Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers

Chapter 10 The Chemistry of Alcohols and Thiols

Solutions to In-Text Problems

10.1 (b)

10.2 The OH group of the alcohol is protonated in a Brønsted acid–base reaction to form the conjugate acid of the

alcohol. This loses water to form a carbocation in a Lewis acid–base dissociation reaction. Finally, in a Brønsted

acid–base reaction, water acts as a Brønsted base to remove a b-proton from the carbocation, which acts as a

Brønsted acid, to give the alkene. The formation of product A by removal of proton (a) is shown here; the formation

of products B and C occurs in an analogous manner by removal of b-protons (b) and (c), respectively.

10.5 (b) Both 3-methyl-3-pentanol and 3-methyl-2-pentanol should give 3-methyl-2-pentene as the major product.

The tertiary alcohol 3-methyl-3-pentanol should dehydrate more rapidly.

10.6 (a) Protonation of the OH group and loss of water as shown in several of the previous solutions, as well as in

Eqs. 10.3a–b on p. 437 of the text, gives a secondary carbocation. As the text discussion of Eq. 10.6 suggests,

INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 10 2

the mechanism involves a rearrangement of the initially formed secondary carbocation to a more stable

tertiary carbocation.

Loss of the two possible b-protons gives the two alkene products.

(b) Rearrangement occurs because a more strained secondary carbocation is converted into a less strained, and

therefore more stable, secondary carbocation.

10.8 (b)

10.9 This reaction involves a carbocation rearrangement. We use HBr as the acid, although, because water is generated as

a product, H3O+ could also be used.

The product in part (c) results from a carbocation rearrangement.

10.10 (b) The product is I—CH2CH2CH2—I.

(d) The compound, neopentyl alcohol, is a primary alkyl halide and cannot react by the SN1 mechanism; and it

has too many b-substituents to react by the SN2 mechanism. Consequently, there is no reaction.

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10.11 (b) (d)

10.12 (b)

10.13 (b)

Although a polar aprotic solvent would accelerate the last step, it would probably work in an alcohol solvent.

The nucleophile, CH3S –, can be generated by allowing the thiol CH3—SH to react with one equivalent of

sodium ethoxide in ethanol.

10.14 (a) Cyanide ion displaces the tosylate ester formed in the first step.

10.15 (b)

10.16 (b) (c)

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10.18 (b) Eq. 10.26 of the text shows that the nucleophilic reaction of the bromide ion on the reactive intermediate

occurs by a concerted (SN2) substitution reaction. Therefore, the reaction should occur with inversion of

stereochemistry, and the product would then be (S)-2-bromopentane. The SN2 reaction occurs at an

acceptable rate on a secondary carbon in the absence of b substituents. In addition, the reaction is fast

because the leaving group is a very weak base. However, it is possible that some SN1 mechanism could occur;

this is hard to predict. To the extent that the SN1 reaction occurs, some racemization might also take place.

10.19 (b) The simplest method for effecting the conversion shown is to treat the alcohol with thionyl chloride and

pyridine. Conversion of the alcohol to a sulfonate ester and treatment of the ester with sodium chloride in a

polar aprotic solvent would also work, but involves more steps.

(d) Because a carbocation intermediate and hence rearrangements are a distinct possibility if this alcohol is

exposed to acidic reagents such as HBr, the sulfonate ester method should be used. Thus, treat the alcohol

with tosyl chloride and pyridine, and treat the resulting tosylate with sodium bromide in a polar aprotic

solvent. This type of solvent suppresses carbocation formation. Also, PBr3 often gives satisfactory results

with unbranched secondary alcohols.

10.21 (b) The conversion of toluene into benzoic acid is a six-electron oxidation.

(c) The oxidation of a secondary alcohol to a ketone is a two-electron oxidation.

(e) The dihydroxylation of an alkene by KMnO4 is a two-electron oxidation.

(g) The addition of HBr to an alkene is neither an oxidation nor a reduction. (One carbon of the alkene is

formally oxidized and the other is reduced by the same amount.)

10.22 (e) The half-reaction of Problem 10.21, part (e):

10.23 (b) This is an oxidation–reduction reaction; the organic compound is reduced, and the –AlH4 is oxidized.

(c) This is an oxidation–reduction reaction; the alkene is oxidized, and the Br2 is reduced.

