organic chemistry: an acid-base approach, second edition

333 Pages·2015·24.75 MB·English

SOLUTIONS MANUAL FOR
ORGANIC
CHEMISTRY:
AN ACID-BASE
APPROACH, SECOND
EDITION
by
Michael B. Smith
SOLUTIONS MANUAL FOR
ORGANIC CHEMISTRY:
AN ACID-BASE
APPROACH, SECOND
EDITION
by
Michael B. Smith
Boca Raton London New York
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Chapter 02
Solutions Manual
Organic Chemistry. An Acid-Base Approach 2nd Edition

27. (a)
O O
H
N O OH2 N O H3O+
O O
K = [NO –] [H O+]
a 3 3
[HNO ] [H O]
3 2
Nitric acid is a stronger acid than water, so water is the base and nitric acid is the acid in this
reaction to generate the nitrate anion (conjugate base) and the hydronium ion (conjugate acid).
(b)
H H
H H-NH
NH O 3
O 3
K = [HO–] NH +]
a 4
[H O] [NH ]
2 3
Water is a stronger acid than ammonia, so ammonia is the base and water is the acid in this
reaction. The reaction products are hydroxide anion (conjugate base) and the ammonium ion (conjugate
acid).
(c)
Br H-NH
Br H 3
NH
3
K = [Br–] [NH +]
a 4
[HBr] [NH ]
3
Hydrobromic acid is a stronger acid than ammonia, so ammonia is the base and HBr is the acid
in this reaction to generate the bromide anion (conjugate base) and the ammonium ion (conjugate acid).
(d)
Cl H OH2 Cl- H3O+
[Cl-] [H O+]
3
K =
a
[HCl] [H O]
2
Hydrochloric acid is a stronger acid than water, so water is the base and HCl is the acid in this
reaction to generate the chloride anion (conjugate base) and the hydronium ion (conjugate acid).
(e)
Cl
Cl
Cl C NH
Cl C H 3
Cl NH2 Cl
Na
K = [CCl –] [NH ]
a 3 3
[HCCl ] NaNH ]
3 2
Note that the acidic hydrogen atom is attached to a carbon atom. The amide anion is clearly a
base, which a very strong base. Therefore, the strong base can react wit the weakly acidic “carbon
acid”. In other words, NH – is the base and chloroform is the acid in this reaction that generates the
2
anion –CCl (the conjugate base) and ammonia, NH , which is the conjugate acid.
3 3

28. The main reason is likely the relative size of the bromide ion (182 pm or 1.82 Å) versus the chloride
ion (100 pm, or 1.00 Å). Greater charge dispersal for the bromide ion leads to greater stability of that
conjugate base, and a larger K .
a

29.
[H O+] [Cl–] [H O+] [Cl–]
3 3
K = K =
a a
[HCl] [HCl] [H O]
2

In this reaction, water is the base that reacts with the HCl. If water is omitted, the base has been
excluded from the equilibrium expression for an acid-base reaction.

30. (a) NH (b) H CO– (c) –NO (d) Br– (e) –:NH (f) –CH
3 3 3 2 3

31. The ‘ate. complex in the “box” is the reaction product, where N of ammonia is the electron donor
(the base).
H N: BF
3 3 F B NH
3 3

32. Oxygen is more electronegative than nitrogen, so nitrogen is more electron-rich, and will be a better
electron-donor. In addition, the ‘ate’ complex from ammonia is an ammonium salt whereas water will
react to form an oxonium salt. The ammonium salt is more stable, which contributes to the overall
increased Lewis basicity of the nitrogen atom in ammonia.

33. In these neutral molecules, phosphorus is larger than nitrogen, with covalent radii of 106 and 71 pm
(1.06 Å and 0.71 Å), respectively. The charge density of nitrogen is greater. Therefore, ammonia is
expected to be the stronger Lewis base.

34. An ate complex is the salt generated by reaction of a Lewis acid with a Lewis base. The atom
derived from the Lewis base expands its valence and assumes a positive charge, whereas the atom
derived from the Lewis base expand its valence and assumes a negative charge.

