8 Suggested solutions for Chapter 8 PROBLEM 1 How would you separate a mixture of these three compounds? CO2H
N
naphthalene
pyridine
para-toluic acid
Purpose of the problem Revision of simple acidity and basicity in a practical situation. Suggested solution Pyridine is a weak base (pKa of the pyridinium ion is about 5.5) and can be dissolved in aqueous acid. Naphthalene is neither an acid nor a base and is not soluble in water at any pH. p-­‐Toluic acid is a weak acid (pKa about 4.5) and can be dissolved in aqueous base. So dissolve the mixture in an organic solvent immiscible with water (say ether Et2O or dichloromethane CH2Cl2) and extract with aqueous acid. This will dissolve the pyridine as its cation. Then extract the remaining organic layer with aqueous base such as NaHCO3 which will remove the toluic acid as its water-­‐soluble anion. You now have three solutions. Evaporate the organic solution to give crystalline naphthalene. Acidify the basic solution of p-­‐toluic acid and the free acid will precipitate out and can be recrystallised. Add base to the pyridine solution, extract the pyridine with an organic solvent, and distil the pyridine. It doesn’t matter if you extract the original solution with base first and acid second. 2 Solutions Manual to accompany Organic Chemistry 2e PROBLEM 2 In the separation of benzoic acid from toluene on p. 164 of the textbook we suggested using KOH solution. How concentrated a solution would be necessary to ensure that the pH was above the pKa of benzoic acid (4.2)? How would you estimate how much KOH solution to use? Purpose of the problem To ensure you understand the relationship between pH and concentration. Suggested solution Even a very weak solution of KOH has a pH>4.2. If we want a pH of 5 (just above the pKa of benzoic acid) we must ensure that we have [H3O+] = 10–5 mol dm–1. The ionic product of water is [H3O+] x[HO–] = 10–14 and so we need 10–9 mol dm–1 of KOH. This is very dilute! The trouble would be that you need one hydroxide ion for each molecule of benzoic acid and so if you had, say, 1.22 g PhCO2H (= 0.01 equiv) you would need 1000 litres (dm–1) of KOH solution. It makes more sense to use a much more concentrated solution, say 0.1M. This would give an unnecessarily high pH (13) but you would need only 100 ml (dm–3) to extract your benzoic acid. PROBLEM 3 What species would be present in a solution of this hydroxy-­‐acid in (a) water at pH 7, (b) aqueous alkali at pH 12, and (c) in concentrated mineral acid? CO2H
HO
Purpose of the problem To get you thinking about what really is present in solution using rough pKa as a guide in a practical situation. Suggested solution The CO2H group will have a pKa of about 4-­‐5 and the phenolic OH a pKa of about 10. So the carboxylic acid but not the phenol will be ionized at pH 7, they will both be ionized at pH 12, and there will be a mixture of Solutions for Chapter 8 – Acidity, Basicity and pKa free acid and protonated acid at very low pH. The proton will be on the carbonyl oxygen atom as this gives a delocalised cation. at pH 7
O
O
at pH 12
O
O
HO
O
OH
O
at low pH
OH
OH
HO
HO
PROBLEM 4 What would you expect to be the site of (a) protonation and (b) deprotonation if these compounds were treated with the appropriate acid or base? In each case suggest a suitable acid or base and give the structure of the products. N
OH
H
N
N
H
N
OH
Purpose of the problem Progressing to more taxing judgements on more interesting molecules. Suggested solution The simple amine piperidine will easily be protonated by even weak acids as the conjugate base has a pKa of about 11. Any mineral acid such as HCl will do the job as would weaker acids such as RCO2H. Deprotonation will remove the NH proton as nitrogen is more electronegative than carbon but a very strong base such as BuLi will be needed as the pKa will be about 30-­‐35. You could represent the product with an N–Li bond or as an anion. RCO2H
N
H
or HCl
H
BuLi
N
H
or
N
Li
N
