IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

tarinbansal

I think the answer is m-nitrobenzoic acid.

First of all,
@Dream
ur explanation is wrong. NO2 is not electron repelling, it is electron withdrawing group by both -I, -R.

Now to decide acidic character I am using the stability of conjugate base. Means in the case in which the -ve charge on O wud be more delocalised, it wud be more acidic.

Now, COOH is at meta position of NO2, means its -R effect wont work here and only -I wud work. And in case of F, COOH is at para position of F. So we can consider its very feeble +R effect and -I effect of F.
Now in case of inductive effect, we see the distance. More the distance, less the inductive effect.
Now, NO2 is nearer to COOH than F. It means its -I > -I of F. Means it wud help decreasing electron density from O in the Conjugate base more than F.
Secondly if we consider a feeble +R effect of F, it wud increase the electron density on O rather than decreasing it.

SO in either way, whether u consider F's +R effect or not, NO2 is acting as a better electron pulling group than F and hence m-nitrobenzoic acid's conjugate will be more stable as compared to the other one.

So, M-nitro benzoic acid is more acidic.

Hope U get it and plz tell me if I hav made some mistake.

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