[RedsAreRaw]

Hey everyone,
I came across this question: "(a) When neopentyl alcohol, (CH₃)₃CCH₂OH, is heated with acid, it is slowly converted into an 85:15 mixture of two alkenes of formula C₅H₁₀. What are these alkenes, and how are they formed? Which one would you think is the major product, and why? (b) Would you expect neopentyl bromide, (CH₃)₃CCH₂Br, to undergo the kind of dehydrohalogenation described in Sec 5.13? (This is the addition of KOH in alcohol to produce an alkene) Actually, when heated in aqueous alcohol, neopentyl bromide slowly reacts to yield, among other products, the same alkenes as those in (a). Suggest a mechanism for this particular kind of dehydrohalogenation. Why does this reaction, unlike that in (a), not require acid catalysts?"

I was able to answer (a), but wasn't sure if i had the correct mechanism for (b). If you guide me if I made any mistakes it would be appreciated. Thanks.

[phth]

What is the mechanism for that reaction (addition of KOH to an alkene) E1 or E2?  Does the mechanism in part a go through an E1 or E2 mechanism?

[RedsAreRaw]

I had (a) via E1 mechanism. Then for (b) I used S_N2 and E1.

[Babcock_Hall]

Your initial post is not easy to decipher.  I can't figure out which is part (a) and which is part (b).  Are you proposing mechanism for neopentyl bromide that converts it to neopentyl alcohol first?

[RedsAreRaw]

Sorry, I should have said that the mechanisms are just for part (b). That's my fault. The first line is the alcohol and water reacting to give small amounts of OH^-. This is what initiates the proposed following reaction. This is why I'm guessing it does not require the acid catalyst.

[Babcock_Hall]

I think that you are making problem (b) more complicated than it needs to be.  What chemical property do good leaving groups generally possess?

[RedsAreRaw]

What chemical property do good leaving groups generally possess?

The ability the leaving group has to accommodate the pair of electrons/negative charge.

[Babcock_Hall]

Right.  Now apply this to the leaving group in neopentyl bromide.

[RedsAreRaw]

I thought that's what I did with the OH^- attacking the carbon with the Br substituent via an S_N2 mechanism. Because there is no H at the quaternary carbon there has to be an alkyl shift after the substitution I thought.

[Babcock_Hall]

I suggest writing a mechanism for elimination that does not involve an S_N2 reaction with hydroxide as a nucleophile.  An E1 might be slow, but I don't see obvious alternatives.

[RedsAreRaw]

So just like this?

[Babcock_Hall]

That looks basically reasonable to me, although you might want to make the last step formally balance.  Also, your original question said that it was a slow reaction.  What do you think is the reason for that?

[RedsAreRaw]

I was originally thinking it was a slow reaction because of the amount of OH^- produced is small, but once it attacks the neopentyl bromide the reaction would progress forward. I wasn't sure how readily a 2^° bromine would progress the reaction via an E1 mechanism. I didn't think it would be a strong enough leaving group to do that. Then with both the Br^- and carbocation being in very low concentrations, I wasn't sure if it would amount to much of a reaction.

[Babcock_Hall]

It's not a 2° bromide.  What sort of carbocation does bromine leave behind when it departs as bromide ion?

[RedsAreRaw]

Sorry, I meant 1^°. And I've read where it is very unlikely for E1 to happen there.