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I. (18 pts). Predict the 1H NMR spectrum for the molecule M, assuming a very pure sample with no solvent effects.

On the structure draw in and label each type of proton clearly (A, B, C, etc.).

A. (08 pts).  Then on the chart below, draw the peaks corresponding to these protons (again clearly labeled A, B, C, etc.),

showing the appropriate chemical shift, multiplicity (s, d, t, quar, quin, sext, etc), and integration (note the # of H's above the peak). 

For each peak, provide an estimate of each coupling constant that contributes to the splitting observed. Ignore long-range coupling. 

NOTE: Do NOT attempt to draw peak shapes. Simply draw the appropriate

number of lines (like the TMS peak shown) that correspond to the multiplicity

with approximately relative peak heights.

B. (04 pts).  Explain how you got to the chemical shift positions you are

using; use the additivity table where appropriate and show your work. 

Again, refer to protons type, A, B, C, etc. 

HA: alkene H, typically in the range ! 5-6 ppm. Triplet coupled to HB , J = 7 Hz

HB: methylene perturbed by alpha C=C and alpha C=O 1.20 + 0.75 + 1.10 = 3.05

Doublet coupled to HA-  J=7

HC: methylene coupled to -OHD perturbed by alpha –OH and alpha C=O

1.20 + 2.30 + 1.10 = 4.60 Doublet J=7 Hz (approx).

-OH: not too predictable, 2-5 ppm. Triplet coupled to HC  Arbitrarily: 3.5 ppm J = 7 Hz approx

01234567891011 -1

ppm

TMS

HA

HBHC

-OH

2H

1H

2H

1H

(relative areas)Integration numbers could bedoubled to reflect actual numberof protons

__________________________________________________________________________________________________________

C. (04 pts).  Explain carefully how the spectrum would change if a small amount of HCl were added to the sample. Draw a

mechanism for any chemical reaction that accounts for the changes. 

R CH2O

Ha

+ Hb+

R CH2O

Ha

Hb

-Ha+ R CH2

O Hb

Protons Ha and Hb exchange rapidly via the oxonium ion formed from protonation

This exchange now averages the spin state of the H attach to O, and the coupling

to the -CH2- group is now zero, no coupling, a singlet. The -OH proton also showsa singlet due to averaging of the environment, and no net coupling to the -CH2-.

Doublet at 4.60 becomes singlet Triplet at 3.5 becomes singlet  

D. (02 pts).  How many peaks do you predict for the 13C NMR spectrum (broad band decoupled) of M ? __4_____

(plane of symmetry in the molecule) 

HO

O

OH

O

M

HA   HA

HB

HB

Hc   Hc

HB

HB

Hc   Hc

 

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II. (22 pts).  Consider the spectral data for molecule X on the data sheets in the Data Supplement. [You may not

“buy” a structure for this problem]

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 CR H

C NR

C CH2

R

R

C O

R

R

R NH2

R Cl

R OH

(non-conj)

sp C-H at 3300 not observed

peak 1700-1800 not observed

N-H at 3100-3400 not observed

O-H 3200-3300 not observed

could be present due to peak at 1622

could be present due to peak at 2250

could be present due to no peak expected in functional group stretching region

 

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

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

(check the best answer; explain any ambiguity)

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

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

(check the best answer; explain any ambiguity)

D. (04 pts). Consider the mass spectrum.

1. What is the molecular weight of the molecule?  ___93_______

2. What is your estimate of the number of carbon atoms in the molecule?  ___6________

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

O? N? F? Cl? S? [both would show significant M+2 peaks

E. (01 pts). 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. (06 pts).  Draw LARGE in the box your best proposal

 for the structure of X. 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 follow table. In the column “coupling connections”,

list the other H(s) to which the proton in question is coupled

Chemical shift (ppm) Pattern (s, d, t, quar, quin, sext) Area Coupling connections

 A 1.80 quintet 2 B, C

 B 2.50 triplet 2 A

C 2.30 quartet 2 A,D

 D 6.75 Triplet 1 C

 E

F

G. (03 pts). Note the peak at ! 6.75 ppm in the 1H NMR spectrum of X and the proton to which you assign it in your

structure.  Explain carefully how this chemical shift position is consistent with your structure. [If you are not confident of a structure,

 just discuss this feature in general terms]

C

H

NC

H

N

electron poor,deshielding

The alkene H is deshielded relative to a simple alkene due to electronwithdrawal by the CN group, both resonance (large) and inductive.  

