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

I encountered this question while solving an objective paper.

$\ce{BuLi}$ 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?

• – Klaus-Dieter Warzecha Mar 31 '17 at 4:22
• Given the choices, C is the answer. As A --> B is written, the halide is added to the dithiane anion. In the real world this mode of addition, if added slowly, could eventually give dialkylation with the eventual formation of glutaraldehyde. To optimize the formation of cyclobutanone, the dithiane anion should be added slowly to the halide. – user55119 Jul 25 at 1:25

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.