1.
Using line-bond notation, draw a neutral molecule that meets the following criteria:
[8 marks]
(a) An aromatic heterocyclic compound that contains 1 oxygen atom and 1 nitrogen atom in a five membered
ring.
plus other possibilities
N
O
(b) An aromatic heterocyclic compound that contains 1 boron atom.
H
B
B
Borabenzene
B
H
plus other possibilities
c) An anti-aromatic heterocyclic compound that contains 1 oxygen atom and 1 nitrogen atom in a six
membered ring.
O
plus other possibilities
N
H
d) A non-aromatic heterocyclic compound that contains 1 oxygen atom and 1 nitrogen atom in a six membered
ring.
O
N
plus other possibilities
2. The following is an acid-base reaction between two conjugate acid/base pairs. Does this equilibrium favour
reactants or products? How can you tell?
[3 marks]
O
O
+
OH
CN
+ HCN
O
pKa = 10
pKa = 5
A Bronsted acid/base reaction between two conjugate acid/base pairs favours the side of the
weakest acid, HCN in this case. So this is product favoured. Marks for the pKas and the
statement.
3.
(a) Using the polygon-in-a-circle method (Frost circles), sketch the pi MO
energy level diagram for the cyclopentadienyl anion. In your
diagram:
• show the relative energy levels of the orbitals
• include the appropriate number of electrons
• label each or bital as bonding, non-bonding, or antibonding.
**NOTE: For this part of the question (part a), you do NOT need to
sketch what the actual MOs look like, just give their relative energies.**
[10 marks]
cyclopentadienyl anion
antibonding
bonding
bonding
(b) Sketch the π–MO for the HOMO of the cyclopentadienyl anion. If there is more than one HOMO, just give
one possible example.
or
(c) Is the cyclopentadienyl anion aromatic, anti-aromatic, or non-aromatic? Explain your reasoning.
It is aromatic because it is a cyclic, planar, π-system (a continuous cyclic arrangement of p-orbitals)
with 6 π electons (4n + 2 π electrons).
4.
Pyridine is a modest Bronsted base at nitrogen whereas pyrrole is
not. In fact, pyrrole is protonated only in strong acid, and
protonation only occurs at C(2), not at the nitrogen atom.
(a) Explain why pyridine is basic and pyrrole is not.
[6 marks]
C(2)
N
N
H
pyridine
pyrrole
Both pyridine and pyrrole are aromatic molecules. If the nitrogen atom on pyridine acts as a base and accepts a
proton the resultant molecule is still aromatic because the lone pair is not part of the π system. Thus, the lone
pair on pyridine is available. The lone pair on pyrrole is part of the π system and therefore is much less
available to react with an acid. If it did react with an acid, aromaticity would be lost.
(b) Using curly arrows to show the movement of electrons, draw the protonation reaction of pyrrole and the
resultant product. Provide a reasonable explanation for why protonation of pyrrole occurs at carbon and
not nitrogen. If your argument uses resonance structures, DRAW them, don’t just describe them.
H
H
N
H
N
N
H
H
Protonation at C2 gives a carbocation that can be
resonance stabilized giving a resonance structure
where all atoms obey the octet rule. Protonation at
nitrogen gives a cation with no opportunity of
resonance stabilization.
H
N
N
H
H
H
5. Classify each of the following species as an electrophile, nucleophile, both, or neither.
[5 marks]
Cl
(a)
Br
(b)
electrophile
(d)
F
(fluorine atom)
(c)
Al
Cl
S
Cl
nucleophile
electrophile
(e)
O
neither (accepted electrophile also)
electrophile (accepted both also)
H 3C
H
6. For each of the following pairs of reagents, suggest a possible first step in the mechanism using curved
arrows to show the movement of electrons and draw the intermediate(s) formed.
[4 marks]
(a)
O
O
H
+
OH
H2O
+
N
H
N
H
H
H
(b)
O
N
O
O
+
N
O
7. You wish to make a sulfide using the following nucleophilic substitution approach.
[5 marks]
pKa = 11
SH
SH
SH
CH3CH2Br
Base
pKa = 7
HS
i)
S
S
What is the purpose of the first step?
