CHEM 233, Fall 2012
Midterm #1
answer
PRINTED
FIRST NAME
PRINTED
LAST NAME
Ian R. Gould
ASU ID or
Posting ID
key
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• PRINT YOUR NAME ON EACH PAGE!
M.O.s
2___________/27
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• READ THE DIRECTIONS CAREFULLY!
concept
3__________/6
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• USE BLANK PAGES AS SCRATCH PAPER
structures
4__________/22
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work on blank pages will not be graded...
dipoles
5__________/14
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•WRITE CLEARLY!
hybrid
6__________/44
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• MOLECULAR MODELS ARE ALLOWED
energies
7__________/24
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• DO NOT USE RED INK
BDE
8__________/24
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• DON'T CHEAT, USE COMMON SENSE!
Total (incl Extra)________/175+5
Extra Credit_____/5
H
He
Li Be
B
N
O
F
Ne
Na Mg
Al Si P
S
Cl
Ar
H/H
~1.0
Me/Me
~0.9
Ga Ge As Se Br
Kr
H/Me
~1.4
Et/Me
~0.95
In Sn Sb Te I
Xe
Me/Me
~2.6
i-Pr/Me
~1.1
Tl Pb Bi Po At
Rn
Me/Et
~2.9
t-Bu/Me
~2.7
K
Ca
Sc Ti V
Cr Mn Fe Co Ni Cu Zn
Rb Sr
Y
Zr Nb Mo Tc Ru Rh Pd Ag Cd
Cs Ba
Lu Hf Ta W
small range
range of values
broad peak
Re Os Ir Pt Au Hg
O H
C N
N H
C O
C
Interaction Energies, kcal/mol
C
H
O
C H
H
C
N
O
OR
amine R NH2 variable and condition
alcohol R OH dependent, ca. 2 - 6 δ
11
220 O
10
200
Aromatic Ar H
mainly 8 - 6.5
C CH3
R C OH
–H2C NR2
C CH2
9
180
O
C
8
160
H
O
–OCH2–
NMR Correlation Charts
7
140
6
120
5
100
R2C
Aromatic
CR2
C CH
4
80
3
60
2
40
–OCH2–
R C N
RC
H
~15
C C
1500
–H2C X
(δ, ppm)
~2
H
NR2
1650
2000
H
H
C
1710
2500
O
C H
~8
~2
C C
O
2200
3000
O
R C OH
H
~10
H
1600
O
C
O
C O H
3500
H
C C
1735
CH
broad ~3000
(cm-1)
~7
H
C
2850–2960
broad ~3300
H
1680
C
H
broad with spikes ~3300
O H
H H
C C
C
2200
C
N
O
2720–2820
2 peaks
3000–
3100
N H
C
1600–1660
H
C
3300
Approximate Coupling
Constants, J (Hz), for
1H NMR Spectra
Infrared Correlation Chart
usually
strong
C
Gauche
Eclipsing
CR
Alkyl
3Y > 2Y > 1Y
1
20
0
0
Alkyl 3Y > 2Y > 1Y
C X
C NR2
-2CHEMISTRY 233, Fall 2012, MIDTERM #1
NAME
Question 1 (14 pts.)
a) Give a line-angle structure for the following condensed formula. Do not forget to add all nonbonding electrons where appropriate.
(CH3CH2)2CC(CH2CH3)CH2COCH(CH3)CCCHO
the graders do not get an absolute grading rubric for my tests, there are always too many
possible ways to get partial credit in my exams, the guidelines in blue therefore are guidelines
ONLY
~3 pts each for the functional groups
-2pts for incorrect alkyl groups
-3pts for this at the end:
O
O
H
O
H
Question 2 (27 pts.) Directly ON TOP of the structures below, draw a picture of the Ψ or Ψ2 as
requested, for the indicated orbitals, AND, in each case write down the atomic orbital or orbitals
that you used, as appropriate
forgot A.O.s = -2pts
did the wrong question (wrong bond) = 7 pts maximum
incorrect size here -2pts
did bonding instread of anti-bonding or vica versa 7pts max
H
H3C
C
O
H
p A.O. (carbon)
p A.O. (oxygen)
H
C
Cl
sp3 A.O. (carbon)
p A.O. (chlorine)
H
Ψ for the C-O π∗ M.O.
