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I am supposed to identify the products A and B in the following scheme:

Epoxide opening with NH3 and HN3

The three options I have been given are:

Possible combinations of products

I think that since $\ce{HN3}$ is a stronger nucleophile than $\ce{NH3}$, B should form via an SN2 mechanism. So, cleavage should take place in such a way that $\ce{OH}$ group remains with the carbon attached to the $\ce{CH3}$ group. Hence I think option 1 is the answer, but I am not sure about it.

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2
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In fact, these are at most chemoselective reactions, so there will probably be a mixture of all possible products.

Ignoring stereochemistry and assuming that we want to know the 'most probable' products, as far as I can tell the main difference is that $\ce{NH3}$ is a rather nucleophilic basic compound, whereas $\ce{HN3}$ is an acid and not very nucleophilic. We are not given the solvent, so we cannot know what side reactions can happen, or what solvation effects we have. So we may assume an 'inert' solvent is used.

Within these hypotheses, I'd say that $\ce{NH3}$ will react by $\mathrm{S_N}$2, attacking mostly on the least substituted side ($\ce{CH2}$), whereas $\ce{HN3}$ will protonate the oxygen first, and the opening by the azide anion will happen by $\mathrm{S_N}$1, i.e. attacking the most stable carbocation (the one with the positive charge on the most substituted carbon. So for me, the 'correct' answer would be 2.

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