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Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Practice the Skill 06.01a

Using the data in the table below, predict the sign and magnitude of ΔH° for the following reaction. Identify whether the reaction is expected to be endothermic or exothermic:

Part 1

Your answer is correct.

Identify the bond(s) broken and the bond(s) formed in the reaction shown. Select all that apply.

H–Br formed (BDE 368 kJ/mol)

Br–Br broken (BDE 193 kJ/mol)

C–Br formed (BDE 285 kJ/mol)

C–H broken (BDE 381 kJ/mol)

C–H broken (BDE 397 kJ/mol)

C–Br formed (BDE 272 kJ/mol)

C–Br formed (BDE 293 kJ/mol)

C–H broken (BDE 435 kJ/mol)

Attempts: 1 of 2 used

Part 2

Correct. The total energy released by bond formation is subtracted from the total energy required to break bonds. The net sum is – 63 kJ/mol. ΔH⁰ for this reaction is negative, which means that the system is losing energy. It is giving off energy to the environment, so the reaction is exothermic.

Predict the sign and magnitude of ΔH⁰, and determine whether the reaction is expected to be endothermic or exothermic. Select all that apply.

The reaction is endothermic.

ΔH⁰ = (397 + 193) – (285 + 368) = – 63 kJ/mol

The reaction is exothermic.

ΔH⁰ = (285 + 368) – (397 + 193) = + 63 kJ/mol

ΔH⁰ = (285 – 368) + (397 – 193) = + 121 kJ/mol

ΔH⁰ = (193 – 397) + (368 – 285) = – 121 kJ/mol

Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Review of Skills - Skill Builder 06.02 Classify each of the highlighted regions below as either a nucleophilic center or an electrophilic center.

Correct. Carbon is more electronegative than lithium and withdraws electron density via induction from the lithium atom, rendering the carbon atom highly nucleophilic.

By looking for inductive effect , the highlighted region above is a nucleophilic center . Correct. Lone pairs are the regions of high electron density making the atom nucleophilic.

By looking for lone pair , the highlighted region above is a nucleophilic center . Correct. The π bond is a nucleophilic center.

By looking for pi-bond , the highlighted region above is a nucleophilic center . Correct. The carbon atom is deficient in electron density and it is an electrophilic center.

By looking for inductive effect , the highlighted region above is a electrophilic center . Correct. The central carbon atom has empty p orbital and is deficient in electron density, therefore be an electrophilic center.

By looking for empty p orbital , the highlighted region above is a electrophilic center .

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Practice the Skill 06.09 Identify all of the electrophilic centers in each of the following compounds:

Your answer is correct.

Correct. The carbon atom of the carbonyl group (C=O) is electrophilic. This is the only electrophilic center in each compound.

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Conceptual Checkpoint 06.07a-b Consider the relative energy diagrams for four different processes:

6.07a

Your answer is correct. Compare energy diagrams A and D. Assuming all other factors (such as concentrations and temperature) are identical for the two processes, identify which process will occur more rapidly. Explain.

Process D will occur more rapidly because it is exergonic, while A is endergonic.

Process D will occur more rapidly because it has a lower energy of activation.

Process A will occur more rapidly because it has a lower energy of activation.

Process A will occur more rapidly because it is exergonic, while D is endergonic.

Attempts: 1 of 2 used

6.07b Your answer is correct. Compare energy diagrams A and B. Which process will more greatly favor products at equilibrium? Explain.

Process B more greatly favors products at equilibrium because it has a lower energy of activation.

Process A more greatly favors products at equilibrium because it is exergonic.

Process B more greatly favors products at equilibrium because it is endergonic.

Process A more greatly favors products at equilibrium because it has a higher energy of activation.

Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Practice the Skill 06.11a-b For each of the following cases, read the curved arrows and identify which arrow pushing pattern is utilized:

6.11a Correct. The curved arrow indicates a hydride shift, which is a type of carbocation rearrangement.

nucleophilic attack

loss of a leaving group

rearrangement

proton transfer

Attempts: 1 of 2 used

6.11b Correct. The curved arrow indicates a nucleophilic attack. In this case, water functions as a nucleophile and attacks the carbocation.

proton transfer

rearrangement

loss of a leaving group

nucleophilic attack

Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Integrated Problem 06.42e

Correct. Consider the following reaction:

The following rate equation has been experimentally established for this process: Rate = k[HO-][CH3CH2Br] The energy diagram for this process is shown below:

Would you expect an increase in temperature to have a significant impact on the position of equilibrium (equilibrium concentrations)? Explain. The position of equilibrium is dependent on the sign and value of ΔG , which is comprised of two terms. The effect of temperature appears in (- TΔS), which is not significant because ΔS is approximately zero . Therefore, an increase (or decrease) in temperature is not expected

to have a significant impact on the position of equilibrium.

Question Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Testbank, Question 088 Your answer is correct. Identify the nucleophilic and electrophilic sites in the reactants of the following reaction.

2

3

4

5

6

1

Question Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Practice Problem 06.28b Correct. Rank the three carbocations shown in terms of increasing stability:

A

B

C

C

B

A

(least stable)

(most stable) Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Skill Building Exercise: MO basis of nucleophiles and electrophiles/ Problem 1 Correct. There are three nucleophilic centers in this compound: the lone pair on the nitrogen, the lone pairs on the oxygen, and the triple bond. Choose all of the nucleophilic centers in the following compound. Select all that apply.

1

2

3

4

5

6

7

8

9

10

11

12

Question Attempts: 1 of 2 used

Review of Concepts Answer the questions below about concepts and vocabulary from Chapter 6.

Correct. The system gives energy to the surroundings in an exothermic process and receives energy from surroundings in an endothermic process. exothermic reactions involve a transfer of energy from the system to the surroundings, while endothermic reactions involve a transfer of energy from the surroundings to the system. Correct. Bond dissociation energy is the energy required to break a covalent bond via homolytic bond cleavage.

Each type of bond has a unique bond dissociation energy, which is the amount of energy necessary to accomplish homolytic bond cleavage. Correct. Entropy is loosely defined as the disorder of a system. Entropy is loosely defined as the disorder of a system. Correct. For a process to be spontaneous, the change in Gibbs free energy must be negative. In order for a process to be spontaneous, the change in Gibbs free energy must be negative. Correct. Thermodynamics is the study of relative energy levels and equilibrium concentrations. The study of relative energy levels and equilibrium concentrations is called thermodynamics . Correct. Kinetics is the study of reaction rates. Kinetics is the study of reaction rates. Correct. Catalysts speed up the rate of reaction. catalysts speed up the rate of reaction by providing an alternate pathway with a lower energy of activation. Correct. Each peak represents a transition state and each valley represents intermediate. On an energy diagram, each peak represents a transition state , while each valley represents intermediate . Correct. Nucleophile has an electron-rich atom that is capable of donating a pair of electrons.

A nucleophile has an electron-rich atom that is capable of donating a pair of electrons. Correct. Electrophile has an electron-deficient atom that is capable of accepting a pair of electrons. An electrophile has an electron-deficient atom that is capable of accepting a pair of electrons. Correct. Nucleophilic attack, loss of leaving group, proton transfer and rearrangements are the four arrow-pushing patterns of ionic reactions. Identify the four characteristic arrow-pushing patterns of ionic reactions. (Several choices may be correct.)

Electrophilic attack

Loss of leaving group

Rearrangement

Nucleophilic attack

Addition

Proton transfer

Substitution

Electron transfer

Correct. Tertiary carbocations are more stable than secondary carbocations, which are more stable than primary carbocations. As a result of hyper conjugation, tertiary carbocations are more stable than secondary carbocations, which are more stable than primary carbocations.

Question Attempts: 1 of 2 used

Integrated Problem 06.42a Correct. Consider the following reaction:

The following rate equation has been experimentally established for this process: Rate = k[HO-][CH3CH2Br]

The energy diagram for this process is shown below:

Identify the two characteristic arrow-pushing patterns that are required for this mechanism. Arrow pushing patterns:

nucleophilic attack

proton transfer

carbocation rearrangement

loss of a leaving group

Question Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Practice Problem 06.20 Consider the following reaction:

Use the data in table below for your answer.

