CHEMISTRY 213
MID-TERM EXAM A
FALL 2006
ANSWERS (IN RED)
Answer questions on the green computer sheet provided with a PENCIL
YOU WILL DRAW THE STRUCTURES OF MOLECULES AA AND AB
ON THE LAST PAGE OF THIS EXAM - REMOVE IT NOW AND
FILL IN YOUR NAME AND REGISTRATION NUMBER. ALSO DO THIS
ON THE GREEN COMPUTER SHEET NOW (Where it says identification).
Ignore course, section, etc., BUT CODE BOTH YOUR NAME & NUMBER.
THERE SHOULD BE 7 PAGES OF QUESTIONS IN THIS EXAM, PLUS
THIS PAGE, PLUS THE STRUCTURE PAGE TO BE HANDED IN.
There are 35 (+Q#1) QUESTIONS WORTH 35 MARKS AND 15 MARKS FOR
THE TWO STRUCTURES and 50 minutes available, so ration your time
accordingly. Questions 2-36 have equal marks.
NO BOOKS, DATA SHEETS OR OTHER MATERIALS ALLOWED
EXCEPT CALCULATORS
----------------------------------------------------------------------------------------------------DATA YOU MAY NEED DURING THIS EXAM:
N = 6 x 1023
h = 6.6 x 10-34 J s
c = 3 x 108 m s-1
R = 8.3 J K-1 mole-1
k = 1.383 x 10-23 J K-1
Chem 213 Mid-Term Exam 2006
A
1
1.
This is exam version A, mark A on the computer sheet as the answer to question 1.
2.
If one of our local radio stations broadcasts waves of energy 0.04035 J.mole-1, what
frequency are they broadcasting on
A: 1070 kHz (CFAX)
C: 90.5 MHz (CBC)
E: 103.1 MHz (JACK-FM)
G: 100.3 MHz (THE Q)
I: 98.5 MHz (OCEAN)
B: 900 kHz (The Village)
D: 101.9 MHz (CFUV)
F: 107.3 MHz (KOOL)
H: 91.3 MHz (THE ZONE)
J: 99.5 MHz (CBC SOOKE)
E = Nhν so ν = E/Nh = 0.04035/(6 x 1023 x 6.6 x 10-34) = 1.019 x 108 Hz =
3.
D
When radiation from the "OCEAN" interacts with a molecule of skin, what is the most
likely change that would occur in the skin molecule
A: bond dissociation
D: bond rotation
4.
A
B: bond stretching/bending
E: electronic transition
C: skin breakdown
F: spin change
F
Sometimes spectroscopists use wave number rather than wave length. What wave
number would “CFAX” radiation have
A: 35667 cm-1
B: 3.57 x 10-5 cm-1
C: 3.57 x 10-6 cm-1
D: 3.57 x 10-8 cm-1
E: 3.57 x 10-3 cm-1
F: 3567 cm-1
G:321.0 cm-1
H: 3.021 x 10-2 cm-1
I: 3.021 x 10+6 cm-1 J: 3.021 x 10+16 cm-1
wn = 1/λ (cm) = ν/c (cm sec-1)
= (1070 x 103)/(3 x 1010)
= 3.57 x 10-5 cm-1
=B
Chem 213 Mid-Term Exam 2006
A
2
The diagram below shows the first four vibrational levels for a bond in a real molecule (note the
vertical scale is exaggerated).
E2
Note: E1
E0=12.5
Of the choices (kJ mole-1):
A: 12.5
B: 24.8
C: 25.0
D: 25.2
E: 49.8
F: 50.0
G: 50.2
H: 62.3
I: 62.5
J: 62.7
E0= 1/2 hν = 12.5, so hν = 25 E1= 3/2 hν = 37.5
5.
What is the energy of the fundamental frequency hν = 25 = C
6.
What is the likely energy for E2 = 5/2 hν - a bit = 62.5-a bit = 62.3 = H
7.
What is the energy of the first overtone 2 hν - bit = 50-a bit = 49.8 = E
8.
