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I need to propose a route for the synthesis of 2-(bromomethyl)benzonitrile from 1-bromo-2-(bromomethyl)benzene. In the attached figure, the structure of the starting compound and the species to be obtained are indicated.

enter image description here

I thought it could be done by treatment with NaCN in acetone. But, under those conditions, my teacher has told me that you replace the bromine attached to the methyl, since aryl halides hardly undergo substitution reactions.

So I really don't know very well how it could be done. I hope you can help me.

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  • $\begingroup$ This question is likely to be closed because you have not made any effort to solve the problem. Therefore, I will not offer a solution but make you work for it. Instead of worrying about how to make the less reactive site react selectively, ask yourself what if both of them react with cyanide. This product can lead to a "vinylogous malonic acid". What might homophthalic acid do on thermolysis? So how do you convert this aromatic carboxylic acid to the product? Let's see what you come up with. @Waylander $\endgroup$
    – user55119
    Commented Oct 30, 2021 at 18:42
  • $\begingroup$ @user55119 If we select conditions to give the bis-nitrile then a straightforward acid hydrolysis will give homophthalic acid. I cannot find a reference to it, but I presume from your comment that it decarboxylates to give o-toluic acid. Formation of the primary amide followed by dehydration with POCl3 or SOCl2 gives the 2-cyanotoluene. A straightforward benzylic bromination (NBS/AIBN) gives the required. Steppy but reliable. $\endgroup$
    – Waylander
    Commented Oct 31, 2021 at 9:26
  • $\begingroup$ I'm afraid it has to be steppy. I found a photochemical decarboxylation with low quantum yield isolated in 7% yield [DOI:10.1248/cpb.40.2188]. The problem with thermolysis may be anhydride formation. Perhaps prolonged heating in acid. The bromination of o-toluic acid is known. A simpler route:? Oxidize homophthalic and form phthalic anhydride. NaBH4 reduction to the lactone and proceed from there. In reality, who would do this? $\endgroup$
    – user55119
    Commented Oct 31, 2021 at 15:32
  • $\begingroup$ Related: en.wikipedia.org/wiki/Letts_nitrile_synthesis $\endgroup$ Commented Oct 31, 2021 at 16:49
  • $\begingroup$ See the Addendum to my answer subsequent to your accept. $\endgroup$
    – user55119
    Commented Dec 21, 2021 at 21:12

2 Answers 2

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As a complement to @Waylander's solution to this problem and after some exchange of Comments on other possible solutions, other methods for arylcyanation provide an alternative methodology. The benzylic bromide in dibromide 1 is problematic because it is the more reactive of the two bromides. Accordingly, dibromide 1 is temporarily converted into the methyl ether 2. Traditional formation of internally solvated Grignard reagent 3 followed by implementation of one of two possible transnitrilation reagents.

2-Pyridyl cyanate 4 has been shown [1] to transfer cyano groups upon reaction with aryl Grignard reagents. Alternatively, a number of dialkylated malononitriles have been utilized as cyano group transfer reagents with Grignard and organolithium reagents. Typically, dimethyl malononitrile 5 [2] functions admirably in this regard. The success of each of these reagents depends upon the lose of of a stabilized leaving group --- the anion of 2-pyridone in the former case and the anion of isobutyronitrile in the latter. With arylnitrile 6 in hand, reincorporation of the benzyl bromide moiety to form target nitrile 7 is accomplished through the agency of hydrogen bromide with the concurrent lose of methyl bromide and water.



Addendum (21 Dec 2021; user55119): The transformation 6 $\rightarrow$ 7 needs clarification. If $\ce{CH3Br}$ is formed by protonation of the ether followed by cleavage to form $\ce{CH3Br}$ then the resulting benzylic alcohol could add to the nitrile faster than the alcohol is converted into the desired bromide. If cyclization occurs, a pathway to 7 may be lost. Ideally, protonated 6 must preferentially cleave at the benzylic site to form 7 and methanol, which is then converted to $\ce{CH3Br}$.@Waylander's suggestion of $\ce{BBr3}$ is a possible, viable alternative.

  1. J. S. Koo, J. I. Lee, Synthetic Communications, 1996, 26, 3709.

  2. J. T. Reeves, C. A. Malapit, F. G. Buono, K. P. Sidhu, M. A. Marsini, C. A. Sader, K. R. Fandrick, C. A. Bussacca, C. H. Senanayake, J. Am. Chem. Soc., 2015, 137, 9481.

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  • $\begingroup$ is the nitrile group going to survive the HBr treatment? $\endgroup$
    – Waylander
    Commented Nov 1, 2021 at 13:22
  • $\begingroup$ @Waylander--I didn't mean to imply 48% HBr. I had HBr in inert solvent -- benzene, perhaps. If Me is cleaved first rather than the benzylic site, then cyclization could be a problem. But the intermediate iminolactone could cleave to the product. $\endgroup$
    – user55119
    Commented Nov 1, 2021 at 14:59
  • $\begingroup$ @Waylander: The Grignard reagent with orthocresol in place of methoxide is known. This should remove the issue of any cyclization via a benzylic alcohol. $\endgroup$
    – user55119
    Commented Nov 1, 2021 at 16:19
  • $\begingroup$ BBr3 may be a better option for the demethylation than HBr as it does not react with nitriles - there are examples of its use in MeCN $\endgroup$
    – Waylander
    Commented Nov 1, 2021 at 18:08
  • $\begingroup$ @Waylander--Certainly BBr3 is an option. Its use in MeCN may not be relevant if nothing comes of complexation. In nitrile 6, N-complexation may lead to oxygen addition to the nitrile with Me and/or benzylic site reacting with bromide. It could be complicated as you mentioned earlier. Nothing ventured; nothing gained. $\endgroup$
    – user55119
    Commented Nov 1, 2021 at 23:08
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This is a non-trivial problem. As your teacher has pointed out the bromomethyl group is very reactive to nucleophiles, so the issue is to find conditions for the reaction of the aryl nitrile that do not affect it.

The standard way of transforming an aryl bromide into an aryl nitrile is Pd-catalysed cyanation with a source of CN such as Zn(CN)2. This Organic Letters paper offers a set of mild conditions for the transformation, but I cannot say for sure if the bromomethyl group would survive. If it does not then you are faced with transforming it into another group that would survive and can be taken back to the benzyl bromide afterwards; -OAc might be a good choice if this is required.

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