# 7.E: Solutions to Foundations of Chemical Reactivity and Mechanisms Problems


## Chemical Reaction Components

### Reactants and substrates

1. Reactant/Substrate - a substance that takes part in and is changed in the course of a chemical reaction. These are found on the left side of the arrow in a chemical reaction equation.

2.

a. Negatively charged, more nucleophilic

b. No resonance with nitrogen, more nucleophilic

c. Nitrogen is less electronegative, more nucleophilic

d. Less steric hindrance, more nucleophilic

3.

4. Steric effects: the consequences on a reaction or a molecule due to the size of atoms or groups (e.g. size; t-Bu > $$\ce{CH_3}$$ > $$\ce{H}$$ and steric effects: t-Bu > $$\ce{CH_3}$$ > $$\ce{H}$$)

Steric hindrance: the slower reaction rate or prevention of a chemical reaction due to the increase in energy from having larger/bulkier groups (e.g. $$S_\text{N}2$$ reactivity: 3$$^\text{o}$$ < 2$$^\text{o}$$ < 1$$^\text{o}$$)

5.

a.

b.

c.

6. Lactose would be considered both a reactant and a substrate. It is changed in the reaction and is the species converted by an enzyme. Water is needed as a reactant to balance the reaction.

### Reagents

7. Reagent - substance added to cause a chemical reaction to occur

Reagents are usually indicated above the reaction arrow in the chemical reaction equation.

Reactants are typically the substance that the desired transformation is being performed on, while reagents are used to achieve those transformations.

8.

a. $$\ce{NaBH_4}$$ can be used to reduce ketones and aldehydes. Lithium aluminum hydride can be used to reduce aldehydes, ketones, esters, and carboxylic acids. $$\ce{H_2}$$ and palladium can be used to reduce alkenes, aldehydes, and ketones.

b. For the first transformation, we want a reagent that will convert the ketone without affecting the ester or alkenes. For this reason we would want to use $$\ce{NaBH_4}$$ to accomplish this reduction. For the second, we want to reduce the ketone and the ester, but not the alkenes. Therefore, we will use lithium aluminum hydride.

9. The main difference between the two reactions is in the size of the base used. In the first reaction, a smaller base is used allowing for easier deprotonation to give the more substituted 2-hexene products. In the second, a more bulky base is used, lowering the yield of the 2-hexene products in favor of the 1-hexene.

10.

a. alkylating agent: transfers an alkyl group from one molecule to another (alkylation; e.g. alkyl halides can be used to alkylate aromatic substrates, as in Friedel-Crafts reactions)

b. fluorinating agent: adds fluorine to the molecule (e.g. $$\ce{HF}$$-pyridine can be used to convert alcohols into alkyl fluorides)

c. dehydrating agent: results in a loss of $$\ce{H_2O}$$ from the molecule (e.g. sulfuric acid can be used as a dehydrating agent to remove $$\ce{H_2O}$$ from sucrose)

d. oxidizing agent: oxidizes the molecule (e.g. bleach, sodium hypochlorite, can be used with a catalyst to produce an epoxide from an alkene)

e. reducing agent: reduces the molecule (e.g. lithium aluminum hydride can be used to reduce a nitrile to an amine)

11.

The reagent is ethylmagnesium bromide, and it reduced the ketone to an alcohol.

The reagent is pyridinium chlorochromate, and it oxidized the alcohol to a ketone.

12.

Prop-1-ene, but-1-ene, and pent-1-ene would each be considered a reactant and a substrate. Isobutane would be considered a reactant as well as a reagent. The new $$\ce{C-C}$$ bond formed is shown in blue, and the added isobutene unit is shown in red.

### Product

13. Product - the chemical compound that results from a chemical reaction. Products are indicated on the right side of the arrow in a chemical reaction equation.

14. Kinetic product - the product of a chemical reaction that is favored by the reaction rate.

Thermodynamic product - the product of a chemical reaction that is favored by the lower energy of the product

15.

