[theanonymous]

5. Please provide a mechanistic explanation for the following observation:



This is what I wrote down (I took it from my notes, but I'm not 100% sure if it's entirely accurate...also, I'm guessing my teacher also wants me to briefly explain what's happening in words)



How would I go about explaining in words what's going on?

Second structure I've drawn: Hydrogen atom (H⁺) moves at the bottom right corner to make way for the incoming nucleophile Br^-; a carbocation (C⁺) is formed in the process
Third structure I've drawn: You now have 2 hydrogen atoms at the bottom (one pointing towards you and one pointing away from you); nucleophile attacks and a C-Br bond is made
Fourth structure of drawn: The result

^

Is that fine (and accurate)?

[BetaAmyloid]

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.

[theanonymous]

Your reaction mechanism isn't correct. You need to show the alcohol on carbon 1

Wait what?
I don't understand
Can you show me?

[BetaAmyloid]

Sorry, I accidentally hit enter, look at my post above!!

[theanonymous]

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.

Uhhh...
I'll try

[theanonymous]

ughghghgh
I'm not doing this right...
What do you mean by protonate and making the C1 and C2 bonds form a pi bond??
Is there a way for you to draw it on here?

[BetaAmyloid]

So your alcohol doesn't have negative charge, it is neutral originally. The hydrogen in HBr leaves Br (leaving a Br anion and a H cation or proton) and is accepted by the alcohol.

So using your drawing, show an arrow from the oxygen on the alcohol to the H in HBr, while showing that the H in HBr is leaving Br. This will give you OH₂ as you have shown, but the oxygen will be protonated.

What step would occur next to cause the water to leave (dehydration) and a pi bond to form between carbons 1 and 2 at the same time?

[theanonymous]

So your alcohol doesn't have negative charge, it is neutral originally. The hydrogen in HBr leaves Br (leaving a Br anion and a H cation or proton) and is accepted by the alcohol.

So using your drawing, show an arrow from the oxygen on the alcohol to the H in HBr, while showing that the H in HBr is leaving Br. This will give you OH₂ as you have shown, but the oxygen will be protonated.

What step would occur next to cause the water to leave (dehydration) and a pi bond to form between carbons 1 and 2 at the same time?

^
The methy groups will move/shift to the bottom (?)

[BetaAmyloid]

Nope! What is on carbon 2 besides methyl? Hydrogen, right? Could the bonding electrons of that hydrogen form the pi bond and then that hydrogen become a proton in solution? Show how that would work.

Note: Make sure to show the positive charge on the oxygen of the -OH₂ group while it is still attached to the parent structure!

[theanonymous]

Nope! What is on carbon 2 besides methyl? Hydrogen, right? Could the bonding electrons of that hydrogen form the pi bond and then that hydrogen become a proton in solution? Show how that would work.

Note: Make sure to show the positive charge on the oxygen of the -OH₂ group while it is still attached to the parent structure!

Ok, I have a quick question... "Carbon 2"... is that the carbon on the bottom of the vertical line.. you know where  ––––– OH₂ is.... Carbon 2 would be located here right?: C₂ ––––––– OH₂

[BetaAmyloid]

If you were naming this compound, the alcohol would be considered more electronegative or polar than the methyl group, thus the carbon where the alcohol is would be considered #1. Then you rank other R-groups as necessary, here the methyl group is attached to carbon #2.

[theanonymous]

If you were naming this compound, the alcohol would be considered more electronegative or polar than the methyl group, thus the carbon where the alcohol is would be considered #1. Then you rank other R-groups as necessary, here the methyl group is attached to carbon #2.

Ok...



Ughh...

This is hopeless.
I'm not good at this yet and this is probably taking longer than it should because my teacher never elaborated on /how/ to do these problems

[theanonymous]

You still there?

[BetaAmyloid]

You are closer!

In your drawing, you have the bonding electrons moving to make that pi bond. But, as that pi bond is forming there would be 5 bonds on carbon 1, which means that the water would have to be leaving as the pi bond is forming.

Your first drawing should have an arrow showing water leaving. Your second drawing should not have water, but should have a pi bond and no charge on carbon 2.

Now, if you draw what I said correctly, you can show the pi bond electrons attacking a proton in solution (which formed from the first drawing you have here), which will put the hydrogen on carbon 1. This will leave a positive charge on carbon 2, and will show that the last segment of your drawing here is correct giving the final product.

[theanonymous]

You are closer!

In your drawing, you have the bonding electrons moving to make that pi bond. But, as that pi bond is forming there would be 5 bonds on carbon 1, which means that the water would have to be leaving as the pi bond is forming.

Your first drawing should have an arrow showing water leaving. Your second drawing should not have water, but should have a pi bond and no charge on carbon 2.

Now, if you draw what I said correctly, you can show the pi bond electrons attacking a proton in solution (which formed from the first drawing you have here), which will put the hydrogen on carbon 1. This will leave a positive charge on carbon 2, and will show that the last segment of your drawing here is correct giving the final product.

Ohhh 



Awesome!!

Thanks!!

I have one more question like this - Imma try to do it on my own first and get back to you if youlll still be here!