CH 334
First Midterm Exam
FORM_B
Wednesday, October 17, 2012
Name:____Key____________________________________________
You may use molecular model kits but no other material with chemical
information on it.
Please do not use MP3 players or other electronic gadgets other than
calculators.
Useful Information:
1. (15 points) Draw correct bond-line structures (show all lone pairs) for the
following compounds. Please include hydrogens and include formal charges
where appropriate.
a. (CH3)3NO
b. C2H6O (any isomer)
c. CH3C(H)O
2. (15 points) Draw the major resonance forms for the following molecule.
Indicate which atom(s) will be Lewis basic.
4 points per resonance form.
The two oxygens will be Lewis basic. (3 points)
There is a minor form that places a lone pair at C-2 that illustrates
some Lewis basicity for that carbon.
3. (10 points) Label the hybridization of each carbon in the molecule shown
below. -2 points for each error/omission up to 10 points.
4. (15 points) The three highest-energy, occupied molecular orbitals of
methyleneimine, H2CNH, are sketched in simplified form below. Energies of the
MO are included in eV.
Lone pair
π-bonding
σ-bonding
a. Label each MO above as either (mostly) lone-pair, sigma-bonding or pibonding.
See above. -2 for each error (max -5)
b. This molecule reacts with strong Bronsted acids as a proton acceptor. Draw
the structure (and any important additional resonance structures) for the
protonated product that forms. Be sure to clearly express the bond angle the
new proton forms with other atoms.
All atoms are coplanar; the H-N-C angles and the H-N-H angles are
close to 120°.
2 points for each resonance structure; 1 pt. for bond angle
c. Explain why protonation occurs where you show it, based on the MO
description provided above.
The most reactive electrons are the lone pair. The resulting cation is
resonance stabilized so long as all hydrogens are coplanar with C and
N.
5. (15 points) Consider the alkane 2,3-dimethylhexane.
a. Draw this molecule in bond-line notation.
5 points
b. Draw a Newman projection of the most stable arrangement around the C2C3 bond. (You may abbreviate any longer chain using its formula.)
5 points
c. How many gauche interactions are there? Circle them on your diagram.
2; circled above. Note that any other rotational isomer will have 3
gauche interactions.
5 points. -2 for each incorrect interaction circled or each one missed.
6. (30 points) Compound B below (shown in bond-line notation) has 5 distinct
C-H bonds.
a. Draw in one of each distinct kind, and label each one as primary, secondary
or tertiary.
5 points
b. Compound B reacts with Br2 (using light to initiate the reaction) to form a
single major product. Draw that product.
5 points
c. Using bond dissociation energies from the table on the last page, calculate
the net ΔH° for the reaction you drew in (b).
Break Br-Br:
Break 3° C-H:
+46 kcal/mol
+96.5 kcal/mol
Form H-Br:
Form 3° C-Br:
-87 kcal/mol
-71 kcal/mol
Net ΔH°r
-15.5 kcal/mol
(You may also properly split this into the separate reaction mechanism
steps; the H-atom transfer to Br has a net ΔH°r of +9.5 kcal/mol, and
the Br atom transfer has a net ΔH°r of -25.0 kcal/mol. The sum of the
two is still -15.5 kcal/mol.)
10 points
d. We photochemically cleave Br2 to form two Br atoms as the initiation step of
the reaction. Show, using the correct curved arrows, the two steps involved in
radical chain propagation that gives the product you claim is formed in (b).
10 points; 5 points each reaction
Bond strengths (kcal/mol):
F-F
Cl-Cl
Br-Br
I-I
H-F
H-Cl
H-Br
H-I
CH3-H
CH3CH2-H
(CH3)2CH-H
(CH3)3C-H
CH3-F
CH3-Cl
CH3-Br
CH3-I
CH3CH2-F
CH3CH2-Cl
CH3CH2-Br
CH3CH2-I
(CH3)2CH-F
(CH3)2CH-Cl
(CH3)2CH-Br
(CH3)2CH-I
(CH3)3C-F
(CH3)3C-Cl
(CH3)3C-Br
(CH3)3C-I
38
58
46
36
136
103
87
71
105
101
98.5
96.5
110
85
70
57
111
84
70
56
111
84
71
56
110
85
71
55