Q69P When 2-bromo-3-phenylbutane is t... [FREE SOLUTION]

Chapter 7: Q69P (page 393)

When 2-bromo-3-phenylbutane is treated with sodium methoxide, two alkenes result (by E2 elimination). The Zaitsev product predominates.
Draw the reaction, showing the major and minor products.
When one pure stereoisomer of 2-bromo-3-phenylbutane reacts, one pure stereoisomer of the major product results. For example, when (2R,3R)-2-bromo-3-phenylbutane reacts, the product is the stereoisomer with the methyl group cis. Use your models to draw a Newmanprojection of the transition state to show why this stereospecificity is observed.
Use a Newman projection of the transition state to predict the major product of elimination of (2S,3R)-2-bromo-3-phenylbutane.
Predict the major product from elimination of (2S,3S)-2-bromo-3-phenylbutane. This prediction can be made without drawing any structures, by considering the results in part (b).

Short Answer

Expert verified

d) 2S,3S is the mirror image of 2R,3Rand it would give a mirror image of the alkene produced by 2R,3R with two methyl groups which arecis. The alkene product is planar and not chiral, so its mirror image is the same, i.e., 2S,3S, and 2R,3R would give the same alkene.

Step by step solution

Explanation of part (a)

E2 elimination is a one-step mechanism and carbon-hydrogen, and carbon-halogen bonds mostly break off to form a new double bond. The base used in this reaction has a huge influence on the mechanism and the reaction rate depends on both the substrate and the base involved. The beta-hydrogen and leaving group must be anti-coplanar with each other.

When 2-bromo-3-phenylbutane undergoes E2 elimination, then the Zaitsev product is a major product. Both phenyl and bromine are anti-coplanar with each other which is the required condition for E2 elimination to occur.

Formation of major and minor products via E2 elimination

Explanation of part (b)

When (2R,3R)-2-bromo-3-phenylbutane reacts, the product is stereoisomer with the methyl group cis. The hydrogen and bromine atoms are anti-coplanar in the transition state and that gives rise to the cis product in which both methyl groups are cis to each other. On drawing the Newman projection of the given alkene, placing hydrogen and bromine anti to each other, and when leaving group eliminates, the double bond is formed. Converting Newman to wedged-dashed form, both methyl groups are cis to each other.

Formation of cis alkene

Explanation of part (c)

When (2S,3R)-2-bromo-3-phenylbutane undergoes E2 elimination, trans product forms. In the Newman projection, hydrogen and bromine are anti to each other and on converting the Newman form to wedged-dashed form, both methyl groups are trans to each other.

Formation of trans alkene

Explanation of part (d)

The 2S,3S is the mirror image of 2R,3Rand it would give a mirror image of the alkene produced by 2R,3R with two methyl groups which arecis. The alkene product is planar and not chiral, so its mirror image is the same, i.e., 2S, 3S, and 2R,3R would give the same alkene.