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Chapter 9: Problem 39

Following are diastereomers (A) and (B) of 3-bromo-3,4-dimethylhexane. On treatment with sodium ethoxide in ethanol, each gives 3,4 -dimethyl-3-hexene as the major product. One diastereomer gives the \(E\) alkene, and the other gives the \(Z\) alkene. Which diastereomer gives which alkene? Account for the stereoselectivity of each \(\beta\)-elimination.

Short Answer

Expert verified
Explain the stereoselectivity of each β-elimination based on the transition states. Answer: Diastereomer A gives the E alkene, while diastereomer B gives the Z alkene as the major product. The stereoselectivity of each β-elimination is determined by the orientation of the ethoxide ion (base) and the groups to be eliminated in the transition state, with the E configuration resulting when the groups being eliminated are on opposite sides of the double bond, and the Z configuration resulting when the groups being eliminated are on the same side of the double bond.

Step by step solution

01

Draw the structures of the diastereomers

Draw the structures of diastereomers (A) and (B) of 3-bromo-3,4-dimethylhexane.
02

Draw the structures of E and Z alkenes

Draw the structures of 3,4-dimethyl-3-hexene in E and Z configurations.
03

Analyze the elimination in diastereomer A

In diastereomer A, analyze the orientation of the ethoxide ion (base) and the groups to be eliminated: the bromide ion (leaving group) and the proton on the β-carbon. Draw the transition state.
04

Determine the stereochemistry of the major product in diastereomer A

Based on the transition state, determine if the resulting alkene is in the E or Z configuration.
05

Analyze the elimination in diastereomer B

In diastereomer B, analyze the orientation of the ethoxide ion (base) and the groups to be eliminated: the bromide ion (leaving group) and the proton on the β-carbon. Draw the transition state.
06

Determine the stereochemistry of the major product in diastereomer B

Based on the transition state, determine if the resulting alkene is in the E or Z configuration.
07

Assign E and Z alkenes to corresponding diastereomers

Combine the results from steps 4 and 6 to assign the E and Z alkenes to the corresponding diastereomers A and B. Explain the stereoselectivity of each β-elimination based on the transition states.

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Most popular questions from this chapter

Predict the \(\beta\)-elimination product(s) formed when each chloroalkane is treated with sodium ethoxide in ethanol. If two or more products might be formed, predict which is the major product. (a) CC1(Cl)CCCCC1 (b) ClCC1CCCCC1 (c) CC1(C)CCC(Cl)CC1

Draw a structural formula for the product of each \(\mathrm{S}_{\mathrm{N}} 2\) reaction. Where configuration of the starting material is given, show the configuration of the product.

Alkenyl halides such as vinyl bromide, \(\mathrm{CH}_{2}=\mathrm{CHBr}\), undergo neither \(\mathrm{S}_{\mathrm{N}} 1\) nor \(\mathrm{S}_{\mathrm{N}} 2\) reactions. What factors account for this lack of reactivity?

Another important pattern in organic synthesis is the construction of \(\mathrm{C}-\mathrm{C}\) bonds. Using your roadmap as a guide, show how to convert propane into hex-1-en-4-yne. You must use propane as the source of all of the carbon atoms in the hex-1-en-4-yne product. Show all reagents needed and all molecules synthesized along the way.

Consider the following statements in reference to \(\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2, \mathrm{E} 1\), and \(\mathrm{E} 2\) reactions of haloalkanes. To which mechanism (s), if any, does each statement apply? (a) Involves a carbocation intermediate. (b) Is first order in haloalkane and first order in nucleophile. (c) Involves inversion of configuration at the site of substitution. (d) Involves retention of configuration at the site of substitution. (e) Substitution at a stereocenter gives predominantly a racemic product. (f) Is first order in haloalkane and zero order in base. (g) Is first order in haloalkane and first order in base. (h) Is greatly accelerated in protic solvents of increasing polarity. (i) Rearrangements are common. (j) Order of reactivity of haloalkanes is \(3^{\circ}>2^{\circ}>1^{\circ}\). (k) Order of reactivity of haloalkanes is methyl \(>1^{\circ}>2^{\circ}>3^{\circ}\).
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