What is the energy of the first hot line 62.3-37.5 = 24.8 = B
9.
What is the dissociation energy of the bond (kJ mole-1)
A: 937.5
B: 925
925-12.5 = 912.5 = C
10.
C: 912.5
D: 900
E: 875
F: 862.3
What is the fundamental frequency of this bond (cm-1)
A: 4040
F: 1690
B: 3330
G: 1100
C: 2850
H: 970
D: 2100
I: 825
ΔE = Nhc/λ so 1/λ = wn = ΔE/(Nhc)
c is in cm/sec
23
-34
10
= 25,000 / (6 x 10 x 6.6 x 10 x 3 x 10 ) = 2104 cm-1 = D
E: 1740
J: 700
Chem 213 Mid-Term Exam 2006
11.
3
Which of the following bonds is this most likely to be
A: S-F
F: C/C
12.
A
B: C-H
G: C=O
C: H-O
H: C-O
D: H-F
I: C-C
E: Cl-Cl
J: O=C-H
If there are 10 million of these molecules in E0, how many will be in E1 at 22oC
A: 9.9 x 105
B: 0
C: 436
D: 7785
4
F: 26
G: 1.85 x 10
H: 368
I: 78
-1
-(ΔE/RT)
-(25,000/8.3 x 295)
-10.21
For 25 kJ mol , Nu/Nl =e
= e
=e
= 3.678 x 10-5
so since NL = 107, NU = 368 = H
E: 1030
J: 106
In the above bond, if one of the atoms were replaced by an atom of 10% lesser mass, and the
force constant increased by 10%
using the choices A: would be smaller
B: no change
C: would be larger
13.
How would the frequency of the bond change
ν ~ /(k/μ) so if k is larger and m is larger, μ is smaller so ν is larger = C
14.
How would the zero point energy change
ZPE = 1/2hν, so if ν is larger ZPE is larger = C
Chem 213 Mid-Term Exam 2006
A
4
The compound AA, C4H7NO, gives the ir spectrum shown below.
|
|
1100 cm-1
2250
nitrile C/N
15.
C-O
How many DBE’s (double bond equivalents) are in AA C4H11-C4H7 = H4 = 2DBE=C
A: 0
F: 0.5
B: 1
G: 1.5
C: 2
H: 2.5
D: 3
I: 3.5
E: 4
J: 4.5
C: =C-H str
H: C=C str
D: O-H str
I: C-O str
E: C=O str
J: -C-H bend
Of the choices:
A: /C-H str B: C/C str
F: C/N str
G: N-H str
For compound AA, which corresponds to absorption at
F
16.
2250 cm-1
-1
17.
1100 cm
I
Of the choices:
A: alcohol
F: nitrile
18.
B: alkene
G: acid
C: ketone
D: aldehyde
H: acetylene I: amide
Which functional group is present in AA
F
E: amine
J: ester
Chem 213 Mid-Term Exam 2006
A
5
The 1H NMR spectrum of compound AA is shown below:
δ
multiplicities
19
4
3
t s
2
1
t
What are the integrations of the three nmr peaks:
A: 1:1:1
F: 2:3:2
B:1:2:1
G:2:2:3
C:2:1:1
H:2:3:3
D:1:2:2
I:3:3:2
E: 2:2:2
J:1:3:2
F
We have a C/N and an -O- which is not an -OH
The CH2 and the CH3 at about 3.5 ppm must both be on this O, so -CH2-O-CH3
This CH2 is a triplet so is next to the other CH2 so have -CH2-CH2-O-CH3
so only the nitrile to add thus N/C-CH2-CH2-O-CH3
DETERMINE THE STRUCTURE OF COMPOUND AA AND ENTER IT ON THE LAST
PAGE OF THIS PAPER IN THE SPACE PROVIDED
A mixture of CF4 and SiF4 showed fundamental absorptions at (cm-1)
A: 908
B: 435
C: 1281
D: 628
E: 800
F: 268
G: 1010
H: 390
Heavy elements < light elements; assym str> sym str> bends
20.
