Organic Chemistry
Organic Chemistry
5th Edition
ISBN: 9780078021558
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 8, Problem 8.53P

For which reaction mechanisms, S N 1 , S N 2 , E1 or E2 -are each of the following statements true? A statement may be true for one or more mechanisms.

a. The mechanism involves carbocation intermediates.

b. The mechanism has two steps.

c. The reaction rate increases with better leaving groups.

d. The reaction rate increases when the solvent is changed from CH 3 OH to ( CH 3 ) 2 SO .

e. The reaction rate depends on the concentration of the alkyl halide only.

f. The mechanism is concerted.

g. The reaction of CH 3 CH 2 Br with NaOH occurs by this mechanism.

h. Racemization of a stereogenic center occurs.

i. Tertiary ( 3 o ) alkyl halides react faster than 2 o or 1 o alkyl halides.

j. The reaction follows a second-order rate equation.

Expert Solution
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Interpretation Introduction

(a)

Interpretation: The validation of the given statement is to be stated according to the mechanism involved in the reaction.

Concept introduction: The two-step unimolecular elimination reaction that favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. In the second step of the reaction, the carbocation forms a double bond. This type of reaction is termed E1 elimination reaction.

The two-step unimolecular reaction which favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. Then, in the second step, the carbocation undergoes substitution. This type of reaction is termed as SN1 reaction.

Answer to Problem 8.53P

The given statement is true for SN1 and E1 reaction mechanisms.

Explanation of Solution

The two-steps reactions are SN1 and E1 in which the first step is the formation of carbocation intermediate. In the second step, if the base is strong, negatively charged, then elimination takes place and if the base is weak, then substitution takes place.

Organic Chemistry, Chapter 8, Problem 8.53P

Figure 1

Conclusion

SN1 and E1 reactions mechanism involves carbocation intermediates.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation: The reaction in which the mechanism has two steps is to be stated.

Concept introduction: The two-step unimolecular elimination reaction that favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. In the second step of the reaction, the carbocation forms a double bond. This type of reaction is termed E1 elimination reaction.

The two-step unimolecular reaction which favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. Then, in the second step, the carbocation undergoes substitution. This type of reaction is termed as SN1 reaction.

Answer to Problem 8.53P

SN1 and E1 reactions have two steps.

Explanation of Solution

The two-steps reactions are SN1 and E1 in which the first step is the formation of carbocation intermediate. In the second step if the base is strong, negatively charged than elimination takes place and if the base is weak then substitution takes place. If the given base is a good nucleophile then it can undergo nucleophilic substitution reaction or if a strong base then elimination reaction.

Conclusion

SN1 and E1 reactions have two steps.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation: The reactions for which the reaction rate increases with better leaving groups is to be stated.

Concept introduction: A good leaving groups is preferred by all substitution and elimination reactions as they increase the rate of the reaction by lowering the transition state energy.

Answer to Problem 8.53P

The rate of SN1, SN2, E1 or E2 reactions increases with better leaving groups.

Explanation of Solution

The more the bigger the atom the better it will be as a leaving group and more efficiently it will leave the group to which they are attached and hence increasing the reaction rate. They do so by lowering the transition state energy.

Conclusion

The rate of SN1, SN2, E1 or E2 reactions increases with better leaving groups.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation: The reactions for which the reaction rate increases when the solvent is changed from CH3OH to (CH3)2SO.is to be stated.

Concept introduction: The choice of the solvent being used determines the mechanism being followed.

Answer to Problem 8.53P

The rate of SN2 and E2 reactions increases when the solvent is changed from CH3OH to (CH3)2SO.

Explanation of Solution

CH3OH is a polar protic solvent while (CH3)2SO is a polar aprotic solvent. The choice of solvent for SN2 and E2 reactions is a polar aprotic solvent as they stabilize the transition state more efficiently than polar protic solvents.

Conclusion

The rate of SN2 and E2 reactions rate increases when the solvent is changed from CH3OH to (CH3)2SO.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation: The reactions for which the reaction rate depends on the concentration of the alkyl halide only is to be stated.

Concept introduction: The rate law for SN1 and E1 reactions is expressed as,

Rate=k[R-X]

The rate law depends only on the concentration of alkyl halide.

Answer to Problem 8.53P

For SN1 and E1 reactions, the reaction rate depends on the concentration of the alkyl halide only.

