OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Chapter 11 – Reactions of Alkyl Halides: Nucleophilic Substitutions and
Eliminations
Solutions to Problems
11.1 As described in Worked Example 11.1, identify the leaving group and the chirality
center. Draw the product carbon skeleton, inverting the configuration at the chirality
center, and replace the leaving group (bromide) with the nucleophilic reactant (acetate).
11.2 Use the suggestions in the previous problem to draw the correct product.
11.3
11.4 All of the nucleophiles in this problem are relatively reactive. See Table 11.1.
(a) CH3CH2CH2CH2Br + NaI ⎯⎯→ CH3CH2CH2CH2I + NaBr
(b) CH3CH2CH2CH2Br + KOH ⎯⎯→ CH3CH2CH2CH2OH+ KBr
(c) CH3CH2CH2CH2Br + HC≡C– Li+ ⎯⎯→ CH3CH2CH2CH2C≡CH + LiBr
(d) CH3CH2CH2CH2Br + NH3 ⎯⎯→ CH3CH2CH2CH2NH3 + Br–
11.5 (a) (CH3)2N– is more nucleophilic because it is more basic than (CH3)2NH and because
a negatively charged nucleophile is more nucleophilic than a neutral nucleophile.
(b) (CH3)3N is more nucleophilic than (CH3)3B. (CH3)3B is non-nucleophilic because it
has no lone electron pair.
(c) H2S is more nucleophilic than H2O because nucleophilicity increases in going down
a column of the periodic table.
11.6 In this problem, we are comparing two effects – the effect of the substrate and the effect
of the leaving group. Tertiary substrates are less reactive than secondary substrates,
which are less reactive than primary substrates.
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Least reactive ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Most reactive
(CH3)3CCl < (CH3)2CHCl < CH3Br < CH3OTos
tertiary secondary good excellent
carbon carbon leaving leaving
group
group
11.7
Polar protic solvents (curve 1) stabilize the charged transition state by solvation and also
stabilize the nucleophile by hydrogen bonding.
Polar aprotic solvents (curve 2) stabilize the charged transition state by solvation, but do
not hydrogen-bond to the nucleophile. Since the energy level of the nucleophile is higher,
ΔG‡ is smaller and the reaction is faster in polar aprotic solvents than in polar protic
solvents.
Nonpolar solvents (curve 3) stabilize neither the nucleophile nor the transition state. ΔG‡
is therefore higher in nonpolar solvents than in polar solvents, and the reaction rate is
slower. Benzene, ether, and chloroform are in this category.
11.8
In this SN1 reaction, attack by acetate can occur on either side of the planar, achiral
carbocation intermediate, resulting in a mixture of both the R and S enantiomeric
acetates. The ratio of enantiomers is probably close to 50:50.
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,OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
11.9 If reaction had proceeded with complete inversion, the product would have had a specific
rotation of +53.6°. If complete racemization had occurred, [α]D would have been zero. The
+5.3
observed rotation was +5.3°. Since = 0.099 , 9.9% of the original tosylate
+53.6
was inverted. The remaining 90.1% of the product must have been racemized.
11.10 The S substrate reacts with water to form a mixture of R and S alcohols. The ratio of
enantiomers is close to 50:50.
11.11 SN1 reactivity is related to carbocation stability. Thus, substrates that form the most
stable carbocations are the most reactive in SN1 reactions.
11.12
The two bromobutenes form the same allylic carbocation in the rate-limiting step.
11.13 Both substrates have allylic groups and might react either by an SN1 or an SN2 route. The
reaction mechanism is determined by the leaving group, the solvent, or the nucleophile.
(a) This reaction probably occurs by an SN1 mechanism. HCl converts the poor –OH
leaving group into an excellent –OH2+ leaving group, and the polar solvent
stabilizes the carbocation intermediate.
(b) This reaction takes place with a negatively charged nucleophile in a polar, aprotic
solvent. It is very likely that the reaction occurs by an SN2 mechanism.
11.14 Redraw linalyl diphosphate so that it has the same orientation as limonene.
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After dissociation of PPi, the cation cyclizes by attack of the double bond π electrons.
Removal of –H by base yields limonene.
11.15 Form the double bond by removing HX from the alkyl halide reactant in as many ways as
possible. The major elimination product in each case has the most substituted double
bond (Zaitsev’s rule).
