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Chemistry 303, fall, 2011

SECOND EXAMINATION

7:30 PM, NOVEMBER 14th, 2011

Duration: 2.0 hr Name_______________________KEY_____________________________________ (Print neatly and use your official PU name, the one on your ID. This is an "open book" examination; you may use anything that is not alive or connected to the Internet. Note: if you do not know the complete or specific answer, give a partial or general answer-- WRITE SOMETHING If you are using a resonance argument in your answer, draw the relevant resonance structures. Write only in the space provided for each question. Score: p2______/18 p3_______/12 p4_______/14 p5____ _/12 p6_______/12 p7_______/16 p8_______/16 Lab:________/14 Lecture total: _________/100 There are 11 pages in this exam; the last page includes (a) tables of nuclear spin values and for common isotopes, (b) isotope distributions, and (c) typical coupling constants for 19F, with 13C and 1H. PLEDGE:_________________________________________________________________

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I. (10 pts). Graded by X. H. and L. Z. Consider the molecule M. A. (02 pts). What would be the pattern (% height, to 2 significant figures) of peaks in the molecular ion region of the mass spectrum (M, M+1, M+2, M+3, etc)? You need not calculate the molecular weight; assume M = 100%. Ignore peaks <1%. M (100%); M+1 (10%); M+2 (4.4%) 1 pt for M+1, 1 pt. for M+2 B. (02 pts). What would be the approximate position (in cm-1) of the single most diagnostic peak in the IR spectrum? 1650 cm-1 1720 cm-1 1780 cm-1 1840 cm-1 circle single best answer What group would give rise to this peak? C-H C=O CH3-S circle single best answer C. (02 pts). What do you predict for the UV spectrum (approximately)? No observable λmax λmax observed with low ε λmax observed with high ε circle single best answer D. (02 pts). How many peaks would you expect in the broad band decoupled 13C NMR spectrum? ___7______ E. (02 pts). What would be the predicted chemical shift in the 1H NMR spectrum for the CH2 group labeled (a) in structure M? Show your work. δ = 1.20 (base for CH2) + 0.5 (alpha S) + 1.10 (beta C=O ketone) = 2.80 ppm No partial credit for B-E _________________________________________________________________________________________________________ II. (20 pts). Graded by J. Conway and J. Mighion Use the spectral techniques (UV, IR, MS, 1H NMR, 13C NMR) to distinguish the pairs of molecules shown. You may use each technique only once. For each pair, specify one method and clearly describe the single most important difference that could be used to distinguish the members of each pair. With mass spec, you may not use the high resolution technique to give the exact molecular formula, and you may not rely on differences in fragmentation pattern. With IR, you may refer only to the region from 1500-4000 cm-1. 2 pts. For correct choice 2 pts. For explanation of choice X 5 parts graded from the top, you can only use a technique one time

No predictable difference in the H NMR, C NMR, or IR spectra. (b) might be expected to show longer wavelength UV absorption than (a). [But there is a better UV case later] Mass spec would easily distinguish them, either by the mass of the molecular ions or the pattern of peaks in the molecular ion region. The molecule with Br (b) would show an M+2 equal in intensity to the M ion. (a) would have an insignificant M+2.

______________________________________________________________________________________________________

No predictable difference in the UV, H NMR. The C NMR spectrum for (a) would show a quartet for the CF3 group, while (b) would be just three singlets. The mass spec would have different molecular ions, and isotope pattern in the molecular ion region.

The IR is particularly useful, because the electron-withdrawing effect of the CF3 group would move the C=O stretch to high energy, higher cm-1. This is because it destabilizes the dipolar resonance structure:

O

SCH3

M

a

Br

ab

O

CF3

O

CCl3a b

3

O

CF3

O

CCl3a b

O

CF3

destabilizedOverall, less single bond character

O

CCl3

less destabilized,bigger contributor

Continued next page--

No predictable difference in the UV, IR, or mass spec. H NMR for (a) would show two 2H triplets for the aryl H; (b) would show 3 distinct peaks, one singlet, one doublet (2H), and one triplet (1H). C NMR for (a) would show 4 peaks while for (b) there would be 5 peaks

______________________________________________________________________________________________________

No predictable difference in the mass spec. IR would show several differences: different C=O stretch is perhaps most obvious. H NMR would show several differences. C NMR would show one C=O for (a) and two C=O for (b). One could find other differences. UV is the most obvious technique: (a) has an sp3 carbon insulating the three benzene rings from each other, andin (b) there are in conjugation.

