Question

1)Characterize the following alkene as having the E or Z configuration. Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any). Use wedge-and-dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one.

2)

Alkenes can be converted to alcohols by hydroboration–oxidation. Draw the structure of the alcohol(s) formed in the following reaction sequence. If applicable, draw hydrogen at a chirality center and use wedge-and-dash bonds to designate the stereochemistry.

3) Then, select all that apply about the product(s) for the each of the following 2 reactions: R,R ; R,S (or S,R) ; S,S ; achiral ; racemic ; diasteromers ; R ; S

Answer 1

(1) 2,3-Dimethyl-2-pentene does not have any configuration (E or Z-configurations) because in this compound, one of the double bonded carbon have two identical (methyl) substituents. 2,3-Dimethyl-2-pentene reacts with bromine to form 2,3-dibromo-2,3-dimethylpentane.

This compound has only one chiral carbon. So, it gives only 2 stereoisomers. They are enantiomers.

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(2) 1-Methylcyclohex-1-ene undergoes hydroboration-oxidation reaction to form less substituted alcohol, 2-methylcyclohexanol. This reaction is favorable for syn-addition. Here, the possible stereoisomers 2. The configurations of 2 chiral centers are opposite in 2 compounds. Therefore, theses are enantiomers.

Answer 2

1) This characterization is based solely on the molecular mass of the groups attached to the alkene. Since the alkene contains both -CH3 groups on one side of the double bond and the other side of the bond has one -CH3 and other -CH2-CH3 groups. Since it cannot be determined from the structure whether its Z or E conformer therefore the answer is that the conformer cannot be determined.

Structure of the product is

2)Hydroboration oxidation will give alcohol at the position of less substitution and the stereochemistry will be such that the Hydride and alcohol attached will be on the opposite ends so the the product will be like.