The purpose of the first step is to convert the ArSH in its conjugate base which is a BETTER nucleophile.
Many said that this step makes it a nucleophile, but really, it is nucleophilic to begin with. Deprotonation
makes it more reactive.
ii)
Note that there are two acidic SH groups in this molecule. Label them with their approximate pKa
values.
When I say label, please label.
iii)
Assuming that the base you use is used in excess (i.e. more than 1 equivalent) what base might you
use to selectively deprotonate only the required SH? Explain your reasoning.
You need to use a base whose conjugate acid is found between 7 and 11. There are several such bases found on
your table, ammonia being the simplest example.
8. Shown below is a mechanism for a solvolysis reaction.
[10 marks]
(a) Use curved arrows to show the electron movement in the following mechanism (Hint: you should start by
adding all missing lone pairs):
!
O
O
+
H
+
Br
H
O
+
Br -
O
O
+
O
H
O
H
O
O
+
O
B
+
BH
H
O
O
(b) Use the graph below to sketch the reaction profile diagram for this product favoured reaction. Be sure to
label all transition states, intermediates, and to identify the rate determining step (RDS).
Step 1 is rate determining
therefore it should have
the highest activation
energy
Transition States
E
Intermediates
The product is lower in energy than
the reactants because it is a product
favoured reaction.
reaction progress
9. Rank the following four molecules in order of acidity. Explain your reasoning.
O
O
SH
OH
OH
[4 marks]
OH
F
The last two both have conjugate bases in which the negative charge is delocalized over two O atoms making
them more acidic than the first two. The third example has an F atom which further inductively stabilizes the
conjugate base making it the most acidic. Of the first two, the thiol is the more acidic because the larger S atom
has orbitals that can diffuse the charge on its conjugate. I was impressed by many concise answers that covered
all the salient points. And for the record, when I say explain, I mean explain or rationalize. Looking up the
pKas is not an explanation.
10.
[5 marks]
i)
In terms of the relevant thermodynamic parameter(s), what does it mean to say that a
reaction is product favoured?
It means that ∆G<0. That’s all you need to say. ∆H and ∆S contribute to this, of course, but
Their signs can vary and still give a negative ∆G.
ii)
What is the sign of ∆S for the following reaction? Explain your reasoning.
O
O
+ SOCl2 (l)
OH (l)
Cl (l)
+ SO2 (g)
+ HCl (g)
The stoichiometry is 2 to 3 which usually gives a positive ∆S. In addition, two of the products
are gases which have a lot of entropy.
Two marks.
iii)
This reaction is product-favoured at all temperatures. What does this tell you about
the sign of ∆H for this reaction? Explain.
It must be negative. If a reaction is spontaneaous at all temperatures then ∆H must be negative and ∆S must be
positive. Some students noted that this statement actually provides an answer to part ii, which is true.
Many noted that some reactions that have a positive ∆S will still be reactant-favoured at low T if ∆H is positive
therefore in this case ∆H must be negative. This is a different way of putting the statement in the second
sentence, but is perfectly reasonable.
11. For each of the following two schemes, classify each reactions as Bronsted acid/base, substitution, addition,
elimination, oxidation, reduction or rearrangement. The reactions are numbered for your convenience. Please
put your answers on the lines provided.
[7 marks]
H3C
H
H
Cl2, H2O
CH3
H3C
Cl
H
H
HO
1
NaOH
H3C
H
(CH3)3SiI
H3C
H
H
CH3
If you gave two answers and one was wrong, you didn’t get any marks.
1)
addition and/or oxidation
2)
substitution (it’s an intramolecular substitution)
3)
elimination and/or reduction
O
HO
H
H+,CH3OH
OH
C
H
1
OH
O
H3CO
OCH3
C
H
2
OH
C
C
O
O
O
1)
addition
2)
substitution
3)
reduction
4)
oxidation
H3CO
OCH3
C
OCH3
H+,CH3OH
3
H3CO
H
PCC
H
4
H
3
2
CH3
O
OCH3
H
CH2OH
1) LiAlH4, THF
2) HCl
CH3