(b) would show peaks at much longer wavelength than (a). ______________________________________________________________________________________________________

No predictable difference UV, C NMR, IR, or mass spec. H NMR would show difference in the coupling constants for the alkene H. The Z isomer (a) would have smaller coupling compared to the E isomer

(b).

OMe

OMe

OMe

OMea b

a b

OO

HO

HO CO2H

O

OH

a b

MePh Me

Ph

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III. (14 pts). Graded by Dennis and J. Bratz When dissolved in CDCl3, a compound (Y) with the molecular formula C4H8O2 yields the 1H NMR spectrum below [Hint: CDCl3 can be slightly acidic]. The IR spectrum of Y shows a strong absorption peak at 1720 and a broad peak centered at 3350 cm-1. Note that the peaks in the 1H NMR spectrum are labeled A,B,C,D.

012345PPM

1H 1H

3H

3HA B

CD

A. (06 pts). Draw a structure for this compound, drawing in all H's. Correlate your structure with the NMR spectrum shown by labeling each H or group of equivalent H's with the appropriate letter, A, B, C, D. 2 pts for correct structure 1 pt for correct labeling A-D (4 total points) B. (02 pts). Explain how your structure is consistent with the peaks at 1720 and 3350 cm-1 in the IR spectrum. The structure shows a normal ketone, typically at 1720 cm-1

The peak at 3350 is due to the –OH stretch 1 pt for each correct assignment C. (02 pts). Explain why the proton A shows up with the pattern shown on the spectrum. Proton A is a quartet because it is coupled to the methyl group D, and not to the –OH unit. 2 pts for correct explanation D. (05 pts). When Y is stirred in D2O, then separated, and the 1H NMR spectrum recorded again in CDCl3, the broad singlet at δ 2.80 ppm has disappeared. The rest of the spectrum is the same as before. Explain carefully with pictures why the broad singlet at δ 2.80 (B) disappears when the molecule is treated with D2O. You need not show a detailed mechanism. Treatment with D2O allows for exchange of the –OH proton with the deuterium in the D2O giving H-O-D and –OD in the molecule. No signal for D is observed in an 1H NMR spectrum.

H3C CH3

O OH

H

+ D-O-D

H3C CH3

O OH

D

+ H-O-D

2 pts for drawing the deuterated alcohol, 2 pts for explaining the peak disappears

H3C CH3

O O

HA

B

CD

H

Y

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IV. (04 pts). Graded by M. VanderWal, E. Welin, M. Wu Consider o-hydroxybenzaldehyde (A) and o-methoxybenzaldehyde (B). Note that A shows a C=O stretch at 1670 cm-1 and B has the C=O stretch at 1695 cm-1. Explain why there is the difference in C=O stretching frequency between the two molecules. In molecule A there is an opportunity for internal H-bonding between the –OH and the carbonyl oxygen. This will particularly stabilize the dipolar resonance form for the carbonyl, introducing more single bond character and resulting in a lower energy C=O stretch. This is not possible for B.

OH

O

A

OH

O

B

CH3

H OH

OH

OH

OCH3

A' B'

2 pts for H-bond in A’ 1 pt for dipole stabilization in A’ 1 pts for explaining that this does not occur in B’ V. (04 pts). It often comes up in 1H NMR analysis that you find two peaks of roughly equal intensity near each other in the spectrum. You are not certain whether these two peaks are singlets arising from two separate and uncoupled protons, or if they are two peaks of a doublet, arising from two equivalent protons coupled to one other proton. In general, what experiment could you perform to distinguish clearly between these two possibilities? Explain. The best answer will be an unambiguous and be generally applicable. Change the operating frequency for the machine, for example from 300 MHz to 600 MHz. Then if the two peaks are due to two different H, both singlets, the distance between the peaks in Hz will double (the difference in ppm will remain the same). If the two peaks are due to a doublet from splitting by an adjacent H, the distance between the peaks in Hz will stay constant. 2 pts for correct answer 2 points for explanation or just two points for an alternate and valid answer VI. (04 pts). Explain why the 1H NMR spectrum of dimethylformamide (M) shows two distinct signals for the methyl groups while the related structure N shows the two methyl groups as a single peak.

O

NH3C

CH3

HCl

NH3C

CH3

M N

HH

O

NH3C

CH3

H

free rotation, the methyl groupsare averaged to the same environment

Very significant resonance structure; leads to restricted rotationso that the methyl groups appear in different environments, differentchemical shifts.

