CHEM 302
Organic Chemistry I
Exam I
16-September-2005
Answers
1) Provide proper IUPAC names for the following compounds:
a) (CH3)2CHCH2CH2CH(CH2CH3)CH2C(CH3)3
4-ethyl-2,2,7-trimethyloctane
b)
H
H
trans-1-(1,1-dimethylethyl)-4-ethylcyclohexane
c)
3-methyl-5-(1-methylethyl)octane
2) Draw structures for the following names
a) cis-1-(2-butyl)-2-ethylcyclopentane
b) trans-1-sec-butyl-2-ethylcyclohexane
c) 4-(2,2-dibromoethyl)-3,5-dichloroheptane
Br
Br
Cl
Cl
d) 5-(2-butyl)-4-ethyl-methylcycloheptane
(maybe a better name would have been: 4-(2-butyl)-5-ethyl-methylcycloheptane)
e) 3-(trans-3,4-dimethylcyclohexyl)-2,2,4-trimethyloctane
3) The rates of hydration of the following isomeric alkenes differ markedly. Which
one reacts faster? Explain briefly but concisely.
The second one would be expected to react faster, since the cation which is
formed is 3°, which is more stable than either 1° or 2°. Cation stability goes in the order
3°>2°>1°>methyl.
A less good answer (which is incomplete) speaks to the first isomer being more
stable (which is most likely is), but ignores the transition state/intermediate through
which the reaction must proceed. This answer would have been made complete by also
speaking to the intermediate cations formed, and the relative stability of these. (Had I
merely given you the (E) and (Z) isomers of the first one, then the stability of the starting
material would have been considerably more relevant.)
4) How would you differentiate between the two compounds in each of the following
pairs by 13C-NMR spectroscopy?
The easiest way to answer this question is to address the number of different
carbon atoms, which is what I will do.
and
a)
There are 4 different carbons in the first one; 8 in the second one.
Cl
Cl
and
Cl
b) Cl
There are 3 different carbons in the first one; 2 in the second.
and
c)
There are 4 different carbons in the first one; 8 in the second.
5) In preparation for this test, you drew a conformational analysis diagram for 3,4dimethylhexane. I gave you the isomer, below left, to use. Now I want you to draw
the conformational analysis diagram for the other isomer of this compound, given
below right. Same assumptions apply i.e. two ethyls yield more steric interaction
than a methyl and an ethyl, and two hydrogens are on the bottom of the steric
interaction scale. Draw it looking down the C3-C4 bond.
H
(old)
H
H
H
(new--do this one)
Here are the six conformers you need to examine. For clarity, the “eclipsed” conformers
are drawn slightly staggered—these are #1, 3, and 5.
H H
H C2H5
H
C2H5
C2H5
CH3
C2H5
H3C
H
H3C
C2H5
CH3
1
3
2
H CH3
H
H3C
C2H5
H3C
C2H5
C2H5
H
H3C
H3C
H
H
H
H3C
H
C2H5
C2H5
CH3
H3C
C2H5
C2H5
6
5
4
By almost any measure chosen, conformer 1 is the highest energy (due to CH3C2H5 eclipsing), and structure 2 or 4 are the lowest (these are very close, so it is really a
matter of choice here.) My organic chemistry instinct gave me the order: 1>5≈3>2≈6>4.
[A quick check via calculations gives an order: 1>5≈3>4>2≈6, so I wasn’t too far off.]
1
2
3
4
5
6
(This is the best diagram I could draw with my software; the idea is what I am
after.)
6) For homework, I gave you a problem in which HCl was added to butadiene-1,3,
and two products were obtained: 3-chlorobutene-1, and (E)-1-chlorobutene-2. An
interesting thing happens when this reaction is run at elevated temperatures: the
product mixture has the same two compounds but in different ratios. Furthermore,
placing a pure sample of either one in a flask and heating to an elevated
temperature produces the same mixture of products in the end. Thinking carefully
about this datum, propose a reasonable mechanism to account for this observation.
Remember than a mechanism shows clearly the movement of electrons, and goes
stepwise.
Below, I’ve shown the two isomers, and the mechanistic arrows which lead to a
resonance-stabilized cationic intermediate. Note that the cationic intermediate arises from
both products, and can therefore be used to account for the formation of both products
from either pure product. Thus, the equilibrium arrows.
Cl
+Cl-
-Cl-
Cl
+Cl- -Cl-
both of which can be represented as