H. (02 pts). Consider the peak at ! 148 and the peak at ! 22 ppm in the 13C NMR spectrum of X. Draw your structure

here again, and then circle the carbon which appears at 148 and put a square around the carbon which appears at 22 ppm.  [If you

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

C

H

N  

C

H

N

H  H

H

H

HH

A

B

C

D

 

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III. (06 pts). The lowest energy conformer of a compound can

vary with pH. The process shown here involves addition of a proton to A and also a

change in the preferred orientation of the carboxyl substituent. 

A. (03 pts). On the chair templates provided below, draw the two chair

conformations that are in equilibrium for compound A. Explain which

conformation is lower in energy, noting all relevant interactions.

O

OO

O

conformers of A(circle most stable)CO2

O2C

axial group, destabilized by small gauche interactions with O

equatorial group, no torsional strain

 

B. (03 pts).  Do the same for B (on templates below) and explain why the -CO2- group in A prefers a different orientation

than the -CO2 H group in B.

O

OO

O

conformers of B(circle most stable)O

HO

great geometry for H-bond stabilization; axial conformation stabilized and becomes preferred.

O

O

H cannot participate in intramolecular H-bonding

no torsional strain, no special stabilization

In A, the carboxylate anion cannot provide any special stabilization, and is destabilized by gauche interactions.In B, the carboxyl group is a great H-bond donor and the axial conformation allows strong stabilization  

___________________________________________________________________________________________________________

IV. (05 pts). Compound C can adopt chair conformations very similar

to those observed for substituted cyclohexane rings.

A. (02 pts). On the chair templates provided below,

draw the two chair conformations that are in equilibrium for compoundC.

O

O

a

b  a

a

1   2

H

H 

B. (03 pts). One of the conformers from part (A) is considerably lower in energy than the other.  Explain which conformer

is least stable, noting all relevant interactions.

Conformer 1 shows one typica l gauche interaction with a CH2 group (a), and a smaller interactionwith the carbonyl group (b).

In the ring fli p isomer, 2, the axial group shows two typical gauche interactions (a).Conformation 2 is less stable due to greater gauche in teract ions . 

O

B

HO

O

O

A

O

OO

H+

O

C

 

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V. (11 pts). Using a catalyst, the epoxide L may be opened by water to form the diol M.

A. (07 pts). Considering only the data below and no other interactions, draw the three staggered conformations of M using the

 Newman projection templates below, looking along the C 1-C 2 bond. Calculate the torsional strain in each, write the total below the

drawing, and circle the most stable conformer.

O

Me MeH H

L

H2O

catalyst

OH

HO Me

MeH

H

M

H–H gauche 0.0 kcal/mol H–CH3 gauche 0.0 kcal/mo

CH3 –CH3 gauche 0.9 kcal/mol H–OH gauche 0.0 kcal/mo

CH3 –OH gauche 0.3 kcal/mol HO–OH gauche 0.1 kcal/mol1 2

OH

H Me

OH

MeH

OH

H Me

Me

HHO

OH

H Me

H

OHMe

1   1 10.9

0.9

0.1   0.10.3

0.3

1.5 kcal/mol

torsional strain

1.0

0.3

0.3

0.7

 

B. (02 pts).  In protic solvents, would the energy of gauche interactions with OH increase, decrease, or be unchanged? Why? 