16. A transition state is a high-energy configuration through which the reaction coordinate progresses, while an intermediate forms and then reacts as the reaction coordinate progresses. An intermediate does not have any bond forming or breaking, but is highly reactive and so will have a short lifespan (e.g. a carbocation). Neither the transition state nor the intermediate would be considered a product, as they are not the final outcome of the reaction.

17.

18.

### Reaction conditions

19. Three variables that could be considered reaction conditions are temperature, solvent used, and reaction time.

20. Conditions that could be changed are the substrate used, the nucleophile used, and the leaving group. In order to favor an $$S_\text{N}2$$ reaction, one could use a methyl or primary substrate if possible and a good, charged nucleophile.

21. Based on Le Châtelier's Principle, I would expect that additional water, a product of the reaction, would cause reactants to be favored and lower the yield. If the product is insoluble in water, however, this would change as the concentration of product will be low due to its being insoluble, causing products to be more favored.

22.

a. The thermodynamic (bottom) product is lower in energy.

b. A larger base has more difficulty deprotonating the carbon necessary to form the thermodynamic enolate, causing the kinetic enolate to be favored. This proton is more open and easily taken away by a base.

c. The kinetic enolate should be formed to give the desired product.

23. Only one condition - the variable - should be changed at a time to give the best indication of what is causing the change. The sets of reaction conditions can be compared by determining which has the best percent yield.

24. A total synthesis of penicillin V is shown below.

Reaction conditions are shown in red. These indicate the solvent(s) used, the temperature, and the reaction time. The percent yield of each reaction is also shown below the arrow, but this is a result of the reaction and not a condition. The reagents used in the reaction are indicated above the arrow.

25. Pressure is very important for reactions that involve gases (as reactants, products, or both), but is relatively unimportant for reactions involving only liquids and solids (such as the example of acetic acid and sodium hydroxide). Increasing the temperature will favor the reverse, or endothermic, reaction because the additional heat will affect the equilibrium (Le Châtelier's Principle).

26. Overall yield $$= 1.00 \times 0.83 \times 0.93 = 77\%$$

Improved yield $$= 1.00 \times 0.89 \times 0.95 = 85\%$$

The improved yields for the second and third steps result in an increase of $$9.5\%$$ for the overall yield.

### Catalyst

27. Catalyst - an additive in a reaction that makes the reaction occur more quickly; catalysts are not consumed in the reaction; catalysts serve to lower the activation energy, causing the rate of the reaction to be increased.

28.

Yes, a catalyst is present. The catalyst is the hydronium ions present from the dissociation of $$\ce{HCl}$$ in water.

29. In the case of the hydrochloric acid, product forms immediately. This is due to the fact that $$\ce{HCl}$$ is a strong acid and therefore dissociated completely in water, allowing for more catalyst to be present for the reaction. In the case of acetic acid, a weak acid, there is less catalyst present and therefore the rate of reaction is slower.

30. The presence of a catalyst has no effect on the difference in free energy of the reactants or products; it lowers the energy of activation needed to reach the transition state in a reaction.

31. A homogeneous catalyst is in the same phase as the reactants, while a heterogeneous catalyst is in a different phase than the reactants.

32. Enzymes lower the activation energy and allow the reaction to proceed more quickly, but they do not affect equilibrium. In tissues where the concentration of carbon dioxide is high, the enzyme will catalyze the reverse reaction (i.e. the reaction between carbon dioxide and water to form carbonic acid).

33. The bubbling is due to the production of oxygen gas from the reaction. Uncut skin means the cells are intact, so there is no catalase present to catalyze the reaction.

A catalase-positive bacterium would be indicated by the bubbles formed from the decomposition reaction, while catalase-negative microorganisms would not show the characteristic bubbling.

