solomons organic chemistry solution manual - chapter 3

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JWCL234-03 JWCL234-Solomons-v1 December 8, 2009 21:29

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3AN INTRODUCTION TO ORGANIC REACTIONS

AND THEIR MECHANISMS: ACIDS AND BASES

SOLUTIONS TO PROBLEMS

3.1 CH3 O F

F

F

H B+ F

F

CH3 O

H

B

+

−

F

F

F

BCH3 CH3O F+ F

F

CH3

CH3

O B

+

−

F

Cl

Cl

CH3 Cl Al

+

−

ClCl

Cl

CH3 Cl ClAl+

(a)

(b)

(c)

3.2 (a) Lewis base

(b) Lewis acid

(c) Lewis base

(d) Lewis base

(e) Lewis acid

(f) Lewis base

3.3 F

F

Lewis base Lewis acid

CH3

CH3

N

H

CH3

CH3

N B+

+ −

FH

F

F

F

B

3.4 (a) Ka =[H3O+][HCO −

2 ]

[HCO2H]= 1.77 × 10−4

Let x = [H3O+] = [HCO −

2 ] at equilibrium

then, 0.1 − x = [HCO2H] at equilibrium

but, since the Ka is very small, x will be very small and 0.1 − x ' 0.1

35



CONFIRMING PAGES

36 AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS

Therefore,

(x)(x)

0.1= 1.77 × 10−4

x2= 1.77 × 10−5

x = 0.0042 = [H3O+] = [HCO −

2 ]

(b) % Ionized =[H3O+]

0.1× 100 or

[HCO −

2 ]

0.1× 100

=.0042

0.1× 100 = 4.2%

3.5 (a) pKa = − log 10−7= −(−7) = 7

(b) pKa = − log 5.0 = −0.669

(c) Since the acid with a Ka = 5 has a larger Ka, it is the stronger acid.

3.6 When H3O+ acts as an acid in aqueous solution, the equation is

H3O++ H2O ® H2O + H3O+

and Ka is

Ka =[H2O][H3O+]

[H3O+]= [H2O]

The molar concentration of H2O in pure H2O, that is [H2O] = 55.5; therefore, Ka = 55.5

The pKa is

pKa = − log 55.5 = −1.74

3.7(a)

(d)

(c) N

H

O−

O−

−(b)

HO

O

3.8 The pKa of the methylaminium ion is equal to 10.6 (Section 3.6C). Since the pKa of the

anilinium ion is equal to 4.6, the anilinium ion is a stronger acid than the methylaminium

ion, and aniline (C6H5NH2) is a weaker base than methylamine (CH3NH2).

3.9 + NH2−

+ NH3−

3.10 CR

H

O

O

CR+ +Na+ Na+

OHO

−

O

O−

CHO OH

C

O O



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ACIDS AND BASES 37

3.11 (a) Negative. Because the atoms are constrained to one molecule in the product, they have

to become more ordered.

(b) Approximately zero.

(c) Positive. Because the atoms are in two separate product molecules, they become more

disordered.

3.12 (a) If Keq = 1

then,

log Keq = 0 =−1G◦

2.303RT1G◦

= 0

(b) If Keq = 10

then,

log Keq = 1 =−1G◦

2.303RT

1G◦= −(2.303)(0.008314 kJ mol−1 K−1)(298 K) = −5.71 kJ mol−1

(c) 1G◦= 1H ◦

− T 1S◦

1G◦= 1H ◦

= −5.71 kJ mol−1 if 1S◦= 0

3.13 Structures A and B make equal contributions to the overall hybrid. This means that the

carbon-oxygen bonds should be the same length and that the oxygens should bear equal

negative charges.

A B

CCH3

O

O −

CH3 C

O

O

−

hybrid

CCH3

O

O

δ−

δ−

3.14 (a) CHCl2CO2H would be the stronger acid because the electron-withdrawing inductive

effect of two chlorine atoms would make its hydroxyl proton more positive. The electron-

withdrawing effect of the two chlorine atoms would also stabilize the dichloroacetate ion

more effectively by dispersing its negative charge more extensively.

(b) CCl3CO2H would be the stronger acid for reasons similar to those given in (a), except

here there are three versus two electron-withdrawing chlorine atoms involved.

(c) CH2FCO2H would be the stronger acid because the electron-withdrawing effect of a

fluorine atom is greater than that of a bromine atom (fluorine is more electronegative).

(d) CH2FCO2H is the stronger acid because the fluorine atom is nearer the carboxyl group

and is, therefore, better able to exert its electron-withdrawing inductive effect. (Remember:

Inductive effects weaken steadily as the distance between the substituent and the acidic

group increases.)