10.25 The oxidation state of each Cr in Cr2O72– is +6, and it changes to +3 in Cr3+. Therefore, six electrons are gained per

mole of dichromate. Two electrons are lost in the oxidation of ethanol to acetaldehyde. To reconcile electrons lost

and electrons gained, three molecules of ethanol are oxidized by one of dichromate; or, it takes one-third mole of

dichromate to oxidize one mole of ethanol. We leave it to you to prove this point (if necessary) by balancing the

complete reaction.

10.26 (b) On the assumption that sufficient PCC has been added, both primary alcohols are oxidized:

10.27 (b) This compound (3-pentanone) can be prepared by a PCC oxidation of the corresponding alcohol,

3-pentanol. (Aqueous dichromate could also be used.)

(d) This aldehyde can be prepared by a PCC oxidation of the corresponding alcohol.

10.29 (b) Hydrogens a and b are constitutionally equivalent and enantiotopic. (The analysis of this case is essentially

identical to the analysis of the a-hydrogens of ethanol; see Eq. 10.49, text p. 466.)

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(e) Replacing Ha and Hb in turn with a “circled H” shows that these hydrogens are constitutionally equivalent

and diastereotopic, as are Hc and Hd. Ha and Hc are constitutionally equivalent and enantiotopic, as are Hb and

Hd. Finally, Ha and Hd are constitutionally equivalent and Diastereotopic, as are Hb and Hc.

10.30 (b) Because deuterium is delivered, the a-carbon of the resulting ethanol bears two deuteriums—that is, the

product is CH3CD2OH—and it therefore has no asymmetric carbon; hence, the molecule is achiral.

10.31 (b) This is a 2-electron oxidation, because a hydrogen (which contributes –1 to the oxidation number of sulfur) is

replaced by an OH (which contributes +1).

10.33 (b) The deuterium-containing alkane can be prepared by protonolysis of a Grignard reagent in D2O; the Grignard

reagent can be prepared from an alkyl halide; and the alkyl halide can be prepared from an alcohol.

(d) The aldehyde can be prepared by oxidation of a primary alcohol; the required primary alcohol can be

prepared by hydroboration–oxidation of an alkene; and the required alkene can be prepared by an E2 reaction

of a primary alkyl halide using a branched base.

10.34 (b) The final target is to enter college. The step prior to this is to pay your tuition. The step prior to this is to

obtain the money for the tuition. The steps prior to this might be …

1. Ask your parents.

2. Get a loan.

3. Get a temporary job.

4. Win the lottery.

Each of these possibilities then suggests courses of action. For example, possibility 2 requires you to make an

appointment at the bank. Possibility 3 requires you to buy a newspaper or to look online at classified

advertising … and so on.

INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 10 6

Solutions to Additional Problems

10.36 (a) (b) (c) (d) (e)

(f) (g) (h)

10.37 (b)

10.39 (b) Fluorines a are constitutionally equivalent and diastereotopic; fluorines a are constitutionally nonequivalent

to fluorine b.

10.40 (b)

10.42 (a) This exchange occurs essentially through a series of Brønsted acid–base reactions. As shown below, once the

deuterium is incorporated into the solvent, it is significantly diluted, so that its probability of reaction with the

alkoxide is very small. In addition, such a reaction is retarded by a significant primary deuterium isotope

effect and competes less effectively with the corresponding reaction of water.

(b) To prepare CH3CH2CH2—OD from CH3CH2CH2—OH, use the same reaction with D2O/NaOD as the

solvent.

10.43 (b) This reaction is a two-electron reduction.

(d) This reaction is a two-electron oxidation.

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10.44 (b) In this case, convert the alcohol into a bromide using a method that involves an inversion of configuration.

(Either PBr3 or the two-step alcohol T tosylate T alkyl bromide sequence shown below will work.) Then, in

a second inversion step, displace the bromide with –O18—H to provide the alcohol with the desired

configuration.

10.45 (b)

(d) In this part, we have to “throw away” a carbon; ozonolysis comes to mind:

(e)

10.46 (a) The sulfonate ester serves as a leaving group in either case:

(b) The triflate anion is a weaker base than the mesylate anion because the polar effect of the fluorines stabilizes

the negative charge in the triflate anion and thereby lowers the pKa of the conjugate sulfonic acid. (See Sec.

3.6C of the text.)

(c) The principle to apply is that the better leaving group is the weaker base. This is true because the leaving

group is accepting a negative charge and breaking a covalent bond in both the Brønsted acid–base reaction

with a base and an electron-pair displacement (SN2) reaction with a nucleophile. The polar effect of the

fluorines should operate in the same way on both processes, because the processes are so similar.

