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Chemistry 303, fall, 2010

SECOND EXAMINATION

7:30 PM, NOVEMBER 15th, 2010

Duration: 2.0 hr

Name____________________KEY________________________________________

This is an "open book" examination; you may use anything that is not alive.

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.

Be aware that in Problem VII, you may "purchase" the unknown structure (Z) for a 10-pt penalty. Go to the

front and request the structure of the proctor and he/she will deduct the appropriate points.

Write only in the space provided for each question. 

Score:

p2______/13 p3_______/08 p4_______/08

p5_______/07 p6_______/07 p7_______/12

p8_______/15 p9_______/12 p10______/18 Lab:________/14

Lecture total: _________/100 Penalty?_______ Adjusted total:__________/100

There are 11 pages in this exam. In addition, there is a separate spectra supplement which includes tables of (a)nuclear spin values and for common isotopes, (b) isotope distributions, (c) Chemical shift additivity table (7

pages). Turn in only the exam.

PLEDGE:_________________________________________________________________

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I.  (07 pts).  Consider the isomers, A, B, and C. H   CH3

A   B   C  

A. (4 pts)  How would you use IR spectroscopy to differentiate between A and C? Give two clear-cut differences.

1. Molecule A has an sp-CH bond with characteristic high energy absorption ca. 3300 cm-1

. C does not.

2. Molecule A is an unsymmetrical (terminal) alkyne, and shows a strong stretch at 2250 cm-1

 for the triple bond stretch

Molecule C is symmetrical and therefore the absorption for the triple bond stretch is not observed

B. (03 pts).  How would you use broad band decoupled 13C NMR spectroscopy to differentiate between B and C?  Give the

single most clear-cut difference.

Broad band decoupled13

C NMR shows one peak for each non-equivalent carbon. In B, all of the carbons are distinct, 8

peaks. In C, there is a plane of symmetry and therefore only 4 different carbons are present, 4 peaks.

___________________________________________________________________________________________________________

II. (06 pts). Consider the methane derivative, V.

A. (03 pts). What do you predict for the pattern that will be observed in the 1 H NMR spectrum?

 Explain carefully. 

The H will be coupled to the F and to the P. Doublet of doublets.

If the coupling constants were equal (unlikely), the H would appear as a triplet.

B. (03 pts). What do you predict for the pattern that will be observed in the 13C NMR spectrum (H broad band decoupled)

The C (13

C) would be coupled to the F and the P. Again doublet of doublets is most likely; triplet is possible.

CCl2P

F

Cl

H

V  

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III. (08 pts).  Consider compound P and its 1H NMR spectrum.

A. (05 pts)  Draw the hydrogens on compound P, and assign each type of hydrogen (H a, H b, etc.) to the corresponding NMR peaks,

and the pattern (s, d, t , quartet, quintet, sextet, etc).

PPM

e

d

c

b

a

(1H) (1H)

(4H)

(6H)   (6H)

OO

CH3

H3C

CH2

H3C

CH2CH3

P

H

H

(a)

(a)

(c)

(c)

(b)

(b)   (d)

(e)

(t)

(q)

(d)

(septet)

(s)

 

B. (03 pts).  How many peaks do you predict for the13

C NMR (broad band decoupled)? 4 5 6 7 8 9 10 11  12

(circle single best answer)

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IV. (08 pts). Isomers K  and L show distinctly different IR spectra in CCl4 solution.

K: sharp peak at 3300 cm-1

, strong sharp peak at 1700 cm-1

.L:  broad peak at 3300-3500 cm-1; strong sharp peak at 1715 cm-1.

A. (04 pts).  Explain the difference in pattern (sharp vs broad) for the peaksaround 3300 cm-1. Include in your answer why the sharp peak for K  is at the lowerend of the 3300-3500 cm

-1 range.

O

OMe

O

O

K

L

O

OMe

H

OH

O

L can be H-bonded to itself, leading to a range of O-H bond energies and a rangeof -OH stretching frequencies, a broad peak.

K is internally H-bonded with a fixed -OH stretching frequency, a sharppeak. It is also a weak bond, low energy, low frequency as the H is partlyshared with another O. H

OMe

 

B. (04 pts).  Explain the difference in position (frequency) for the peaks at around 1700-1715 cm=1.

O

OMeK

O

H

O

OMe

O

H

L is a fairly normal ketone, at 1715 cm-1. However, the fixed and favorable H-bonding in K makes the dipolar resonance

structure more favorable, leading to more single bond character for

  the C=O, and lower stretching frequency.

