[KVargasBakas]

hey all,
I'm doing a write up for a lab explaining the preparation of 2-chloro2-methylbutane and I'm doing alright with establishing that it is an Sn1 and also the overall reaction along with the reaction mechanisms as well. However, I am extremely confused with what the side reactions would be and would also really appreciate someone explaining to me why the reaction favours the Sn1 pathway versus say one of the side reactions which I believe to be an elimination??

So basically I'm asking for the side reactions for the preparation of 2-chloro2-methylbutane in HCl. Overall reaction is listed below. Thank you very much!!

2-methyl2-butanol + HCL -----> 2-chloro2-methylbutane + H20 (which is then dried out to leave only the 2-chloro2-methylbutane)

[KVargasBakas]

Also, I  could provide the mechanisms of the reaction if that would help as well.

[NahiEw]

So the experiment you are doing involves protonation on the -OH and then followed by one of the two following ways:
Cl- attack on the carbon baring OH2+ and it leaves.
OH2 leaves the carbon creating a carbocation
The first way is not very favorable because Cl- is a relatively weak nucleophile. The second way will be more favored as it undergoes rearrangements to achieve tertiary carbocation. Now that you have a carbocation intermediate, it's either going to be E1 or SN1. It's not going to be E1 because, again, you are dealing with pretty weak nucleophiles that cannot extract the protons next to the charged carbon very well. Therefore, it will just be a chloride capture at the end which is essentially an SN1 reaction.

[Doc Oc]

Just as a point of clarification, the Cl- couldn't attack the carbon until the H2O left because there isn't a hydrogen on that carbon.  It is true that Cl- is a weak nucleophile, but this substrate was specifically chosen so there wouldn't be competing SN1/SN2 reactions, only SN1 is possible.

Also, if you look at the E1 elimination sequence, you would start with the carbocation, H2O, and Cl-.  If the Cl- pulls off a proton, you will have the alkene, H2O, and HCl.  Now consider what happens to alkenes in the presence of strong acids and see what you get from that reaction.

[orgopete]

In an SN1 reaction, the slow step is formation of the carbocation. The products are determined by the further reactions of that carbocation.

"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." http://www.cem.msu.edu/~reusch/VirtualText/alhalrx3.htm#hal8

However, I do not believe that is all to be said about the reaction. If the elimination can react with HCl in the reaction mixture, then the elimination intermediate can also be driven to the same substitution product. Thus, even though elimination may compete well with the substitution reaction, this elimination reaction may be masked by its further reaction.

Also, I am not verifying this statement of Reusch, though I do believe it is essentially correct. However, as it indicates product ratios are difficult to control, therefore, I find it difficult to predict whether a particular reaction is going to give an SN1 or an E1 product.