CHEMISTRY 235 – FINAL EXAMINATION – SUMMER 2005 – ANSWERS
1. Answers are given for the 20 parts, in order:
MeO
O
O
OH
Br
O
Cl
O
O
O
OEt
OEt
Br
COOH
O
H
HO
OH
COOH
CN
O
no reaction
NH2
NH2
COOH
N
+ NH4+
+ CH3CO2H
HO
NO2
O
no reaction
Ph
O
Ph
N
S
NHCH3
2. A single substance can often be converted into more than one closely related synthetic intermediate
by appropriate selection of experimental conditions. For the following pairs, provide experimental
conditions to carry out each of the indicated conversions [the answer requires the starting material
to be converted independently to each of the two products]. Each conversion may be done in one
step, assuming an aqueous work-up.
[8 marks]
OH
a)
CH3Li
H2O
Et2O
O
O
(CH3)2CuLi
H2O
Et2O
O
b)
DIBAL-H
H
H3O+
toluene
-78º
CN
O
H3O+
OH
∆
3. Answer the following questions about the Diels Alder reaction. Remember, stereochemistry is
important. [8 marks]
a) Each of the following pairs of starting materials provides a single major Diels Alder product.
Give the formula of each one.
OMe
CO2Me
+
+
CHO
MeO
H H
CO2Me
CHO
b) Give the starting materials which would best yield the Diels Alder product shown here:
CO2Me
CO2Me
Best would be the pair:
+
[the opposing pair, acetylene and the dienoate, is a reverse polarity DA and not as likely.]
c) 2,3-Dimethyl-1,3-butadiene readily undergoes Diels-Alder reactions. 2,3-Di-(t-butyl)-1,3butadiene (shown below) does not. Explain this difference.
(CH3)3C
(CH3)3C
In order for a diene to undergo the DA reaction, it must be able to adopt the s-cis configuration,
shown above. The two t-butyl groups are too large to permit this conformation to exist (severe steric
interactions) and thus the compound exists solely in the s-trans form, shown here:
4. A chemist wanted the product shown below as a possible precursor for a steroid synthesis. Provide
a structure for each of the lettered intermediates [A - E]. [8 marks]
O
MeO
+
C11H12O4
O
AlCl3
NH2NH2
D
C13H14O4
MeO
COOH
KOH
DMSO
∆
OH
TsOH
benzene
∆
H3O+
E
COOH
O
B
O
MeO
O
O
D
5. Consider the compound shown here.......
SOCl2
C
C13H16O5
MeO
O
O
MeO
B
C13H16O3
MeO
A
O
HO
A
AlCl3
O
O
O
E
MeO
COCl
O
O
C
O
O
O
D
H
O
D
Br
a) excess D2O
O
O
H
D
reaction with:
O
O
H
Ph
H
b) 1 mole Br2
c) 1 mole PhCHO
d) Explain why optically active piperitone ("D = -32/), shown below, is converted to racemic
piperitone upon standing in a solution of sodium ethoxide in ethanol.
EtO
O
O
H
O
H
deprotonation of (-)-piperitone by ethoxide produces a planar enolate; when this is reprotonated,
addition of H+ can occur from below [regenerating (-)-piperitone], or from above [giving (+)-piperitone].
Since the protonation of the enolate must occur with equal ease from above or below, a racemate results.
6. There are fatal flaws in each step in the following sequence of reactions. Clearly identify what is
wrong with each reaction.
NO2
HNO2
CF3CO3H
H2SO4
CH2Cl2
NH2
O
NH2
Cl
AlCl3
O
Br
NaNO2
Br2
H2SO4
0-5°
Fe
CO2Et
O
Mg
H
H3O+
OEt
LiAlH4
Et2O
dry Et2O
O
O
CH2OH
CH2OH
MeLi
H2O
H3O+
dry Et2O
O
i) must be HNO3 not HNO2 for a nitration
OH
ii) this reagent oxidizes NH2 to NO2 - here a reducing agent is needed
iii) with the unprotected NH2, this reaction will not go
iv) Br2 and Fe are electrophilic substitution conditions; here we need CuBr, HBr
v) two problems – cannot do a Grignard reaction in a compound which contains a keto group (it
will react internally); reaction of a Grignard with an ester does not yield an ester product!
vi) LAH will also reduce the ketone
vii) MeLi will quench with the OH which is already in the compound.
