KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only
KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only
4th Edition
ISBN: 9781308429878
Author: SMITH
Publisher: MCG
Question
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Chapter 8, Problem 8.57P
Interpretation Introduction

(a)

Interpretation:

To draw the mixture of enantiomers A and B obtained from cis and Trans isomers of 1-bromo-4-tert-butylcyclohexane with KOC (CH3)3.

Concept introduction:

The most common mechanism for dehydrohalogenation is the E2 mechanism. An E2 reaction exhibits a second-order kinetics, in which the reaction is bimolecular and both the alkyl halide and the base appear in the rate equation. The most straightforward explanation for the second-order kinetics is a concerted reaction: all bonds are broken and formed in a single step. In other words simultaneously all bonds are broken and formed.

The transition state of an E2 reaction consists of four atoms from the alkyl halide-one hydrogen atom, two carbon atoms, and the leaving group (X)-all aligned in the same plane. There are two ways for the C-H and C-x bonds to be coplanar. When the H and X atoms oriented on the same side of the molecule. This geometry is called syn periplanar. While the H and X atoms oriented on opposite sides of the molecule. This geometry is called anti periplanar. The dihedral angle for the C-H and C-X bonds equals 0° for the syn periplanar arrangement and 180° for the anti periplanar arrangement.

Whereas the E1 reaction is a twostep reaction of first order kinetics. The two steps reaction is the bond to the leaving group breaks first before the π bond is formed.

Expert Solution
Check Mark

Answer to Problem 8.57P

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  1

Explanation of Solution

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  2

The elimination of the Br from the reactant using base to form the respective alkene.

Conclusion

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  3

Interpretation Introduction

(b)

Interpretation:

To determine the isomer that reacts faster with KOC(CH3)3.

Concept introduction:

The most common mechanism for dehydrohalogenation is the E2 mechanism. An E2 reaction exhibits a second-order kinetics, in which the reaction is bimolecular and both the alkyl halide and the base appear in the rate equation. The most straightforward explanation for the second-order kinetics is a concerted reaction: all bonds are broken and formed in a single step. In other words simultaneously all bonds are broken and formed.

The transition state of an E2 reaction consists of four atoms from the alkyl halide-one hydrogen atom, two carbon atoms, and the leaving group (X)-all aligned in the same plane. There are two ways for the C-H and C-x bonds to be coplanar. When the H and X atoms oriented on the same side of the molecule. This geometry is called syn periplanar. While the H and X atoms oriented on opposite sides of the molecule. This geometry is called anti periplanar. The dihedral angle for the C-H and C-X bonds equals 0° for the syn periplanar arrangement and 180° for the anti periplanar arrangement.

Whereas the E1 reaction is a twostep reaction of first order kinetics. The two steps reaction is the bond to the leaving group breaks first before the π bond is formed.

Expert Solution
Check Mark

Answer to Problem 8.57P

The bulky tert-butyl group in the cyclohexane ring is in the equatorial position.

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  4

In the cis-isomer, the Br is in the axial position. So cis-isomer is faster than the trans-isomer.

Explanation of Solution

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  5

The bulky tert-butyl group in the cyclohexane ring is in the equatorial position. The cis isomer has the Br in the axial position while in the trans isomer the Br is in the equatorial position. For the dehydrogenation to occur, the halogen must be in the axial position to afford trans diaxial elimination of H and X. Thus the cis isomer react faster than trans where the Br lies in the equatorial position.

Conclusion

Thus cis-isomer is faster than the trans-isomer due to axial position of Br in the structure.

Interpretation Introduction

(c)

Interpretation:

To draw the structures for C and D from cis and trans-1-bromo-4-tert-butylcyclohexane with NaOCH3.

Concept introduction:

The most common mechanism for dehydrohalogenation is the E2 mechanism. An E2 reaction exhibits a second-order kinetics, in which the reaction is bimolecular and both the alkyl halide and the base appear in the rate equation. The most straightforward explanation for the second-order kinetics is a concerted reaction: all bonds are broken and formed in a single step. In other words simultaneously all bonds are broken and formed.

The transition state of an E2 reaction consists of four atoms from the alkyl halide-one hydrogen atom, two carbon atoms, and the leaving group (X)-all aligned in the same plane. There are two ways for the C-H and C-x bonds to be coplanar. When the H and X atoms oriented on the same side of the molecule. This geometry is called syn periplanar. While the H and X atoms oriented on opposite sides of the molecule. This geometry is called anti periplanar. The dihedral angle for the C-H and C-X bonds equals 0° for the syn periplanar arrangement and 180° for the anti periplanar arrangement.