### The role of energy

34. Your values for bond dissociation energies may vary some.

a. Bonds broken: $$\ce{C-Cl}$$; $$\ce{O-H}$$ = $$339 \: \text{kJ/mol} + 498 \: \text{kJ/mol} = +837 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C-O}$$; $$\ce{H-Cl} = -381 \: \text{kJ/mol} + -431 \: \text{kJ/mol} = -812 \: \text{kJ/mol}$$

Enthalpy of reaction = $$+837 \: \text{kJ/mol} - 812 \: \text{kJ/mol} = +25 \: \text{kJ/mol}$$

Reaction is endothermic.

Entropy is approximately 0.

Not spontaneous at all temperatures.

b. Bonds broken: $$\ce{C=C}$$ (3); $$\ce{C-C} = +602$$ (3) $$+ \: 356 \: \text{kJ/mol} = +2162 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C-C}$$ (5); $$\ce{C=C} = -356$$ (5) $$+ -602 \: \text{kJ/mol} = -2382 \: \text{kJ/mol}$$

Enthalpy of reaction = $$-220 \: \text{kJ/mol}$$

Reaction is exothermic.

Entropy of reaction is negative.

Spontaneous at some (lower) temperatures.

c. Bonds broken: $$\ce{C=C}$$; $$\ce{C=O}$$; $$\ce{C-O} = +602 + 799 + 358 \: \text{kJ/mol} = +1759 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C-C}$$ (2); $$\ce{C=O}$$; $$\ce{C-O} = -356$$ (2) $$+ -799 + -358 \: \text{kJ/mol} = -1869 \: \text{kJ/mol}$$

Enthalpy of reaction = $$-110 \: \text{kJ/mol}$$

Reaction is exothermic.

Entropy is negative.

Spontaneous at some (lower) temperatures.

35.

a. Products are favored.

b. Reactants are favored.

c. Neither is favored.

d. Reactants are favored.

e. Products are favored.

36. Bonds broken: $$\ce{C=O}$$; $$\ce{O-H} = +799 + 459 \: \text{kJ/mol} = 1258 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C-O}$$ (2); $$\ce{O-H} = -358$$ (2) $$+ -459 \: \text{kJ/mol} = -1175 \: \text{kJ/mol}$$

Enthalpy of reaction = $$+83 \: \text{kJ/mol}$$

This reaction is endothermic. Therefore, the equilibrium will favor products if more heat is put into the reaction (higher temperature).

37.

a. Bonds broken: $$\ce{C-C}$$ triple bond $$= +962 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C=C}$$; $$\ce{C-H}$$ (2) $$= -682 +$$ (2) $$-427 \: \text{kJ/mol} = -1536 \: \text{kJ/mol}$$

Enthalpy = $$+962 - 1536 \: \text{kJ/mol} = -574 \: \text{kJ/mol}$$

b. Bonds broken: $$\ce{C=C} = +682 \: \text{kJ/mol}$$

Bonds formed: $$\ce{C-C}$$; $$\ce{C-H}$$ (2) $$= -356 +$$ (2) $$-410 \: \text{kJ/mol} = -1176 \: \text{kJ/mol}$$

Enthalpy = $$682 - 1176 \: \text{kJ/mol} = -494 \: \text{kJ/mol}$$

c. Palladium acts as a catalyst in this reaction, meaning that it lowers the activation energy. This combined with the fact that the reaction is exothermic, makes the reduction of the alkene very favorable with the typical hydrogenation conditions.