35.
H C Cl
3
O Al Cl
H C Cl
3

36. The C—H bond is much stronger than the N—H bond, so it is more difficult to break. Nitrogen is
larger and better able to accommodate charge relative to carbon, so H C- is significantly less stable
3
(more reactive) than H N-. In addition, nitrogen is more electronegative than carbon, so the amide anion
2
is more stable, which means that it is less reactive (it is a stronger base). If the -NH conjugate base is
2
more stable, K is larger, and ammonia is a stronger acid.
a

37. (a) CH OH is the strongest acid in this series. The O—H bond is more polarized and easier to
3
break, and the methoxide anion, H CO-, is more stable than the anions from CH or CH NH . NaF does
3 4 3 3
not have an acidic proton, and it is not a Brønsted–Lowry acid.
(b) As explained in section 2.4, the size of the conjugate base increases from fluoride towards iodide, so
the iodide in is more stable. This means that K is larger for HI and decreases going towards HF. Since
a
iodide is much larger, the H—I bond is longer, and weaker, s it is easier to break relative to the others.

38. The iodide in is much larger, and the charge is dispersed over a greater area. Therefore, it is more
difficult for iodide to donate electrons to an acid relative to fluoride. In other words, iodide is a weaker
base.

39.
O H O O O
N O BASE H-BASE N O N O N O
O O O O

As shown, nitric acid generates the resonance stabilized nitrate anion. In the nitrate anion, the
charge is dispersed over several atoms, which makes it more difficult for that species to donate electrons
to an acid. For hydroxide ion, HO-, the charge is concentrated on the oxygen atom, and it is much easier
to donate electrons. The charge is not dispersed as in the nitrate anion, and hydroxide is more basic.

40. The fluoride is much more stable relative to the methide anion, H C:-, because fluoride is more
3
electronegative. Therefore, it retains electrons whereas the less electronegative carbon does not.
Therefore, carbon will donate electron more easily and the methide anion is the stronger base.

41. Determine the pK for each of the following.
a
(a) K = 1.45x105 (b) K = 3.6x10–12 (c) K = 6.7x10–31 (d) K = 18 (e) K = 3.8x1014
a a a a a
pK = -log K . (a) 5.16 (b) 11.44 (c) 30.17 (d) –1.26 (e) –14.6
a 10 a

42. The more acidic acid will have the smaller pK . Of this series, HCl is the strongest acid (pK -7)
a a
relative to HF (pK 3.17). Water has a pK of 15.7 and ammonia has a pK of about 25. Clearly, HCl
a a a
has the smallest pK .
a
43. The least acidic acid will have the largest pK . Of this series, HCl is the strongest acid (pK -7)
a a
relative to HF (pK 3.17). Water has a pK of 15.7 and ammonia has a pK of about 25. Clearly,
a a a
ammonia is the least acidic and has the largest pK .
a

44. NaF is an ionic salt, Na+ and F-. The electron rich fluoride ion is the only atom of these two that
can donate electrons, so F is the basic atom.

45. Clearly, water is more acidic. Figure 2.5 shows that oxygen is more electronegative than sulfur, so
the O—H bond is more polarized than the C—S bond, and should be weaker. This is consistent with a
-
larger K for water (smaller pK ). Further, the more electronegative oxygen atom in the HO anion
a a
-
makes it a poorer electron donor relative to HS , so hydroxide is the stronger base. This is also
consistent with a larger K for water.
a

46.
• •
• •O• • • •O• • • •O• •
A
C C
C H • • H C
H3C • •O• • H3C • •O• • 3 • •O• •
• •O• • • •O• • • •O• •• •
C
C H B C H
H C H
H C C 3 H C
3 H 3
H
H H
The reaction of A gives the conjugate base shown, and B gives the conjugate base indicated. In
both cases, the charge is dispersed over three atoms (resonance). The OH bond is weaker than the NH
bond, and that proton is easier to remove. Oxygen holds onto electron better than nitrogen (it is more
electronegative), so the conjugate base from A is less likely to donate electrons (it is more stable, which
shifts the equilibrium towards the conjugate base). Although it is not obvious from the diagram, the
charge dispersal is more efficient in the conjugate base from A. All of these combine to make A much
more acidic (pK of 4.8 versus 46).
a

47.
O O
O
+ NaOH C C + HOH
C H
H C
H C O H3C O 3 O
3
H H
H H
+ NaOH + HOH
H
H O H O
(a) The conjugate base derived from formic acid is resonance stabilized by charge dispersal over several
atoms, as shown. The conjugate base from methanol has the charge concentrated on oxygen, and no
charge dispersal is possible.
(b) If there is a larger concentration of the conjugate base, the equilibrium is shifted toward the right
(towards the conjugate base), and K is larger.
a
(c) If K for formic acid is much larger, it will be the stronger acid, and will react better with NaOH.
a
48. Since F is more electronegative than B, one predicts that C—F is the more polarized covalent bond.
49. Aluminum has d orbitals and boron does not. According to the rules presented for Lewis acid
strength, the d orbitals are more available for donation, which makes AlCl the weaker Lewis acid
3