Whereas the E1 reaction is a twostep reaction of first order kinetics. The two steps reaction is the bond to the leaving group breaks first before the π bond is formed.

Expert Solution
Check Mark

Answer to Problem 8.57P

cis -1-bromo-4-tert-butylcyclohexane with NaOCH3 forms the product

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  6

trans-1-bromo-4-tert-butylcyclohexane with NaOCH3 forms the product

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  7

Explanation of Solution

The major products C and D formed when cis and trans-1-bromo-4-tert-butylcyclohexane reacts with NaOCH3 is shown. The respective reaction takes place by SN2 reaction mechanism that is by backside attack of the nucleophile.

.

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  8

Conclusion

cis -1-bromo-4-tert-butylcyclohexane with NaOCH3 forms the product.

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  9

trans-1-bromo-4-tert-butylcyclohexane with NaOCH3 forms the product.

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  10

Interpretation Introduction

(d)

Interpretation:

To determine the isomer that reacts faster with NaOCH3.

Concept introduction:

The most common mechanism for dehydrohalogenation is the E2 mechanism. An E2 reaction exhibits a second-order kinetics, in which the reaction is bimolecular and both the alkyl halide and the base appear in the rate equation. The most straightforward explanation for the second-order kinetics is a concerted reaction: all bonds are broken and formed in a single step. In other words simultaneously all bonds are broken and formed.

The transition state of an E2 reaction consists of four atoms from the alkyl halide-one hydrogen atom, two carbon atoms, and the leaving group (X)-all aligned in the same plane. There are two ways for the C-H and C-x bonds to be coplanar. When the H and X atoms oriented on the same side of the molecule. This geometry is called syn periplanar. While the H and X atoms oriented on opposite sides of the molecule. This geometry is called anti periplanar. The dihedral angle for the C-H and C-X bonds equals 0° for the syn periplanar arrangement and 180° for the anti periplanar arrangement.

Whereas the E1 reaction is a twostep reaction of first order kinetics. The two steps reaction is the bond to the leaving group breaks first before the π bond is formed.

Expert Solution
Check Mark

Answer to Problem 8.57P

The cis -1-bromo-4-tert-butylcyclohexane reacts faster when compared to trans-1-bromo-4-tert-butylcyclohexane.

cis -1-bromo-4-tert-butylcyclohexane with NaOCH3.

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  11

trans-1-bromo-4-tert-butylcyclohexane with NaOCH3

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  12

Explanation of Solution

NaOCH3, is a strong nucleophile. The reaction follows the SN2 in which the backside attack occurs. With the cis isomer the nucleophile can attack by the equatorial position to avoid the 1, 3-diaxial interaction. Moreover only the cis isomer has the Br and Β carbon in the anti periplanar that is in trans diaxial arrangement when compared to trans isomer. Thus the cis isomer reacts faster than trans isomer.

cis -1-bromo-4-tert-butylcyclohexane with NaOCH3

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  13

trans-1-bromo-4-tert-butylcyclohexane with NaOCH3

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only, Chapter 8, Problem 8.57P , additional homework tip  14

Conclusion

Thus the cis -1-bromo-4-tert-butylcyclohexane reacts faster when compared to trans-1-bromo-4-tert-butylcyclohexane

Interpretation Introduction

(e)

Interpretation:

To explain why there is different products formed when the alkyl halides uses two different alkoxides.

Concept introduction:

The most common mechanism for dehydrohalogenation is the E2 mechanism. An E2 reaction exhibits a second-order kinetics, in which the reaction is bimolecular and both the alkyl halide and the base appear in the rate equation. The most straightforward explanation for the second-order kinetics is a concerted reaction: all bonds are broken and formed in a single step. In other words simultaneously all bonds are broken and formed.

The transition state of an E2 reaction consists of four atoms from the alkyl halide-one hydrogen atom, two carbon atoms, and the leaving group (X)-all aligned in the same plane. There are two ways for the C-H and C-x bonds to be coplanar. When the H and X atoms oriented on the same side of the molecule. This geometry is called syn periplanar. While the H and X atoms oriented on opposite sides of the molecule. This geometry is called anti periplanar. The dihedral angle for the C-H and C-X bonds equals 0° for the syn periplanar arrangement and 180° for the anti periplanar arrangement.

