Sciencemadness Discussion Board - Tricyclopentyl alcohol treated with HCL

Tricyclopentyl alcohol treated with HCL

benzophenone - 7-11-2018 at 06:49

Hello. I have been having a lot of trouble with the reaction shown in this image.

I just can't picture how it would lead to a tetra-substituted alkene with formula C17H28. Can anyone come up with a representation of the product?

DraconicAcid - 7-11-2018 at 07:45

The first thing that's going to happen is that you'll get rid of the hydroxyl to form a primary carbocation. Primary carbocations are unstable as all get out, so you'll immediately have a rearrangement where one of the cyclopentyls shifts to that primary carbon.

That gives you a tertiary carbocation. Remove an H+ from one of the carbons next to it, and viola! An alkene! If you take off the right one, it will be tetrasubstituted.

benzophenone - 7-11-2018 at 09:44

Thank you so much. You have made a perfect explanation. Following your instructions, I would end with this molecule, correct?

fusso - 7-11-2018 at 10:03

How would it dehydrate when the solvent is H2O?

Loptr - 7-11-2018 at 10:18

Quote: Originally posted by DraconicAcid

The first thing that's going to happen is that you'll get rid of the hydroxyl to form a primary carbocation. Primary carbocations are unstable as all get out, so you'll immediately have a rearrangement where one of the cyclopentyls shifts to that primary carbon.

That gives you a tertiary carbocation. Remove an H+ from one of the carbons next to it, and viola! An alkene! If you take off the right one, it will be tetrasubstituted.

Wouldn't you also get substitution, albeit to a lesser extent? Or is the rearrangement too fast for that to happen? If a more stable secondary/tertiary carbocation can be formed, will it be formed compared to other possibilities?

Quote:

Other common strong acids such as HCl, HBr or HI are less suitable catalysts as nucleophilic substitution reactions will probably interfere

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch15/ch15-4...

EDIT: Ah, I also see that no deprotonation can occur due to that being a quaterniary carbon, so the alkyl shift occurs instead.

Another question, how do you get the removal of the hydrogen to an alkene? The carbocation has been resolved at this point, so you just have an unsymmetrical alkane.

EDIT: Also, with the water present, wouldn't this reaction take the more favorable E2 route, rather than an E1 (I supposed you meant this because of the immediate carbocation formation)? Water acting as the base? With the E2 you wouldn't get the intermediate carbocation. Unless because of that carbon already being quaterniary, a carbocation is formed?

[Edited on 7-11-2018 by Loptr]

CuReUS - 7-11-2018 at 22:26

Quote: Originally posted by benzophenone

Thank you so much. You have made a perfect explanation. Following your instructions, I would end with this molecule, correct?

you would get this.This is the reaction that is happening -https://en.wikipedia.org/wiki/Wagner%E2%80%93Meerwein_rearra...

wagner.png - 1kB

Metacelsus - 7-11-2018 at 23:52

OP said tetrasubstituted alkene, that one is trisubstituted.

This would be the correct tetrasubstituted product:

Screen Shot 2018-11-08 at 7.51.49 AM.png - 23kB

CuReUS - 8-11-2018 at 06:42

Quote: Originally posted by Metacelsus

OP said tetrasubstituted alkene, that one is trisubstituted

Ya I know that.I thought the OP made a mistake in the structure

Quote:

This would be the correct tetrasubstituted product

but there would be a lot of torsional strain in that molecule.It doesn't look like it would form naturally or freely.

DraconicAcid - 8-11-2018 at 11:23

Quote:

As I understand it, the primary carbocation is so unstable that rearrangement is virtually simultaneous with ionization.

Quote:

Another question, how do you get the removal of the hydrogen to an alkene? The carbocation has been resolved at this point, so you just have an unsymmetrical alkane.

EDIT: Also, with the water present, wouldn't this reaction take the more favorable E2 route, rather than an E1 (I supposed you meant this because of the immediate carbocation formation)? Water acting as the base? With the E2 you wouldn't get the intermediate carbocation. Unless because of that carbon already being quaterniary, a carbocation is formed?

After rearrangment, you still have a carbocation, and water will deprotonate it to give an alkene. It's not going to be E2, because water's not a strong base, and you can't get the rearrangement without forming a carbocation.