7. Aromaticity [8 marks]
Use the Hückel rule to estimate whether each of the following compounds will exhibit aromaticity
or not. Give your reasons. The correct answer for the wrong reason will not get any marks.
N
N
N
N
O
O
aromatic
10 π electrons
2 per double bond
2 per oxygen
not aromatic
saturated carbons
in the periphery
not conjugated
aromatic
14 π electrons
2 per double bond
none of the N lone
pair electrons count
not aromatic
8 π electrons
4 per double bond
8. For each section of this question, pairs of structures or pairs of hydrogens within a structure are
compared with respect to a property or process. In each section, indicate which choice is correct,
and briefly indicate why.
a) Which hydrogen is more rapidly abstracted by base, Ha or Hb?
O
Hb
Ha since the anion which is produced is situated between two
carbonyls and can therefore be delocalized by both. Hb has only one
adjacent carbonyl.
Ha
O
b) Which is the stronger acid?
O
O
OH
F
F F
OH
CF3COOH is stronger; the fluorines are extremely electronegative and inductively withdraw electron density away from
the carboxylate anion, thus stabilizing it. Methyl groups are
electron donors, and the effect is opposite.
c) Which compound will undergo addition with cyanide ion (-CN) at a faster rate?
O
the aldehyde: steric constraints in the ketone slow down -CN
attack, and cause greater steric strain or compression in the
product. Also, the extra methyl substituent, being electron
donating, also decreases the ease of attack at the C=O by
diminishing the partial positive charge.
O
H
d) Which proton is more acidic, Ha or Hb?
O
O
O
Ha
Hb
Ha: the carbonyl can participate freely in the charge
delocalization through enolate formation; in the ester, the
second oxygen competes for the carbonyl group’s attention
through resonance interactions, decreasing the enolate
stability.
9. The reaction of N-phenylbenzamide, shown below, with fuming sulfuric acid [H2SO4 + SO3] gives
predominantly one product. On the basis of the structures of the intermediates involved, explain
- which ring is attacked, and why
H
- why the other ring is not attacked
N
- why only one product is seen.
O
As we look at the molecule, the right hand ring is attacked.
H
N
Ph
O
H
N
Ph
SO3
H
N
Ph
O
H
SO3
O
H
H
N
Ph
O
H
SO3
SO3
H
N
Ph
O
H
SO3
This is due to the delocalization possible within the Wheland intermediate, particularly the fourth
resonance structure which results from the donation of the nitrogen lone pair, shown above in red.
The other ring is not attacked because the intermediate which results is destabilized by the presence
of the electron-withdrawing carbonyl group:
H
O3S
δ+
NHPh
O
δ+
intermediate for ortho- or paraattack has to place two partly
positive charges adjacent -destabilizing!
δ+δ+ NHPh
O
δ-
or
intermediate for meta- attack has two
partly positive charges adjacent to
the partly positive substituent-also destabilizing!
δ NHPh
+
H
O3S
O δ
10. Amines can act as acids when protonated; the pKa is then defined as loss of the proton from the
ammonium species. You can also think of this in the reverse fashion -- basicity is a measure of the
ease with which the proton is picked up, and this will depend on the availability of the lone pair.
Using this reasoning, match each of the compounds below with one of the pKa values on the right,
and give a very brief explanation of your assignment (i.e. why the acidities vary in the way you
chose)
H
N
NH2
NH2
H
N
NH2
O
8.33
4.63
NO2
1.00
11.12
NH2
6.15
All the anilinium ions will be more acidic than the others; resonance delocalization of the nitrogen
lone pair into the ring makes proton loss from the salt very easy (since no resonance can occur in the
quaternary ammonium system). Conversely, you could say that the lone pair is tightly held in the
aromatic amines, meaning they will pick up a proton less easily.
