[BetaAmyloid]

95% Correct.

Make sure that arrow in the first drawing is from the BOND (electrons) to the oxygen, showing that those electrons went to the oxygen, not ambiguously that the -OH₂ just left.

The second drawing should show those pi electrons reaching out and grabbing a H⁺, not that the pi electrons just jumped to what was carbon 1 and a hydrogen just all the sudden appeared (show where from).

Also, after that first drawing there is no need to show stereochemistry (no dashed line on the methyl) because that methyl is now in a trigonal planar electronic geometry. The bromine anion could attack from the front or the back, which is why the product given in the question is just solid lines, since you cannot be completely confident as to whether the bromine or methyl is in the front or the back (there will be a racemic mixture).

I will check back periodically to help you out.

And don't give up, it isn't that you won't ever understand. O-chem takes time, never feel dumb or that you shouldn't ask questions (especially ask a lot to your professor), because I can bet you that there are many many others that feel the same.

[theanonymous]

95% Correct.

Make sure that arrow in the first drawing is from the BOND (electrons) to the oxygen, showing that those electrons went to the oxygen, not ambiguously that the -OH₂ just left.

The second drawing should show those pi electrons reaching out and grabbing a H⁺, not that the pi electrons just jumped to what was carbon 1 and a hydrogen just all the sudden appeared (show where from).

Also, after that first drawing there is no need to show stereochemistry (no dashed line on the methyl) because that methyl is now in a trigonal planar electronic geometry. The bromine anion could attack from the front or the back, which is why the product given in the question is just solid lines, since you cannot be completely confident as to whether the bromine or methyl is in the front or the back (there will be a racemic mixture).

I will check back periodically to help you out.

And don't give up, it isn't that you won't ever understand. O-chem takes time, never feel dumb or that you shouldn't ask questions (especially ask a lot to your professor), because I can bet you that there are many many others that feel the same.

Ohh ok!
Like this?

[BetaAmyloid]

That's it!

[theanonymous]

Thanks!!!!

Last question:

6. 6. The reaction of 2,2-dimethyl-1-propanol with HBr is very slow and gives 2-bromo-2-methylbutane as the major product.



Give a mechanistic explanation for these observations.

EDIT:

OH WAIT!!!!
I think I got the answer!



Does that look right?? 

I'm so happy right now!

[BetaAmyloid]

Always remember that you have the arrowheads pointing towards the positive charges, so the oxygen should be accepting that proton (not the proton accepting the oxygen, since which is bigger and more electronegative? - oxygen is).

That oxygen becomes protonated. The bond (electrons) between the oxygen and carbon go to the oxygen to give water and a positive charge on that carbon.

A carbocation rearrangement occurs where the "bottom" methyl shifts to the positive carbon and ends up giving a positive charge on the "middle" carbon.

This allows for the bromine anion to attack that "middle" carbon and form the final product.

Show what I just said in words with pictures.

[theanonymous]

Always remember that you have the arrowheads pointing towards the positive charges, so the oxygen should be accepting that proton (not the proton accepting the oxygen, since which is bigger and more electronegative? - oxygen is).

That oxygen becomes protonated. The bond (electrons) between the oxygen and carbon go to the oxygen to give water and a positive charge on that carbon.

A carbocation rearrangement occurs where the "bottom" methyl shifts to the positive carbon and ends up giving a positive charge on the "middle" carbon.

This allows for the bromine anion to attack that "middle" carbon and form the final product.

Show what I just said in words with pictures.

^ Like that?
Should the CH₂ still be there though?

[BetaAmyloid]

1) No -- in your first picture the arrow should go from the oxygen to the proton because the oxygen is accepting the proton because of it's electronegativity.

There are five carbons in a "wheel" like formation, just for understanding, lets call them top, left, bottom, right, and center for the names of the carbon (you should be able to see which is which).

2) Now that the oxygen (alcohol) is protonated, the bond (electrons) between the right carbon and the alcohol go to the oxygen to form water and a positive charge on the right carbon.

3) The bond (electrons) between the center and bottom carbons now move in a carbocation rearrangement to the right carbon (the one with the positive charge).

4) There should now be a positive charge on the middle carbon that the bromine anion attacks (like you have shown in your last mechanistic step).

[theanonymous]

1) No -- in your first picture the arrow should go from the oxygen to the proton because the oxygen is accepting the proton because of it's electronegativity.

There are five carbons in a "wheel" like formation, just for understanding, lets call them top, left, bottom, right, and center for the names of the carbon (you should be able to see which is which).

2) Now that the oxygen (alcohol) is protonated, the bond (electrons) between the right carbon and the alcohol go to the oxygen to form water and a positive charge on the right carbon.

3) The bond (electrons) between the center and bottom carbons now move in a carbocation rearrangement to the right carbon (the one with the positive charge).

4) There should now be a positive charge on the middle carbon that the bromine anion attacks (like you have shown in your last mechanistic step).

Like this??

[BetaAmyloid]

Almost, In your second picture there should be no positive charge on the carbon and that bottom methyl should not be doing anything yet -- you should only show the water leaving.

One extra step between your second picture and third picture would be the formation of a positive charge on that right carbon due to the water leaving. Then, you should show the electrons of that bottom methyl moving to the positive charge on the right carbon. Do not show it going from carbon to carbon, it is the ELECTRONS (the bond!!!) going to carbon.

See if you can do that.

[theanonymous]

Almost, In your second picture there should be no positive charge on the carbon and that bottom methyl should not be doing anything yet -- you should only show the water leaving.

One extra step between your second picture and third picture would be the formation of a positive charge on that right carbon due to the water leaving. Then, you should show the electrons of that bottom methyl moving to the positive charge on the right carbon. Do not show it going from carbon to carbon, it is the ELECTRONS (the bond!!!) going to carbon.

See if you can do that.

Like this?

[BetaAmyloid]

You got it!

*Note: It would be more accurate if you showed the bond (the electrons) in HBr going to bromine leaving just the proton.

So H-Br show an arrow from the dash to bromine giving a bromine anion and a proton which is then used in the mechanism.*

[theanonymous]

You got it!

*Note: It would be more accurate if you showed the bond (the electrons) in HBr going to bromine leaving just the proton.

So H-Br show an arrow from the dash to bromine giving a bromine anion and a proton which is then used in the mechanism.*

Awesome!!!! 
Thanks a bunch!!!

[BetaAmyloid]

Anytime!

[orgopete]

Your reaction mechanism isn't correct. I'll give you the first step, protonate the alcohol which will ultimately allow dehydration and the formation of a pi bond between carbons 1 and 2. See what you can do from there.

So, you are saying that if 2-deutero-2-methylcyclohexanol were used, the product would lose all deuterium atoms and no migration would occur?

[BetaAmyloid]

So, you are saying that if 2-deutero-2-methylcyclohexanol were used, the product would lose all deuterium atoms and no migration would occur?

No, in the step where the pi electrons undergo electrophilic attack to a proton, this could have been a deuterium if the original reactant had deuterium. Deuterium wasn't present though. Isn't that right?