CHEM 2312: Homework 3
Due Sept. 15, 2003 (1:00 pm)
Student Name_________________________
Student ID:____________________________
Note: An answered hardcopy must be submitted to Dr. AJ Ragauskas; all 7 questions must be
answered. All chemical structures must be illustrated with molecular drawing software.
Question 1. Provide a multi-step synthesis of the molecule on the left starting with the corresponding
structure on the right. Do not provide mechanisms, just the synthetic steps (including
reagents, and solvents and conditions when applicable). Note: the product can not be made
in a single step. (5 points)
CN
from this:
a. make this:
Br
CN
NaCN
NBS
CCl4
light or heat
Question 2. Calicene (structure below) has a surprisingly large dipole for a hydrocarbon. (a) Provide
the structure of a major (i.e. stable) resonance contributor that explains the polarity. (b)
Provide a brief (one sentence) explanation that accounts for the stability of the resonance
structure that you have drawn. (5 points)
This resonance structure is especially stable because both of the rings
(cyclopentadienyl anion and cyclopropenyl cation) are aromatic {since
n= 1 for cyclopentadiene anion and 0 for cyclopropenyl cation with
respect to Huckels rule: 4n+2} (
or other resonance structure
CHEM 2312: Homework 3
Due Sept. 15, 2003(1:00 pm)
Question 3.
Student Name_________________________
Student ID:____________________________
Both cyclooctatetraene and styrene have the molecular formula C8H8 and undergo
combustion according to the equation:
C8H8 + 10 O2 -> 8 CO2 + 4 H2O
The measured heats of combustion are 4393 and 4543 kJ/mol. Which heat of combustion
belongs to which compound and why? (4 points)
Answer>> Styrene contains an aromatic ring and will be appreciable stabilized by resonance, which
makes it lower in energy than cyclooctatetraene.
Styrene>> 4393 kJ/mol (# π e-= 6 > 4n+2 n=1)
Cyclooctatetraene >> 4543 kJ/mol mol (# π e-= 8 does not satisfy Huckels rule)
Question 4.
Classify each compound as aromatic or not and explain via Huckels’ rule why? (6 points)
O
H2N
N
N
A
Aromatic>>
H
H
B
N
B
H
B
N H
B
N
H
HC
O
O
N
D
N H
O
E
F
A (6 π electron, satisfy Huckel rule)
C (6 electron from 3 N atoms, satisfy Huckel rule)
D (4 π electron, 2 electrons from O, satisfy Huckel rule)
F (8 π electron,2 electrons from N, n=2 with respect to 4n+2, satisfy Huckel rule)
Not aromatic >>B (6 π electrons + 2 nonbonding electrons from oxygen does not satisfy Huckel rule)
E (4 π electrons + 4 nonbonding electrons from the two oxygen does not satisfy
Huckel rule)
CHEM 2312: Homework 3
Student Name_________________________
Due Sept. 15, 2003 (1:00 pm)
Student ID:____________________________
Question 5. Classify each compound as aromatic or not and explain why? (4 points)
H
H
H
H
H
A
B
C
D
A is not aromatic because of CH2 and hence π is not a cyclic conjugated system
B is not aromatic because it contains 11 π electrons and does not satisfy Huckel rule
C is aromatic because it contains 10 π electrons and does satisfy Huckel rule (i.e., n=2; 4n+2)
D is not aromatic because it contains 12 π electrons and does not satisfy Huckel rule
Question 6.
Provide a molecular orbital energy diagram description for the ground state and lowest
excited state of benzene. (4 points)
Ground State
π
*
E
π
π
Molecular orbital
π
Excited State
π
*
E
π
π
π
Molecular orbital
CHEM 2312: Homework 3
Due Sept. 15, 2003 (1:00 pm)
Question 7.
Student Name_________________________
Student ID:____________________________
On heating 1,2,4,5-tetramethylbenzene with fuming sulfuric acid one product is isolated
with the molecular formula of C10H14O3S in 94% yield. Identify the structure of the
product and provide a reasonable reaction mechanism for the product. (6 points)
O
HO S OH
H-O-SO3H
O
H
O
O S O H
H
O
-
OSO3H
-
OSO3H
H
O
S OO
O S
O
+ H2SO4
O
O
S OO
O
HO S OH
O
O
S O H
O
C10H14O3S