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: 1-chloro-2-(trifluoromethyl)benzene has \( Cl \) on C1 and \( CF_3 \) on C2. Sodamide in liquid ammonia is a very strong, small base that reacts with aryl halides through the benzyne (elimination-addition) mechanism.

Step 2: The first step is E2-type removal of \( HCl \): the base removes a proton from a carbon ortho to the chlorine. The two ortho positions to C1 are C2 and C6. C2 already carries \( CF_3 \) and has no removable hydrogen there, so the proton must come from C6.

Step 3: Loss of \( H \) from C6 and \( Cl^- \) from C1 generates a benzyne (aryne) with a strained triple bond character across the C1 to C6 bond.

Step 4: The amide ion \( NH_2^- \) can add to either end of this benzyne triple bond, C1 or C6, generating a carbanion at the carbon not attacked. \( CF_3 \) is strongly electron withdrawing by induction and stabilizes a negative charge best when the charge sits on the nearest (ortho) carbon, which is C1.

Step 5: Therefore \( NH_2^- \) preferentially attacks C6, leaving the more stable carbanion at C1, adjacent to \( CF_3 \), and that carbanion is then protonated by ammonia. The final product carries the amino group at what was C6, which sits meta to the \( CF_3 \) group at C2.

\[ \boxed{\text{3-(trifluoromethyl)aniline}} \]