3 4 Solutions Manual to accompany Organic Chemistry 2e The second example is more complicated but contains a normal tertiary amine so protonation will occur there with most acids as as the conjugate base has a pKa of about 11. We use TsOH this time but that has no special significance. The tertiary amine cannot be deprotonated and in any case the alcohol is more acidic and a strong base will be needed, say NaH. OH
H
H
SO2OH
OH
NaH
+
N
N
Me
H
O
H
N
The third example is more complicated still. There is a normal OH group (pKa of about 16) and a slightly acidic alkyne (pKa of about 32). The basic group is not a simple amine but a delocalised amidine. Protonation occurs on the top (imine) nitrogen as the positive charge is then delocalised over both nitrogens. Protonation on the other nitrogen does not occur. The pKa of the conjugate base is about 12. N
N
OH
H
×
H
H
H
N
N
N
N
N
OH
N
OH
OH
The first proton to be removed by base will be from the alcohol and this will need a reasonably strong base such as NaH. Removal of the alkyne proton requires a much stronger base such as BuLi. You might represent the product as an alkyne anion or a covalently bonded alkyllithium. N
N
OH
N
N
NaH
N
BuLi
N
O
O
Solutions for Chapter 8 – Acidity, Basicity and pKa PROBLEM 5 Suggest what species would be formed by each of these combinations of reagents. You are advised to use estimated pKa values to help you and to beware of those cases where nothing happens. (a)
O
(b)
O
+
(c)
O
HN
O
NH +
+
OH
N
H
O
F3C
O
Purpose of the problem Learning to compare species of similar acidity or basicity. Suggested solution In each case one of the reagents might take a proton from the other. In example (a), would the phenolate anion remove a proton from acetic acid? The answer is yes because acetic acid is a much stronger acid than phenol The difference is five pH units so the equilibrium constant would be about 105 and the equilibrium would lie far across to the right. (a)
O
OH
O
O
+
+
O
OH
pKa about 5
pKa about 10
Example (b) has a similar possible reaction but this time the pKa difference is much smaller and the other way so the equilibrium constant is 100 and favours the starting materials. ■ See page 178 of the textbook (b)
HN
NH +
N
O
pKa about 7
for the effect of fluorine on pKa. O
O
NH +
OH
pKa about 5
Example (c) is rather different. We do have another carboxylic acid but this is a much stronger acid because of the three fluorine atoms and the equilibrium is far over to the left. 5 6 Solutions Manual to accompany Organic Chemistry 2e (c)
O
O
+
N
H
+
F3C
N
O
F3 C
pKa about 5
OH
pKa about –1
PROBLEM 6 What is the relationship between these two molecules? Discuss the structure of the anion that would be formed by the deprotonation of each compound. N
H
O
N
OH
Purpose of the problem To help you recognise that conjugation may be closely related to tautomerism. Suggested solution They are tautomers: they differ only in the position of one hydrogen atom. It is on nitrogen in the first structure and on oxygen in the second. As it happens the first structure is more stable. They are both aromatic (check that you see why) but the first has a strong carbonyl group while the second has a weaker imine. Deprotonation may appear to give two different anions but they are actually the same because of delocalization. Note the different ‘reaction’ arrows: equilibrium sign for deprotonation and double headed arrow for delocalization. –H+
N
H
O
+H+
N
O
N
O
N
OH
Solutions for Chapter 8 – Acidity, Basicity and pKa PROBLEM 7 The carbon NMR spectrum of these compounds can be run in D2O under the conditions shown. Why are these conditions necessary? What spectrum would you expect to observe? 13C NMR
spectrum
run in
DCl/D2O
N
H2N
N
H
O
OH
13C NMR
spectrum
run in
NaOD/D2O
Purpose of the problem NMR revision and practice at judging the states of compounds at different pHs. Observation of hidden symmetry from conjugation. Suggested solution Both compounds are quite polar and not very soluble in the usual NMR solvents. In addition thay have NH or OH protons that exchange in solution and broaden the spectrum. With acid or base catalysis the NH and OH protons are exchanged with deuterium and sharp signals appear. But in the strong acid or base used here, ions are formed. The first compound, a strongly basic guanidine (see p. 167 of the textbook) forms a cation in DCl. The cation is symmetrical, unlike the original guanidine, and a very simple spectrum results: just three types of carbon in the benzene ring and one very low field carbon (at large δ) for the carbon in the middle of the cation. N
