# Conversion of 4-nitrotoluene to 2-bromobenzoic acid [closed]

In the above conversion why doesn't bromine add to methyl group? Is the first reaction following free radical or polar reaction (Friedel-Crafts). If it's following free radical mechanism then Br should add to CH3, if it's F-C substitution then why there is no catalyst when NO2 is deactivating group?

What would be the change if I first do seconds step (i.e reduction to amine) and then bromination? Would it be wrong to do bromination at last?

Is the intermediate (2-bromo-4-nitrotoluene) the only major product of first step?

## closed as off-topic by bon, Klaus-Dieter Warzecha, Todd Minehardt, ringo, FreddyApr 16 '16 at 5:46

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• Too many questions in one! 1) You are right, ring substitution needs a Lewis acid catalyst. Without the catalyst nothing is going to happen. For free radical substitution of the methyl group you will need either UV light or a radical initiator. 2) and 3) You need to look up directing effects of substituents. 4) Yes. – orthocresol Apr 15 '16 at 12:34
• At what temp or conditions was the bromine reacted in the first step? The methyl group may require a higher temp or specific conditions to be activated I'm only guessing tho , good luck. – Technetium Apr 15 '16 at 12:36
• @Swastik okay no problem – user26831 Apr 15 '16 at 13:37
• I hope my answer will clarify your doubts @physics – Swastik Apr 15 '16 at 13:58

Sorry but I would like to answer your questions from last to first,

(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place

(2) -COOH is deactivating meta directing group (due to presence of >C=O) so friedel-crafts bromination would take place at meta position.

(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2

(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced one of the -H at ortho (to -CH3) of benzene.

(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)

• preference of -CH3(activating) OVER -NO2(deactivating) will take place - I doubt this. $\ce{NO2}$ is a very strongly deactivating group whereas methyl is only weakly activating. – bon Apr 15 '16 at 14:31
• The directing effects of the methyl and nitro groups in p-nitrotoluene reinforce each other so that's not a problem in this specific case, but as bon said, the strongly deactivating nitro group is likely to exert a greater influence on the regioselectivity. – orthocresol Apr 15 '16 at 17:43
• [for your reference}(utdallas.edu/~biewerm/17.pdf) is the site which will answer your question.You can refer "Benzene and its derivative" chapter of any standard organic chemistry books like J.D. March or Paula Yukarinis or P.Bahadur or O.P Tandonor Solomons(name of eminent chemistry book writer) @ bon,orthocresol – Swastik Apr 16 '16 at 16:38