[LuckyDude35]

Got a couple of questions, I have a question that is: bromomethane + hydrogen sulfide in DMF (at 25 degrees C). I know the S acts the nucleophile and the Br is the leaving group. My question is what happens to the hydrogen that was on the S? Does it bond to the Br now? Thanks for your *delete me*

[movies]

Yeah, you get HBr as a side product.  It will be dissociated into H+ and Br- ions, most likely.  In these reactions there is typically some kind of non-reactive base to absorb the acid that is produced.

[LuckyDude35]

alrighty, then I have another question. I am trying to work on 1-methylcyclopentanol + HCL (at 0 degrees C), Im not sure where to start with this one, any help?

[movies]

Well, give me your thoughts on it.

[LuckyDude35]

The  0 degrees seems odd to me. I would think that the Cl from the HCL would be the nucleophile, would the OH come off of the 1-methylcyclopentanol? Would this be SN2 then?

[movies]

Rember the rules of substitution and elimination.  What do those suggest?

[LuckyDude35]

I think it would be elimination since the base is larger

[movies]

What is the base?  How does it compare in size to the bases used in typical elimination reactions?

[LuckyDude35]

It would have to be the 1-methylcyclopentane since we only have HCL other then that. It seems larger then most other bases we run into

[movies]

How is 1-methycyclopentane basic?  There aren't any basic atoms or negative charges.

[LuckyDude35]

I'm sorry, I got confused. So this would be a substitution. Wouldn't the Cl from the HCL take thespot of the OH on the larger compound to produce water when it comes off?

[movies]

That certainly sounds reasonable.

So what kind of substitution mechanism would this suggest?  You might want to check back on one of my posts from last night.

[LuckyDude35]

I think it would be SN1 since the HCL is uncharged and SN2 prefers a charged nucleophile?

[movies]

I think that's the correct answer, but I think your reasoning is flawed.

You have to think of the mechanism in steps.  After each step the species that are present change and that changes what the likely next step is.

So to go through an SN1 mechanism step by step, using this problem as an example:

1) You have a strong acid (HCl, which is likely dissociated into H+ and Cl- ions) and a somewhat basic substrate (the alcohol).  When you have a strong acid/base and a weaker base/acid, it's usually a good starting point to bring these two together.  So, protonate the alcohol.

2) Now there is a protonated alcohol and a Cl- ion in solution.  The alcohol in this case is at a very hindered position (tertiary carbon) so SN2 is very unlikely.  You do, however, have an exceptionally good leaving group (water), which can leave spontaneously.  So have the water leave.

3) Now there are three species in solution: water, Cl-, and the carbocation from where the water left.  This carbocation is quite stable since it is tertiary, so it can linger in solution for a while, but the positive charge still wants to be quenched.  At this point there are two things that could happen in order to quench the positive charge: elimination or SN1 substitution.  Since Cl- isn't a very strong base, elimination isn't likely.  Cl- is, however, a pretty decent nucleophile, so it can attack the carbocation and form 1-chloro-1-methylcyclopentane.

So things are more subtle than just whether or not the nucleophile is charged.

[LuckyDude35]

Oh, Ok. I see. Thank you!