Correct. (a) Estimate ΔH for this reaction. Bonds broken kJ/mol Bonds formed kJ/mol RCH2—Br

285

RCH2—OR

-381

RCH2O—H

435

H—Br

-368

Sum = -29 kJ/mol

Answer *1: the tolerance is +/-2% Answer *2: the tolerance is +/-2% Answer *3: the tolerance is +/-2%

Answer *4: the tolerance is +/-2% Answer *5: the tolerance is +/-2%

Correct. (b) Is ΔS for this reaction positive or negative. Explain. ΔS of this reaction is positive because one mole of reactant is converted into two moles of product.

Correct. (c) Determine the sign of ΔG. negative

Correct. (d) Is the sign of ΔG dependent on temperature? Answer: No

Correct. (e) Is the magnitude of ΔG dependent on temperature? Answer: Yes

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved. Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Practice Problem 06.36

Correct. For the following mechanism, identify the sequence of arrow-pushing patterns:

Step A: proton transfer

Step B: nucleophilic attack

Step C: proton transfer

Step D: proton transfer

Step E: loss of leaving group

Step F: nucleophilic attack

Step G: proton transfer

Question Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Integrated Problem 06.45

Correct. Consider the following reaction. Predict whether an increase in temperature will favor reactants or products. Justify your prediction.

ΔS for this process is negative , which means that -TΔS is positive . Thus -TΔS will dominate over the ΔH at high temperatures, generating a positive value for ΔG. Therefore, the reaction will favor the reactants at high temperatures.

Question Attempts: 1 of 2 used

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Testbank, Question 088 Your answer is correct. Identify the nucleophilic and electrophilic sites in the reactants of the following reaction.

2

3

4

5

6

1

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 06

Integrated Problem 06.50

Compound 1 has been prepared and studied to investigate a novel type of intramolecular elimination mechanism (J. Org. Chem. 2007, 72, 793–798). The proposed mechanistic pathway for this transformation is presented below.

6.50a Correct. The hydroxide ion functions as a nucleophile and attacks the carbon atom of the ester group, pushing the π electrons up to the oxygen atom.

Provide curved arrows consistent with the change in bonding for the reaction of compound 1 with hydroxide.

Edit

Attempts: 1 of 2 used

6.50b Correct. The anionic oxygen atom serves as a base and removes a proton in an intramolecular fashion. The electrons in the C—H bond form a C=C π bond; the C—O σ bond breaks thus converting the adjacent C—O bond to a double bond, expelling HO¯ as a leaving group.

Provide curved arrows consistent with the change in bonding for the second step.

Edit

Attempts: 1 of 2 used

6.50c Correct. Step 3 is a proton transfer to create a resonance-stabilized anion.

Provide curved arrows consistent with the change in bonding for the third step.

Edit

Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 07

Apply the Skill 07.39 Correct. The reagent is ethoxide, which is both a strong base and a strong nucleophile. E2 elimination will predominate if the substrate is secondary or tertiary. Of the possible isomers of C4H9Cl, there is only one that is a secondary halide and one that is a tertiary halide. Compound A and compound B are constitutional isomers with molecular formula C 4H9Cl. Treatment of compound A with sodium methoxide gives trans-2-butene as the major product, while treatment of compound B with sodium methoxide gives a different disubstituted alkene as the major product.

Draw the structure of compound A.

Edit

Propose the structure compound B.

Edit

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Print by: Tori Sutton CHM2210-21Spring 0001:CHM2210-21Spring 0001 / HW: Chapter 07

Apply the Skill 07.38 Compound A and compound B are constitutional isomers with molecular formula C 3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates. Propose structures A and B.

Correct. There are only two constitutional isomers with the molecular formula C 3H7Cl. Sodium methoxide is both a strong nucleophile and a strong base. When compound A is treated with sodium methoxide, a substitution reaction predominates. Therefore, compound A must be the primary alkyl chloride. When compound B is treated with sodium methoxide, an elimination reaction predominates. Therefore, compound B must be the secondary alkyl chloride. compound A:

Edit

Your answer is correct.

compound B:

Edit

Question Attempts: 1 of 2 used

Copyright © 2000-2021 by John Wiley & Sons, Inc. or related companies. All rights reserved....