 ________________________________________________________________________________________________

O

OMe

O

OH

K   L

OH OMe

 

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V. (21 pts). A reaction that you will see next semester involves the conversion of acetanilide (1) into TWO of

the three isomeric structures (2, 3, 4).

Upon running the reaction you are pleased to find that the two products (call them X and Y) produced are easily separated,

and you are left to identify the structures by spectroscopy. X and Y correspond to two of the three possible structures, 2-4.

A. (02 pts). Would IR spectroscopy be useful in assigning the structures of X and Y, assuming you use the region from

1500-4000 cm-1 only. Explain briefly. For your reference, the spectrum of X is given here:

B. (05 pts). The13

C NMR data were also collected and given here:

Isomer X 13C NMR: (! 169.2, 136.5, 135.8, 134.8, 125.6, 123.4, 122.3, 25.5)

Isomer Y 13C NMR: (! 169.2, 145.9, 142.0, 124.4, 118.4, 25.4)

Can you distinguish the isomeric products using 13C NMR alone? Describe specifically the single most clear-cut distinction you can

determine about the isomers X  and Y from the 13C NMR spectra and clearly define any ambiguities.

Structure 4 has a plane of symmetry (assuming rapid rotation about the phenyl-N single bond) and therefore should show

only 6 peaks. Consistent with the data for Y, and inconsistent with the data for X.

Both 2 and 3 would have 8 peaks in the13

C spectrum due to 8 non-equivalent carbons. It would be difficult to

unambiguously decide which is isomer X based on the13

C NMR data alone.

No. The peaks in the region 1500-4000

cm-1

 are due to the C-H, N-H, and C=O.

Essentially the same in the isomers 2-4. 

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C. (07 pts). The 1H NMR spectrum of Y is shown here.  Draw the structure of Y  in the box and explain carefully how it is

consistent with the patterns of peaks between 7 and 9 ppm in this spectrum. You need not deal with relative chemical shifts. Assume

all coupling to H on adjacent carbons is 9 Hz. Ignore long-range coupling.

Ha is coupled only to Hb and is therefore one of the doublets, at 7.7 or 8.1 ppm.

Hb is coupled only to Ha, and is the other doublet.

Hc is identical by symmetry to Hb.

Hd is identical by symmetry to Ha.

Y

NH

O

Ha

Hb

NO2

Hc

Hd

 

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D. (12 pts). The 1H NMR spectrum of X is shown here.

1. (08 pts).  Draw the structure of X  in the box and explain carefully how the patterns between 7 and 9 ppm are consistent with this

spectrum. All of the multiplets in the region show area 1H. You need not analyze relative chemical shifts. Assume all primary

coupling constants are 9 Hz. 

Ha is coupled only to Hb and is therefore one of the doublets, at 8.2 or 8.8 ppm.

Hb is coupled to Ha and Hc, and is therefore a triplet (assuming equal J).

At either 7.2 or 7.7 ppm.

Hc is coupled to Hb and Hd, and is therefore a triplet (assuming equal J).

At either 7.2 or 7.7 ppm.

Hd is coupled only to Hc and is therefore one of the doublets, at 8.2 or 8.8 ppm.

2. (04 pts).  Explain briefly how the other candidate structure (not Y ) is not compatible with these patterns. 

NH

O

NO2

Hb

Hc

HdHa

isomer 3

This isomer would show a singlet for Ha, along with two doublets (Hd and Hb)and a triplet for Hc, and therefore does not fit the spectrum for X

 

X

NH

O

NO2

Ha

Hb

Hc

Hd

 

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VI. (15 pts). Use the techniques of spectroscopy (UV, IR, MS, 1H NMR, 13C NMR) to distinguish the pairs of molecules

shown. You may use each technique only once. Clearly describe the features that would distinguish the pairs of molecules using the

method selected. With mass spec, you may not use the high resolution technique to give exact molecular formula.

The more obvious methods are discussed here. Other good answers are possible.

