Wittig reaction with an alkyl dibromide

Recently, I have come across a problem, in which a ketone needs to be converted into an alkene. However, in the alkene product, there is also another bromide functional group, at the end of the alkyl chain. I was thinking of using a dibromide (an alkyl chain with bromine atoms at both ends) and performing the Wittig reaction. However, I foresee that both bromide functional groups would participate in the reaction. Would one solution to this be to use a protecting group? Or would the problem be solved if I manipulate the stoichiometry of the reactants (e.g. ensure that the dibromide is in excess etc.)?

• Use a bromo alcohol with the alcohol protected as THP ether. After the Witting, deprotect and form the required bromide. – Waylander Nov 9 '18 at 7:34
• @Waylander Would manipulation of the stoichiometry of the reactants work as well? Just wondering. – Tan Yong Boon Nov 9 '18 at 7:38
• There is a danger that when you deprotonate the Phosphonium bromide to do the Wittig that the anion will react with the remote bromide, either to cyclise or intermolecularly. – Waylander Nov 9 '18 at 8:04
• A structure would have been helpful. You might not need a protecting group. Possible that the alkoxy ylid is fine. – user55119 May 2 at 12:32

1 Answer

It seems that the method to circumvent this problem is indeed to use a protecting group to protect one end of the alkyl chain, which is the end to be converted to a bromide after the Wittig reaction.

As Waylander aptly mentioned, another problem which comes with using an alkyl dibromide is that there may be an intramolecular substitution reaction, in which the anionic carbon in the phosphonium ylide reacts with the other end of the alkyl dibromide.

To circumvent this problem, we could use an alkyl bromide with an alcohol functional group at the other end of the chain. This alcohol group can then be protected using various protecting groups, such as a tetrahydropyran ether or trimethylsilyl ether. After the Wittig reaction, the compound can then be deprotected using acidic hydrolysis. After which, the alcohol functional group can then be converted to a bromide using $$\ce {PBr3}$$. The desired compound is thus obtained.