CHEMISTRY 213

MID-TERM EXAM B

FALL 2007

ANSWERS

Answer questions on the green computer sheet provided with a PENCIL

YOU WILL DRAW THE STRUCTURES OF MOLECULES BA AND BB

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 31 (+Q#1) QUESTIONS WORTH 31 MARKS AND 15 MARKS FOR

THE TWO STRUCTURE PROBLEMS (BA and BB) and 50 minutes available,

so ration your time accordingly. Questions 2-32 have equal marks.

NO BOOKS, DATA SHEETS OR OTHER MATERIALS ALLOWED

EXCEPT CALCULATORS

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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 k = 1.383 x 10-23 J K-1

R = 8.3 J K-1 mole-1

Chem 213 Mid-Term Exam 2007 B 1

1. This is exam version B, mark B on the computer sheet as the answer to question 1. B

The compound BA, C5H8O, contains two functional groups and gives the ir spectrum shown below.

| | | | ignore ~2750 1692 942 cm-1

aldehyde conj-aldehyde geminal or trans-alkene

2. How many DBE’s (double bond equivalents) are in BA

A: 0 B: 1 C: 2 D: 3 E: 4F: 0.5 G: 1.5 H: 2.5 I: 3.5 J: 4.5

C5H12-C5H8 = 4H = 2DBE = C

Of the choices:

A: /C-H str B: =C-H str C: O-H str D: C=O str E: -C-H strF: C=C str G: C-C str H: C-O str I: -C-H bend J: =C-H bend

For compound BA, which corresponds to absorption at3. 2750 cm-1 B4. 1692 cm-1 D5. 942 cm-1 J

Chem 213 Mid-Term Exam 2007 B 2

Of the choices:

A: ether/alkene B: acetylene/aldehyde C: alcohol/alkeneD: alcohol/aldehyde E: ketone/alkene F: ketone/acetyleneG: alkene/aldehyde H: acetylene/ether I: acetylene/alcoholJ: acetylene/alkene

6. Which two functional group(s) are present in BA

DETERMINE TWO POSSIBLE STRUCTURES FOR COMPOUND BA AND ENTER THEMON THE LAST PAGE OF THIS PAPER IN THE SPACE PROVIDED

contains a conjugated aldehyde C=C-CHO = C3Oso we have another 2C to find942 is between a trans double bond (970) and a gem double bond (900)so is either H2C=C(CH2CH3)CHO or CH3CH2-CH=CH-CHOcannot be the cis compound (700) or mono substituted (not then a conj aldehyde)Could maybe be CH3CH=C(CH3)CHO

Atmospheric S16O2, has IR active fundamental vibrations at A:1362 B: 1358 C: 1151 D: 1147 E: 518 F: 515 cm-1

[there are two isotopes of sulfur, 32S and 34S ]

7. Which would be the symmetric stretching vibration of 34SO2

asym>sym and light mass> heavy massso sym/heavy is lowest of stretching freqencies = 1147 = D

8. Since these vibrations were ir active, is the molecule linear =G or bent = Hfor symmetric stretch to be active, molecule must be bent = H

9. What is the total number of fundamentals for 34S16O2

A: 2 B: 3 C: 4 D: 5 E: 6F: 7 G: 8 H: 30 I: 32 J: 24 x 1023

3N-6 where N=3, so have 3 fundamentals = B

Chem 213 Mid-Term Exam 2007 B 3

10. Chlorine (Cl2) is a pale green gas at room temperature (300K). Basing your answer onBoltzmann, how many molecules are in the first excited electronic state at roomtemperature - assume 105 molecules are in the ground state. [hint: you will need toestimate the energy gap using data we discussed in the first week of term!]

A: 0 B: 1 C: 10 D: 100 E: 1000F: 3 G: 34 H: 340 I: 3400 J: 104

green gas means absorbs in visible, where ΔE = ~ 200-300kJ/moleNU/NL = e-(DE/RT) is then 0 ( cannot have a fraction of a molecule in Nu)

11. If you had a mixture of PH3 and 13CF4 in an NMR spectrometer operating at fixedfrequency but which could vary the magnetic field, and you scanned from low field tohigh field, in which order would the nuclei achieve resonance.