38. Enthalpy: total heat content of a system

Entropy: degree of disorder or randomness in a system

Gibbs free energy: enthalpy minus the product of entropy and absolute temperature

Exothermic: a process in which the net transfer of energy is to the surroundings from the system ($$\Delta H$$ is negative)

Endothermic: a process in which the net transfer of energy is from the surroundings to the system ($$\Delta H$$ is positive)

Exergonic: a process which has a negative $$\Delta G$$, and so is spontaneous

Endergonic: a process which has a positive $$\Delta G$$, and so is nonspontaneous

39. Rate $$\text{M/s}$$

First order: $$\text{s}^{-1}$$

Second order: $$\text{M}^{-1} \text{s}^{-1}$$

Third order: $$\text{M}^{-2} \text{s}^{-1}$$

40. Bonds broken: 2 $$\ce{C-H}$$ + 1 $$\ce{C-C} = \left( 2*464 \: \text{kJ/mol} \right) + \left( 1*368 \: \text{kJ/mol} \right) = 1296 \: \text{kJ/mol}$$

Bonds formed: 1 $$\ce{C=C}$$ + 1 $$\ce{H-H} = \left( 1*632 \: \text{kJ/mol} \right) + \left( 1*435 \: \text{kJ/mol} \right) = 1067 \: \text{kJ/mol}$$

$$\Delta H = 1296 \: \text{kJ/mol} - 1067 \: \text{kJ/mol} = +229 \: \text{kJ/mol}$$

$$\Delta G_{294 \: \text{K}} = 229 \: \text{kJ/mol} - 294 \: \text{K} \left( 0.120 \: \text{kJ/mol} \cdot \text{K} \right) = +193 \: \text{kJ/mol}$$, so this reaction is nonspontaneous at room temperature.

The reaction will be spontaneous when $$\Delta G < 0$$:

$$0 < 229 \: \text{kJ/mol} - T \left( 0.120 \: \text{kJ/mol} \cdot \text{K} \right)$$

$$-229 \: \text{kJ/mol} < \left( 0.120 \: \text{kJ/mol} \cdot \text{K} \right)$$ T\)

$$1908 \: \text{K} < T$$

The reaction will be spontaneous when the temperature is greater than $$1908 \: \text{K}$$ $$\left( 1635^\text{o} \text{C} \right)$$.

41.

For an exothermic reaction, the transition state will more closely resemble the reactants, while it will more closely resemble the products for an endothermic reaction.

## Basic Classes of Reactions

42.

43.

44.

a. Proton transfer

b. Nucleophilic attack

c. Proton transfer

45. 1st step = $$-124 \: \text{kJ/mol}$$; 2nd step = $$-73 \: \text{kJ/mol}$$

Energy diagram should show a difference of $$124 \: \text{kJ/mol}$$ between the reactant and the 1st product, followed by a difference of $$73 \: \text{kJ/mol}$$ between the 1st and 2nd product.

46.

a.

b. The first step is a nucleophilic attack. This is followed by two proton transfer steps.

c. The nucleophile is the amine of the second reagent. The electrophile is the carbon of the nitrile functional group of the first reagent.

47.

The possible locations of an addition reaction are shown in red. Cyclohexane and 2,4-dimethylpentane do not have a double bond, so there cannot be an addition reaction performed on these compounds.

48.

Oleic acid to stearic acid, cyclohexa-1,4-diene to cyclohexane, and buta-1,3-diene to butane are hydrogenation reactions. Ethane to ethanol is a hydration reaction. Isopentane to 2-methylbut-2-ene is an elimination reaction.

49. Markovnikov addition is the observation that in addition reactions with $$\ce{HBr}$$, the $$\ce{H}$$ will end up on the less substituted carbon while the $$\ce{Br}$$ (or other group) will be on the more substituted carbon. Anti-Markovnikov addition is where the $$\ce{H}$$ is observed to be on the more substituted carbon and the $$\ce{Br}$$ on the less substituted carbon. In these reactions, a carbocation is formed when the alkene is protonated. Since the stability of the carbocation increases with substitution, it is more likely that the more substituted carbon will be a carbocation while the less substituted carbon will have the $$\ce{H}$$. The $$\ce{Br}$$ anion will then attack the carbocation, resulting in the observed trend.

50.

