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Explain mustard gas hydrolysis at a very high rate:

$\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$

I tried searching online but did not find this reaction anywhere. Can someone explain me the steps/mechanism of this reaction? Even some hints might help.

My attempt:

According to Wikipedia

Solvolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) or elimination where the nucleophile is a solvent molecule. For certain nucleophiles, there are specific terms for the type of solvolysis reaction. For water, the term is hydrolysis; for alcohols, it is alcoholysis; for ammonia, it is ammonolysis; for glycols, it is glycolysis; for amines, it is aminolysis.

It clearly states that hydrolysis/solvolysis can only be $\mathrm{S_N1}$ or elimination. However I'm wondering how can $\mathrm{S_N1}$ occur on a primary carbon! Morever water doesnt seem to be a strong enough base for elimination. So what will be the preferred reaction mechanism?

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    $\begingroup$ The cation is somehow stabilized via S. Without S, it would be just ethyl chloride which hydrolyzes pretty slowly, as expected. $\endgroup$
    – Ivan Neretin
    Commented Sep 21, 2016 at 20:39
  • $\begingroup$ You mean SN1 occurs?What will be the end product?Can you link the mechanism? $\endgroup$
    – user14857
    Commented Sep 21, 2016 at 20:41

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Luckily I got the mechanism via Google Image Search.

Mechanism of hydrolysis

Internal attack by sulphur atom takes place during the reaction. This is called anchimeric assistance, neighbouring group effect or $\mathrm{S_N{NGP}}$. Wikipedia describes it well.

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  • $\begingroup$ Yes, an $S_N1$ reaction takes place. As, @Ivan pointed out, if it weren't for the it would just be like ethyl chloride. The key to this whole process is the presence of lone pair of electrons on sulfur which it uses to stabilise the primary carbocation (by forming the triangular cation depicted). $\endgroup$
    – getafix
    Commented Sep 22, 2016 at 1:56
  • $\begingroup$ @getafix I would rather say it an extended version of $\ce{S_N1}$ i.e. $\ce{SN_{NGP}}$.It's not strictly $\ce{S_N1}$. $\endgroup$
    – user14857
    Commented Sep 22, 2016 at 5:11
  • $\begingroup$ Of course I agree. There is clear neighbouring group participation. Sorry, my comment did not make it explicit. However, I never knew $SN_{NGP}$ was a term..learn something new everyday $\endgroup$
    – getafix
    Commented Sep 22, 2016 at 5:17
  • $\begingroup$ @getafix Never mind.We're all here to learn :-)! $\endgroup$
    – user14857
    Commented Sep 22, 2016 at 5:20
  • $\begingroup$ Indeed. What I meant is that the rate determine step is unimolecular, i.e $S_N1$ $\endgroup$
    – getafix
    Commented Sep 22, 2016 at 5:26

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