Among the anilines, p-nitro is most acidic: NO2 is strongly electron withdrawing by resonance, and
thus aids in delocalization of the nitrogen lone pair into the ring. This makes it less basic (less likely
to pick up a hydrogen ion) and therefore more acidic. p-Amino is going to be more basic than aniline,
since the substituent is an electron donor (making the other N more likely to pick up a proton).
Morpholine (with the oxygen in the ring) is considerably more acidic than piperidine. Oxygen is
electronegative, and thus by the inductive effect causes the electron pair to be held more tightly on the
nitrogen. This makes it less basic (less likely to pick up a hydrogen ion) and therefore more acidic.
11. Answer the questions below which related to mechanism of reaction. Remember to include all
intermediate structures and use proper arrows. [8 marks]
a) Give the mechanism for the ester hydrolysis shown here, clearly showing where the labelled O18 is.
18
O
H2O
OEt
H+ ∆
18 O
H+
H2O 18
H2O
OH
OEt
18
HO
OH
+H+
OEt
HO
18
18
HO 18
OH
-H+
OEt
18
18
18
OH
+H+
18
OH
+ EtOH
b) Give the mechanism for one of the following two reactions:
O
OCH2CH3
H 3 O+
LiAlH4
+
AlH3
O
H
OCH2CH3
OCH2CH3
H
H
H
O
H
CH3OH
H
H
AlH2OCH2CH3
AlH2OCH2CH3
O
HO
HOCH2CH3
O
O
AlH3
or:
CH2OH
ether
∆
H+
CH2OH
H3O+
O
+ CH3CH2OH
+ Al(OH)3
OCH3
OH
+
OEt H
O
-H+
OEt
H+
18
CH3OH
O
HO
H
H+
OH
H+
O
OH
O
OH
OH
H
CH3OH
H
O
OCH3
O
OCH3
O
CH3OH
O
OH2
12 The reaction of the diene shown below with dry HBr can lead to three different products (do not
worry about temperature effects). Give a formula for each of these possible products, and draw a
mechanism which accounts for their formation.
HBr (one mole)
Br
H
Br
Br
Br
Br
same
product
Br
H
Br
Br
Br
Br
D. SYNTHESIS [36 MARKS]
13. Complete THREE of the four syntheses shown below, using only the indicated starting materials
and any other necessary reagents, solvents, etc., remembering that you must make any organometallic
reagents you with to use [this refers to RMgX or RLi, not to LiAlH4]. Each synthesis may be carried
out in five or less steps.
a)
OCH3
from benzene and any
other necessary reagents
OH
Br
Br
Br
Br2
SO3
Fe
H2SO4
CH3I
KOH
heat
OH
Br
HO-
OH
SO3H
OMe
HCHO H3O+
Li
Et2O
OMe
b)
F
from benzene and any
necessary reagents
Br
NO2
NO2
HNO3
Br2
H2SO4
Fe
NaNO2
H2SO4
0 - 5º
HPF6
NH2
NH3
H2S
Br
EtOH
Br
F
mineral oil
heat
Br
c)
O
using one of
CH2(COOEt)2 ,
COOEt
or
plus benzene any other necessary
reagents excluding benzene compounds
H2C
COCH3
CH3I
NBS
AlCl3
hν
Br
O
COOEt
Br
O
NaOEt
EtOH
CH2(COCH3)2 ,
COOEt
H3O+
heat
from compounds of four
carbons or less
d)
O
Br
two moles
Mg
OEt
H
Et2O
OH
PCC
Ph3P=CH2
CH2Cl2
and
H3O+
THF
O
CH3I
Ph3P
THF
BuLi
Ph3P=CH2
O