I want to convert terephthalic acid (TPA) to an alcohol (however I think the correct term would be "a phenol").

Terephthalic acid

I've read that lithium tetrahydridoaluminate ($\ce{LiAlH4}$) can produce an alcohol from a carboxylic acid. And, according to Wikipedia, dicarboxylic acids (such as TPA) react around the same. So does $\ce{LiAlH4}$ reduce TPA to a phenol (or an other product)?


I want to make sure that you understand the different between two commonly confused molecules (their names don't help).

Phenol is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.


Benzyl alcohol is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.

benzyl alcohol

Conversion of benzoic acid to phenol requires loss of a carbon atom.

$$\ce{Ph{\bf C}O2H -> PhOH + {\bf CO}}$$

Lithium aluminum hydride converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, stop here. If you want the phenol, read on.

$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{->}}}\ce {PhCH2OH}$$

To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a Swern oxidation. Then benzaldehyde can be converted to phenyl formate using a Baeyer-Villiger Oxidation. The formate could be hydrolyzed in aqueous base.

$$\ce{PhCH2OH \overset{Swern}{->} PhCHO}$$ $$\ce{PhCHO \overset{Baeyer-Villiger}{->} PhOCHO}$$ $$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \ H2O}}{\underset{2. \ \ce{H3O+}}{\ce{->}}} \ce{PhOH}$$

In practice, it would be easier to buy the phenol. In your case, you want hydroquinone.


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    $\begingroup$ Thanks for the very good answer - but is it also possible to remove two methanol molecules from the 1,4benzenedimethanol molecule to also yield benzene? $\endgroup$ – user2117 Jan 19 '14 at 15:56

As nice and thoroughly described by Ben Norris, 1,4-(bishydroxymemethyl)benzene is available via reduction of the starting material with $\ce{LiAlH4}$. In order to minimize the amount of alanate used, one might consider to first convert the terephthalic acid to a corresponding ester.

If you're heading for the phenol, which doesn't make a lot of sense since hydroquinone is cheap as dirt, the following "strategy" might work:

  1. brominate terephthalic acid to yield 2,5-dibromo terephthalic acid
  2. heat the dibromo acid in an aqueous solution of sodium hydroxide, cupric bromide and 1,2-diaminoethane. You'd end up with 2,5-dihydroxy terephthalic acid, probably through a copper-catalyzed nucleophilic substitution
  3. decarboxylate the 2,5-dihydroxy terephthalic acid (or its disodium salt) by heating with calcium oxide.

You'll get hydroquinone, and unless your arene was labeled, you won't notice that you haven't replaced the $\ce{COOH}$ with $\ce{OH}$ but actually changed the positions of the substituents.

But all of this is nothing but a mind game ;)

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