BELLARMINE COLLEGE
DEPARTMENT OF CHEMISTRY & PHYSICS
CHEM217 INTERMEDIATE CHEMISTRY II
SECOND MIDTERM EXAM
3/6/ 99
TIME: 50 MINUTES
ANSWER KEY
PART ONE: RELATIVE REACTIVITIES – Arrange any five (6) of the following in decreasing order (i.e., greatest
first) for the indicated reactivity. Use the following code to identify your answers. Do only five (6) of questions 1 – 8.
No mechanisms are required. (15)
A. i > ii > iii
B. i > iii > ii
1.
C. ii > i > iii
D. ii > iii > i
E. iii > ii > i
F. iii > i > ii
Rate of E2 elimination (NaOEt/EtOH/):
i) CH3CH2Cl
ii) CH3CH2I
iii) CH3CH2OH
ANSWER: C
The variable here is leaving group ability, Cl- vs. I- vs. OH-. Leaving group ability follows conjugate
base stability. Cl and I are in the same group. I - is more stable than Cl- so I- is a better leaving group
than Cl-. I- is larger than Cl- so that the valence shell electrons in I- are more delocalized and hence
repel each other less. OH- is a much stronger base than either Cl- or I- (or H2O is a much weaker acid
than either HI or HCl) and so is the worst leaving group.
2.
Nucleophilicity in SN2 reaction:
ii) CH3O-
i) CH3OH
iii) PhO-
ANSWER: D
Since all these bases are oxygen-based, nucleophilicity is proportional to basicity. Conjugate bases are
obviously more basic (nucleophilic) than their acids so CH 3O- (methoxide) will be a stronger nucleophile
than CH3OH. PhO- (phenoxide) is less basic than CH3O- because the benzene ring is able to delocalize
(stabilize) the O negative charge by resonance. PhO - is, however, a stronger base than the neutral
CH3OH.
O-
O
O
-
O
-
3.
Leaving group ability in SN1 reaction:
i) Br-
ii) PhO-
iii) Cl-
ANSWER: B
Leaving group ability is inversely proportional to basicity. Br - is more stable than Cl- (see Q1 above)
which in turn is more stable than PhO-. Another way of looking at this is to examine the conjugate
acids. Both HBr and HCl are strong acids (pK a's < 0) whilst PhOH (phenol) is considered to be a weak
acid with a pKa around 10. Strong acids have weak conjugate bases, which are good leaving groups.
4.
Rate of E1 elimination:
i) Ph2CICH3
ii) CH3CHICH3
iii) (CH3)3CI
ANSWER: B
The variable here is carbocation stability. Both i) and iii) give 3 o carbocations whilst ii) produces a 2o
carbocation. Carbocation stability decreases in the order 3o > 2o > 1o > Me due to the electron-donating
capability (inductive and resonance - hyperconjugation) of alkyl groups attached to the carbocation
center. Phenyl groups are better able to stabilize a positive charge (resonance) than alkyl groups (more
delocalized) which makes the carbocation from i) more stable than the carbocation from iii).
+
CH 3
CH 3
CH 3
+
etc.
+
CH 3
CH 3
5.
Rate of SN2 reaction with -CN:
i) (CH3)3CCH2Cl
ii) CH3CH2CH2Cl
iii) (CH3)2CHCH2Cl
ANSWER: D
Though all three alkyl chlorides are nominally 1o substrates, they each suffer from differing amounts of
secondary steric hindrance. Alkyl groups connected not to the elelctrophilic center but further down
the chain hinder backside attack of the nucleophile. The more methyl groups on carbon 2, the more
hindered the nucleophilic attack.
CH 3
CH 3
CH 3
CN
C
C
H
H
Cl
1o carbon
methyl groups block
attack of nucleophile
6.
Rate of SN2 reaction with CH3I:
i)
C6H5O-
ii) p-NO2C6H4O-
iii) p-CH3OC6H4O-
ANSWER: F
The more stable the phenoxide ion the less reactive it is as a nucleophile and vice versa. Electronwithdrawing groups (EWG's) will stabilize the negative charge through resonance whilst electrondonating groups (EDG's) will destabilize the ion.
O-
O-
O
O
etc.
N+
-
O
N+
-
O
O
OCH 3
OCH 3
O-
unstable resonance
structure
7.
Rate of SN1 reaction with H2O
i) p-NO2C6H4CH2Cl
ii) C6H5CH2Cl
iii) p-CH3OC6H4CH2Cl
ANSWER: E
The variable is carbocation stability. EDG's will stabilize a carbocation whilst EWG's will destabilize a
carbocation. Note this is exactly the opposite of phenoxide stability as shown in Q6 above.
