[Lindsay<3]

http://i219.photobucket.com/albums/cc138/saveourpens/securedownload.jpg

Hey guys, in the first part, am I right that the first reaction would be
 sn1 , giving mainly substitution products and some elimination

and the 2nd reaction would be e1 because of the sterically hindered nucleophile? 

[nj_bartel]

for the second one I'd say e2 because of the strength of the base

[Telamond]

I agree with nj_bartel. I also think that the second reaction should be an E2. Also, high temperature generally favours elimination reactions.

However, I think that the first one would be an E1 since the nucleophile isn't that good (and the high temperature).

But after reading the heading, it seems like there's only SN1 and E1 to chose from...?

[azmanam]

high temp for #1 does usually indicate elimination... but only if you have a competent base.  ethanol's not that hot of a nucleophile... but I'd take it as a nucleophile over an electrophile in this situation.  Sn1 for me.

[Lindsay<3]

Hey thanks guys!

Ive figured out that the 1st reaction actually has both sn1 and e1 products with the sn1 being the major product.
the 2nd reaction is definetly sn2.

 Can someone please tell me how to draw the e2 product for the 2nd reaction? It is definetly e2. Im confused as which hydrogen the nucleophile would attack.

[azmanam]

1) it attacks a proton, not a hydrogen (trivial, but important).

2) KOtBu is generally too large to take sterically hindered protons, so it usually gives the Hoffmann elimination product.  In this case, do you think it will matter?

[nj_bartel]

Just a note, S_N1 and E1 always occur together, competitively.  If you're making a distinction as to which is occurring, what you're actually saying is which occurs to a greater extent.