50.
NH Cl + NaOH NH + HOH + NaCl
4 3
conj. conj.
base acid
Chapter 03
Solutions Manual
Organic Chemistry. An Acid-Base Approach 2nd Edition

22. Al is 1s22s22p63s23p1 He is 1s2 Be is 1s22s2 Mg is 1s22s22p63s2
Cl is 1s22s22p63s23p5 Br is 1s22s22p63s23p63d104s24p5 Ti is 1s22s22p63s23p63d24s2
Cu is 1s22s22p63s23p63d104s1

23.
1s 2s

24. Both are group 1 elements. Potassium is a larger atom, which means that the 4s1 electron is held
less tightly than the 3s1 electron on sodium. Using a very simple rationale, it should be easier to lose the
electron from potassium, and the resulting K+ ion is larger than the Na+ ion and will be more stable.

25. No single orbital may hold more than two electrons, and if there is more than one orbital of the
same energy (degenerate orbitals), no one orbital may hold two electrons until all orbitals hold one. For
the second row, there is one s-orbital and three degenerate p-orbitals. Therefore, adding one electron to
the 2s-orbital yields lithium (Li) and adding the second to that orbital yields beryllium (Be). Adding one
electron to a 2p-orbitals gives boron (B), and the second electron goes to a different 2p-orbital to give
carbon (C), and the third fills the last 2p-orbitals to yield nitrogen (N). The next electron will spin pair
in a 2p-orbital to give oxygen (O) and the net electron will spin pair to give fluorine (F).

26. The four hydrogen atoms about carbon in methane arrange as far as possible from each other, with
carbon in the middle: a tetrahedron. Ammonia has three atoms and a lone electron pair, with nitrogen in
the middle of a tetrahedron. Since the electron pair cannot be seen, the shape of the three hydrogen
atoms and the nitrogen constitutes a pyramidal structure.

27.
sp3 sp3
sp3 sp3 sp3 sp3 sp3
Br Br sp3 O sp3 Cl CH2CH3
(a) C (b) CH C CH (c) CH C N (d) C
Br 3 3 3 Br
CH CH
Br 2 2 CH=CH
2
The sp3 hybridized carbon atoms are indicated by the arrows.

28.
H
Br
(a) CH3O- Na+ (b) C (c) CH3C≡C-H (d) CH3C≡C-Na (e) NaF
H
H
Ionic Covalent Covalent Ionic Ionic
CH
O 2
O
(f) CH —CH (g) (h) (i)
3 3 C H
C
Covalent Covalent Covalent Covalent

29. The numbers indicate that it requires less energy to remove an electron from K (4.341 eV, 124.3
kcal mol-1), so it is expected to be the most reactive. Since Li (5.392 eV, 113.5 kcal mol-1) has the
highest ionization potential, it is more difficult to lose an electron and it will be less reactive than Na
(5.139 eV, 100.4 kcal mol-1), the least reactive of the three. Therefore, K is more reactive than Na,
which is more reactive than Li.

30.
H H
Cl H H H C
(a) C Cl (b) C O (c) H O
Cl H C
Cl H H
H
H
H H H H
C H
H H
C C
(d) H H (e) H (f)
N C C C H
C H
H C H Cl
C H Cl
H H H H
H
H
. H H

31. ΔH° = H°products - H°reactants
(a) ΔH° = (H°C-Br + H°H-O) - (H°H-Br + H°C-O)
ΔH° = (67 + 104.2) - (87.4 + 257.3) = 171.2 - 344.7 = -173.5 kcal mol-1.
(b) ΔH° = 2xH°C-I - (H°C-C + H°I-I)
ΔH° = 2x50 - (145 + 36.5) = 100 - 181.5 = -81.5 kcal mol-1.
(c) ΔH° = (H°C-O + H°C-H) - (H°O-H + H°C-C)
(257.3 + 80.6) - (104.2+ 145) = 337.9 - 249.2 = 88.7 kcal mol-1.

(d) ΔH° = (H°C-I + H°N-H) - (H°C-N + H°H-I)
ΔH° = (50 + 75) - (184 + 71.4) = 125.0 - 255.4 = -130.45 kcal mol-1.

32. I—CH (C—I = 56 kcal or 234 kJ mol-1); Br—CH (C—Br = 70 kcal or 293 kJ mol-1);
3 3
Cl—CH (C—Cl = 84 kcal or 351 kJ mol-1.
3