H
Ψ2 for the C-Cl σ* M.O.
C
N
H sp A.O. (nitrogen)
1s A.O. (hydrogen)
Ψ for the N-H σ M.O.
Question 3 (6 pts.) Give one property of electrons that can only be explained by taking into
account their wave behavior
this was DIRECTLY from the problem set mentioned nodes with no mentioned positive and
explanation 3/4 pts
negtive with no
explanation 3/4 pts
there are several, the most obvious for us are:
• we can only explain the formation of both bonding and anti-bonding molecular orbitals this way
• the wave behavior requires quantization of electron energies, which forces them to occupy
distinct orbitals, this is why we have orbitals, if we have no wave behavior, we have no orbitals!
• the fact that we can only give a probability of finding en electron in a particular position is a
consequence of the wave behavior
Extra Credit (5 pts.) In the "Organic Chemistry in Real Life" pages, anti-bonding molecular
orbitals were described as playing an important role in which of the following
color
burning hydrocarbons
ATP energy conversion
CHEMISTRY 233, Fall 2012, MIDTERM #1
-3-
NAME
Question 4 (22 pts.) For the structures A and B below:
a) State (in words, do not draw on top of the structures) which is the largest BOND DIPOLE
moment(s) in A and which is the largest bond dipole moment(s) in B.
b) Between these two largest dipole moments (one from A and one from B), which would be
larger? Give a brief explanation.
c) Draw the MOLECULAR DIPOLE MOMENTS ON TOP OF THE STRUCTURES. Your drawing
does not need to illustrate the size of the dipole, only the direction. If there is no molecular dipole,
indicate so.
d) Indicate which structure would have the larger MOLECULAR dipole moment, give a brief
explanation
N
N
F
Br
C
C
A
B
the largest bond dipole moment in A is that associated with the C-N triple bonds
the largest dipole moment in B is that associated with the C-F bond
the C-N triple bond dipole moment is larger because even though F is more electronegative
than N, the electrons in the pi-bonds in the triple bond are highly polarizable
A has a larger MOLECULAR dipole moment because it has two larger bond dipole
moments that add in (roughly) the same direction as the two smaller bond dipole moments
in B
too many options for a rubric, but if 2 of the 4 parts are correct then 12 pts, if 3 of the 4 parts are
correct then 17 pts, but generally half correct = half points
Question 5 (14pts.) For the vinyl anion shown to the right:
a) give the hybridization for the circled carbon atom
b) state, in words, exactly what the various pairs of electrons around the
circled atom "do", e.g. in a σ-bond to atom X, in a π-bond to atom Z, are
non-bonding, etc.
c) list the hybridized valence orbitals for the circled carbon, and how
these are used to accomodate the electron pairs mentioned in part b)
sp2
H
C
C
H
H
vinyl anion
too many options for a rubric, generally half correct = half points
there were 4 questions JUST LIKE THIS ONE on the problem set
the C has an sp2 hybrid A.O. containing one pair of non-bonding electrons
the C has an sp2 hybrid A.O. that is used to build the σ-bond to the hydrogen atom
the C uses an sp2 hybrid A.O. to build the σ-bond to carbon, which contains one of the pairs
of electrons in the C=C double bond
the C uses an unhybirdized p A.O. to build the π-bond to carbon, which contains the
second of the pairs of electrons in the C=C double bond
-4-
CHEMISTRY 233, Fall 2012, MIDTERM #1
NAME
Question 6 (44 pts.) For the molecular formula C4H8O
a) Give the degrees of unsaturation 1 degree of unsaturation 2pts
b) Draw SIX structural isomers that obey the normal rules of valence for each atom. Include all nonbonding electrons. You can draw Lewis structures or line-angle structures (your choice). If you
draw line-angle structures, don't forget to include the H atoms that are normally included as part of
the functional groups. CIRCLE AND IDENTIFY ALL FUNCTIONAL GROUPS IN YOUR
-1 missing H atoms in funct. groups
STRUCTURES!!