A: C-F-H-P B: H-P-F-C C: H-F-P-C D: H-C-F-PE: P-C-F-H F: C-P-F-H G: C-P-H-F H: H-F-C-PI: P-C-H-F J: C-F-P-Hthe relative γ values are H=1, F=0.94, P=0.4 and C=0.25 (you don’t need to know values, butmust know order!) so for 2.3 Tesla, H = 100MHz but C = 25 MHz, to get C at 100MHz wouldtake 4x the fieldso order as you scan from low to high field is H<F<P<C = C

Of the choices:A: Avogadro’s number (N)B: nuclear spin quantum number (I)C: induced fieldD: applied fieldE: spin-spin couplingF: rotational quantum number (J)G: magnetogyric ratio (()

Which is chiefly responsible for12. the different frequencies needed to observe P, F ,13C and H in the same magnetic fieldG: magnetogyric ratio (()

Chem 213 Mid-Term Exam 2007 B 4

13. the difference in Hz between CHCl3 and CHI3 in a 60MHz and 300MHz spectrumD: applied field14. the fact that the 13C spectra of CH2Cl2 and CDCl3 both show three linesB: nuclear spin quantum number (I) = 1 for D (3 lines) and 1/2 for H (3 lines because 2 protons)

Of the choices:A: singlet B: doublet C: quartet D: sextetE: septet F: doublet of quartets G: quartet of quartetsH: doublet of doublet of triplets I: quartet of doublets J: doublet of doublet of quartets

What pattern would you observe for the 15. F in PF6 d B16. P in PF6 septet E17. H in 13CH3CF3 dq F18. H in 13CF3-PH2 ddq J

19. What pattern of intensities would you expect for the 13C of CF313CH3

A: 1:3:3:1 B: 1:1:1:1 C: 1:2:2:1 D: 1:3:3:1:1:3:3:1E: 1:4:4:6:4:4:1 F: 1:6:15:20:15:6:1 G: 1:3:3:1:3:9:9:3:3:9:9:3:1:3:3:1H: 3:9:9:3 I: 1:3:3:1:3:9:9:3:1:3:3:1 J: 1:3:3:1:2:6:6:2:1:3:3:1

qq so 1x(1:3:3:1) 3x(1:3:3:1) 3x(1:3:3:1) 1x(1:3:3:1)= G: 1:3:3:1:3:9:9:3:3:9:9:3:1:3:3:1

Chem 213 Mid-Term Exam 2007 B 5

Compound BB has formula C10H15NO and shows the IR and 1H NMR spectrum shown below:

| | | 3400 cm-1 1050 825 cm-1

alcohol (too large for 2o amine) C-O para-benzene

multiplicities: d d t s t s para benzeneThe peak at δ 1.9 exchanges with D2O = alcohol.

20. How many DBE has compound BB

A: 0 B: 0.5 C: 1 D: 2 E: 3F: 4 G: 5 H: 6 I: 7 J: 8

C10H22+1 - C10H15 = H8 = 4DBE = aromatic =F

Chem 213 Mid-Term Exam 2007 B 6

21. From the IR spectrum, what is the functional group in compound BB

A: alcohol B: nitrile C: ether D: aldehyde E: ketoneF: tertiary amine G: alkene H: acetylene I: primary amine J: amide

22. From the NMR spectrum, what are the integration ratios for the 6 sets of peaks

A: 2:2:2:3:2:1 B: 1:1:1:3:2:1 C: 2:2:2:4:2:1 D: 1:1:1:4:1:1E: 2:2:2:6:2:1 F: 1:1:1:1:2:3 G: 3:3:3:6:3:1 H: 2:2:2:1:2:3I: 2:2:2:6:2:2 J: 2:3:2:6:2:1

