[a confused chiral girl]

Hi there!

for this question, the starting material is given, as well as the reagents and products formed. These are drawn in red in the diagram, and the question asks us to draw the detailed mechanism of this formation. I have attempted this question, and my work is shown in black...although this is how far I've gotten. I don't know what to do next 
I have a feeling that my work is not correct so far anyway...can someone please show me?

thank you!! 

[Dan]

First of all, you must show the stereochemistry of the C-Br bonds you are forming.

Secondly, the product you got to has a nucleophilic -SH substituent, where might it react?

You might find this question easier if you redraw the products, eg below

[kremar]

what you've got so far is correct. like Dan said, look at what you've got as your product. it has both a bromide (a good leaving group) and a thiol (a good nucleophile).

to get the stereochemistry right, draw the proper isomer mix you get from the bromine adition, and remember that the last step requires a proper conformer to give the leaving group the right orientation.

[a confused chiral girl]

thanks for the suggestions.

I've tried to understand what you all have suggested, and this is one way that I think it would happen....
but what I've got is a prodcut so far with 2 negative charges, which I think is unlikely. I think I went the wrong way...
can someone please show me how to finish the reaction with 2 enantiiomers? I am stuck.

thank you all!

[kremar]

you are making things more complicated than they are...
try to think of it like this:
how would R-Br and R'-SH react?

the only difference is that in this case the reaction is intramolecular.

[a confused chiral girl]

you are making things more complicated than they are...
try to think of it like this:
how would R-Br and R'-SH react?

the only difference is that in this case the reaction is intramolecular.

the Br would attack the H on the S? that's the only possibility that I can think of.... can you please show me the rest of what I've gotten so far? thank you so much!

[sjb]

In my opinion, you're almost there. Now in the first "blue" step, how can the bromide be displaced? You can't really protonate it, as you've said you're working under basic conditions. Consider things like ether synthesis (as you're forming a thioether) - do you know any ways of making them?

S

[a confused chiral girl]

Br attacking the S?

[Dan]

No.

The R-SH group is nucleophilic. Br^- is a good leaving group. If you have a good nucleophile and a center with a good leaving group present, what can happen?

For example, if you had R-CH₂OTf (^-OTf is an excellent leaving group) and you threw in some NaI (I^- is an excellent nucleophile) you should expect the nucleophilic I^- to attack R-CH₂OTf in an Sn2 fashion and form R-CH₂I.


The blue compound you drew after eliminating HBr is wrong, it would be a + charge left behind, ie a secondary carbocation, not anion - but I think you should be mainly looking at other substitution mechanisms... nudge nudge wink wink...

[organoman]

Water will not attack the carbonyl carbon. Rather in alkaline conditition alfa carbon becomes heighly acidic.
try with this information.