CHM238
1.
Fall
2002
Practice Exam 2
Complete 5 out of the 6 following reactions, showing the main organic products. Cross one out or graded in order.
CH3
O
CH 3
O
AlCl3
O
CH3 H3C
+ H3C
CH3
Cl
(a) Friedel-Crafts Acylation, CH3 is o,p director
Br
H3C
CH
H3C
CH 3
CH3
+ Br2 NBS
(b) benzylic substitution
CH3
CH3
CH3
OH
Br
o
350 C + high P
+ NaOH
H2O
OH
(c) reaction thru benzyne intermediate
NO 2
Cl
+
NO2
O
OH
NaOH
-
N
+
O
-
OH
Cl
from
Meisenheimer
(d) Nucleophilic Aromatic Substitution
CH3
CH 3
H3C
+ Br2, FeBr3
Br
Br
(e) o,p directed electrophilic aromatic substitution.
N
C
N
+ Br2, FeBr3
(f) meta directed, EAS
Br
C
2.
Fill in the proper reagents A- F
Br
Br
+A
O
CH3
+C
+B
O2N
A
Br2, FeBr3
B
O2N
HNO3/H2SO4
C
Na+ CH3O-
O
O
+D
+E
+F
OH
O
CH3
AlCl 3
DCl
3.
E
H2, Pt
F
KMnO4
In the following reaction of chlorobenzene with nitric/sulfuric acid, complete the resonance structures below with the
correct bonds. Then answer the questions underneath.
Cl
Cl
Cl
Cl
+
Cl
HNO 3
H2SO4
H
NO 2
H
NO 2
H
NO 2
H
NO 2
.
(a) Why is the sulfuric acid necessary? Nitric acid is already acidic, isn’t it? (Note : pKa (Nitric Acid) = -1.5, pKa
(Sulfuric Acid) = -10).
O
H
O
O
+
H
+
+
+
+
N
O
H2O
N
N
O
OH
O
H
O
(b) What is the reactive electrophile in the above reaction?
NO2+, nitronium ion.
(c) If we used only pure (fuming) sulfuric acid, what would be the product(s)?
mostly sulfonation of Cl benzene, both o and p, because SO3H+ becomes the superelectrophile and there is not as much
protons for the dehydration of nitric acid.
(d) Chlorine is o,p directing group but chlorobenzene is slower to react with nitric/sulfuric acid than benzene is. Why?
Chlorine is o,p directing because of resonance. Note the bonus structure above. 4 resonance structures vs 3 for meta subst.
Chlorine is EWG through sigma bonds because of its high electronegativity. Chloro-substituted benzene has less electron
density to donate towards the NO2+ group.
+
4.
Starting with toluene, design syntheses, providing the correct reagents for the following transformations. Both
processes can be accomplished with 2 steps, but there is more than one correct answer for each. Assume that ortho and
para isomers can be separated.
OH
O
CH3
Br
Toluene is reacted with KMnO4 to make benzoic acid. Benzoic acid is then brominated, Br 2, FeBr3 to give metabromobenzoic acid. Carboxylic acids are EWG.
OH
O
CH 3
Br
Toluene is first brominated with Br2, FeBr3. Then KMnO4 turns the methyl group into a carboxylic acid. This is the only
way to make para substituition, because Br is o,p director.
5. Assign the NMR spectrum below to the molecule with the formula C 8H8O2 that reacts with nitric/sulfuric acid to make
the meta product. Draw arrows to peaks. This is the reactant not the product. IR peak near 1700 cm -1, 1620 cm-1 and C-H
sp2
Molecule is single substituted benzene (5H) with a C=O (1700 cm-1 IR peak) and EWG-meta director.
O
H
CH3
O
H
H
3H
3H
2H
8
7
6
5
4
6. Draw the following alcohols:
HO
H
C
H3C
CH
CH3
CH2
(S)-2-butanol
H3C
OH
OH
CH
CH2
CH2
H2C
H3C
allyl alcohol
isobutyl alcohol
7. Rank the order of acidity from l most acidic  3 least acidic.
1
2
3
phenol
methyl alcohol
isopropyl alcohol
2
3
1
phenol
p-methylphenol
p-chlorophenol
8. Write the reagent that you would use to reduce the following carbonyl compounds over the arrow. Predict the product.
O
CH3
LAH
O
O
CH
HC
C
HC
CH2
OH
NaBH4
LAH
OH
CH
CH
9. Write reagents over the arrows for these transformations:
CH 3
O
KMnO 4
C
O
PCC
LAH
OH
CH CH 3MgBr
CH 2OH
CH
OH
CH 3
Acid Workup
10. What would be the product of
Br
(a) 1-propanol + PBr3 
(b) cyclohexanol + POCl3 
H3C
H
H3C
(c) (S)-2-butanol + TosCl/pyridine followed by NaBr in DMF?
Br
C
CH2
CH3
O
H3C
CH2 C
O
H3C CH2 C
OH (e) 1-propanol + PCC 
H
(d) 1-propanol + Jones Reagent (CrO3-H2SO4) 
11. Predict the products of the reaction of Phenyl MgBr with the carbonyl compounds below (assume acid workup):
OH
O
H3C
H
H3C
O
OH
O
NR
Benzene + alkoxide
HO
12. The alcohol in 11.c would cause trouble. Why?
How could we get around this problem?
12. The problem is the alcohol will react with Grignard to make benzene and alkoxide.
R
pKa
OH
+
R
MgBr
R
O
-
+
R
16
H
45
Keq = 1030
The solution is to protect the alcohol, perform the Grignard and then deprotect.
OH
OH
O
HO
TMSCl
pyridine
CH3
TMSO
O
PhMgBr
CH3
TMSO
CH3
HO
CH3