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Topic: C12H15Br ---> C13H18O  (Read 1718 times)

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Offline jlanrdy287

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C12H15Br ---> C13H18O
« on: November 29, 2013, 05:30:25 PM »



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I am unclear what it means going from +22.1o to 0o. Does that mean it is achiral? This is the mechanism I have come up with..

« Last Edit: November 29, 2013, 07:20:29 PM by jlanrdy287 »

Offline orgopete

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Re: C12H15Br ---> C13H18O
« Reply #1 on: November 29, 2013, 08:46:39 PM »
If there are no errors in this problem, then I am not pleased with how it has been formulated. Because the structure has two chirality centers, loss of one center (the first intermediate) could give a complete loss of the chirality at the bromide center. That would give a new diastereomeric mixture, but the chirality at the methyl group should remain unchanged. If a single methyl isomer, the product should be a diastereomeric mixture, but because it retained one chirality center, it would be be a surprise for the product to be optically inactive.

The product could be optically inactive if the original mixture were an exactly 50:50 mixture of diastereomers. Then loss of the bromide chirality could give an optically inactive mixture of diastereomers.

However, the drawing is also incorrect as the planar carbocation should be attacked from both sides. This has not been illustrated.
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