[orgoclear]

Which method will this substrate react? SN1 or E1

(CH3)3-C-CH2Br + OH- -->

Can we decide the product without information about the temperature conditions?

[Dan]

The important thing here is the structure (the position is neopentyl). Try to draw the mechanism for E1 on this substrate...

I find it unlikely that this substrate will go by either E1 or S_N1...

I would suggest slow S_N2, as is typical for neopentyl centres, or, given the appropriate conditions, an intermediate carbocation could undergo rearrangement via a methyl migration. I think hydroxide is probably too good a nucleophile for the latter case though - if it was a poorer nucleophile, such as water, then carbocation rearrangement would probably be my bet.

[orgoclear]

As the substrate is sterically hindered on one side, shouldnt we expect it to undergo SN1/E1 in this case?

Same reasons apply for my not thinking about SN2/E2

[eureka123]

Yeah Sn2 will be difficult
becoz in SN2 attack of Nucleophile takes place from rear side of Halide 

[Dan]

Elimination, E1 or E2, on this system has a problem. Which hydrogen are you proposing the base takes off? There are no beta hydrogens. It's impossible. This is why I asked you to try to draw the mechanism.

Neopentyl carbocations are still primary and highly unstable as a result. In the presence of a good nucleophile, they will react by S_N2, but the reaction will be slow due to steric hindrance. This is classic neopentyl reactivity, it's not my own crazy idea, check any organic chemistry textbook, for an example see the google return below.

[orgopete]

I understand and agree with orgoclear. It is difficult to predict whether one should get SN1 or E1 products from a reaction of neopentyl bromide. It is also possible to perform an SN2 reaction but not an E2 reaction on it. In my opinion, it can be difficult to predict the answers to this type of question a priori. If one wished to perform an SN2 reaction, one would use high concentrations of reagents in SN2 solvents. The same reaction diluted with a polar protic solvent can change mechanisms. I would be unable to predict at which concentrations that would be expected to occur.

However, as Dan points out from Carey and Sandberg, neopentyl bromide also gives the products of rearrangement to t-pentyl system. This is obviously the rearrangement that must have occurred from an SN1 reaction. That was the OP's original question. To this point, I note that Reusch in his online textbook, he states something like it is difficult to predict the ratios of SN1 to E1 products. I would concur.

[azmanam]

I suspect we'll get E1 only AFTER rearrangement.  It's not likely to be straight elimination, because, as Dan pointed out, there are no beta-protons.  It's unlikely to be SN1 because primary carbocations are too unstable to form in any appreciable amount.  For the same reason, it's unlikely to form the carbocation, then rearrange, then eliminate.  It's unlikely to be SN2 because of the severe steric hindrance of beta branching.  I just lectured on beta branching this morning, and neopentyl electrophiles react some 100,000 times slower than ethyl bromide in SN2 reactions.  So if you let it go long enough, yeah, you'd probably get some SN2... eventually.

So I predict one of the rare cases where rearrangement and LG expusion occur simultaneously.  This gives the stable tertiary carbocation.  We're in good base, so E1 elimination should occur after carbocation formation.  That's my guess.  My sketch of the arrow-pushing is below.

Below that is a scheme from the alcohols chapter of Brown Foote and Iverson's Organic text.  It shows precedence for this type of rearrangement/carbocation formation.  That system is in weak base (actually acidic conditions, so net SN1 results), but the theory is the same?

[orgoclear]

@ Dan: Yeah I know there is no beta H to eliminate but upon rearrangement, it forms a stable tertiary carbocation. (although the rate of formation of the initial primary carbocation is very little)

That was the reason why E2 wont take place but E1 can

Again as Azmanam pointed out, SN2 will be very very slow indeed..

Again the carbocation is tertiary so, wouldnt we expect E1 to be more dominant that SN1 ?

[orgopete]

From Reusch, www.cem.msu.edu:

To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:

     1. The cation may bond to a nucleophile to give a substitution product.
     2. The cation may transfer a beta-proton to a base, giving an alkene product.
     3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.

Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place.

I don't know that I agree completely. Some examples can be driven to elimination. I have wondered what the products are for the solvolysis of t-butyl halides in NaOH. When we did this experiment, the kinetic clearly show it is first order (its not E2) and because we are using an indicator, we detect the pH change from basic to acidic. Will the product be isobutylene as azamanam suggests or will it be a mixture of isobutylene and t-butanol as implied by Reusch?