No, we would not expect the anti-Markovnikov addition to have the same intermediate because it would result in a different product.

51.

The carbons in red became chiral centers, although the cyclopentene compound's product is meso. The enantiomers would need to be separated in order to obtain a single product.

### Elimination

52.

53. $$E1$$ eliminations proceed by loss of a leaving group, followed by deprotonation and formation of the alkene. $$E2$$ eliminations are concerted, meaning that the loss of the leaving group and the deprotonation occur simultaneously.

54. $$\ce{NaOH}$$ and $$\ce{NaOMe}$$ are both bases and will result in eliminations. $$\ce{NaBr}$$ and $$\ce{NaN_3}$$ do not act as bases and therefore do not result in elimination.

55.

Bonds broken: $$\ce{C-H}$$; $$\ce{C-Cl}$$; $$\ce{C-C} = +411 + 327 + 346 \: \text{kJ/mol} = 1084 \: \text{kJ/mol}$$

Bonds formed: $$\ce{O-H}$$; $$\ce{C=C} = -459 + -602 \: \text{kJ/mol} = -1061 \: \text{kJ/mol}$$

Enthalpy of reaction = $$1084 - 1061 = +23 \: \text{kJ/mol}$$

Entropy will be positive.

56.

a. The second step is the elimination step of the reaction.

b. The first reaction will not result in a condensation product. The product does not have a proton that can give the elimination step.

The second reaction will result in a condensation product. There are protons present to allow the elimination to occur.

c. The resulting alkene could then undergo a variety of addition reactions to the carbon-carbon double bond.

57. Zaitsev's rule predicts that the product of an elimination reaction will favor the more substituted alkene. Hofmann's rule predicts that the product of an elimination reaction will favor the less substituted alkene.

58.

59.

This is a dehydrohalogenation proceeding through an $$E2$$ reaction mechanism.

60. Hydride shift:

Alkyl shift:

Hydride shifts and alkyl shifts are called 1,2-hydride/alkyl shifts to denote that the group moved to an adjacent carbon. This would only be observed with the $$E1$$ reaction mechanism because there is no carbocation intermediate in the $$E2$$ mechanism.

61.

If a cis alkene is the desired product, you would be able to use the racemic mixture since the enantiomer gave the same produce. You would get the same result with an optically pure sample of the starting material, but the additional cost would not be worth it. The diastereomers would give the trans alkene.

### Substitution

62.

63.

64. For this problem, the reaction of the acid halide and acid anhydride should be more exothermic than that of the amide. Entropy for these reactions should be approximately zero for these reactions. Ultimately, the reaction with the acid halide and acid anhydride should be more likely to occur spontaneously.

65.

66.

a.

b. First reaction: Nucleophile is amine functional group. Electrophile is carbonyl carbon of acid chloride.

Second reaction: Nucleophile is oxygen of carboxylate functional group. Electrophile is alkyl iodide.

c. The amine should be more nucleophilic. Nitrogen is less electronegative than oxygen, as well as the fact that the electrons on the oxygen are distributed more by resonance, while those of the nitrogen do not participate in resonance.

67.

68. The $$S_\text{N}1$$ reaction mechanism is a stepwise mechanism where a leaving group is lost then there is a nucleophilic attack on the carbocation intermediate. Carbocation rearrangement may occur before the nucleophilic attack.

The $$S_\text{N}2$$ reaction mechanism is a concerted mechanism where the nucleophilic attack and loss of the leaving group occur simultaneously, resulting in an inversion of configuration. "Umbrella flip" is a term sometimes used to describe the inversion of stereochemistry observed in the $$S_\text{N}2$$ reaction: the "umbrella" made of $$\ce{H}$$, $$\ce{Me}$$, and $$\ce{Et}$$ in the example below "flips" to the opposite configuration, much like an umbrella may be turned inside out by strong winds.