10.48 The oxidation of a secondary alcohol to a ketone is a 2-electron oxidation. [See the solution to Problem 10.21(c).] In

the process, CrO3, a form of Cr(VI), is converted into Cr3+, a form of Cr(III); hence the chromium half-reaction is a

3-electron reduction. Therefore, 2/3 mole of CrO3 is required to oxidize 1 mole of the alcohol.

The molecular mass of the alcohol is 116; therefore, 10.0 g = 0.0862 mole. Consequently, (0.667)(0.0862) =

0.0575 mole of CrO3 is required for the oxidation. The molecular mass of CrO3 = 100; therefore, 5.75 g of CrO3 is

required for the oxidation.

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10.50 (a) The glycol is oxidized. This follows from the fact that a bond to carbon is replaced by a bond to oxygen at

each carbon of the glycol. The other participant in the reaction, periodate (IO4–) must therefore be reduced.

Indeed, the ionic product iodate (IO3–) contains one less oxygen bound to the iodine.

(b) The number of electrons involved in the oxidation half-reaction is determined from the oxidation numbers of

the carbons that change:

The number of electrons lost is [(+1) + (+1)] – [0 + 0] = +2. (This result could also be determined from a

balanced half-reaction.)

The iodine can be assigned an oxidation number of +7 in periodate and +5 in iodate. How do we know

this? Assign +2 to every oxygen because oxygen is divalent and presumably has two bonds to the iodine. (See

the top of text p. 456 for a similar case.) Assign a –1 for every negative charge. Hence, the reduction of

periodate is a two-electron reduction. (You can verify this with a balanced half-reaction.) Another way to

reach the same conclusion is to note that one mole of periodate is required per mole of diol. Because the diol

undergoes a two-electron oxidation, periodate must undergo a two-electron reduction.

(c) From the balanced equation, shown in the problem, 0.1 mole of periodate is required to oxidize 0.1 mole of

the diol.

10.52 The reactivity data and the molecular formula of A indicate that compound A is an alkene with one double bond. The

identity of compound D follows from the oxidation of 3-hexanol; it can only be 3-hexanone (see following

equation). It is given that 3-hexanone is an ozonolysis product of alkene A (along with H—CO2H (formic acid), not

shown in the following equation). Since alkene A has seven carbons and one double bond, and 3-hexanone has six

carbons, the carbon of alkene A not accounted for by 3-hexanone must be part of a ACH2 group. Therefore, the

identity of alkene A is established as 2-ethyl-1-pentene. The identities of compounds B and C follow from the

reactions of A.

10.54 (b) (d) (f)

(h)

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10.57 The two branches of citrate at the central carbon are enantiotopic. Hence, they are chemically distinguishable to a

chiral catalyst such as an enzyme. Evidently, the difference is such that the dehydration occurs into the unlabeled

branch, as shown in Fig. P10.57.

One difference in H2SO4 solution is that the carboxylate groups are not ionized, but this is not the key

difference. The point of the problem is that an achiral laboratory reagent will not make the distinction between

enantiotopic groups. Hence, equal amounts of dehydration should occur into each branch, and there is no reason to

expect exclusively the Z stereochemistry observed in the product of the enzyme-catalyzed reaction.

10.58 First, draw the 2S,3R stereoisomer of the product so that the stereochemistry of the addition can be deduced.

Only a malate stereoisomer with the 2S configuration will dehydrate; if the enzyme is stereospecific in one direction,

it must be stereospecific in the other. (See Eqs. 7.29a–b, text p. 300.) The pro-R hydrogen at carbon-3 comes from

the solvent; if the reaction had been run in H2O, this would be a hydrogen. This is an anti-addition; we leave it to

you to confirm this point. (See Problem 7.52(a), text p. 318.)

(b) The fumarate stereoisomer obtained in this reaction is the same as in part (a).

The same comments apply to the dynamic reversal of the reaction; the starting material, to the extent that it

remains at equilibrium, will be a mixture of the 2S,3S and 2S,3R diastereomers, and the deuterium will not

wash out.

10.61 (b) The carbocation rearrangement in this mechanism also involves a ring expansion. In this case, a tertiary

carbocation is converted into another tertiary carbocation.

(d) This is an addition to the alkene that is conceptually similar to hydration or HBr addition.

Trifluoromethanesulfonic acid (triflic acid), a strong acid, protonates the alkene double bond to give a

carbocation, which then undergoes a Lewis acid–base association reaction to give the product.

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