3.15 All compounds containing oxygen and most compounds containing nitrogen will have an

unshared electron pair on their oxygen or nitrogen atom. These compounds can, therefore,

act as bases and accept a proton from concentrated sulfuric acid. When they accept a proton,

these compounds become either oxonium ions or ammonium ions, and having become ionic,

they are soluble in the polar medium of sulfuric acid. The only nitrogen compounds that do



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not have an electron pair on their nitrogen atom are quaternary ammonium compounds, and

these, already being ionic, also dissolve in the polar medium of concentrated sulfuric acid.

3.16 (a) + +HCH3O

Stronger acid

pKa = 16

methanolH2

Weaker acid

pKa = 35

H−

Stronger base

(from NaH)

−CH3O

Weaker

base

(b) + +HCH3CH2O

Stronger acid

pKa = 16

ethanolNH2

−

Stronger base

(from NaNH2)

NH3

Weaker acid

pKa = 38

−CH3CH2O

Weaker

base

(c) H + +N

H

H

Stronger acid

pKa = 38

CH2CH3

Stronger base

(from CH3CH2Li)

hexaneCH3CH3

Weaker acid

pKa = 50

NH2−−

Weaker

base

(d) NH2−

Stronger base

(from NaNH2)

H + +N

H

H

H+

Stronger acid

pKa = 9.2

(from NH4Cl)

liq. NH3NH3

Weaker acid

pKa = 38

NH3

Weaker

base

(e) H ++O H

Stronger acid

pKa = 15.7

OC(CH3)3

Stronger base

[from (CH3)3CONa]

H2OHOC(CH3)3

Weaker acid

pKa = 18

H O• •

• •

••−

Weaker

base

−

(f) No appreciable acid-base reaction would occur because HO− is not a strong enough base

to remove a proton from (CH3)3COH.

3.17 HC CH NaHhexane

+ HC CNa H2+(a)

HC CNa D2O+ HC CD NaOD+(b)

CH3CH2Li CH3CH2DD2O+ LiOD+(c)

CH3CH2OH CH3CH2ONaNaH+ H2+(d)

CH3CH2ONa CH3CH2OTT2O+ NaOT+(e)

CH3CH2CH2Li CH3CH2CH2DD2O+ LiOD+(f )

hexane

hexane

hexane



CONFIRMING PAGES

ACIDS AND BASES 39

Problems

Brønsted-Lowry Acids and Bases

3.18 NH2− (the amide ion)(a) −

H C C (the ethynide ion)(d)

−−OH (the hydroxide ion)(b) CH3O (the methoxide ion)(e)

H− (the hydride ion)(c) H2O (water)(f )

3.19−OHNH2 H H C C> > > >

− − − −CH3O H2O≈

3.20 (a) H2SO4

(b) H3O+

(c) CH3NH3+

(d) NH3

(e) CH3CH3

(f) CH3CO2H

3.21 H2SO4 > H3O+> CH3CO2H > CH3NH3

+> NH3 > CH3CH3

Lewis Acids and Bases

3.22 CH3CH2 Cl

Lewis

base

Lewis

acid

ClCl

Cl

AlCl3 CH3CH2 Al++ −

Cl

Lewis

acid

Lewis

base

CH3 OH

Lewis

base

Lewis

acid

BF3+ F

F

CH3

H

O B+

−

F

++ CH3 OH2

CH3

CH3

CCH3

CH3

CH3

C H2O+

(a)

(b)

(c)



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Curved-Arrow Notation

3.23 (a) CH3 OH + H I I+CH3

H

O H+

−

(b)

(c)

CH3 NH2 + H Cl Cl+CH3

H

H

N H+

−

H+C

H

H

H

H

C F + C H

H

H

+ F−C

H

H

3.24 (a)

+

+ BF3

BF3O−

O

(b)

+

+ BF3BF3−

O O

(c) +

H

Cl−

O

OH

OH

+ O

H Cl

(d) CH3CH2CH2CH2 CH3CH2CH2CH3−

O+ +O H Li Li+

3.25

No appreciable acid-base reaction takes place because CH3CH2ONa is too weak a

base to remove a proton from ethyne.

(c)

(e) O H +−

CH3 CH2CH3hexane

(from

LiCH2CH3)

CH2 O + CH3CH3−

CH3 CH2

S

O

O

O H H+−

C6H5 S

O

O

O OH H+C6H5−

O(b)

(d) C C H +−

H CH2CH3 C C +H−

CH3CH3hexane

(from

LiCH2CH3)

C

O

O H H+−

CH3CH2 C

O

O O H+CH3CH2 H−

O(a)



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ACIDS AND BASES 41

Acid-Base Strength and Equilibrium

3.26 Because the proton attached to the highly electronegative oxygen atom of CH3OH is much

more acidic than the protons attached to the much less electronegative carbon atom.