69. The carbocation is planar, but can interact with the leaving group (anion) and solvent molecules. In very polar solvents, ions are completely separated, e.g. $$\ce{NaCl}$$ in an aqueous solution. In less polar solvents, the cation and anion may retain some interaction as a solvent-separated ion pair or a tight ion pair. This concept, known as the intimate ion pair, can explain the tendency for the nucleophile to attack the opposite side of the leaving group and the slight preference for inversion.

70.

The reactants are favored ($$\ce{SH^-}$$ is a better leaving group than $$\ce{OH^-}$$)

The products are favored ($$\ce{OH^-}$$ is a better leaving group than $$\ce{NH_2^-}$$)

The products are favored ($$\ce{SH^-}$$ is a better leaving group than $$\ce{NH_2^-}$$)

71.

### Rearrangement

72.

73.

74.

a. Yes.

b. Yes.

c. Yes.

d. No.

75.

76.

a.

b.

77. The most common hybridization of a carbocation is $$sp^2$$, with the third $$p$$ orbital vacant. Tertiary carbocations are more stable than primary carbocations due to the inductive effects of the alkyl groups (also meaning that a methenium ion, $$\ce{CH_3^+}$$, is less stable than a primary carbocation). An allylic carbocation is relatively stable due to the delocalization (resonance stability) of the positive charge, and the same is true of benzylic carbocations. Vinylic (and by extension aryl) carbocations are not stabilized by delocalization (the orbitals do not overlap) and the carbocation has $$sp$$ hybridization (more $$s$$ character is less stable).

78.

79.

The first rearrangement is a 1,2-hydride shift, but the indicated rearrangement is unfavorable because the carbocation on the left is more stable. The second rearrangement is a 1,2-alkyl or 1,2-methyl shift, and the rearrangement is favorable because the carbocation on the right is more stable.

80.

81.

### Acids and Bases

82.

83.

a.

b.

84. A Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. A carbocation would not donate nor accept an $$\ce{H^+}$$. A carbocation may accept a pair of electrons, so it can be considered a Lewis acid.

85.

86.

Yes, this reaction would involve a proton transfer from the $$\ce{HCl}$$ (Brønsted-Lowry acid) to the alkene (Brønsted-Lowry base).

No, this substitution reaction $$\left( S_\text{N}2 \right)$$ does not involve a proton transfer step, so neither compound is acting as a Brønsted-Lowry acid/base.

Yes, this elimination reaction $$\left( E2 \right)$$ would involve a proton transfer from the brominated alkane (Brønsted-Lowry acid) to the potassium hydroxide (Brønsted-Lowry base).

87.

88.

The protonated carboxylic acid is more electrophilic than the non-protonated carboxylic acid due to the presence of the positive charge. The acid acts as a catalyst, making the carboxylic acid more readily react with the alcohol to form the ester.

89.

The difference in p$$K_a$$ is about 1 unit, so the ratio will be 10:1 in favor of the reactants. Since the difference is 1 unit, the reaction is reversible.

90. The Henderson-Hasselbalch equation is:

$\text{pH} = \text{p} K_a + \text{log} \left( \frac{ \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \right)$

The p$$K_a$$ of formic acid is 3.75, so

\begin{align} 3.75 &= 3.75 + \text{log} \left( \frac{ \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \right) \\ 0 &= \text{log} \left( \frac{ \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \right) \\ 1 &= \frac{\left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \\ \left[ \ce{HA} \right] &= \left[ \ce{A^-} \right] \end{align}

The p$$K_a$$ of acetic acid is 4.75, so

\begin{align} 3.75 &= 4.75 + \text{log} \left( \frac{ \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \right) \\ -1 &= \text{log} \left( \frac{ \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \right) \\ 0.1 &= \frac{\left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \\ \left[ \ce{HA} \right] &\neq \left[ \ce{A^-} \right] \end{align}

The concentration of acetic acid and acetate would be equal when pH = p$$K_a$$ = 4.75.