(a)
(b)
(c)
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Chapter 11 – Reactions of Alkyl Halides: Nucleophilic Substitutions and
Eliminations
Solutions to Problems
11.1 As described in Worked Example 11.1, identify the leaving group and the chirality
center. Draw the product carbon skeleton, inverting the configuration at the chirality
center, and replace the leaving group (bromide) with the nucleophilic reactant (acetate).
11.2 Use the suggestions in the previous problem to draw the correct product.
11.3
11.4 All of the nucleophiles in this problem are relatively reactive. See Table 11.1.
(a) CH3CH2CH2CH2Br + NaI ⎯⎯→ CH3CH2CH2CH2I + NaBr
(b) CH3CH2CH2CH2Br + KOH ⎯⎯→ CH3CH2CH2CH2OH+ KBr
(c) CH3CH2CH2CH2Br + HC≡C– Li+ ⎯⎯→ CH3CH2CH2CH2C≡CH + LiBr
(d) CH3CH2CH2CH2Br + NH3 ⎯⎯→ CH3CH2CH2CH2NH3 + Br–
11.5 (a) (CH3)2N– is more nucleophilic because it is more basic than (CH3)2NH and because
a negatively charged nucleophile is more nucleophilic than a neutral nucleophile.
(b) (CH3)3N is more nucleophilic than (CH3)3B. (CH3)3B is non-nucleophilic because it
has no lone electron pair.
(c) H2S is more nucleophilic than H2O because nucleophilicity increases in going down
a column of the periodic table.
11.6 In this problem, we are comparing two effects – the effect of the substrate and the effect
of the leaving group. Tertiary substrates are less reactive than secondary substrates,
which are less reactive than primary substrates.
1 10/26/2023
, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Least reactive ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Most reactive
(CH3)3CCl < (CH3)2CHCl < CH3Br < CH3OTos
tertiary secondary good excellent
carbon carbon leaving leaving
group
group
11.7
Polar protic solvents (curve 1) stabilize the charged transition state by solvation and also
stabilize the nucleophile by hydrogen bonding.
Polar aprotic solvents (curve 2) stabilize the charged transition state by solvation, but do
not hydrogen-bond to the nucleophile. Since the energy level of the nucleophile is higher,
ΔG‡ is smaller and the reaction is faster in polar aprotic solvents than in polar protic
solvents.
Nonpolar solvents (curve 3) stabilize neither the nucleophile nor the transition state. ΔG‡
is therefore higher in nonpolar solvents than in polar solvents, and the reaction rate is
slower. Benzene, ether, and chloroform are in this category.
11.8
In this SN1 reaction, attack by acetate can occur on either side of the planar, achiral
carbocation intermediate, resulting in a mixture of both the R and S enantiomeric
acetates. The ratio of enantiomers is probably close to 50:50.
11/16/2023 2
,OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
11.9 If reaction had proceeded with complete inversion, the product would have had a specific
rotation of +53.6°. If complete racemization had occurred, [α]D would have been zero. The
+5.3
observed rotation was +5.3°. Since = 0.099 , 9.9% of the original tosylate
+53.6
was inverted. The remaining 90.1% of the product must have been racemized.
11.10 The S substrate reacts with water to form a mixture of R and S alcohols. The ratio of
enantiomers is close to 50:50.
11.11 SN1 reactivity is related to carbocation stability. Thus, substrates that form the most
stable carbocations are the most reactive in SN1 reactions.
11.12
The two bromobutenes form the same allylic carbocation in the rate-limiting step.
11.13 Both substrates have allylic groups and might react either by an SN1 or an SN2 route. The
reaction mechanism is determined by the leaving group, the solvent, or the nucleophile.
(a) This reaction probably occurs by an SN1 mechanism. HCl converts the poor –OH
leaving group into an excellent –OH2+ leaving group, and the polar solvent
stabilizes the carbocation intermediate.
(b) This reaction takes place with a negatively charged nucleophile in a polar, aprotic
solvent. It is very likely that the reaction occurs by an SN2 mechanism.
11.14 Redraw linalyl diphosphate so that it has the same orientation as limonene.
3 10/26/2023
, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
After dissociation of PPi, the cation cyclizes by attack of the double bond π electrons.
Removal of –H by base yields limonene.
11.15 Form the double bond by removing HX from the alkyl halide reactant in as many ways as
possible. The major elimination product in each case has the most substituted double
bond (Zaitsev’s rule).
(a)
(b)
(c)
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