CHEM 233, Fall 2014
Midterm #1
answer
PRINTED
FIRST NAME
Ian R. Gould
key
PRINTED
LAST NAME
ASU ID or
Posting ID
Person on your LEFT (or Aisle)
Person on your RIGHT (or Aisle)
energies
1___________/30
........
• PRINT YOUR NAME ON EACH PAGE!
f groups
2___________/18
........
• READ THE DIRECTIONS CAREFULLY!
dipoles
3__________/22
........
• USE BLANK PAGES AS SCRATCH PAPER
bde
4__________/43
........
work on blank pages will not be graded...
isomers
5__________/20
........
•WRITE CLEARLY!
M.O.
6__________/20
........
• MOLECULAR MODELS ARE ALLOWED
hybrid
7__________/22
........
• DO NOT USE RED INK
• 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
Ga Ge As Se Br
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
~1.0
Kr
H/Me
In Sn Sb Te I
Xe
Me/Me
Tl Pb Bi Po At
Rn
Me/Et
H
C
H
N H
C
N
C
N
10
200
OR
R C OH
~8
8
160
~2
H
NR2
1650
H
O
–OCH2–
C CH3
–H2C NR2
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
NMR Correlation Charts
H
H
C
2000
~2
C C
O
C CH2
O
C
H
~10
H
1600
O
Aromatic Ar H
mainly 8 - 6.5
9
180
~7
H
C C
–H2C X
11
220 O
H H
H
1710
amine R NH2 variable and condition
alcohol R OH dependent, ca. 2 - 6 δ
(δ, ppm)
Approximate Coupling
Constants, J (Hz), for
1H NMR Spectra
1735
CH
2500
O
C H
~2.7
1680
C
2200
3000
O
R C OH
t-Bu/Me
C C
C
O
C O H
3500
~2.9
O
broad ~3000
(cm-1)
~1.1
H
2850–2960
broad ~3300
~0.95
i-Pr/Me
C
1600–1660
H
broad with spikes ~3300
O H
Et/Me
~2.6
C
2200
C
~1.4
O
2720–2820
2 peaks
3000–
3100
~0.9
Infrared Correlation Chart
H
C
3300
Me/Me
usually
strong
C
Gauche
Eclipsing
H/H
O
C H
Interaction Energies, kcal/mol
CR
Alkyl
3Y > 2Y > 1Y
1
20
0
0
Alkyl 3Y > 2Y > 1Y
C X
C NR2
-2-
CHEMISTRY 233, Fall 2014, MIDTERM #1
NAME
Question 1 (20 pts) Rank the indicated pairs of electrons A, B, C and D in order of INCREASING
energy. Give a BRIEF explanation. (the points are for the explanation, not for getting the order
correct)
A non-bonding
H
N
C
N
H
D σ-bonding
B non-bonding
C <
D <
B <
lowest
energy
C σ-bonding
A
highest
energy
non-bonding electrons are higher energy than bonding electrons, so A + B > C + D, A
are in sp3 hybridized A.O., B in an sp hybrid A.O., sp3 more p character thus A are
highest energy. The electrons in C are in a M.O. that was "built" from an sp3 and an sp
hybrid A.O., compared to D which is sp3 + sp3, thus C are lowest of all
grading, mostly common sense, 1/2 correct for 1/2 points etc, 4 pts for correct order
Question 2 (14 pts) Circle and identify all of the functional groups in the provided structure.
The structure is shown in its neutral form, there are no formal charges. (ignore alkyl chains,
and you do not have to specify primary, secondary or tertiary if relevant)
thyroxine
a thyroid hormone that
regulates metabolism
iodide
ether aromatic
I
aromatic
O
carboxylic acid
NH2
OH
iodide
O
I
I
OH
iodide
alcohol
I
iodide
amine
halide can be used in place of iodide
Extra Credit (5 pts.) BRIEFLY give ONE way in which anti-bonding orbitals are used by
organic molecules
they accept electrons in chemical reactions AND they are where the electrons "go to" upon
photochemical excitation
CHEMISTRY 233, Fall 2014, MIDTERM #1
-3-
NAME
Question 3 (22 pts) For the structures A, B and C, indicate the molecular dipole moments ON
TOP OF EACH STRUCTURE and RANK them from smallest to largest (do not try to indicate
their relative size using arrow length). Give a BRIEF explanation. YOUR EXPLANATION
SHOULD INCLUDE THE TERM "BOND DIPOLE MOMENT(S)"
O
O
S
A
B
A
C
+
S
+
+
S
<
smallest
B <
C
largest
O
oxygen is more electronegative than S, will polarize the electrons in the π-bond more, bonds to
O will have larger BOND dipole moments
In A the 2 C=O bond dipoles add to give the largest molecular dipole moment, in C the 2 C=S
bond dipoles add to make the smallest molecular dipole and B is in the middle
Question 4 (24 pts.) For the structure below
a) Add the curved arrow-pushing that describes homolytically cleavage of the O-Ha bond
(ONLY), and draw the products of bond cleavage (include all non-bonding electrons).
