[MarkOfBrock]

I recently had a quiz in class. I was unable to complete this question... The question stated to propose a complete reaction mechanism (similar to the attached picture)
Im really lost..i got stuck writing it and i am totally stuck now. The TAs kept our test.

I have attached the image of a sample question. Can anyone help me with it.

The question also mentions is it initially formed carbication primary secondary tertiary...
assuming with what i tried, i did tertiary. I am not sure if i am right on this.

This picture is not the exact reaction due to the TAs keeping our test/quizes , however, a friend of mine at a different university has a similar question. I feel if i could figure it out i would be able to apply it to mine, however, he is stuck on it just as much as i am.

Thanks for any help that you can provide,

Many thnx Mbrock

[movies]

Well, how far did you get?  It sounds like you got the first part (protonate the alkene to make a tertiary carbocation).  Next you have to convert a four membered ring to a five membered ring.  This will be a good thing because you release a lot of strain, even if it means making a somewhat less stable carbocation.

[english]

Ring shift is always energetically feasible when expanding to a 5 or 6 carbon ring.

This means that even with an initial tertiary carbocation, thr ring will expand to a less strained ring to form another carbocation, which is secondary or tertiary.  In this case it's secondary, which is normally less stable.

Ask yourself whether it would be favorable to expand a ring or not.  Would a 5 membered ring be more stable than a four membered with a tertiary carbocation?  Cyclobutane is very sterically strained, so yes it would. 

Ring expansion always has to do with 5 or 6 membered rings.  If we were to expand to a 7 membered ring from a 6 membered ring, it would take more energy than if we didn't.  So this wouldn't happen under normal circumstances.

[MarkOfBrock]

Well, how far did you get?  It sounds like you got the first part (protonate the alkene to make a tertiary carbocation).  Next you have to convert a four membered ring to a five membered ring.  This will be a good thing because you release a lot of strain, even if it means making a somewhat less stable carbocation.

Alright i understand your statement of the ring strain. I have thought of this before hand.
However, let me proceed to tell you my first step. Is this really right..

i think i did pronate the alkene , but i ended up with + (Br-)

                                  +
So it looks like this   >-H  + Br-    (the R stands for the 4carbon ring still)
                                 R

(hope that makes sence if i must ill draw a picture)

Now I assume that the electron on the Bromine will attack the + charge on that carbon no?

i think this is where i make my mistake...when i do this i draw an arrow from the - charge on the Br to the + charge on the carbon and i result in

(excuse the attempt thru dashes but its the best i can do)

    |_br
   / \
  R    H     (so roughly a carbon in the middle bonded to another C (or methyl) to a Bromine to a H and        

to a cyclobutane

[MarkOfBrock]

I am running in circles i do not know why i cant turn it into a ring...

[english]

I am running in circles i do not know why i cant turn it into a ring...

OK.  I have drawn out this reaction (as neat as possible):

This is how ring shift is going on.  The numbers 1 and 2 will give you reference.  This will be your minor product because you have a 2° carbocation after ring shift—



You can also have a methyl shift occur after the ring shift to form a major product because this will give you a 3° carbocation.  This is a more stable carbocation and thus is easier to form than the 2°—



Try to see what's going on here according to what I've said.  You've already attempted and are probably getting rather irritated, so I only mean to help.

[movies]

Mark, your first step is correct and you do produce a bromide ion in the process.  Your instinct to bring the two charges together is a good one, but the alkyl shift in this case is preferred because you have an opportunity to lower the overall energy of the molecule before you trap the cation with the Br-.

[MarkOfBrock]

Mark, your first step is correct and you do produce a bromide ion in the process.  Your instinct to bring the two charges together is a good one, but the alkyl shift in this case is preferred because you have an opportunity to lower the overall energy of the molecule before you trap the cation with the Br-.

I think i got it. I had to use different color of pens. Also when i did it, i i had the V differntly i guess it epends how you open the ring.

Also- my name isnt mark ... or is it

~XCL

ps. thanks for all the help, imanaged to get a few more questions in my workbook of organic chemistry.

im gonna try some more otherwise ill ask you guys if i get stuck, till then many thanks