-(trifluoromethyl) aniline

Hint:

Work out which ortho hydrogen is removed to form the benzyne, then see which carbanion the CF3 group stabilizes better.

Answer 1

The Correct Option is C

Step 1: Set up the ring positions explicitly: C1 bears $Cl$, C2 bears $CF_3$, and C3 through C6 bear $H$. This is a benzyne-forming substrate because $NaNH_2$/liquid $NH_3$ is the classic reagent combination for aryl halide elimination-addition.

Step 2: Elimination needs an ortho hydrogen relative to the leaving $Cl$ at C1. Of the two ortho carbons, C2 is blocked by $CF_3$, so only C6-H can be removed, fixing the triple bond of the benzyne intermediate between C1 and C6.

Step 3: Model the benzyne as offering two electrophilic carbons, C1 and C6, for the incoming $NH_2^-$. Whichever carbon is NOT attacked ends up as a carbanion that must be stabilized before protonation.

Step 4: Rank the two possible carbanions by proximity to the electron withdrawing $CF_3$ at C2: a carbanion at C1 is directly adjacent (ortho) to $CF_3$ and receives strong inductive stabilization, while a carbanion at C6 is two bonds further away (meta to $CF_3$) and is stabilized far less.

Step 5: The pathway that generates the better stabilized carbanion (at C1) is favored kinetically, which means the nucleophile itself lands at C6. Mapping C6 relative to the $CF_3$ at C2 around the ring shows a 1,3 (meta) relationship.

\[ \boxed{\text{NH}_2 \text{ ends up meta to CF}_3 \Rightarrow \text{3-(trifluoromethyl)aniline}} \]