Chem 14C Lecture 1
Spring 2011
Final Exam Part B
Solutions
Page 1
Statistics
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A note about exam keys
: The answers presented here may be significantly longer than expected from a
student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as
well.
To see the final course grade cutoffs, consult the grading scale on the Chem 14C course web page.
1.
Benzene ring (or phenyl), amine, and ester.
2.
Molecule
A
has five
lone pairs and three
atoms that are more electronegative than carbon.
3.
C–O
This bond is composed of the atoms with the largest electronegativity difference. Note that not all
the answer choices are present in molecule
A
.
4.
H
N
O
O
CH
3
H
N
O
O
CH
3
All other choices violate at least one
resonance contributor preference rule,
and are therefore less significant
.
5.
(c) Both structures are of approximately equal significance.
In fact, they are degenerate
.
6.
(a) Conjugation
: Stabilization due to delocalization of pi electrons in three or more parallel, adjacent,
overlapping
p
orbitals.
(b) Aromaticity
: Stabilization due to delocalization of 4n+2 (n = any integer) pi electrons in a closed
loop of three or more parallel, adjacent, overlapping
p
orbitals.
(c) Anomeric carbon
: In a cyclic carbohydrate, the carbon atom that becomes the carbonyl carbon in the
carbohydrate's acyclic form.
7.
Many answers are possible. For example:
O
OH
HO
OH
8.
Any molecule adopts a shape, geometry, or conformation that allows conjugation because conjugation
increases molecular stability.
9.
(a) I and III; (b) I; and (c) I and III.
10.
H
N
H
CO
2
CH
3
H
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Chem 14C Lecture 1
Spring 2011
Final Exam Part B
Solutions
Page 2
11.
In the set of molecules that includes Concerta and all its stereoisomers, there are four
molecules that are
enantiomers and four
molecules that are diastereomers.
Each stereoisomer in the set is an enantiomer
as well as a diastereomer of some other stereoisomer in the set
.

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- Fall '04
- Whiteshell
- Organic chemistry, Atom, CHCH2, CHCHCH
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