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In this question, I am confused whether the Grignard reagent will attack the epoxide or the episulfide (thiirane) first. The solution says ring opening of the episulfide will occur due to greater strain.

enter image description here

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    $\begingroup$ I'm not convinced by the strain argument, but you could think about leaving group ability and which bond is easier to break... $\endgroup$
    – orthocresol
    Apr 13 '20 at 7:22
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    $\begingroup$ This is kind of stupid set of answers. I mean any idiot can point to right answer by elimination. $\endgroup$ Apr 13 '20 at 9:10
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There are quite a few reasons why the episulfide might open before the epoxide:

  • Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the $\ce{C-S-C}$ bond angle as compared to the $\ce{C-O-C}$ angle, and thereby increasing the angle strain in the episulfide relative to the epoxide
  • Sulphur will also tend to be a better leaving group, as basicity decreases down the group(in this case, group 16)
  • The carbon-sulphur linkage is weaker than the carbon-oxygen one

So, the mechanism should look like this:

enter image description here

Some points to note:

  • Sulphur compounds also tend to be pretty good nucleophiles due to their softness as per HSAB as elaborated here. Therefore, the $\ce{RS-}$ formed after ring opening will again attack on the epoxide to open the latter.
  • This will also lead to the formation of an ethereal linkage in the final product, which will be quite favorable due to the high stability of $\ce{C-O}$ ether bonds
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I'm not sure I buy the strain argument either. Is there enough difference between the reactivities that you would get selectivity? In the real world I think you would get a mixture, possible complex enough to be called a mess. However according to this reference here

Grignard reagents attack the least hindered carbon atom of thiiranes.

So the sequence is as follows: RMgX attacks the least hindered carbon of the thiirane in preference to the least hindered carbon of the epoxide as the thiirane is more reactive due to the ring strain. The resulting sulfide anion then attacks the epoxide at the alpha position giving a primary alkoxide and a thiirane, this step is reversible but as RS- is the better nucleophile the thiirane predominates.

When the MeI is added it methylates the primary alkoxide giving product c

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    $\begingroup$ I feel like I might have seen this kind of sulfur Payne rearrangement before, with a different (softer) nucleophile, maybe an alkoxide (I can't remember)... but am a bit busy now to look it up $\endgroup$
    – orthocresol
    Apr 13 '20 at 8:46

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