Section:
A-Hudson
B-Katz
Name__Answer Key
Question 1.
a) Draw a Lewis structure for the molecule CO2.
O C O
b) Add dipole arrows above, showing all polar bonds.
c) Is CO2 a polar molecule? Yes / No (circle your answer)
d) What is the hybridization state of the carbon atom in CO2?
sp
e) Draw a picture that clearly represents the interacting orbitals of all the sigma bonds.
O
C
O
f) Draw a picture that clearly represents the interacting orbitals of all the pi bonds.
O
C
O
CH241 Exam 1 – Fall 2016
Section:
A-Hudson
B-Katz
Name__Answer Key
Question 2.
a) On the template below, draw two valid Lewis structures for the formate anion
(HCO2–). Make sure to add all lone pairs and identify any non-zero formal charges.
H
O
O
C
C
H
O
O
b) Add curved arrows to the structure on the left (above) showing how it can be
converted into the structure on the right.
c) What is the hybridization state of the carbon atom in formate?
sp2
d) Is the formate anion polar? Yes / No (circle your answer)
In the space below, briefly explain your answer.
The (large) bond dipoles of the C–O bonds do not fully oppose each other (120° angle
between them, not 180°). Thus the molecule cannot fully cancel all the bond dipoles
and is polar.
Question 3. Circle the structure with the most stable pi bond:
CH 3
H 3C
Question 4. Using a Newman projection, draw butane in the eclipsed conformation as
viewed down the C2-C3 bond axis.
CH
CH 3 3
H
H
H
H
CH241 Exam 1 – Fall 2016
Section:
A-Hudson
B-Katz
Name__Answer Key
Question 5. Consider the isomer of dibromobutane shown in the Newman projection A
below.
Br
H 3C
Br
H
CH 3
H
A
a) Three isomers of dibromobutane are shown (structures B-D). Compare each structure
to Newman projection A, and describe the relationship of B-D as either: structural
isomers, conformational isomers, enantiomers, or diasteriomers.
Note that conformational isomers must be able to interconvert by single bond rotations.
CH 3
H
Br
H
Br
H 3C
B
H
H 3C
Br
Br
CH 3
H
C
Br
H
CH 3
H
Br
CH 3
D
Relationship between:
A and B: ___Diastereomers____________________________________
A and C: ___Diastereomers____________________________________
A and D: ___Structural Isomers__________________________________
b) For structures B-D, circle any that are chiral. (None are chiral!)
Question 6. Draw (S)-4-tert-butyl-1-heptene
CH241 Exam 1 – Fall 2016
Section:
A-Hudson
B-Katz
Name__Answer Key
Question 7. Identify the number of stereoisomers for each of the following compounds.
F
a)
Br
Cl
b)
H
2
4
Question 8.
a) Provide the IUPAC name for each of the following compounds A-C:
A
(E)-3-methyl-2-pentene
B
2-ethyl-1-butene
C
Cl
(3R,4R)-4-chloro-6-ethyl-3-methyloctane
b) How many signals will appear in the 1H NMR spectrum of A (in other words, how
many hydrogen atom environment/types)?
5 signals
c) How many signals will appear in the 1H NMR spectrum of B (in other words, how
many hydrogen atom environment/types)?
3 signals
CH241 Exam 1 – Fall 2016
Section:
A-Hudson
B-Katz
Name__Answer Key
Question 9. Circle ‘soluble’ or ‘insoluble’ to make the following statement accurate:
When performing a recrystallization, the compound that you wish to isolate as a solid
crystal should be [soluble/insoluble] in the boiling solvent of your choice, and
[soluble/insoluble] in that solvent when cooled.
Question 10. When separating two compounds X and Y by thin layer chromatography
(TLC) with hexanes as the mobile phase, you find that compound X has an Rf of 0.8 and
compound Y has an Rf of 0.1.
a) Which compound is more polar? X / Y (circle your answer)
b) Would you expect the Rf of compound Y to be greater in ethyl acetate than in hexanes?
Yes / No
(circle your answer)
Give a brief explanation for your answer below:
Ethyl acetate is a more polar solvent than hexanes. Using a more polar eluent will
cause all species to travel more rapidly through silica gel, thus increasing the Rf of
both X and Y.
CH241 Exam 1 – Fall 2016
Section:
A-Hudson
B-Katz
Name__Answer Key
Extra Credit. [finish the rest of the test first!] Although the average C–H bond
dissociation energy (BDE) is about 100 kcal/mol, some trends emerge when we compare
specific compounds. Give a clear and complete explanation for the following trend in
BDE’s for the C–H bonds of ethane, ethylene, and acetylene:
101 kcal/mol
111 kcal/mol
H
H3C C H
H
130 kcal/mol
H
HC C H
H2C C
H
Better orbital overlap will result in a stronger bond from the standpoint of homolytic
cleavage (not the same as heterolytic cleavage, which has exactly the opposite trend!).
The hydrogen 1s orbital has better orbital overlap with more spherical orbitals, such as
sp-hybrids (50% S) and overlaps more poorly with elongated orbitals, such as sp3hybrids (25% S). The sp-hybrid is also lower in energy, closest in energy to the hydrogen
1s, and the opposite is true for the sp3-hybrid (highest energy, farthest in energy from the
hydrogen 1s). The overlap of orbitals with similar sizes, shapes, and energies give the
strongest bonds.
Csp3-H1s
Csp2-H1s
H
H
H
H3C C
Csp-H1s
H
worst for the σ-bond, atomic
orbitals farthest in energy
HC C
H2C C
H
H
better overlap/energy matching
than sp3, but worse than sp
CH241 Exam 1 – Fall 2016
best overlap for the σ-bond,
atomic orbitals closest in energy