Ring contraction in a diazotisation reaction

[Physics lover]

Homework Statement

Question is in My attempt section

Relevant Equations

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I know HNO₂ favours diazotisation.So amine reacted and formed diazorised product with N⁺=NCl^- which leaves and gives a carbocation to which OH^- attacks.So according to me the answer should be cyclohexane-1,2-diol.
But in the answer key,cyclopentane carboxyaldehyde was given as answer.
I know it will be only formed when ring contraction takes place.But I can't guess why it would take place.Please help.

 

[mjc123]

gives a carbocation to which OH- attacks

Can you think of anything else that carbocation could do? In particular, could it form a more stable carbocation by rearranging?

 

[Physics lover]

Can you think of anything else that carbocation could do? In particular, could it form a more stable carbocation by rearranging?

yes the carbocation could do 1,2 H^- shift with adjacent carbon to which other OH group is attached.

 

[Physics lover]

yes the carbocation could do 1,2 H^- shift with adjacent carbon to which other OH group is attached.

I think the sane thing can also be achieved through a ring contraction.That will also give the same carbocation as of 1,2 H^- shift.But according to me shift would be a better option because 6 membered ring is more stable than 5 membered ring.

 

[chemisttree]

Review the pinacol rearrangement.

 

[Physics lover]

Review the pinacol rearrangement.

ok i get your point.
So now,H+ would attact on any OH group and will form a carbocation.Further there will be rearrangement by 1,2 H^-
shift.Now OH group removes its H⁺ and forms a ketone.
But i know that that exocyclic double bonds are unstable.So ring contraction will take place to give aldehyde.Am i right?

 

[chemisttree]

Almost. The pinacol rearrangement classically starts with a 1,2-diol and strong acid. The acid protonates one of the alcohols and makes it a good leaving group. The protonated alcohol leaves and we have a carbocation adjacent to an alcohol. That intermediate is what you should focus on. By whatever mechanism, be it protonated alcohol leaving or a diazotized amine leaving, you arrive at the same thing... carbocation adjacent to a secondary alcohol. Pinacol rearrangement follows.

The diazotization route allows you to control which carbon becomes the carbocation. Might be important someday.

 

[Physics lover]

Almost. The pinacol rearrangement classically starts with a 1,2-diol and strong acid. The acid protonates one of the alcohols and makes it a good leaving group. The protonated alcohol leaves and we have a carbocation adjacent to an alcohol. That intermediate is what you should focus on. By whatever mechanism, be it protonated alcohol leaving or a diazotized amine leaving, you arrive at the same thing... carbocation adjacent to a secondary alcohol. Pinacol rearrangement follows.

The diazotization route allows you to control which carbon becomes the carbocation. Might be important someday.

Thank you :)
By the way I just wanted to know whether my explanation for ring contraction to take place is correct or not.

 

[chemisttree]

I don’t think so. Starting with the newly-formed carbocation, allow the carbon on the other side of the alcohol to migrate over to the cation. In this process, it is likely the bond breaking and bond forming process occurs simultaneously via a three-member (C-C-C) ring structure where the carbocation is smoothly converted to a neutral entity and the positive charge develops on the oxygen as a protonated oxonium ion which is MUCH more stable and the driving force for this rearrangement. The free lone pair of electrons on the alcohol oxygen starts the process by forming a carbonyl which weakens the C-C bond allowing it to migrate efficiently. What is formed is the enol of an aldehyde which is in equilibrium with formyl group of course.

 

[Physics lover]

I don’t think so. Starting with the newly-formed carbocation, allow the carbon on the other side of the alcohol to migrate over to the cation. In this process, it is likely the bond breaking and bond forming process occurs simultaneously via a three-member (C-C-C) ring structure where the carbocation is smoothly converted to a neutral entity and the positive charge develops on the oxygen as a protonated oxonium ion which is MUCH more stable and the driving force for this rearrangement. The free lone pair of electrons on the alcohol oxygen starts the process by forming a carbonyl which weakens the C-C bond allowing it to migrate efficiently. What is formed is the enol of an aldehyde which is in equilibrium with formyl group of course.

Thanks for your efforts but can you please tell me in simple words.I cannot understand what really happened.
Can you please tell me what's wrong in my solution.We know that exocyclic double bonds are unstable so the reaction would take other path after the formation of carbocation.

 

[chemisttree]

There is NO HYDRIDE SHIFT in this reaction.

 

[Physics lover]

View attachment 260489
There is NO HYDRIDE SHIFT in this reaction.

Thanks a lot.I got it.

 

[chemisttree]

Great. I believe I may have mislead you a bit though. What is formed is NOT an enol (my post#9) but simply a protonated aldehyde. ::faceslap::

 

[Physics lover]

Great. I believe I may have mislead you a bit though. What is formed is NOT an enol (my post#9) but simply a protonated aldehyde. ::faceslap::

yes by that statement i was confused but now it's all clear.

 

[Physics lover]

Great. I believe I may have mislead you a bit though. What is formed is NOT an enol (my post#9) but simply a protonated aldehyde. ::faceslap::

By the way i wanted to PM you.But why that option is not available

 

[chemisttree]

I just reviewed my settings. Reset to “Members Only”. Fire away!

 

[Physics lover]

View attachment 260489
There is NO HYDRIDE SHIFT in this reaction.

I am a little confused again.
Here as we see 1,2 methyl shift takes place since there is no hydrogen on any carbon attached with OH group.
But if hydrogen is there will not hydride shift take place,as hydrogen has better shifting tendency.
Please explain.

 

[chemisttree]

Hydrogen does not have a better tendency to shift in this case. The extra stability of the oxonium ion vs the carbocation drives this. Hydride shifts are governed by small differences in carbocation stability. Carbocation vs oxonium is a large difference in stability. You cannot generate a carbocation easily. Look at the special conditions required for an E1 elimination, for example.
This 1,3-carbon shift occurs because the alcohol is protonated, a free lone pair from oxygen forms a carbonyl which weakens the C-C bond on the other side of the carbonyl (than the carbocation). The cation adjacent to carbonyl is stabilized somewhat by the newly forming pi system kind of like an enol stabilization.

 

[Physics lover]

Hydrogen does not have a better tendency to shift in this case. The extra stability of the oxonium ion vs the carbocation drives this. Hydride shifts are governed by small differences in carbocation stability. Carbocation vs oxonium is a large difference in stability. You cannot generate a carbocation easily. Look at the special conditions required for an E1 elimination, for example.
This 1,3-carbon shift occurs because the alcohol is protonated, a free lone pair from oxygen forms a carbonyl which weakens the C-C bond on the other side of the carbonyl (than the carbocation). The cation adjacent to carbonyl is stabilized somewhat by the newly forming pi system kind of like an enol stabilization.

ok thanks for the help.