In a protic solvent, such as water, the –OH groups would enter into H-bond donation and H-bond acceptance with with the

solvent, becoming effectively bigger and giving more repulsion—larger gauche interactions

C. (02 pts). One can calculate the barrier to rotation about the central carbon-carbon bond in M using the usual steric repulsion

parameters, but it turns out that the barrier for M is actually about 3 kcal/mol higher than estimated. What additional factor can you

imagine which would contribute to a raising of the barrier to rotation about the C 1-C 2 bond?   Explain with words and pictures.

The question is whether the transition state (highest energy eclipsed) is raised by some special interaction or whether the

lowest energy conformation is lowered by some special interaction. Either way, the barrier would be increased.

Two acceptable possibilities:

a. in one eclipsed structure, the –OH groups are syn and might suffer

from strong dipole-dipole repulsion [although the same conformation might

be stabilized by H-bonding]

b. in two staggered structures (center and right, above), the –OH groups

could be H-bonded and make it more difficult to rotate by stabilizing the lowest

energy conformer.

HO

H Me

H

OH

Me

 

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VI. (04 pts). Imagine the C=O stretching vibration in the infrared spectrum of the ketone cyclohexanone.

A. (02 pts). Suppose one adds an equimolar amount of BF3 to the cyclohexanone and again measures the IR

spectrum. Would you expect the C=O stretching vibration to: 

Increase Decrease Stay the same

B. (02 pts).  Explain your choice: 

O + BF3

acid/base interaction

OBF3

OBF3

relatively larger resonance contributor (same number of charges)

O significant, but minor resonance contributor More single bond character in theacid/base complex compared tothe simple ketone; lower energystretch of the C-O bond.

 

_________________________________________________________________________________________________

VII. (06 pts). For each pair of compounds below, circle the one that would exhibit the higher C=O stretching frequency in its

infrared spectrum and explain your choice.

A. (03 pts).O

or C

O

sp hybridized; strong C=O bond

[actually around 2300 cm-1 

____________________________________________________________________________________

B. (03 pts).

O

H3C

or 

O

F3C

O

H3C

O

F3C

dipolar resonance structure destabilized by

the inductive effect of the -CF3 group; less

single bond character, higher energy

stretch 

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VIII. (10 pts).  For each of the pairs of molecules listed below, determine whether the given technique could

definitively distinguish between the two. If the technique can differentiate the two compounds, explain how. If the technique cannot

distinguish them, explain why and suggest an alternative technique (IR, UV, Mass spec, 1 H NMR, 13C NMR) that could, making sure to

 justify your answer with one clear point. You may choose each alternative technique only once on this page. 

1. Cl

Cl

and

2. and

1H NMR Spectroscopy:

Distinguishable?  yes no

Alternative technique: IR UV mass 13C-NMR

Explanation:

A. (05 pts)

Distinguishable?  yes no

Alternative technique: IR UV mass 13C-NMR

Explanation:

The alkene H will show coupling to each other. The coupling constant for the Z isomer is smaller than for the E.

H

H

H

H

E isomer  Z

The proton NMR would show a single peak for both, at approx. the same chemical shift. The carbon NMR would also showa single peak. The IR spectra would be different, but not predictably and not in the functional group region. Neither wouldshow UV absorption. The molecular weights are different, so they are easily distinguised by mass spec.

1. and

2. and

13C NMR Spectroscopy:

O

O

O

O

Distinguishable?  yes no

Alternative technique: IR UV mass 1H-NMR

Explanation:

B. (05 pts)

Distinguishable?  yes no

Alternative technique: IR UV mass 1H-NMR

Explanation:

By symmetry, both show 4 peaks in the carbon NMR; not distingishable. IR spectra would not be predictably different.Mass spec would work, but we already used it. Proton NMR also shows the same pattern for both. But the one onthe right has more p orbitals interacting and would absorb at longer wavelength.

