[faust]

Hello,

I have a little question about an execice
Could you tell me why, in the first reaction, the first step (replacing Br by RCOO) is slow? Is it because this is a SN1?
Then why the second step (replacing all Br by RCOO) is faster? Do we have a neigbouring group assistance somewhere?

Then on the second reaction, the exercice says that the kinetic law of this reaction is : kobs = k1 + k2[Nu]

What could be the mechanism? A SN1 followed by a SN2? I don't really know how to link the kinetic expression with a plausible mechanism.

Thank you for your help.

[Dan]

Then on the second reaction, the exercice says that the kinetic law of this reaction is : kobs = k1 + k2[Nu]

That looks familiar, what is the solvent in this reaction?

[faust]

this is : CH3CN : 35°C

[Dan]

Here's an idea:

In the attatched scheme, k(obs) = k[MeCN] + K2[Nu]

MeCN is the solvent, so we can assume it is in great excess, and that [MeCN] is essentially a constant. If we then say that k1 = k[MeCN]

=> k(obs) = k1 + k2[Nu]

[Dan]

For the first one, maybe once you get one RCOO- group on, it can facilitate the loss of further Br- by intramolecular attack by oxygen.

[faust]

Thank you very much for your help. It's clearer now

[movies]

For the first one, maybe once you get one RCOO- group on, it can facilitate the loss of further Br- by intramolecular attack by oxygen.

That is what I was thinking as well.  You could make a six seven-membered ring oxocarbenium ion.

I doubt the acetonitrile plays a big mechanistic role in the first step; it's a poor nucleophile.

[faust]

6 member ring? If you use the oxygen from the carbonyle you have a 7 member ring and if you use the non carbonyle oxygen you have a 5 member ring.

How can I find which oxygen is the better nucleophile?

Thank you again for your precious help.

[movies]

Oops, you're right: 7, not 6.  That isn't quite as good, but it's still okay.

The carbonyl oxygen is much more nucleophilic than the ether-like oxygen because you can stabilize the resulting charge by resonance.