CH 2+
CH 2+
CH 2
CH 2
etc.
etc.
+
OCH 3
+
N
OCH 3
-
+
O
N
-
O
O
+
O
unstable resonance
structure
8.
% of enol form in keto-enol tautomerism:
i) CH3COCH3
ii) CH3COCH2COCH3
iii) CH3COCH2CO2Et
ANSWER: D
-Dicarbonyls tend to have a greater % of the enol form than monocarbonyls because hydrogen bonding
between the enol OH and the remaining carbonyl O stabilizes the enol form.

H
O
O
-Diketones tend to have a greater % of the enol form that -ketoesters because the carbonyl of an ester
is more stable than the carbonyl of a ketone due to the electron-donating nature of the alkoxy group by
resonance. This greater carbonyl stability favors the keto form over the enol form.

O
O
O
O-
+
OEt
OEt
PART TWO: EXPLANATIONS/MECHANISMS - Do any two (2) of the following questions. Remember to keep
verbiage to a minimum and to use structures, drawings, mechanisms, etc. Answers on the following page, please. (20)
9.
For any one (1) of the questions you didn't answer in PART ONE above provide an explanation.
SEE PART ONE ABOVE.
10. Ethers can often be prepared by the reaction of alkoxide ions with alkyl halides (the Williamson ether synthesis).
Suppose you wanted to prepare methoxycyclohexane. Which of the two possible routes shown here would you
choose? Explain (your answer must compare both routes).
Both routes involve the SN2 attack of an alkoxide ion on an alkyl iodide. The key difference is that route 1
involves methyl iodide as the substrate whereas iodocyclohexane, a 2 o alkyl iodide, is involved in the second
route. Rates of SN2 reactions are highly dependent on crowding in the transition state which results in the
following decreasing order of reactivity: Me > 1o > 2o >> 3o. In addition, 2o substrates because of steric
hindrance are also more prone to undergoing E2 reactions. For both of these reasons, the first route would
give the greatest amount of the desired ether product.
O
O-
CH 3
I
CH 3
+ I-
CH3O-
CH3O-
H
H
H
H
I
SN2 - nucleophile approach
hindered
I
E2 - H is much less hindered
11. When 2,2-dimethylcyclohexanol undergoes acid-catalyzed dehydration, the two products formed are 1,2dimethylcyclohexene and isopropylidenecyclopentane rather than the expected 3,3-dimethylcylcohexene. Write
reasonable mechanisms that show the formation of the two products.
This is all about carbocation rearrangements followed by E1 elimination.
H+
OH 2+
OH
CH 3
CH 3
CH 3
CH 3
CH 3
+
CH 3
o
2 carbocation
CH 3
CH 3
CH 3
CH 3
CH 3
CH 3
H
+
OH2
+
H
CH 3
H2O
o
3 carbocation
o
3 carbocation
CH 3
12. Although both cis-4-tbutyl-1-bromocylohexane and trans-4-tbutyl-1-bromocyclohexane react with potassium
t
butoxide to yield 4-tbutylcyclohexene, the cis isomer reacts 500 times faster. Explain.
The trick here is to draw chair conformations, to remember that the tbutyl group is so large that it very
much prefers to be in the equatorial position (conformational locking group), and that the E2 mechanism is
operating (tbutoxide is a strong base) which means anti coplanar stereochemistry.
Br
trans
cis
Br
H
-
most stable
conformation
can't undergo
E2 mechanism
O
Br
least stable conformation
can undergo E2 mechanism
but there isn't very much of it
around.
H
-
O
PART THREE: REACTION PRODUCTS - For any four (4) of the following reactions provide the major organic
product(s). Be sure to include stereochemistry where required. No mechanisms are required. (12)
13.
CH 3O
OH
14.
Ph
Ph
16.
CH 3
Ph
NC
via
H
17.
CH 3O
15.
CH 3
H
TsO
CH 3
Ph
18.
H
Ph
19.
Ph
CH 3
H
OH
(CH3)2CH=CH2
Ph
Ph
CH 3
OH
H
CH 3
CH 3CH 2
CH 3
R/S racemic mixture
PART FOUR: COMPLEXES
20. Draw all possible isomers for the square planar complex PtBr 2(NH3)2. Pt is in its +2 oxidation state. (3)
Br
NH 3
Br
Pt
Br
Pt
NH 3
cis
NH 3
NH 3
Br
trans