b) BRIEFLY explain which of the two bonds O-Ha and C-Hb would have the largest bond
dissociation energy, include the term "energy of the electron(s)" in your explanation
Ha
Hb
O
Hb
O
Ha +
In the O-H bond the electrons are lowest in energy since both "see" the most
electronegative element oxygen. Breaking the O-H bond raises the energy of the electrons
in the bond the most, the BDE fo O-Ha is larger than that for C-Hb
c) Using the axes provided, draw an ENERGY DIAGRAM for cleavage of the O-Ha and C-Hb
bonds ON THE SAME DIAGRAM (you can normalize your diagrams where you like)
d) Indicate the MAGNITUDES of the two bond dissociation energies on your diagrams
Energy
BDE
C-Hb
BDE O-Ha
rR–H
-4-
CHEMISTRY 233, Fall 2014, MIDTERM #1
NAME
Question 5 (40 pts) For the molecular formula C4H8O:
a) Give the degrees of unsaturation.
max # of hyrdogens = (4 x 2) + 2 = 10
actual # of hydrogens = 8
degrees = (10 - 8) / 2 = 1 degree
grading 2pts
b) In EACH of boxes A and B, draw a pair of STEREOISOMERS with molecular formula C4H8O
that obey the normal rules of valence. The two pairs of stereoisomers must be different. Your
structures can be Lewis or line-angle, your choice. INCLUDE ALL NON-BONDING ELECTRONS.
A
stereoisomer PAIR #1
B
stereoisomer PAIR #2
OH
OH
&
&
OH
OH
and many others...
and many others...
c) In box C, draw FIVE more structural isomers of C4H8O that obey the normal rules of valence. Do
not use any structure that appeared in Boxes A or B. Your structures can be Lewis or line-angle,
your choice. INCLUDE ALL NON-BONDING ELECTRONS.
C
5 structural isomers (no structures here can also appear in either box A or box B)
O
O
OH
H
O
O
O
O
and many others...
OH
OH
-5-
CHEMISTRY 233, Fall 2014, MIDTERM #1
NAME
Question 6 (30 pts.) Directly ON TOP of the structures shown, draw a picture of the Ψ or Ψ2 as
requested, for the indicated orbitals. All non-bonding electrons and formal charges are shown.
also give the atomic orbitals that are used to "build" the molecular orbitals as appropriate
in each case indicate the positions of all NODES, or positions where the probability of
finding the electrons is zero, as appropriate
node
sp3 A.O. on oxygen
sp3 A.O. on carbon
O
Ψ for σ M.O.
H
a)
Ψ2 for π∗ M.O.
H3C
b)
H3C
C
p A.O. on C
p A.O. on N
zero probability
N
H
zero probability
Question 7 (25 pts.) For the provided structure:
a) assign the hybridization for the circled carbon atom below
b) draw a picture of the Ψ for the non-bonding electrons on TOP of the structure
c) show that you understand the meaning of the hybridization assignment by making a small table
that summarizes all of the valence hybrid (and any unhybridized) atomic orbitals associated with
this carbon atom, and state how they are used (e.g. used to make a sigma bond to the chlorine, I
know there is no chlorine in the structure, this is just to show you what do to)
~120°
H3C
C
H3C
C
H
this the carbon is sp2 hybridized, the geometry is BENT
sp2 - σ-bond to hydrogen
sp2 - σ-bond to carbon
sp2 - non-bonding electrons
p - π-bond to carbon
d) give the approximate C-C-H bond angle indicated with the arrow, assign the geometry around
the circled carbon atom, AND, give a BRIEF explanation (2-3 sentences MAX.) for your choice of
geometry that includes the terms "energy of the electrons", "VSEPR", "electron domains".
there are 3 domains of electrons around the carbon, VSEPR requires a trigonal planar-like
geometry to minimize the total energy of the electrons, because the carbon only has 2 atoms
attached to it and the position of the electrons cannot be determined with certainty, the geometry
can only be defined as bent