Question

I'm not sure if 4 is correct so I just wanted some clarification, and for 5 I also am not sure as I need more reasoning for why product B was formed!

4. Propose an arrow-pushing mechanism for the reaction below. 1. Cl₂, H₂O 2. NaH HO O-H Loh HOH Na-u 00 27,

Answer 1

It might go like the following---

cz tul H Chloronium ion intermediate. attacks on the less substituted carbon. 14.0 Het THU + = NaH + Nace Me su Me-se BY 67 Product A AS-CH2 +HBY sulfur Ylide + BT By + Br Br Me I Met Br Br2 Bre Bols or sine Product B

Product A is the major kinetically controlled product here. But the reaction doesn't stop after forming A. With the group -SMe attached to the alkene, S_N1 can happen by the opening of bromonium ring, forming a carbocation, which is stabilized by conjugation with the Sulfur loan pair. Now, Bromide ion doesn't generally act as a base by abstracting proton; but with positive charge on the adjacent Sulfur atom the acidity of the -CH₃ proton is enhanced, thus forming a stable Sulfur ylide(shown above). Now, this CH₂^- is a good nucleophile and does a NGP(Neighboring Group Participation) on the carbon bearing positive charge. After that, bromide attacks the less substituted carbon to open the ring(attack on the other carbon generates product A),forming 1,3-dibromide and Sulfide gets methylated either by the starting alkene or the initially formed product A, to form product B.

Actually the formation of product A is reversible, because bromide is a good nucleophile as well as a good leaving group. Thus after a long time there should be a considerable amount of product B along with the "expected" product A.

Thus the formation of B can be explained.