Explanation of Solution

The two step reactions are SN1 and E1 in which the first step is the formation of carbocation intermediate. The rate law for both of these reactions is expressed as,

Rate=k[R-X]

The rate of both the reactions depends only on the concentration of alkyl halide and is independent of the base concentration.

Conclusion

For SN1 and E1 reactions the reaction rate depends on the concentration of the alkyl halide only.

Expert Solution
Check Mark
Interpretation Introduction

(f)

Interpretation: The reaction for which the mechanism is concerted is to be stated.

Concept introduction: The one-step bimolecular elimination reaction that favors the removal of a proton from carbon adjacent to the leaving group by a base that results in the formation of a carbocation. Then, the formation of a double bond takes place simultaneously. This type of reaction is termed E2 elimination reaction.

The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as SN2 reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 8.53P

SN2 and E2 reactions follow concerted mechanism.

Explanation of Solution

In a concerted mechanism, all the bond breaking and bond making takes place simultaneously in a single step. In SN2 reaction, the cleavage of the carbon-leaving group bond and formation of carbon-nucleophile bond takes place in a single step. In E2 reaction, the cleavage of the carbon-leaving group bond and formation of the double bond takes place in a single step. Hence, both of these reactions follow concerted mechanism.

Conclusion

SN2 and E2 reactions follow concerted mechanism.

Expert Solution
Check Mark
Interpretation Introduction

(g)

Interpretation: The mechanism followed by the reaction of CH3CH2Br with NaOH is to be stated.

Concept introduction: Substitution and elimination depends on the base being employed.

Answer to Problem 8.53P

SN2 and E2 reaction mechanism is followed by the reaction of CH3CH2Br with NaOH.

Explanation of Solution

In the given question the alkyl halide is primary and the base is a strong base as well as a strong nucleophile. So, substitution as well as elimination is possible. Since, SN2 reactions favors primary alkyl halide with a strong nucleophile substitution takes place through SN2 mechanism. Since the alkyl halide is primary, E2 elimination pathway will be followed for elimination reaction by a strong base.

Conclusion

SN2 and E2 reaction mechanism is followed by the reaction of CH3CH2Br with NaOH.

Expert Solution
Check Mark
Interpretation Introduction

(h)

Interpretation: The reaction for which racemization of a stereogenic center occurs is to be stated.

Concept introduction: Racemization is possible where a planar carbocation is the intermediate.

Answer to Problem 8.53P

For SN1 reaction, racemization of a stereogenic center occurs.

Explanation of Solution

Racemization can take place where a planar carbocation intermediate is formed and the incoming nucleophile approaches it from both the sides. In a SN1 reaction, a planar carbocation is formed as the intermediate and attack of the nucleophile is possible from both the sides leading to a mixture of enantiomers being formed.

Conclusion

For SN1 reaction, racemization of a stereogenic center occurs.

Expert Solution
Check Mark
Interpretation Introduction

(i)

Interpretation: The reactions for which tertiary (3o) alkyl halides react faster than 2o or 1o alkyl halides is to be stated.

Concept introduction: A tertiary carbocation is more stable then secondary followed by a primary carbocation.

Answer to Problem 8.53P

For SN1 E1 and E2 reactions tertiary (3o) alkyl halides react faster than 2o or 1o alkyl halides.

Explanation of Solution

In SN1 and E1 reaction, the first step is the formation of carbocation and follows the order tertiary>secondary>primary. In an E2 reaction, two sp3 carbons are transferred to sp2 carbon atoms which leads to substituents moving further apart decreasing any steric interactions. So, substituted compounds undergo E2 reactions rapidly.

Conclusion

For SN1, E2 and E1 reactions tertiary (3o) alkyl halides react faster than 2o or 1o alkyl halides.

Expert Solution
Check Mark
Interpretation Introduction

(j)

Interpretation: The reaction which follows a second-order rate equation is to be stated.

Concept introduction: Second order reaction is one in which the reaction rate depends on the concentration of both the reacting species.

Answer to Problem 8.53P

SN2 and E2 reactions follow the second-order rate equation.

Explanation of Solution

In SN2 and E2 reactions, the reaction rate depends on both the alkyl halide and base used. The rate law is expressed as Rate=k[R-X][B]. Second order reaction is one in which the reaction rate depends on the concentration of both the reacting species.

Conclusion

SN2 and E2 reactions follow second-order rate equation.

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Students have asked these similar questions
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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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