Whereas the E1 reaction is a twostep reaction of first order kinetics. The two steps reaction is the bond to the leaving group breaks first before the π bond is formed.

Expert Solution
Check Mark

Answer to Problem 8.57P

The bulky base OC(CH3)3 favours the elimination by an E2 reaction mechanism, to get a mixture of enantiomers A and B. The strong nucleophile OCH3 favours nucleophile substitution by an SN2 mechanism. Inversion of configuration results from backside attack of the nucleophile. Thus different product are formed.

Explanation of Solution

-OCH3, is a strong nucleophile. The reaction follows the SN2 in which the backside attack occurs. With the cis isomer the nucleophile can attack by the equatorial position to avoid the 1, 3-diaxial interaction. Moreover only the cis isomer has the Br and Β carbon in the anti periplanar that is in trans diaxial arrangement when compared to trans isomer. Thus the cis isomer reacts faster than trans isomer. The bulky base OC(CH3)3 favours the elimination by an E2 reaction mechanism, to get a mixture of enantiomers A and B.

Conclusion

Thus different product are formed when the alkyl halides uses two different alkoxides are used.

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Chapter 8 Solutions

KCTCS Organic Chemistry Value Edition (Looseleaf) - Text Only

Ch. 8 - Prob. 8.11PCh. 8 - What alkenes are formed from each alkyl halide by...Ch. 8 - Prob. 8.13PCh. 8 - Prob. 8.14PCh. 8 - Problem 8.15 How does each of the following...Ch. 8 - Prob. 8.16PCh. 8 - Prob. 8.17PCh. 8 - Prob. 8.18PCh. 8 - Prob. 8.19PCh. 8 - Problem 8.19 Explain why...Ch. 8 - Prob. 8.21PCh. 8 - Draw the alkynes formed when each dihalide is...Ch. 8 - Prob. 8.23PCh. 8 - Draw a stepwise mechanism for the following...Ch. 8 - 8.24 Rank the alkenes shown in the ball-and-stick...Ch. 8 - Prob. 8.26PCh. 8 - 8.26 What is the major E2 elimination product...Ch. 8 - Prob. 8.28PCh. 8 - Prob. 8.29PCh. 8 - Prob. 8.30PCh. 8 - Prob. 8.31PCh. 8 - PGF2 is a prostaglandin, a group of compounds that...Ch. 8 - Prob. 8.33PCh. 8 - Prob. 8.34PCh. 8 - Prob. 8.35PCh. 8 - Prob. 8.36PCh. 8 - Prob. 8.37PCh. 8 - Prob. 8.38PCh. 8 - Prob. 8.39PCh. 8 - Prob. 8.40PCh. 8 - What alkene is the major product formed from each...Ch. 8 - Prob. 8.42PCh. 8 - Prob. 8.43PCh. 8 - Prob. 8.44PCh. 8 - 8.42 In the dehydrohalogenation of...Ch. 8 - Prob. 8.46PCh. 8 - Prob. 8.47PCh. 8 - Prob. 8.48PCh. 8 - Prob. 8.49PCh. 8 - Prob. 8.50PCh. 8 - Draw the products of each reaction. a.c. b.d.Ch. 8 - Draw the structure of a dihalide that could be...Ch. 8 - Under certain reaction conditions, 2,...Ch. 8 - For which reaction mechanisms, SN1, SN2, E1 or...Ch. 8 - Draw the organic products formed in each reaction....Ch. 8 - Prob. 8.56PCh. 8 - Prob. 8.57PCh. 8 - Draw all products, including stereoisomers, in...Ch. 8 - Draw all of the substitution and elimination...Ch. 8 - Prob. 8.60PCh. 8 - 8.59 Draw a stepwise, detailed mechanism for each...Ch. 8 - Draw the major product formed when...Ch. 8 - Draw a stepwise, detailed mechanism for the...Ch. 8 - Explain why the reaction of with gives ...Ch. 8 - Draw a stepwise detailed mechanism that...Ch. 8 - Prob. 8.66PCh. 8 - 8.65 Explain the selectivity observed in the...Ch. 8 - Prob. 8.68PCh. 8 - Although dehydrohalogenation occurs with anti...Ch. 8 - 8.68 (a) Draw all products formed by treatment of...
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