1 point for mentioning hindered rotation and double bond character 2 pts for drawing the resonance structure 1 pt for explaining why structure N does not display this

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V. (12 pts). Graded by J. Alleva and I. Ojini The following question engages your knowledge of analyzing resonance structures to explain 1H NMR chemical shifts. Note the related structures A-D. In each case one of the alkene hydrogens is indicated (HA-HD) along with the 1H NMR chemical shift for that H, just above it. There is a wide range of chemical shift values.

Me Me

HA O

Me Me

HB N

Me Me

HDMe

Me

HC

OMeA B C D

4.105.70 7.00 6.20

Me

A. (04 pts). Using resonance arguments, explain the difference in chemical shift for HA vs HB. Cpn A does not have significant dipolar resonance structures (e.g., A'). Cpn B has two reasonable dipolar resonance structures, with the (-) on oxygen. In B', there is a (+) on the carbon bearing HB. Like other examples of the effect of electronegative atoms in deshielding nearby H, HB appears further downfield (higher δ) compared to HA.

Me Me

HA O

Me Me

HB

A B

5.70 7.00

Me Me

HA O

Me Me

HB O

Me Me

HB

B' B"A'

electron-deficient, deshielding

2 pts for drawing B’ 1 pt for mentioning the deshielding of B’ 1 pt for comparison to A’ B. (04 pts). Using resonance arguments, explain the difference in chemical shift for HB vs HD. Both B and D have significant dipolar resonance structures. However, the contribution from D' (and D") is less because N is less electronegative than O, and less well stabilizes the (-) charge.

O

Me Me

HB

B

7.00

O

Me Me

HB O

Me Me

HB

B' B"

N

Me Me

HD

D

6.20

N

Me Me

HD N

Me Me

HD

D' D"

Me

Me Me

2 pts for D’ resonance structure 1 pt for B’ resonance structure 1 pt for explanation of relative difference C. (04 pts). Using resonance arguments, explain the difference in chemical shift for HA vs HC.

Again, A has no significant dipolar resonance structures. With the methoxy group in C, a new dipolar resonance structure appears (C') which adds electron density to the carbon bearing HC. This leads to extra shielding, and HC appears upfield, lower δ, compared to HA. 2 pts for C’ structure 1 pt for A’ structure 1 pt for shielding argument

4.10

Me Me

HA O

Me Me

HC

A C

5.70

Me Me

HA O

Me Me

HC

C'A'

electron-rich, shielding

OMe

OMe

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VI. (16 pts). Graded by Capacci and Garber Summarized below is the 1H NMR spectrum of procaine (P). (a) 1.06 (t, J = 7 Hz, 3H) (b) 2.61 (q, J = 7 Hz, 2H) (c) 2.82 (t, J = 7 Hz, 1H) (d) 4.16 (br s, 1H) (e) 4.33 (t, J = 7 Hz, 1H) (f ) 6.61 (d, J = 9 Hz, 1H) (g) 7.84 (d, J = 9 Hz, 1H) 0.5 pts per correct label

A. (3.5 pts). Assign the H in P with the signals listed above. Label each H on the structure P with the letter (a, b, c…) corresponding to the appropriate chemical shift from the list here. It may be necessary to calculate selected chemical shifts, when possible, to avoid ambiguous chemical shift assignments. You need not show this work.

B. (3.5 pts). Fill in the third column of this table by indicating which nearby H are responsible for the coupling patterns. E.g., for the first entry, you will consider the proton you have assigned as a, and write the letter(s) corresponding to the H(s) it is coupled to.

Label Chemical shift position Which proton(s) coupled to?

a 1.06 b b 2.61 a c 2.82 e d 4.16 -- e 4.33 c f 6.61 g g 7.84 f

0.5 pts for each correct coupling assignment C. (05 pts). Explain carefully with pictures how you made the chemical shift assignments for the H's on the benzene ring.

O O

HH

Hf HfH2N

R

O O

HH

Hf HfH2N

R

electron-rich, shielded

The two Hf are more shielded than the usualbenzene-type protons due to the electron donationfrom the lone pair on N. So the doublet due to Hfis upfield (smaller δ) compared to benzene.

2 pts for drawing structures, 1 pt for explanation

O O

HgHg

Hf HfH2N

R

O O

HgHg

Hf HfH2N

Relectron-poor, deshielded

The two Hg are less shielded than the usualbenzene-type protons; so the doublet due to Hfis downfield (larger δ) compared to benzene.

