ORGANIC CHEMISTRY
ORGANIC CHEMISTRY
5th Edition
ISBN: 9781259977596
Author: SMITH
Publisher: MCG
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Chapter 8, Problem 8.11P
Interpretation Introduction

(a)

Interpretation: The change in the rate of an E2 reaction is to be stated when the concentration of alkyl halide is tripled.

Concept introduction: E2 type of reaction follows second order kinetics in which the rate depends on both the reactants. The rate law equation for E2 reaction is expressed as,

Rate=k[RX][B:]

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of an E2 reaction is tripled when the concentration of alkyl halide is tripled.

Explanation of Solution

E2type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base.

The rate law equation for E2 reaction is expressed as,

Rate=k[RX][B:]

According to the given statement, concentration of alkyl halide is tripled. Hence, the rate of reaction is

Rate=k[3RX][B:]=3(k[RX][B:])

Therefore, the rate of the reaction increases by three times when the concentration of [RX] is tripled.

Conclusion

The rate of an E2 reaction is tripled when the concentration of alkyl halide is tripled.

Interpretation Introduction

(b)

Interpretation: The change in the rate of an E2 reaction is to be stated when the concentration of base is halved.

Concept introduction: E2 type of reaction follows second order kinetics in which the rate depends on both the reactants. The rate law equation for E2 reaction is expressed as,

Rate=k[RX][B:]

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of an E2 reaction is halved when the concentration of base is halved.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base.

The rate law equation for E2 reaction is expressed as,

Rate=k[RX][B:].

According to the given statement, concentration of base is halved. Hence, the rate of reaction is,

Rate=k[RX][12B:]=12(k[RX][B:])

Therefore, the rate of the reaction decreases by half when the concentration of [B:] is halved.

Conclusion

The rate of an E2 reaction is halved when the concentration of base is halved.

Interpretation Introduction

(c)

Interpretation: The change in the rate of an E2 reaction is to be stated when the solvent is changed from CH3OH to DMSO

Concept introduction: E2 reactions are usually preferred in polar aprotic solvents as the transition state involves charge dispersion which is stabilized by polar aprotic solvents.

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of an E2 reaction increases when the solvent is changed from CH3OH to DMSO.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. The choice of solvent for this type of reaction is polar aprotic. DMSO is a polar aprotic solvent and CH3OH is a polar protic solvent. So changing the latter with the former increases the rate of the reaction.

Conclusion

The rate of an E2 reaction increases when the solvent is changed from CH3OH to DMSO.

Interpretation Introduction

(d)

Interpretation: The change in the rate of an E2 reaction is to be stated when the leaving is changed from I- to Br-

Concept introduction: A leaving group bigger in size is preferred as it leaves easily and fast while a smaller leaving group leaves with difficulty and slow.

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of an E2 reaction is decreased when the leaving group is changed from I- to Br-.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. Iodine is bigger in size as compared to bromine. An atom bigger in size is a good leaving group. Thus, iodine is a better leaving group then bromine. Thus, when I- is replaced by a poor leaving group Br-, the rate of elimination decreases.

Conclusion

The rate of an E2 reaction is decreased when the leaving is changed from I- to Br-.

Interpretation Introduction

(e)

Interpretation: The change in the rate of an E2 reaction is to be stated when the base is changed from OH- to H2O

Concept introduction: Strong base is more efficient in proton abstraction and hence, it is more preferred in E2 type of reactions.

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of an E2 reaction is decreased when the base is changed from OH- to H2O.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. A stronger base will facilitate the abstraction of proton and hence will increase the rate of E2 reaction. OH- is a stronger base than H2O and replacing it with H2O will decrease the rate of the reaction.

Conclusion

The rate of an E2 reaction is decreased when the base is changed from OH- to H2O.

Interpretation Introduction

(f)

Interpretation: The change in the rate of an E2 reaction is to be stated when the the alkyl halide is changed from CH3CH2Br to (CH3)2CHBr.

Concept introduction: The rate of E2 reaction depends on the halide being used. A halide in which carbon attached to the leaving group has more number of alkyl groups is usually preferred.

Expert Solution
Check Mark

Answer to Problem 8.11P

The rate of E2 reaction increases when the alkyl halide is changed from CH3CH2Br to (CH3)2CHBr

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. A highly substituted product is preferred in E2 reaction as it is more stable. This can be obtained when the carbon attached to the leaving group has more number of alkyl groups. Since (CH3)2CHBr has more number of alkyl groups, substituting it with CH3CH2Br will increase the rate of the reaction.

Conclusion

The rate of E2 reaction increases when the alkyl halide is changed from CH3CH2Br to (CH3)2CHBr

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Students have asked these similar questions
How does each of the following changes affect the rate of an E2 reaction? a. tripling [RX] b. halving [B:] c. changing the solvent from CH3OH to DMSO d. changing the leaving group from I− to Br e. changing the base from −OH to H2O f. changing the alkyl halide from CH3CH2Br to (CH3)2CHBr
Which of the following will react faster and why
There is at least two steps for each synthetic reaction

Chapter 8 Solutions

ORGANIC CHEMISTRY

Ch. 8 - Prob. 8.11PCh. 8 - Problem 8.12 What alkenes are formed from each...Ch. 8 - Prob. 8.13PCh. 8 - Problem 8.14 What alkenes are formed from each...Ch. 8 - Problem 8.15 How does each of the following...Ch. 8 - Problem 8.16 Draw both the SN1 and E1 products of...Ch. 8 - Prob. 8.17PCh. 8 - Prob. 8.18PCh. 8 - Problem 8.19 Explain why...Ch. 8 - Prob. 8.20PCh. 8 - Problem 8.21 Draw the alkynes formed when each...Ch. 8 - Problem 8.22 Draw the products in each...Ch. 8 - Problem 8.23 Draw a stepwise mechanism for the...Ch. 8 - 8.24 Rank the alkenes shown in the ball-and-stick...Ch. 8 - Prob. 8.25PCh. 8 - 8.26 What is the major E2 elimination product...Ch. 8 - Prob. 8.27PCh. 8 - Prob. 8.28PCh. 8 - Prob. 8.29PCh. 8 - 8.30 Label each pair of alkenes as constitutional...Ch. 8 - Prob. 8.31PCh. 8 - Prob. 8.32PCh. 8 - Prob. 8.33PCh. 8 - For each of the following alkenes, draw the...Ch. 8 - Prob. 8.35PCh. 8 - Prob. 8.36PCh. 8 - Prob. 8.37PCh. 8 - What alkene is the major product formed from each...Ch. 8 - Prob. 8.39PCh. 8 - Prob. 8.40PCh. 8 - Pick the reactant or solvent in each part that...Ch. 8 - 8.42 In the dehydrohalogenation of...Ch. 8 - Prob. 8.43PCh. 8 - Prob. 8.44PCh. 8 - Prob. 8.45PCh. 8 - Prob. 8.46PCh. 8 - Prob. 8.47PCh. 8 - Prob. 8.48PCh. 8 - What alkyl chloride affords the following alkene...Ch. 8 - Draw the products formed when each dihalide is...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...Ch. 8 - Prob. 8.55PCh. 8 - Draw all products, including stereoisomers, in...Ch. 8 - Draw all of the substitution and elimination...Ch. 8 - Prob. 8.58PCh. 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. 8.64PCh. 8 - 8.65 Explain the selectivity observed in the...Ch. 8 - Prob. 8.66PCh. 8 - Prob. 8.67PCh. 8 - 8.68 (a) Draw all products formed by treatment of...
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