The product of a reaction series involving 1,3-dithiane, 1-chloro-3-iodopropane and butyllithium

Chemistry Asked on January 6, 2022

I encountered this question while solving an objective paper.

1,3-dithiane + butyl lithium, then + 3-chloro-1-iodopropane, then + butyl lithium, then + mercury(II) chloride, methanol and water

$ce{BuLi}$ is a strong base and it can be used to abstract a proton from the substrate. Low temperature favours substitution more than elimination. So my idea was that the ring will break up $ce{Li}$ will be substituted in place of two $ce{S}$ atoms. The next product should involve formation of a six membered saturated ring but I am unable to proceed further. What is going to happen?

One Answer

Some right ideas, but unfortunately, not correct. The correct answer is (C).

The protons in between the two sulfurs are more acidic because of hyperconjugative stabilization into the sulfur. Addition of 1-iodo-3-chloropropane results in the substitution product as you guessed. Another equivalent of butyllithium does the same thing again, except, this time, we do an intramolecular substitution for ring closure. Mercury has high affinity for sulfur, so that deprotects what is analogous to a ketal to give the ketone.


The key for these kinds of problems is to approach it from both ends. For example, the products are all carbonyl groups. Where would those groups come from? They can only be from the last step because otherwise butyllithium would add into a carbonyl. This gives you some idea for what 'C' would look, and so on.

For the forward direction, it's really important that you not kill all the reactivity by substituting off the heteroatoms. If you do that, you'll end up with an alkane, and there's no good chemistry unless you want $ce{C-H}$ activation magic. Since butyllithium is not adding into anything, consider acid-base chemistry because it's also a very powerful base.

Answered by Zhe on January 6, 2022

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