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What will be the final major product?

2-(thiiran-2-yl)oxirane + RMgX to 1; 1 + CH3I to 2

The answer is 2-(methoxymethyl)-3-(substituted methyl)thiirane:

2-(methoxymethyl)-3-(substituted methyl)thiirane

I think $\ce{R}$ of $\ce{RMgX}$ attacks thiirane ring because negative charge on sulfur is more stable than that of oxygen.

But why doesn't the negative sulfur subsequently attack on less hindered carbon of the ring? Instead, it attacks the β-carbon which is more hindered and opens the ring so that methyl of $\ce{CH3I}$ can make bond with negative oxygen atom. Why does the reaction proceed that way?

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2 Answers 2

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Sulfur attacking the less-hindered end of the epoxide would generate a [4] ring. This is more strained than a [3] ring and thus energetically less favourable. The attack is reversible, so the more energetically favourable (thermodynamic) product will be the outcome which is trapped by methylation in the final step.

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    $\begingroup$ Actually cyclopropane is a bit more strained than cyclobutane. There is an entropic effect favoring three- over 4-ring formation. See here. $\endgroup$
    – ron
    Jul 30 at 19:18
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One other aspect is after the initial opening of the sulfur-bearing ring, the sulfur ends up closer to the bridgehead carbon than to any other part of the oxirane ring. (You can see this in a molecular model of the intermediate anion, in which the other two atoms are tied back from the sulfur by the small ring size.) So electronic overlap and thus bond rearrangement occurs more readily through the bridegehead.

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