The deshielding could be due to any of three effects: a. the inductive withdrawing effect of the C=O b. the resonance withdrawing effect of the C=O (shown) c. the magnetic anisotropy of the C=O (shielding/deshielding cone)Full credit for any one of these, if nicely explained.

1 pt for drawing structure and 1 pt for explanation D. (04 pts). 1. (02 pts). Procaine is protonated by one equivalent of HCl to give a salt better known as Novocain, which is used as a local anesthetic in dentistry. What would you predict to be the structure of Novocain? Explain your choice. Draw below using the structure of Procaine. 1 pt for protonating correct nitrogen, 1 pt for basicity argument. This is the most basic site in the molecule, comparable to ammonia. The other amino group is conjugated to the benzene ring and the lone pair is somewhat delocalized and therefore stabilized. [Shown in the resonance structures for (C) above.] 2. (02 pts). The 1H NMR spectrum of Novocain shows two distinct resonances for the NH protons: a 2H peak at δ 4.20 ppm and a 1H peak at δ 11.2 ppm. Is this spectrum consistent with your predicted structure? Explain. 1 pt for saying the NMR confirms this structure, 1 pt for explanation The proton NMR confirms the site of protonation as show in the picture. There is now one H on the ammonium ion, with a (+) charge and a strong downfield shift (like an acid). The other –NH2 is relatively unchanged.

O O

HH

H HH2N

CH2H2C

NCH2CH3

CH2CH3

P

(a)

(a)

(b)

(b)(c)

(e)

(f)(f)

(g) (g)

(d)

O O

HH

H HH2N

CH2H2C

NCH2CH3

CH2CH3

H

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VII. (16 pts). J. Shields and P. Deifik The following are spectral data for compound Z. Mass spectrum: m/z = 142 (M+, 100%), 143 (4.7%) IR: 1792 (s) cm-1 (no other significant peaks in the region 1500-4000 cm-1 other than a set of sp3 C-H stretches) 1H NMR (CDCl3): δ 1.40 (t, J = 7.1 Hz, 3H), 4.41 (q, J = 7.1 Hz, 2H) proton-decoupled 13C NMR (CDCl3): δ 13.8, 64.6, 115.1 (q, J = 285 Hz), 157.9 (q, J = 42 Hz) A. (02 pts). What general type of functional group is suggested by the IR spectrum of compound Z? C=O 2 pts for chosing carbonyl, no partial credit, full credit if they chose something with a carbonyl B. (03 pts). What structural fragment is suggested by the 1H NMR spectrum of compound Z? Explain. An ethyl group, CH3-CH2-X. triplet for the CH3 and quartet for the CH2

2 pts for chosing ethyl group, 1 pt for explaining why [the chemical shift for the CH2 suggests that X is strongly electronegative—not required for full credit] C. (06 pts). How many different carbons are contained in compound Z? Please explain the splitting patterns and the importance of the observed coupling in the proton-decoupled 13C NMR spectrum of compound Z. Four sets of signals, four carbons. Two singlets due to carbons bonded to H (or O) 2 pts for for 4 carbons One quartet with large coupling: some NMR-active nucleus other than H. Possibly CF3 group. Coupling of 285 Hz falls in the range for F directly attached to C (geminal) 1 pt for saying there is a –CF3, 1 pt for saying the CF3 is attached to another carbon One quartet with smaller coupling: Coupling of 42 Hz falls in the the range of F attached to C which is attached to the carbon being observed: vicinal coupling. 1 pt for explaining splitting, 1 pt for vicinal coupling D. (05 pts). Propose a structure for compound Z and indicate how the spectral data (mass spectrum, IR, 1H NMR, and 13C NMR) are compatible with your proposed structure.

C CO

H2C

CH3

OFFF

a. mass spectrum molecular wt: M = 142 consistent with C4H5F3O2 ; M+1 = 4.7% consistent with 4 carbons

b. IR: C=O stretch consistent with ester, perturbed by electron-withdrawing group. Dipolar resonance structure destabilized, less single bond character.

c. H(a) is at 1.40 due to beta -O-, a triplet due to coupling to H(b). H(b) is at 4.41 due to alpha -O-, a quartet due to coupling to H(a)

d. C(a) appears at 13.8; C(b) appears at 64.6; C(d) appears at 157.9 as a quartet with beta coupling to CF3; C(c) appears at 115.1 as a quartet with alpha coupling to CF3.

;

C CO

H2C

CH3

OFFF

Z(a)

(b)(c)(d)

3 pts for correct structure, A-D each worth 0.5 pts.

