[ma_he_sal]

Hi. Can someone help me with this reaction? I need to propose the mechanism and the tip I have is the mechanism follows a non-classical carbocation route. The asterisk is an isotopic label.

[vnaraya3]

I am assuming that this is incredibly low yield, and that water is added in a LARGE excess (although still I would have expected the question-writer to put "H3O+", not "H2O"). Even considering all this, this still seems like an unlikely process to me (although I am just a student) - but nonetheless I can attempt to justify most of the steps.
I wrote a number below each arrow, so I can explain/justify it below:

1.) I just rewrote it in a planar conformation. There's no stereochemistry because it's meso.

2.) LiAlH4 is a hydride reagent. Hydride attacks the starred carbon, which opens the epoxide (favorable because of ring strain), forming an alkoxide

3.) The alkoxide abstracts a proton from water to make an alcohol.

4.) *For some unknown reason* water protonates the -OH group again, turning it into OH2+, good leaving group. (This would make a lot more sense if it was H3O+, not H2O, in the problem!)

5.) Loss of OH2 produces a secondary carbocation

6.) An favorable alkyl shift occurs, to produce a benzyl carbocation (stabilized by resonance). I pretend there was stereochemistry in this step to make the next part more clear.

7.) I just rewrote it in a different conformation - now the starred carbon is going "into the page"

8.) Another alkyl shift occurs for unknown reasons (it's meso again so I stopped writing stereochemistry)

9.) A hydride shift produces a highly stabilized tertiary benzylic cation (this is among the most stable types of carbocations you can have...perhaps this step is driving the previous steps forward...?)

10.) Another hydride shift occurs *for unknown reasons*, so that now the starred carbon bears the positive charge.

11.) Hydroxide attacks the carbocation.

12.) This is the product! Rewrite it in the conformation given in the problem.

To be honest none of this really makes any sense to me unless it's under acidic conditions, unless there's such an extreme excess of water. So actually probably in the last step, it's water that attacks the carbocation (followed by deprotonation), and in the second step the alkoxide attacks H3O+, not H2O. (I'm too lazy to change it now lol).

Anyways, that's the most plausible mechanism I could think of.

[critzz]

I am assuming that this is incredibly low yield, and that water is added in a LARGE excess (although still I would have expected the question-writer to put "H3O+", not "H2O"). Even considering all this, this still seems like an unlikely process to me (although I am just a student) - but nonetheless I can attempt to justify most of the steps.
I wrote a number below each arrow, so I can explain/justify it below:

1.) I just rewrote it in a planar conformation. There's no stereochemistry because it's meso.

2.) LiAlH4 is a hydride reagent. Hydride attacks the starred carbon, which opens the epoxide (favorable because of ring strain), forming an alkoxide

3.) The alkoxide abstracts a proton from water to make an alcohol.

4.) *For some unknown reason* water protonates the -OH group again, turning it into OH2+, good leaving group. (This would make a lot more sense if it was H3O+, not H2O, in the problem!)

5.) Loss of OH2 produces a secondary carbocation

6.) An favorable alkyl shift occurs, to produce a benzyl carbocation (stabilized by resonance). I pretend there was stereochemistry in this step to make the next part more clear.

7.) I just rewrote it in a different conformation - now the starred carbon is going "into the page"

8.) Another alkyl shift occurs for unknown reasons (it's meso again so I stopped writing stereochemistry)

9.) A hydride shift produces a highly stabilized tertiary benzylic cation (this is among the most stable types of carbocations you can have...perhaps this step is driving the previous steps forward...?)

10.) Another hydride shift occurs *for unknown reasons*, so that now the starred carbon bears the positive charge.

11.) Hydroxide attacks the carbocation.

12.) This is the product! Rewrite it in the conformation given in the problem.

To be honest none of this really makes any sense to me unless it's under acidic conditions, unless there's such an extreme excess of water. So actually probably in the last step, it's water that attacks the carbocation (followed by deprotonation), and in the second step the alkoxide attacks H3O+, not H2O. (I'm too lazy to change it now lol).

Anyways, that's the most plausible mechanism I could think of.

You are following a classical explanation for the cation-rearrangement, but OP stated that the rearrangement would proceed non-classical. That's why it doesn't appear to make any sense. 

Hi. Can someone help me with this reaction? I need to propose the mechanism and the tip I have is the mechanism follows a non-classical carbocation route. The asterisk is an isotopic label.

Hint: search for norbornyl cation or phenonium ion.

[Babcock_Hall]

@OP,

Welcome to the forum.  Please read the Forum rules (red link).  You must show your attempt or your thoughts before we can help you.

[orgopete]

This is how I think the reaction should take place. I am assuming LAH reduction of an epoxide is not a very facile reaction. Some reduction can form AlH3. I'm using this to justify a Lewis acid catalyzed reaction. The Lewis acid will catalyze opening the epoxide to give a norbornyl cation. This can rearrange to give the interverting or non-classical carbocation. This new carbocation can capture a hydride to give the product and perpetuate the formation of a Lewis acid.

In this case a rapid reduction of the epoxide would block the carbocation reaction.

[orthoformate]

This article:
Some 2- and 7-Derivatives of Benznorbornene
PAUL D. BARTLETT, VILLIAM P. GIDDIXGS
Journal of the American Chemical Society Vol. 82, pgs 1240-6

Discusses this exact problem. The proposed mechanism is on pg. 1243. The reported yield is actually pretty good: 86%

The paper shows exactly what Orgopete described