I am tasked to synthesise para-chloroanisole from methanol and benzene. Other reagents are permitted.

Would it be correct to have the benzene attack the methanol $\ce{H}$ and then attack the carbocation that forms with the methoxy ion? From there use $\ce{Cl2}$ and $\ce{AlCl3}$ which would install para and ortho or a combination?

Also if you write one equivalence, does that suffice to say that you are only having it attach once, so that you would only have a max of two groups at the end for each possible product e.g. addition at the 1,2 positions and 1,4 positions?

  • $\begingroup$ You should, in general, ask only one question per post. $\endgroup$
    – Jan
    Commented Feb 9, 2017 at 13:01

1 Answer 1


Benzene is not basic enough to be protonated by methanol, which is precisely the first step you are proposing.

The use of free methanol to generate an aromatic methyl ether immediately inspired me to take a look at the Buchwald-Hartwig coupling. While the website only mentions the applicability to phenol couplings, the bible of organic chemistry, Strategic Applications of Named Reactions in Organic Synthesis also mentions the possibility of using alcoholates as a coupling partner.[1]

This leaves us the question of how to get to a compound suitable to be subjected to a Buchwald-Hartwig coupling. We would need an aryl halide into which the palladium catalyst may insert. Fortunately, we need halogenation anyway, so we might just make it a two-step procedure. Chlorination of benzene is well-known and well-taught, so I decided to check whether bromination of chlorobenzene is also known. Turns out that Ferguson et al. published $89~\%$ yield of the bromination of chlorobenzene using bromine and $\ce{AlBr3}$ in $\ce{CS2}$ — fully para-selectively.[2] This gives us the following sequence:

$$\ce{C6H6 ->[Cl2/AlCl3] Cl-C6H5 ->[Br2/AlBr3][CS2, $89~\%$] Cl-C6H4-Br ->[Pd(OAc)2, NaOMe][base] Cl-C6H4-OMe}$$

Wherein both the latter two are para-substituted.

The required sodium methanolate is easily synthesised from methanol using sodium hydride:

$$\ce{MeOH + NaH -> NaOMe + H2 ^}$$


[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 70.

[2]: L. N. Ferguson, A. Y. Garner, J. L. Mack, J. Am. Chem. Soc. 1954, 76, 1250–1251. DOI: 10.1021/ja01634a013.


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