NOW ANALYSE THE NMR SPECTRUM AND DEDUCE THE STRUCTURE OF BB

DRAW YOUR STRUCTURE FOR AB ON THE NEXT PAGE IN THE PLACE PROVIDED

para-benzene -OH (probably since we have a 1050 band is a – CH2OH

two CH2 groups are coupled (triplets) so are –CH2CH2– and the 6H singlet has to be 2 CH3 groups since we have 2C not yet used

–benzene– -CH2CH2– –OH -N< -CH3 -CH3 identical

The only place the two methyl groups can be identical is on the -N< and the chemical shift isthen reasonable - so have –N(CH3)2

HO CH2CH2N(CH3)2 (CH3)2N CH2CH2OH

HO- at δ 5-10 not at δ 1 CORRECT-CH2N< not at d 3.8

Chem 213 Mid-Term Exam 2007 B 7

23. If a radiation source is emitting radiation of wavelength 4 μm, what is the energy of theradiation (per mole)

A: 16 J mole-1 B: 29.7 J mole-1 C: 160 J mole-1 D: 297 J mole-1

E: 1.6 kJ mole-1 F: 2.97 kJ mole-1 G: 16 kJ mole-1 H: 29.7 kJ mole-1

I: 160 kJ mole-1 J: 297 kJ mole-1

λ = 4μm = 4 x 10-6m E=(Nhc/λ) = (6 x 1023 x 6.6 x 10-34 x 3 x 108)/4 x 10-6

= 29.7 x 103 J mole-1 =H

24. If a sample of water was placed near to the source in Q#23, what is the most likelychange that would occur in the water - hint: think where water absorbs in cm-1.

A: nothing B: sample boils C: spin change D: bond rotationE: electronic transition F: bond stretching/bending G: bond dissociation

4μm = 2500 cm-1 and water does NOT absorb at 2500 cm-1, so nothing, A

25. What wave number would radiation of frequency 500 MHz have

A: 16.7 cm-1 B: 1.67 cm-1 C: 0.167 cm-1 D: 0.0167 cm-1

E: 0.00167 cm-1 F: 2 cm-1 G: 20 cm-1 H: 200 cm-1

I: 2,000 cm-1 J: 20,000 cm-1

c=νλ so wn = 1/λ = ν/c (in cm.sec-1)= 500 x 106 / 3 x 1010 = 0.0167 cm-1 = D

Chem 213 Mid-Term Exam 2007 B 8

The diagram below shows the first four vibrational levels for a bond in a real molecule (note thevertical scale is exaggerated).

The fundamental frequency of this bond is at 1715 cm-1 .

Of the choices (kJ mole-1):A: 10.2 B: 20.4 C: 30.4 D: 30.6 E: 40.6F: 40.8 G: 50.8 H: 51.0 I: 51.2 J: 71.2

26. What is the energy of the fundamental frequency [also = E1-E0]E = Nhν = Nhc/λ but wn = 1/λ so E = Nhc.wn= 6 x 1023 x 6.6 x 10-34 x 3 x 1010 x 1715 {c in cm.sec-1} = 20.4 kJ/mole = B

27. What is the zero point energyZPE = hν/2 = 20.4/2 = 10.2 kJ/mole = A

28. What is the energy of E1

E1 = hν x 3/2 = 30.6 kJ/mole = D

29. What is the probable energy of E2

a bit less than 5/2 x hν (51) so choose 50.8 = G

Chem 213 Mid-Term Exam 2007 B 9

30. What is the energy of the first overtoneE2-E0 = 50.8-10.2 = 40.6 kJ/mole = E

31. What is the dissociation energy of the bond (kJ mole-1)

A: 608 B: 588 C: 598 D: 577 E: 508 F: 618608-E0 = 608-10.2 = 598 = C

32. If their was only one functional group in this molecule and its ir spectrum also showeda band at 1100cm-1 , what is the bond above likely to be

A: O–H B: C–N C: C– H D: =C–H E: –O– C=O F: C/C G: >C=O H: C– O I: C– F J: C– Cl

1100 = -O--C- so basic frequency is a C=O so is an ester E

END