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XI . (14 pts). Lab-Related Question. Graded by Henry, M. Pirnot and K. Reiber

A. (06 pts). 1. (04 pts). An orgo lab student reported the melting point range of compound A as 46-51 °C and the calibration melting point range of benzoic acid as 115-119 °C. Assuming that the benzoic acid is pure, is the melting point range of compound A reliable, even with an appropriate correction factor? Why or why not? The important clue here is the broad mp range reported for the pure benzoic acid calibration standard. The most critical factor in determining an accurate mp is the rate at which the sample is heated in the mp range. If the temperature is increased too rapidly in the mp range (greater than 1-2 °C/min.), then the thermometer, sample, & heating block will not be in thermal equilibrium, & erroneous mp readings will result (usually low & broad observed mps). Since the mp of benzoic acid was not determined properly, there is no reason to believe that the mp of A was determined properly either. (Mp corrections are only appropriate if the mps are taken correctly.) 2. (02 pts). The observed melting range of compound B is 103-104 °C and the observed melting point range for pure benzoic acid is 123-124 °C. What is the corrected melting point range for compound B? The observed mp midpoint for benzoic acid is 123.5 °C, which is 1.1 °C higher than the accepted mp of benzoic acid (122.4 °C). Thus, the mp apparatus is reading 1.1 °C too high, & this correction factor must be subtracted from the observed mp range for B. corr. mp range for B = 103-104 °C – 1.1 °C = 101.9-102.9 °C (102-103 °C)

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B. (08 pts). Steam distillation of ground cloves affords clove oil, which contains two constituents. The major constituent X, which is soluble in 5% aqueous NaOH solution, is characterized by the following spectral data: mass spectrum: m/z = 164 (M, 100%), 165 (11.1%) IR (liquid film): 3518 (br, s), 3061-2843 (multitude of weak peaks), 1638 (m), 1613 (m), 1607 (m), and 1514 (s) cm-1 [br = broad, m = medium, s = strong] proton-decoupled 13C NMR (CDCl3): 10 peaks Choose (circle) the best structure for constituent X from the four possible structures (1-4) shown below. Please give at least one reason that allowed you to eliminate each of the incorrect structures for X.

Reasons: Structure 1: wrong molecular weight (C10H12O; mw = 148 g/mole; while X has a molecular weight of 164 g/mole from the molecular ion at m/z = 164 in the mass spectrum) Structure 2: correct molecular weight (C10H12O2; mw = 164 g/mole) & M+1 intensity (10 x 1.1% = 11%), but not compatible with the IR spectrum of X. Structure 2 would not show an O-H stretch at 3518 cm-1 & would show a C=O stretch at ~1700-1730 cm-1. Additionally, structure 2 would only exhibit 8 peaks in its proton-decoupled 13C NMR spectrum, & would not be soluble in 5% aqueous NaOH solution. (Of the common function groups, only carboxylic acids & phenols are acidic enough to dissolve in 5% aqueous NaOH solution.) Structure 4: correct molecular weight & M+1 intensity, IR compatibility (O-H stretch), should be soluble in 5% aqueous NaOH solution; but is too symmetrical (i.e., would show only 8 peaks in its proton-decoupled 13C NMR spectrum) Structure 3: this is the best structure for X. It has the correct molecular weight, M+1 intensity, IR compatibility (O-H stretch), 13C NMR symmetry (10 peaks) &, as a phenol, should be soluble in 5% aqueous NaOH solution.

End exam

CH3

OH

OCH3

O

CH3

OCH3

OH

OH

CH3

HO

1 2 3 4

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Table 1. Nuclear spin for common nuclei which are "NMR active".

Nucleus Spin Natural abundance 1H ½ 99.98 2H 1 0.015 13C ½ 1.11 31P ½ 100 19F ½ 100 11B ½ 80

Table 2. Some common isotopes and their abundance Isotope abundance 1H 99.98 2H 0.015 12C 98.9 13C 1.10 14N 99.6 15N 0.37 16O 99.8 17O 0.038 18O 0.200 35Cl 75.8 37Cl 24.23 79Br 50.7 81Br 49.3 127I 100 Table 3. Some typical coupling constants for 19F and 13C H and F on same carbon (geminal) H

C

F

JC-H 100-130 Hz

JC-F 250-350 Hz

JH-F 80-100 Hz F on carbon adjacent to C and H C

H F

JH-F = 15-30 Hz

JC-F = 30-50 Hz