Which would be the asymmetric stretching vibration of CF4.
so 1281 = C
21.
Which would be (without doubt) a bending vibration of SiF4 so 268 = F
22.
Which would be the symmetric stretching vibration of SiF4 so 800 = E
23.
What is the total number of fundamentals for SiF4 3N-6: 15-6=9 = G
A: 3
F: 8
B: 4
G: 9
C: 5
H: 15
D: 6
I: 32
E: 7
J: 24 x 1023
Chem 213 Mid-Term Exam 2006
A
6
Of the choices:
A: Avogadro’s number (N)
B: nuclear spin quantum number (I)
C: induced field
D: applied field
E: spin-spin coupling
F: rotational quantum number (J)
G: magnetogyric ratio (()
Which is chiefly responsible for
24.
the fact that different frequencies are needed to observe 13C and 1H using the same
magnet
G: magnetogyric ratio (()
25.
the difference between the appearance of a 62MHz and 125MHz carbon spectrum
D: applied field
26.
the separation between the C’s in the C spectrum of 13CH313CF213CF3
C: induced field
27.
the fact that the two identical F’s in the 19F spectrum of 13CH313CF213CF3 do not appear
as a singlet E: spin-spin coupling
Of the choices:
A: sexet of quartets
B: septet of doublets
C: quartet of doublets
D: doublet of triplets E: doublet of quartets F: doublet of sextets G: doublet of septets
H: quartet of quartets I: quartet of septets
J: septet of quartets
Assuming normal J values, where nJXH<nJXY, in the appropriate spectra, what pattern would
you observe for the
d of q
=E
28.
CF3 in CH3P(CF3)2
29.
CH3 in CH3P(CF3)2
d of 7
=G
30.
P in CH3P(CF3)2
7 of q
=J
and using the choices:
A: dq7
B:qd7
C:7dq
D: dqq
E: qdq
F: dq6
G:qd6
I: 6dq
J:6qd
13
13
31.
C in CH3P(CF3)2 (assume you can see all normal J values) = A
Chem 213 Mid-Term Exam 2006
A
7
For the molecule
4
5
CH2CO 2Me
3
2
CHO
1
Using the chemical shifts:
A: 13
B: 10
F: 4.8
G: 3.8
32.
33.
34.
C: 7.8
H: 2.3
D: 7.3
I: 1.8
E: 6.8
J: 1
Which will be the closest to the chemical shift for H-3
Which will be the closest to the chemical shift for H-5
Which will be the closest to the chemical shift for H-1
D normal H
G -OCH3 + C=O
B aldehyde
Compound AB, has formula C8H9BrO and shows the IR and 1H NMR spectra shown below:
*
3400
-OH
35.
|
760 cm-1
(-ortho-benzene)
How many DBE has compound AB
C8H18 - C8H10 = H8 = 4DBE = F
A: 0
F: 4
D: 2
I: 7
B: 0.5
G: 5
C: 1
H: 6
E: 3
J: 8
Chem 213 Mid-Term Exam 2006
A
8
36.
From the IR spectrum, what is the principle functional group in compound AB
A
A: alcohol
F: ester
B: nitrile
G: alkene
C: acid
D: aldehyde
H: acetylene I: anhydride
E: ketone
J: amide
6
3
NMR:
10
9
multiplicities:
8
7
5
ddtt
q
1:1:1:1
1
4
1 δ
2
s
d
1 :
3
Integration ratios:
:
Note the peak at 2.4 ppm, exchanged with D2O; so is the -OH
The aromatic protons: not para (d,d) not meta (s, d,t,d), must be ortho (d,t,t,d)
the q at 5.2 ppm and the d at 1.5 ppm MUST be coupled to each other so are a >CH-CH3
so we have an ortho-benzene< and a >CH-CH3 as the bifunctional groups
+ an -OH and a -Br (don’t forget this!)
so must join together the bifunctional groups
CHOH-CH3
Br
CHBr-CH3
OH
correct
shields to <7 ppm
5-10 ppm