[user5433]

Hi I have the following reactions.  I was wondering someone could check my answers.


1) SN2 with the products being the original molecule but inverted and Br replaced with CH3O.  Also NaBr is a product.
2) No reaction because Br is attached to an aryl group and CH3 on the other end is a very bad leaving group.
3) BR SHOULD BE OTS.  SN1 because OTs is a good leaving group and CH3CH2OH is a good base.  The products are a racemic mix with CH3CH2O replacing OTs (one being (R) and one being (S)).  Also HOTs is formed.
4) No reaction because of a bad leaving group and bad nucleophile.
5) SN1 because the nucleophile is a weak base and the resulting product has resonance stabilization.  The starting material should then be the substrate with a very good leading group to compensate for the bad nucleophile.
6) SN1 because of a good leaving group and good but uncharged nucleophile.  Products will be the nucleophile replacing Br with a racemic mix of (R) and (S) configurations.  Also HBr will be formed. 

I've just started to learn this so I'm a bit unsure of my answers.  Any help would be appreciated, thanks.

[discodermolide]

It would help (me at least) if you could provide a better scan, I can't make out some of the compounds. Therefore it is difficult to say what is right and wrong. As far as I can see you need to re-think numbers 3 & 4.

[Dan]

1) SN2 with the products being the original molecule but inverted and Br replaced with CH3O.  Also NaBr is a product.
2) No reaction because Br is attached to an aryl group and CH3 on the other end is a very bad leaving group.

I think these are OK

3) BR SHOULD BE OTS.  SN1 because OTs is a good leaving group and CH3CH2OH is a good base. 

I assume you mean EtNH₂? Can you explain why you have said the basicity of amine will favour S_N1?

4) No reaction because of a bad leaving group and bad nucleophile.

Think about the acid-base reaction between an alcohol and HBr

5) SN1 because the nucleophile is a weak base and the resulting product has resonance stabilization.

I do not understand your resonance argument at all, please explain.

Acetate is not that bad nucleophile, you can do S_N2 reactions with it under the right conditions.

  The starting material should then be the substrate with a very good leading group to compensate for the bad nucleophile.

You should suggest an appropriate leaving group in your answer.

6) SN1 because of a good leaving group and good but uncharged nucleophile.  Products will be the nucleophile replacing Br with a racemic mix of (R) and (S) configurations.  Also HBr will be formed. 

I would describe an alcohol as a poor/weak nucleophile.

[user5433]

I assume you mean EtNH₂? Can you explain why you have said the basicity of amine will favour S_N1?

Sorry this problem was completely different on my sheet; this scan was different for this problem.  The double bond is on the other side, the leaving group is OTs, and the nucleophile is CH3CH2OH, which is an uncharged alcohol and therefore a poor nucleophile.  It should go through SN1.  I am, though, confused about what the Pd/H2 will do. 

4) No reaction because of a bad leaving group and bad nucleophile.

Got it.  It is acid/base and through SN1. 

5) SN1 because the nucleophile is a weak base and the resulting product has resonance stabilization.

I was thinking that the reaction could proceed through SN1 because of the resonance stabilization of the product in the acetate group (around the double bonded O).  But now that you mention it, acetate is a moderately strong nucleophile and this reaction can definitely go through SN2.

Thanks for your help (and discodermolide) and sorry for the poor scan.  I really am starting to grasp this material for which I had no understanding yesterday.