24 pts
-2 for a stereoisomer here
O H alcohol
O ether
O
alcohol
O ether
H
alkene
alkene
ether
O
O
ketone
O
O
ether
aldehyde
H
O
H alcohol
O
alkene
alkene
O
H alcohol
H
O
H
O epoxide
alcohol
O
epoxide
alcohol
alkene
O
H
alcohol H
H
O alcohol
O alcohol
alkene
O
O
aldehyde
ether
here are a few......
O ether
H
-4 incorrect structure
-1 for wrong functional group
-1 for missing non-bonding electrons
c) Draw THREE PAIRS of stereoisomers that obey the normal rules of valence for each atom.
Include all non-bonding electrons. You can draw Lewis structures or line-angle structures (your
choice). If you draw line-angle structures, don't forget to include the H atoms that are normally
included as part of the functional groups. DO NOT IDENTIFY THE FUNCTIONAL GROUPS THIS
TIME! DO NOT INCLUDE ANY STRUCTURES in part c) THAT WERE DRAWN AS PART OF
YOUR ANSWER TO PART b) OF THIS QUESTION
-4 if structure also in part b
O
H
18 pts
O
+
O
+
O
H
here are four pairs
O
H
+
O
H
O
H
+
O
H
CHEMISTRY 233, Fall 2012, MIDTERM #1
-5-
NAME
Question 7 (24 pts.) Rank the pairs of electrons indicated, A, B, C, D and E in order of increasing
energy. Give an explanation for your choice.
switched -3
switched -3
B (σ-bonding)
A (non-bonding)
H
C
(non-bonding)
O
P
H
D
(σ-bonding)
lowest
energy
D
<
B
<
E
<
A
<
no sp2/sp3 description -3pts
no non-bonding description -3pts
no electronegativity description -3pts
C
highest
energy
E (π-bonding)
non-bonding electrons are higher in energy than bonding electrons, therefore A and C are highest
in energy, phosphorus is larger and less electronegative than oxygen, C are thus highest
π-bonding electrons are higher in energy than σ-bonding electrons, thus E is next highest
bonding M.O. B is built by combining sp3 + sp3 hybrid A.O.s, bond D is built from sp2 + sp3
sp2 A.O.s are smaller and electrons in them are lower in energy than electrons in sp3 A.O.s, which
have higher p character, thus the electrons in bond D are lowest in energy of all
in general 1/2 correct = 1/2 poinst etc.
Question 8 (24 pts.)
a) Rank the bonds indicated A, B and C in terms of increasing bond dissociation energy, give a
brief explanation.
A
H
6pts
H
H
H
H
C < A
< B
C
C
C
H
C
smallest
largest
C
C
C
BDE
BDE
B
H H
H
H C
H
H
bond B is from H to an sp hybridized carbon, sp A.O. is smaller lower in energy, this is the
strongest bond and thus has the largest BDE
10pts
both A and C are to sp3 hybridized carbons, but cleavage of C generates a 3° radical whereas
cleavage of A generates a 1° radical, 3° radicals are more stable due to hyperconjugation, it takes
less energy to form the more stable radical, thus C has the smallest BDE
many different answers, 1/2 correct = 1/2 pts
b) Treat homolytic bond dissociation as a simple chemical reaction. For BOND A below, add
the curved arrows to the structure that illustrate bond-breaking and show the products of
homolytic bond breaking (i.e. what you get after breaking the bond) on the "product side" of the
reaction arrow below.
A
8pts
H
H
H
H
H
H
H
H H
C
C
C H
C H
C
C
C
C
C
C
C
C
+ H
H H
H
H H
H
H C
H C
H
H
H
H