What is the best method to prepare meta-nitrobenzoic acid? [closed]








**Closed**. This question is [opinion-based](/help/closed-questions). It is not currently accepting answers.
 
 






---








**Want to improve this question?** Update the question so it can be answered with facts and citations by [editing this post](/posts/91509/edit).




Closed 5 years ago.








[Improve this question](/posts/91509/edit) 



**Question**:



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


[![question](https://i.stack.imgur.com/udy0z.jpg)](https://i.stack.imgur.com/udy0z.jpg)


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](https://www.amazon.in/Elementary-Problems-Organic-Chemistry-Chouhan/dp/9380537247))




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](https://en.wikipedia.org/wiki/Mesomeric_effect).


 1