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We start with compound M:
[Please imagine green bonds as dotted bonds going into the plane]
From compound M, we are told that the 3 membered ring is opened u using sodium azide yielding a mix of isomers N and O.
I drew the following:
and
[Again, please imagine the green bonds as dotted bonds and the blue bonds as wedge bonds]
Given are the following two molecules:
Are my solutions the same as the ones given? If they aren't, where did I go wrong?
Many thanks!
First of all, for both products, you have not given the O atom in the carbonyl position. In organic chemistry, we assume that there is a C atom if no atom is mentioned specifically. Now for this reaction, the carbonyl group will be there as it is.
Now we can consider the reaction. The epoxide is away from the plane, i.e. below the plane. For the steric reason, it is more feasible for azide to attack from the upper side of the plane. Now again, azide can attack both C centre of the 3-membered epoxide ring. In that case, we'll be having two different products. But in both cases, we will have the azide group on the above of the plane and O atom in the below of the page.
Now if I assume that you have mistakenly skipped the O atom in your both products, then your first molecule matches with the first answer given below. But the second answer given by you is not correct and doesn't match with the second answer. You can have the green and blue coloured bond in the same place and then you have to alternate the position of OH group and azide group. Then that will match the answer given below and the answer will be correct.