3.27 CH3CH2 O H CH3CH2 OC C H+ +− −

C HH Cliq. NH3

3.28 (a) pKa = − log 1.77 × 10−4= 4 − 0.248 = 3.75

(b) Ka = 10−13

3.29 (a) HB is the stronger acid because it has the smaller pKa.

(b) Yes. Since A− is the stronger base and HB is the stronger acid, the following acid-base

reaction will take place.

+ B−

Weaker

base

Weaker

acid

pKa =20

A H+A−

Stronger

base

Stronger

acid

pKa =10

H B

3.30 C Cether

then

H NaNH2

NaH

+C6H5 C− +

+C NaC6H5 NH3

C−

C T2O+C6H5+

Na C +C TC6H5 NaOT

CH3 O HCH

then

then

NaH

+ CH3

−O Na+

−O Na+

−O Na+

−O Na+

CH

CH3

H2+(b)

(c)

(a)

NaOD+CH3 OCH

CH3

DCH3

CH3

CH

CH3

D2O+

CH3CH2CH2OH H2+CH3CH2CH2+

D2O+ + NaODCH3CH2CH2ODCH3CH2CH2

3.31 (a) CH3CH2OH > CH3CH2NH2 > CH3CH2CH3

Oxygen is more electronegative than nitrogen, which is more electronegative than carbon.

The O-H bond is most polarized, the N-H bond is next, and the C-H bond is least polarized.

(b) CH3CH2O−< CH3CH2NH−

< CH3CH2CH2−

The weaker the acid, the stronger the conjugate base.



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3.32 CH3C CH CH3CH CH3CH2CH3>

>

>CH2(a)

CH3CHClCO2H CH3CH2CO2H CH3CH2CH2OH>(b)

CH3CH2OH2+

CH3CH2OH CH3OCH3> >(c)

3.33 CH3NH3 CH3NH2 CH3NH< <

CH3O CH3NH−< <

(a)

(b)

CH3C C < <CH3CH(c)

+ −

CH− CH3CH2CH2−

CH3CH2−

−

−

General Problems

3.34 The acidic hydrogens must be attached to oxygen atoms. In H3PO3, one hydrogen is bonded

to a phosphorus atom:

H O O

O

O

H

P H H O

O

O

H

P H

3.35 C

O

HC

O

O H

H O H+ + H O

H

O

C

O

O CH3

H O H+ H O

O CH3

C H

O

+ O CH3CH O

O CH3

C H

O

O H

H

O

H O + II +CH3CH3 OH

(a)

(b)

(c)

(d)

−

−

−

−

−

−

−

−

H O + CH2 CH2

CH3

CH3

CH3

CH3

CH CCl

OH H

+

+

Cl−−

(e)



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ACIDS AND BASES 43

3.36 (a) Assume that the acidic and basic groups of glycine in its two forms have acidities and

basicities similar to those of acetic acid and methylamine. Then consider the equilibrium

between the two forms:

• •

• •O H

•••

•

O

CH N

H

• •

CH2

Stronger

base

Stronger

acid

• •

••

• •O

•••

•

O

CH N

H

H

CH2

Weaker

base

Weaker

acid

+ −

We see that the ionic form contains the groups that are the weaker acid and weaker base.

The equilibrium, therefore, will favor this form.

(b) The high melting point shows that the ionic structure better represents glycine.

3.37 (a) The second carboxyl group of malonic acid acts as an electron-withdrawing group

and stabilizes the conjugate base formed (i.e., HO2CCH2CO −

2 ) when malonic acid loses a

proton. [Any factor that stabilizes the conjugate base of an acid always increases the strength

of the acid (Section 3.11C).] An important factor here may be an entropy effect as explained

in Section 3.10.

(b) When −O2CCH2CO2H loses a proton, it forms a dianion, −O2CCH2CO2−. This dianion

is destabilized by having two negative charges in close proximity.

3.38 HB is the stronger acid.

3.39 ∆G ° = ∆H °− T∆S °

= 6.3 kJ mol−1 − (298 K)(0.0084 kJ mol−1K−1)

= 3.8 kJ mol−1

log Keq = log Ka = −pKa = −

pKa =

pKa =

pKa = 0.66

∆G°

2.303RT

∆G°

2.303RT

3.8 kJ mol

(2.303)(0.008314 kJ mol−1K−1)(298 K)

−1

3.40 The dianion is a hybrid of the following resonance structures:

OO

O OO O

OO O

O O

O OO

O O

−

− − − −

− − −

If we mentally fashion a hybrid of these structures, we see that each carbon-carbon bond

is a single bond in three structures and a double bond in one. Each carbon-oxygen bond is

a double bond in two structures and a single bond in two structures. Therefore, we would

expect all of the carbon-carbon bonds to be equivalent and of the same length, and exactly

the same can be said for the carbon-oxygen bonds.



