Organic Chemistry I Q.1/25 Examination Three 14 November ...
Transcript of Organic Chemistry I Q.1/25 Examination Three 14 November ...
Name:
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Organic Chemistry I
Examination Three
14 November 2012
ANSWERS and NOTES
MPTP – a single low dose of this compound causes permanent symptoms of Parkinson's disease by
destroying dopaminergic neurons in the substantia nigra of the brain.
(Used as a Parkinson’s disease model in animal studies.)
Careless and untidy work will be penalized.
An illegible answer is a wrong answer.
To score full points, all explanations must be accompanied
by appropriate reaction schemes, mechanisms and
intermediates as appropriate.
Q.1/25
Q.2/25
Q.3/25
Q.4/25
Q.5/25
Q.6/25
Q.7/25
Q.8/25
Q.9/25
Q.10/25
Total/250
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1. MPTP is a common (and dangerous) impurity from the illicit synthesis of the designer
drug “MPPP” (shown below).
An unsatisfactory attempt to purify a sample of MPPP using recrystallization from hot
ethanol instead yielded mainly MPTP. Why? Please explain using a mechanism. What
are the other product(s) likely to be found in the recrystallization mixture?
MPPP MPTP
Q.1
3
2. Write a reasonable mechanism for the following reaction, showing all products.
Q.2
4
3. Show all missing starting materials, reagents or products in the following
transformations:
a.
b.
c.
d.
Q.3
5
4. Propose reasonable synthetic sequences to convert the specified starting materials
SM4 into the designated products, TM4: a retrosynthetic approach for each synthesis is
necessary and should arrive in each case at the specified starting material.
Retrosynthesis
Synthesis
Retrosynthesis
Synthesis
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Retrosynthesis
Synthesis
Q.4
7
5. Consider the reaction mechanisms for each of the following: in each case, there is an
unusual or unexpected feature which you are required to identify or explain after
providing the reaction mechanism.
Mechanism
SN1 with the departing Bromide ion hindering ingress of methanol
Feature
Neither 100% yield of trans product (SN2) nor 50:50 if simple SN1 so possibly a mix of both
mechanisms. But see above.
Mechanism
Br
OEt
1. EtOH 2. -H+
OEt
Ho
1. EtOH
2. -H+
Feature
Hydride shift to form a tertiary carbocation whence the second product.
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Mechanism
Feature
The scrambling of the deuteriums means that there must be an allylic carbocation and therefore
a hydride shift too.
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d.EtOH
OEt
I
+ OEt (!)
Mechanisms
I
OEt
o
Hydrideshift
OEt
H
AA
1. EtOH 2. -H+
1. EtOH 2. -H+
II
I I
A simultaneous bond breaking and ring expansion is a non‐standard way to understand this
odd reaction.
Feature
The (forbidden?) formation of a seven membered ring from a six membered ring is clearly (!)
the feature. It looks like a primary carbocation and a hydride shift features too!
Q.5
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6. When the following elimination reaction was conducted in ethanol with a low concentration of hydroxide ion, the following observation was made:
Provide the mechanism(s) for the formation of each of the products, additionally given
that when the overall concentration of the reaction mixture was reduced by a factor of
two, the reaction rate was reduced by a factor of 2.5.
The rate reduction factor suggests that both E1 and E2 reactions occur here. The respective mechanisms
are as follows:
Br
Major
MinorE1
Br
H
H
(more highly substituted)
E2
BrH
HH
B
B
Minor
Br
HH
Major Minor
B
Algebraic analysis of the rate data
Rate(0) = k1[RBr] + k2 [RBr][Base] Rate(/2) = k1[RBr]/2 + k2[RBr][Base]/4
= Rate(0)/2 ‐ k2[RBr][Base]/4
So the reduction in reaction rate is greater than two.
Q.6
11
7. Show the products obtained by treating 3‐chloro‐2,2,3‐trimethylpentane with
concentrated sodium ethoxide in ethanol. Label them as “major” and “minor”, and
provide their systematic names. Explain why the products are produced in this ratio
using appropriate projections to assist.
Elimination via an E2 mechanism
This will involve the loss of the elements of HCl with the hydrogens coming from the positions
shown: Newman projections are shown viewed along the bond indicated.
Cl
t-Bu CH3
H
CH3H
t-Bu
CH3 ClH
CH3H
unfavorable
Antiperiplanar for E2
t-Bu CH3
CH3H
Antiperiplanar for E2
Major
t-Bu
CH3
H
CH3
Minor
(Z)-3,4,4-trimethylpent-2-ene
(E)-3,4,4-trimethylpent-2-ene
Q.7
12
8. Three substitution products and three elimination products are obtained from the
following reaction. Account for the formation of each product shown using a
mechanistic explanation.
P1 P2 P3 P4 P5 P6
As the mechanism unfolds, each carbocation from the initial E1/SN1 step can lose a proton for overall
elimination (and alkene formation (P4, P5, P6)) or react with methanol and then lose a proton to form
an ether (P1, P2, P3). The details have been dealt with in other questions (see Q.5b and Q.6)
Br
H
H
Ho
o
OCH3
OCH3
OCH3
P1
P2
P3
P4
P5
P6
Q.8
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9. Write the mechanism for the ZnCl2/HCl “Lucas Test” reaction with an alcohol. Why
does the solution become cloudy immediately for tertiary alcohols, but only becomes
cloudy after 5 minutes for secondary alcohols?
ZnCl2 can be used to catalyze certain SN reactions: ZnCl2 functions as a Lewis acid that complexes strongly with the lone‐pair electrons on oxygen:
Test results and mechanisms at room temperature:
• Primary: No reaction SN2 not required here………………..
• Secondary: Reaction in ~5 minutes SN2 and SN1
• Tertiary, allylic, and benzylic: Reaction immediate SN1
Primary
Secondary
Q.9
14
10. Please provide mechanistic explanations for each of the following experimental observations:
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No change in stereochemistry in the ring implies that the reaction involves t‐BuOH in an SN1 reaction
The strong base removes one of the more acidic protons in an elimination reaction. Cumulene C can also form.
Br
Br
Br
HaHaHb
HbHa
Ha
The Ha are moreacidic than the Hb
Br
Br
H O
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An intramolecular SN1 reaction after the initial protonation.
Q.10