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Technically there is inversion around that carbon. If you check the rotation, the original compound is R and the correct answer in green is S which means that there is inversion (which is true for acid catalyzed epoxide ring opening). However if you see your answer and assign chirality you'll find that it is R like the original compound and thus did not invert. So the answer in green would be correct...super sneaky problem!

You might wanna double check with Ari about this though 😛
 
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It is inverted. If you look at it closely, your leaving group is the oxygen of the epoxide and your nucleophile is your ethanol, which has attacked from the other side. The groups have switched sides so the chirality changes as well.
 
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helloworld9 said:
Technically there is inversion around that carbon. If you check the rotation, the original compound is R and the correct answer in green is S which means that there is inversion (which is true for acid catalyzed epoxide ring opening). However if you see your answer and assign chirality you'll find that it is R like the original compound and thus did not invert. So the answer in green would be correct...super sneaky problem!

You might wanna double check with Ari about this though 😛
Click to expand...

👍 Always check the configurations, never assume how it's "supposed" to look.
 
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