# Action of H2S and OH- on an alkyl halide?

What would be the action of $$\ce{OH-}$$ and $$\ce{H2S}$$ on 2,6 dichloroheptane? I came across this question in my organic chemistry workbook but I have simply never seen a question of this type before. I think a cyclic compound may be formed, but I am not sure which.

I thought really hard with help from the comments under my question, I came up with this solution. However in my book it says that there should be only one sulphur atom in the cyclic compound. How does this make sense as there are two halogens in the original haloalkane? Is my mechanism incorrect?

• It rather depends on the reaction conditions, but the formation of 5 and 6-member rings is a favourable process and likely to occur, particularly with sulfide as a nucleophile – Waylander Feb 26 '19 at 19:04
• I assumed you mean $\ce{OH-}$ (like alkaline $\ce{H2S}$?) – Mathew Mahindaratne Feb 27 '19 at 3:14
• @Waylander thank you! But why would there be a sulphide? – viktor nikiforov Feb 27 '19 at 11:21
• @Mathew Mahindaratne Yes. – viktor nikiforov Feb 27 '19 at 11:22
• H2S plus OH- gives HS- which will react readily with the haloalkane to give RSH, this will be further deprotonated by OH- to give RS- which will react internally to give the cyclic species. – Waylander Feb 27 '19 at 12:03