H3C

O O

H3C

O

N

O

CH3

CH3

A.

a   b

Method:

Explain:

CH3

CH3

IR

(a) would show one C=O peak. Both C=O are approx. the same environment

(b) would show a normal C=O at 1715 cm-1, and an an amide-type C=O at

  lower frequency (1680 cm-1)

Note: mass spec would also differentiate them (different nominal masses) but this is the only one that works with IR

_____________________________________________________________________________

CH3O

OCH3

H3C

H3C CH3

SH3C

c   d

Method:

Explain:

B MS

Both molecules have the same nominal molecular weight. However, the presence of onesulfur atom is confirmed by an M+2 of 4% the intensity of the parent ion. For (c), the M+2will be very tiny,

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VII. (30 pts). Consider the spectral data for compound Z on the separate data sheets.

A. (04 pts).  Consider the IR spectrum, from 4000-1500 cm-1. Circle the functional groups that you can rule out from the spectrum

and explain in a phrase with the single clearest element of data (or absent data).

C CH2

R

R

C O

R

R

R Cl

R OH

C N

R

R

R

C CHRmissing the typical peaks for sp C-H stretch at 3300 cm-1 and for

terminal   RC   CH triple bond stretch.

missing the typical -OH stretch at 3200-3500 cm-1

 

B. (01 pt). Consider the 1H NMR spectrum.  How many different types of protons are suggested by the spectrum? 

1___ 2___ 3___ 4___ 5_X_ 6___ 7___ cannot tell____

(check the best answer; explain any ambiguity)

C. (01 pt). Consider the 13C NMR spectrum.  How many different carbons are suggested by the spectrum? 

1___ 2___ 3___ 4___ 5___ 6___ 7_X_ cannot tell____

(check the best answer; explain any ambiguity)

D. (05 pts). Consider the mass spectrum.

1. What is the molecular weight of the molecule?  ___123_____

2. What is your estimate of the number of carbon atoms in the molecule?  ___8 or 9_____

10% is approx. The exact % considering one 15N and 8 C is 8.88 + 0.4 = 9.28%

3. Which of the following atoms can be ruled out by the mass spectrum alone? (circle all correct answers)

O? N? F? Cl? S? Br?

E. (01 pt). Consider the UV spectral data.

 Are there conjugated pi bonds in the structure?

Circle single best answer: YES NO CANNOT TELL

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F. (10 pts).  Draw LARGE in the box your best proposal

 for the structure of  Z. On your structure, label each H (or set

of equivalent H) with a letter for reference (A, B, C, D, etc).

 Referring to your structure and labels, please fill in the fol low table.

 In the column “coupling connections”, list the other H(s) to which the proton

in question is coupled (not coupling constants).

 NOTE : you may "buy" the structure Z at a penalty of 10 pts.

_______________________________________________________________

Chemical shift Pattern

(ppm) (s, d, t, etc) Area Coupling connections

 A 1.25 s 6 none

 B 1.92 d 2 To E

C 2.08 s 3 none

 D 5.45 d 1 To E

 E 5.95 d of t 1 To D and B

G. (03 pts). Note the peaks centered at ! 1.92 ppm in the 1H NMR spectrum of Z and the proton(s) to which you assign

them in your structure.  Explain carefully how this chemical shift position and pattern is consistent with your structure. [If you are not

confident of a structure, just discuss this feature in general terms]. Use the additivity tables to support your assignment; show your

work.

This peak is due to HB (2 equivalent H) coupled to HD. HB appears as a doublet.

Calculated chemical shift: methylene 1.20 base"-alkene 0.75

#-imine (C=N) not given. (probably small since beta- C=O is only 0.3)

_________

total: 1.95 (compared to 1.92 observed)

H. (03 pts). Note the peaks centered at ! 5.45 ppm in the 1H NMR spectrum of Z and the proton(s) to which you assign them

in your structure.  Explain carefully how this pattern is consistent with your structure. [If you are not confident of a structure, just

discuss this feature in general terms].

This is a doublet of triplets for 1H, due to HD coupled to HE (producing a doublet) and to the two HB (to convert each peakof the doublet into a triplet).

I. (02 pts). Consider the peak at ! 176 ppm and the peak at ! 31 ppm in the 13C NMR spectrum of X. Draw your structure

here again, and then circle the carbon which appears at 176 and put a square around the carbon which appears at 31 ppm.  [If you

are not confident of a structure, just discuss this feature in general terms]

These two peaks are singlets; therefore there is no H attached to either one. One is

part of the C=N group and is deshielded (176 ppm) while the other is a simple

alkyl carbon slightly perturbed by the alpha-C=N (31 ppm).

