# How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps?

The question asks to carry out the following conversion in not more than 2 steps:

Benzoic acid to m-nitrobenzyl alcohol

I first used nitration (conc. $\ce{HNO3}$ /conc.$\ce{H2SO4}$) and then I'm got stuck, as reducing the $\ce{-COOH}$ group would also reduce $\ce{-NO2}$ group (using $\ce{LiAlH4}$).

I tried to reduce $\ce{-COOH}$ group using $\ce{LiAlH4}$ to $\ce{-CH2OH}$ and then proceed to nitration but got confused whether $\ce{-CH2OH}$ is o/p or m-directing. I think it should be o/p directing.

• Then $\ce{-COOH}$ is meta directing. Does that solve your question? (think about a reducing agent with selective affinities) – Satwik Pasani Feb 21 '14 at 5:07
• @SatwikPasani I can't think of such a reagent which can reduce acid but not nitro... – evil999man Feb 21 '14 at 5:44

1. Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first.
2. Reduce m-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of $\ce{LiAlH4}$ needed, don't forget that you are reacting an acid with a hydride.

Edit 1: Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.

Edit 2: Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't m-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?

Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) Friedel-Crafts alkylations or etherification. I'd dare to say that the result is a mess.

• Wikipedia says that it will reduce nitro group. See reduction to amines. en.wikipedia.org/wiki/Reduction_of_nitro_compounds – evil999man Feb 21 '14 at 6:50
• @Awesome Irrelevant in your case. Check out the link given in that article. – Klaus-Dieter Warzecha Feb 21 '14 at 7:42
• Oh right...Thanks. But please do tell whether $\ce{-CH2OH}$ is o/p directing or m-directing. – evil999man Feb 21 '14 at 7:51
• Isn't the hydroxymethyl substituent like a methyl group with a hydroxy group dangling from its far end? In electrophilic aromatic substitution, it's not m-directing. But would it survive treatment with sulfuric acid/nitric acid? – Klaus-Dieter Warzecha Feb 21 '14 at 7:58
• According to wikipedia $\ce{LiAlH4}$ and similar metal hydrides will reduce nitroarenes to azo compounds. en.wikipedia.org/wiki/… – bon Jan 7 '15 at 18:56

http://pubs.acs.org/doi/abs/10.1021/jo00956a011

Required is a reduction that acts upon the carboxyl as such. Consider borane-THF, or its equivalent sodium borohydride and BF3·Et2O. Next best is borane-dimethyl sulfide, given the odor. Aliphatic nitro groups are reduced.

Use nitration by mixed acid and then $\ce{SOCl2}$ to replace the $\ce{OH}$ of carboxylic group. Then use $\ce{NaBH4}$ to reduce acid chloride to alcohol.

• this is a three stage process not a two stage – bon Jan 7 '15 at 18:51
• bon is right. 2 steps, not three. Please study the question well before answering. That way the quality of the Q and A will be maintained. – M.A.R. Jan 7 '15 at 19:34

You were initially on the right track: After nitration with conc. HNO3 and conc. H2SO4, perform reduction with Diborane (B2H6), under mild conditions as it does not easily reduce functional groups like nitro, halo, ester, etc. Voila, you have m-Nitrobenzyl alcohol.

• If a reducing agent does not attack an ester, I’m having a hard time imagining it attacking an (even more stabilised) acid. – Jan Sep 13 '16 at 20:45