Which is the best method for carrying out the following reaction?


I know that nitro group is a meta directive group in benzene nucleus and so is the carboxyl group. Option (b) cannot be the right answer since it first does the Friedel Crafts alkylation and then it does the nitration which would bring the Nitro group on either ortho or para position which is not desired.

However by going through options (c) and (d) I am getting the same product which is asked in the problem.

If we go via option (c) we will have:

  1. Alkylation resulting in methylbenzene
  2. Oxidation which will produce benzoic acid
  3. Nitration which will give m-nitrobenzoic acid

And if we go by option (d) we will have:

  1. Nitration - giving us nitrobenzene
  2. Alkylation - this will introduce methyl group at meta position
  3. Oxidation of the methyl group will give us again m-nitrobenzoic acid

According to my book the correct option is given as only (d). And that's where I'm confused. I'm not able to decide which of the following will be the best method.

Source:- "Elementary problems in Organic Chemistry" (S. Balaji publication) Chapter - Aldehydes, Ketones and Carboxylic Acids (page 401) (amazon link)

  • 2
    $\begingroup$ Remark: This paper suggests that option (c) can be used. Though it does not comment if its the best method. $\endgroup$ Feb 28, 2018 at 14:54
  • 1
    $\begingroup$ The paper also says “In practice, this nitration reaction can result in the production of quite a bit of the ortho product as well, unless the temperature is kept very cold throughout the reaction“. $\endgroup$
    – MollyCooL
    Feb 28, 2018 at 14:59
  • $\begingroup$ @MollyCooL Oh, good catch, I skipped that! Though I don't see any reason why ortho products would also be obtained.. :/ $\endgroup$ Feb 28, 2018 at 15:01
  • $\begingroup$ True even I don’t get that :( $\endgroup$
    – MollyCooL
    Feb 28, 2018 at 15:03
  • $\begingroup$ The d) method does not work, as nitrobenzenes are usually very unreactive toward Friedel-Crafts Alkylation due to high electron withdrawing effect of -NO2 group. $\endgroup$
    – S R Maiti
    Feb 28, 2018 at 16:06

1 Answer 1


I think the answer should be option C, because in both cases the first group obtained i.e. $\ce{-COOH}$ or $\ce{-NO2}$, is meta directing. But $\ce{-NO2}$ is has greater negative mesomeric effect than $\ce{-COOH}$. Hence if $\ce{-NO2}$ is attached first the reaction rate will be slower than if $\ce{-COOH}$ is added first.

You may check the negative mesomeric effect order here.


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