__________________________________________________________________

Alternate acceptable structures: 

N

H3C

CH3

CH3

H

H HH

A

BD

E

C

NCH3

CH3

H

H HH

A

BD

E

C

CH3

 

NCH3

H3C

H3C

H

HHH

A

B

C

D

E

 

C

NCH3

H3C

H3C

H

HH

H

 

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VII. (14 pts).  Lab-Related Question.

A. (06 pts).  In yet another dyslexic moment in the orgo lab, Joe L. Sapphire set up his Oxidation-Reduction reaction employing 10%aqueous NaOH solution (from the Analgesic Experiment), rather than the recommended 50% aqueous NaOH solution. During the

workup of his reaction, he failed to obtain any carboxylic acid upon acidification of the aqueous layer. (Yes, his aqueous solution wasdefinitely acidic!) In an attempt to isolate some of his carboxylic acid, he concentrated his acidic aqueous layer by boiling off somewater. When enough water had been removed, a colorless solid separated. This colorless solid failed to redissolve upon addingethanol to the hot mixture, but did dissolve upon adding additional water. If isolated, the colorless solid failed to melt below 400 °C.(Melting points above 400 °C cannot be safely taken on our melting point apparatus.) What was this colorless solid, and how did it form?

Presumably Joe L.!s oxidation-reduction reaction either didn!t work or did not go anywhere close to completion as a resultof his blunder. So, presumably most of his material (i.e., unreacted unknown aldehyde) is in the methylene chloride layer.

While it was worth boiling off some of the water from his acidic aqueous layer in an attempt to isolate any of his carboxylic

acid that might be present, the colorless solid doesn !t appear to be his carboxylic acid. All the carboxylic acids listed in

Table 1 melt below 400 °C, and the solubility properties of the colorless solid (i.e., insoluble in hot ethanol, soluble in

water) aren!t compatible with the carboxylic acid either. In fact, the colorless solid sounds like a salt! The colorless solid is

NaCl, mp 801 °C. NaCl is produced by the neutralization of the excess sodium hydroxide with the HCl;

i.e.,

NaCl is, of course, quite water soluble; but, if enough water is removed, the solution becomes saturated, and the NaClwill precipitate out. The NaCl is ionic; and, as such, will not dissolve in hot ethanol, but will re-dissolve upon addition of

more water.

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B. (08 pts). Steam distillation of ocimum basilicum L. affords basil oil, which is composed of about 80% estragole (X). Spectral datafor estragole (X) are summarized below.

mass spectrum: m/ z  = 148 (M, 100%), 149 (11%)

IR (liquid film): 3077-2835 (multitude of weak peaks), 1640 (m), 1611 (m), 1511 (s), 1301 (m), 1247 (s), 1176 (s), 1038 (s), 992(m), 914 (m), and 810 (m) cm-1 [There are no absorptions in the 3100-4000 cm-1 region of the spectrum. m = medium, s = strong]

Choose (circle) the best structure for estragole (  X  ) from the four possible structures ( 1-4 ) shown below. Please briefly explain how

 you eliminated each of the incorrect structures for X .  (Hint: You do not need to utilize any absorptions in the “fingerprint” region ofthe IR spectrum of X to answer this question.)

Structure 1: wrong molecular weight (C10H12; mw = 132 g/mole; while X has a mw of 148 g/mole from the parent ion

at m/ z  = 148 in the mass spectrum)

Structure 2: correct molecular weight (C7H4N2O2; mw = 148 g/mole); but the M+1 peak in the mass spectrum of X at

m/ z  = 149 should only be about 8 % (7 x 1.1 = 7.7 %) of the molecular ion. Also, it could be argued that structure 2 would

show a CN triple bond stretch at ~ 2250 cm-1

 in the IR spectrum, which was not observed.

Structure 4: correct molecular weight (C10H12O); mw = 148 g/mole) and the expected M+1 intensity (10 x 1.1 = 11.0

%). However, structure 4 should exhibit an O-H stretch at ~ 3300 cm-1

 in the IR spectrum. No such absorption was

observed.

Structure 3: As Sherlock Holmes once said “when you have eliminated the impossible, whatever remains, however

improbable, must be the truth”. Thus, structure 3 must be the correct structure for estragole (X). Structure 3 has the

correct molecular weight and M+1 intensity, as well as IR compatibility. (Of course, there are also key spectral feature in

the forbidden “fingerprint” region of the IR spectrum, which could further substantiate this structural assignment, but they

were not required.)

 End exam

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