Question

State the number and kind of stereoisomers formed when \((R)-3\)-methyl-1-pentene is treated with these reagents. Assume that the starting alkene is enantiomerically pure and optically active. Will each product be optically active or inactive? C=CC(C)(C)CC ( \(R\) )-3-Methyl-1-pentene (a) \(\mathrm{Hg}(\mathrm{OAc})_{2}, \mathrm{H}_{2} \mathrm{O}\) followed by \(\mathrm{NaBH}_{4}\) (b) \(\mathrm{H}_{2} / \mathrm{Pt}\) (c) \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2}\) in \(\mathrm{NaOH}\) (d) \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\)

Short answer

Also, determine if the products will be optically active or inactive. Reaction with: (1) Hg(OAc)2, H2O followed by NaBH4 (2) H2/Pt (3) BH3 followed by H2O2 in NaOH (4) Br2 in CCl4 Answer: (1) One product, optically active. (2) One product, optically active. (3) One product, optically active. (4) Two products (enantiomers), both optically active.

Step-by-step solution

(1) Reaction with Hg(OAc)2, H2O followed by NaBH4

This reaction is an oxymercuration-demercuration process. The reagents \(\mathrm{Hg}(\mathrm{OAc})_{2}, \mathrm{H}_{2} \mathrm{O}\) are used for the oxymercuration step; it results in the addition of a hydroxyl group to the double bond. The \(\mathrm{NaBH}_{4}\) is subsequently for the demercuration step, which removes the mercury group. The stereoselectivity of this reaction is Markovnikov's rule, and it proceeds via a syn addition. No stereochemistry change occurs at any other stereocenter. Hence, only one stereoisomer will be formed, and it is expected to remain optically active.

(2) Reaction with H2/Pt

In this reaction, hydrogen gas is added to the double bond in the presence of a platinum catalyst. It results in the syn addition of hydrogen atoms to the double bond, reducing the double bond to a single bond while maintaining the original stereochemistry. As a result, only one stereoisomer will be formed, and it will maintain optical activity.

(3) Reaction with BH3 followed by H2O2 in NaOH

This reaction involves hydroboration-oxidation, which adds a hydroxyl group to the double bond in an anti-Markovnikov fashion. The reaction proceeds via syn addition and only one stereoisomer will be formed, which should remain optically active.

(4) Reaction with Br2 in CCl4

Bromination of alkenes results in the addition of bromine atoms across the double bond, producing a dibromide product. This reaction is an anti-addition, and this produces two stereoisomers, which are enantiomers. These products will be optically active. In summary: (a) One product, optically active. (b) One product, optically active. (c) One product, optically active. (d) Two products (enantiomers), both optically active.