Organic Question TPR Test A...HELLLPPPP!

[MDtoBe777]

Organic Q:
190: You are given the structure of 1,2-epoxycyclohexane (structure x) which is treated with HBr to form compound Y. Compound Y is then treated with NaOH (aq) to give compound X.

Okay, so after treating it with HBr you form an -OH group and a -Br group hanging off of adjacent carbons on the cyclohexane. Then you treat it with NaOH and you go back to your starting product? That is the answer but I do not see why. Why doesnt the Br get displaced and then either elimination can occur or substitution?? I'm confused. Help?

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[blebl34]

yeh i missed that one as well, dont really understand it

 

[dandelion]

What happens is H-Br (nucleophile) attacks the epoxide, forming an alkoxide with Br attached to the adjacent C. Then H+ protonates the alkoxide, and you have the product that you said. When treated with NaOH, the -OH group becomes deprotonated, and the alkoxide forms. The O of the alkoxide displaces the adjacent Br, and you have what you started with.

Carbocations don't usually form in basic solution, I think, so Sn1 and E1 are out. OH- is not a sterically hindered base, so Sn2 is favored over E2. It's possible that OH- from the solution displaces the Br, but it's much more likely that the neighboring alkoxide does this because it's closer.

This is what I think, knowing what the answer is . . . . .

 

[BisMuOxo]

What happens is H-Br (nucleophile) attacks the epoxide, forming an alkoxide with Br attached to the adjacent C. Then H+ protonates the alkoxide, and you have the product that you said. When treated with NaOH, the -OH group becomes deprotonated, and the alkoxide forms. The O of the alkoxide displaces the adjacent Br, and you have what you started with.

Carbocations don't usually form in basic solution, I think, so Sn1 and E1 are out. OH- is not a sterically hindered base, so Sn2 is favored over E2. It's possible that OH- from the solution displaces the Br, but it's much more likely that the neighboring alkoxide does this because it's closer.

This is what I think, knowing what the answer is . . . . .

Your explanation is good except for the first part. The alkoxide does not exist in an acidic solution. The H+ first protonates the epoxide and then Br- attacks to form the bromohydrin.

Also, Sn2 is slowed for hydroxide because of the fact that you have a secondary bromide, but the hindrance is not as important with the alkoxide which is placed perfectly to attack the neighboring carbon.