91. Isoelectric point is the point at which there is no net charge on a molecule. For many acids (e.g. formic acid) and bases (e.g. ammonia) this is when there is no charge at all ($$\ce{HCOOH}$$ and $$\ce{NH_3}$$, respectively), but for some compounds - such as amino acids - this will be the zwitterion form (e.g. $$\ce{NH_3^+} \ce{-R-COO^-}$$).

The isoelectric point of arginine is:

$\text{pI} = \frac{9.00 + 12.10}{2} = 10.55$

The isoelectric point of glutamic acid is:

$\text{pI} = \frac{2.16 + 4.15}{2} = 3.16$

The isoelectric point of valine is:

$\text{pI} = \frac{2.27 + 9.52}{2} = 5.90$

### Oxidation and Reduction

92.

93. Oxidation: a reaction in which a compound undergoes an increase in oxidation state (loss of electrons)

Reduction: a reaction in which a compound undergoes a decrease in oxidation state (gain of electrons)

Oxidizing agents: PCC, $$\ce{OsO_4}$$, $$\ce{KMnO_4}$$

Reducing agents: $$\ce{NaBH_4}$$, LAH, $$\ce{H_2}/\ce{Pd}$$

94.

95.

a. The three given reactions are examples of oxidation reactions.

b.

96. In the first reaction, $$\ce{NAD^+}$$ is being reduced to $$\ce{NADH}$$, and malate is being oxidized to oxaloacetate. In the second reaction, $$\ce{FAD}$$ is being reduced to $$\ce{FADH_2}$$, and $$\ce{NADH}$$ is being oxidized to $$\ce{NAD^+}$$.

97.

Loss of hydrogen: oxidation

Dehydrogenation: oxidation

Hydrogenation: reduction

Loss of oxygen: reduction

99. The most highly oxidized state carbon can obtain is $$+4$$. The most reduced state is $$-4$$.

Most highly oxidized:

Most highly reduced:

100.

101. Glucose-6-phosphate is oxidized to 6-phosphoglucono-$$D$$-lactone, while $$\ce{NADP^+}$$ is reduced to $$\ce{NADPH}$$. $$\ce{NADP^+}$$ acts as an oxidizing agent. Glucose-6-phosphate dehydrogenase is an enzyme, and its role is to catalyze the reaction.

102. Heterolytic bond cleavage: bond breaking that results in the formation of ions. Homolytic results in radical formation.

103. Initiation, propagation, termination. Initiation involves the formation of radical species. In the propagation step, the net chemical reaction takes place. In termination, two radicals join together to give a compound with no unshared electrons.

104.

105.

106.

107.

The order of stability for radicals and carbocations are the same.

108. Coupling results in the termination of the radicals.

$\ce{Br} \cdot + \ce{Br} \cdot \longrightarrow \ce{Br-Br}$

$\ce{CH_3} \cdot + \ce{Br} \cdot \longrightarrow \ce{Br-CH_3}$

$\ce{CH_3} \cdot + \ce{CH_3} \cdot \longrightarrow \ce{H_3C-CH_3}$

109. Chain reaction: chemical reaction in which the mechanism involves at least one step where an intermediate/product (e.g. radical species) may continue the reaction until the reactants are used up or the chain is terminated.

This could be useful in some scenarios, such as energy production in nuclear reactors (nuclear fission chain reaction), but can also be harmful, such as the depletion of the ozone layer by chlorine and bromine free radicals from chlorofluorocarbons and bromofluorocarbons, respectively.

110.

This is the weakest $$\ce{C-H}$$ bond, and the radical is stabilized by inductive effects as well as resonance.

111.

The homolytic cleavage of the $$\ce{C-Cl}$$ bond requires energy (light, $$h \nu$$). Antarctica is dark during their winter (Northern Hemisphere summer), and starts to receive more light as it turns to their summer (Northern Hemisphere winter).

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