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Challenge Problems

3.41 (a) A is CH3CH2S− B is CH3OH

C is CH3CH2SCH2CH2O− D is CH3CH2SCH2CH2OH

E is OH−

(b) CH3CH2 S H CH3+ O CH3CH2 S + HO

CH3CH2 S CH2 CH2+

O

OH

H+CH3CH2 S CH2CH2 O

CH3CH2 S CH2CH2 O

O

H+

H

CH3CH2 S CH2CH2 O

−

−

−

−

−

CH3−

3.42 (a) + +CH3(CH2)8OD CH3(CH2)8Li CH3(CH2)8DCH3(CH2)8O−

Li +

Hexane could be used as solvent. Liquid ammonia and ethanol could not because

they would compete with CH3(CH2)8OD and generate mostly non-deuterio-labelled

CH3(CH2)7CH3.

(b) +NH2 CH3C CH CH3CNH3 + C− −

Hexane or liquid ammonia could be used; ethanol is too acidic and would lead to

CH3CH2O− (ethoxide ion) instead of the desired alkynide ion.

(c) HCl NH2 NH3 Cl++

+−

Hexane or ethanol could be used; liquid ammonia is too strong a base and would lead

to NH +

4 instead of the desired anilinium ion.

3.43 (a,b)

CH N

O

CH3

CH3 C

H

O

+

−

N

CH3

CH3

The uncharged structure on the left is the more important resonance form.

(c) Since DMF does not bind with (solvate) anions, their electron density remains high and

their size small, both of which make nucleophiles more reactive.



CONFIRMING PAGES

ACIDS AND BASES 45

3.44 (a)

(b)

+(c)

OO

C

H3C CH3

+ NH3

O

+

O

D2O + OD−

O

O

−

−O

+

−O

−NH2

−

C

H3C CH2

C

H3C CH2D

−

C

H3C CH2

3.45 The most acidic hydrogen atoms in formamide are bonded to the nitrogen atom. They

are acidic due to the electron-withdrawing effect of the carbonyl group and the fact that

the resulting conjugate base can be stabilized by resonance delocalization of the negative

charge into the carbonyl group. The electrostatic potential map shows deep blue color near

the hydrogen atoms bonded to the nitrogen atom, consistent with their relative acidity.

QUIZ

3.1 Which of the following is the strongest acid?

(a) CH3CH2CO2H (b) CH3CH3 (c) CH3CH2OH (d) CH2 CH2

3.2 Which of the following is the strongest base?

(a) CH3ONa (b) NaNH2 (c) CH3CH2Li (d) NaOH (e) CH3CO2Na

3.3 Dissolving NaNH2 in water will give:

(a) A solution containing solvated Na+ and NH −

2 ions.

(b) A solution containing solvated Na+ ions, OH− ions, and NH3.

(c) NH3 and metallic Na.

(d) Solvated Na+ ions and hydrogen gas.

(e) None of the above.



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3.4 Which base is strong enough to convert (CH3)3COH into (CH3)3CONa in a reaction that

goes to completion?

(a) NaNH2 (b) CH3CH2Na (c) NaOH (d) CH3CO2Na

(e) More than one of the above.

3.5 Which would be the strongest acid?

(a) CH3CH2CH2CO2H (b) CH3CH2CHFCO2H (c) CH3CHFCH2CO2H

(d) CH2FCH2CH2CO2H (e) CH3CH2CH2CH2OH

3.6 Which would be the weakest base?

(a) CH3CO2Na (b) CF3CO2Na (c) CHF2CO2Na (d) CH2FCO2Na

3.7 What acid-base reaction (if any) would occur when NaF is dissolved in H2SO4?

3.8 The pKa of CH3NH +

3 equals 10.6; the pKa of (CH3)2NH2+ equals 10.7. Which is the

stronger base, CH3NH2 or (CH3)2NH?

3.9 Supply the missing reagents.

hexaneCH3CH2C CH +

CH3CH2C CD LiOD+

(b)

(a)

CH3CH2C CH3CH3C Li++

−

3.10 Supply the missing intermediates and reagents.

T2OLiOT+

(c)

(b)

CH3CHCH2OT

CH3

LiBrCH3Br 2 Li+ +

(a)

solomons organic chemistry solution manual - chapter 3

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