H2N
N
H
DCl
D2O
D
N
D
N
D2N
D2N
N
D
N
D
The second compound loses a proton from the OH group to give a delocalized symmetrical anion. There will be five signals in the NMR: the two methyl groups are the same (at small δ) as are the two CH2 groups in the ring (at slightly larger δ). There is one unique carbon joined to the two methyl groups (at small δ) and another in the middle of the anion (at large δ). Finally both carbonyl groups are the same (at even larger δ). 7 8 Solutions Manual to accompany Organic Chemistry 2e NaOD
D2O
O
OH
O
O
O
O PROBLEM 8 These phenols have approximate pKa values of 4, 7, 9, 10 and 11. Suggest with explanations which pKa value belongs to which phenol. OH
O2N
NO2
OH
O2N
OH
Cl
Me
OH
Me
Me
OH
Purpose of the problem Detailed examination of electronic effects to estimate pKa values. Suggested solution Electron-­‐withdrawing effects make phenols more acidic and electron-­‐
donating effects make them less acidic. Phenol itself (the fourth example) has a pKa of 10. The only compound less acidic than phenol must be the third with three weakly electron-­‐donating methyl groups. One chlorine atom has an inductive electron-­‐withdrawing effect so the last compound has pKa 11. The remaining two have the powerful electron-­‐withdrawing nitro group. So the first compound, with two nitro groups, must have pKa of 4 (making it as strong an acid as acetic acid) and the second, with one nitro group, must have pKa of 7. OH
O2N
NO2
OH
O2N
Me
Me
pKa about 7
pKa about 4
OH
pKa about 10
OH
Me
pKa about 11
Cl
OH
pKa about 9
9 Solutions for Chapter 8 – Acidity, Basicity and pKa PROBLEM 9 The pKa values of the amino acid cysteine are 1.8, 8.3, and 10.8. Assign these pKa values to the functional groups and draw the most abundant structure that the molecule will have at pH 1, 5, 9, and 12. CO2H
HS
NH2
cysteine
Purpose of the problem Further revision in thinking about acidity and basicity of functional groups, this time in one molecule. Amino acids are particularly important. Suggested solution At high pH cysteine exists as a dianion as both the thiol and the carboxylic acid are anions. If we now add acid, at about pH 10 (actually 10.8) the amine will get protonated, then the thiol will be protonated at about pH 8 (actually 8.3) and finally the carboxylic acid will be protonated at low pH, rather lower than say MeCO2H, as the electron-­‐
withdrawing ammonium group increases its acidity (actually at 1.8). S
CO2
NH2
at pH 12
add acid
pKa RNH2
about 10
S
CO2
NH3
at pH 9
add acid
pKa RSH
about 8
HS
CO2
NH3
at pH 5
add acid
pKa RCO2H
about 4
HS
CO2H
NH3
at pH <2
10 Solutions Manual to accompany Organic Chemistry 2e PROBLEM 10 Neither of these two methods for making pentan-­‐1,4-­‐diol will work. What will happen instead? OH
O
OH
H
Br
OH
Me
MgBr
Me
Mg
Et2O
OH
H
MeCHO
BrMg
OH
Purpose of the problem To help you appreciate the disastrous effects that innocent-­‐looking groups may have because of their weak acidity. Suggested solution The OH group is the Wicked Witch of the West in this problem. Whoever planned these syntheses expected it to lie quietly and do nothing. All chemists have to learn that things don’t go our way just because we want them to do so. Here, although the OH group is only a weak acid (pKa about 16) it will give up its proton to the very basic Grignard reagents. In the first case, one molecule of Grignard is destroyed but the reaction might succeed if two equivalents were used. O
H
Me
OH
MgBr
OH
Me
OMgBr
Me
MgBr
?
H
pentane1,4-diol
The second case is hopeless as the Grignard reagent destroys itself by intramolecular deprotonation. This synthesis could be rescued by putting a protecting group on the OH. BrMg
Br
OH
Et2O
Me
Mg
H
O
OMgBr