5 peaks 4 peaks

Three pairs of carbons in the left molecule are identical; the carbonyl carbons are different.In the right hand structure, all carbons are paired up by symmetry

 

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IX. (18 pts). A tough structure problem! Consider the tabulated data for Structure Z. [ IR, 13C NMR, and 1H NMR spectra

are in the Data Supplement.]1H NMR (CDCl3): ! 1.44 (3H, t, J=7Hz); 2.00 (2H, quint, J=7Hz); 2.40 (2H, t, J=7Hz), 2.60 (2H, t, J=7Hz); 3.90 (2H, quart,

J=7Hz); 5.35 (s, 1H).13C NMR (CDCl3): ! 17.0, 21.0, 29.0, 37.0, 64.0, 107, 178, 200 ppm.

Infrared (liquid film): key peaks at 1605(s), 1650(s), 3080(w), 2900-3000(m) cm-1  s=strong; w=weak; m= moderate

Ultraviolet: "max 257 nm (# 11,200); 300 nm (# 34)

Mass spectrum molecular ion region: 142 (0.2%), 141 (9.0%), 140 (100%).

A. (02 pts). What is the molecular formula for Z

? _C 8 H 12O2_________

B. (02 pts).  How many degrees of unsaturation are there in Z?  ___3_____

C. (05 pts).  Draw the structure for Z which best fits all the data, showing all

hydrogens. Label the groups of equivalents H’s for reference.

You may “buy” the structure of Z for a penalty of TEN points.

D. (05 pts).  Explain in detail how your structure fits the patterns of signals

(splitting) in the 1 H NMR spectrum. 

Ha appears at 1.44 as a 3H triplet, coupled to Hb

Hb appears at 3.90 as a 2H quartet coupled to Ha

Hc appears at 5.35 as a 1H singlet, with no adjacent H

Hd appears at 2.60 (or 2.40) as a 2H triplet coupled to He

He appears at 2.00 as a 2H quintet coupled to Hd and Hf

Hf appears at 2.40 (or 2.60) as a 2H triplet coupled to He

E. (04 pts). The peaks at 107 and 178 ppm in the 13C NMR spectrum are atypical for the functional group(s) with which they are

normally associated.  Identify the carbons in your spectrum which give rise to these peaks by drawing the chemical shift number next

to the corresponding carbon. Then provide a rationale for these chemical shifts based on our usual parameters.

O

O

O

O

O

O

O

O

O

O

H

deshieldedcarbon

shielded

(178 ppm)

(107 ppm)

The double effect of a resonance withdrawing effect of the C=O and a resonance donor effectof the alkoxy O leads to very different chemical shifts for the alkene carbons.

This carbon also deshielded by electron-withdrawing effect of O  

O

O   CH3

HH

H

HH

H H

HH

a

bc

d

e

f   Z

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IX. (14 pts).  Lab-related questions:

A. (07 pts).  The aldehydes shown to the right,

4-chlorophenoxyacetaldehyde (A) and 2-hydroxy-1-naphthaldehyde (B),

would not make suitable unknowns in the Oxidation-Reduction Experiment.

For each case, explain why the aldehyde would not be a suitable unknown.

(Hint: You should consider not only the structure of the aldehydes,

but also the separation protocol for the products.)

A) Aldehyde A contains two hydrogens alpha to the carbonyl group, shown in bold.

OCH2   H

O

ClA  

The experiment stipulates that none of the aldehyde unknowns would contain alpha hydrogens. (Stay tuned next

semester to find out why alpha hydrogens would cause a problem.) Aldehyde B is a little trickier. The problem is not with

the structure of the aldehyde starting material (assuming that an excess of 50% aqueous NaOH solution was employed in

the reaction), but with the alcohol product. The alcohol product, C, is also a phenol, and, as such, would be soluble in the

aqueous base (but not in the methylene chloride layer). Thus, the alcohol and carboxylic acid products would not be

separated employing our standard separation protocol. 

OH

CH2OH

C  

B.  (07 pts). Compound X gives the IR spectrum summarized below:

IR (neat): 3030 (w), 2915 (s), 2713 (w), 1727 (s), 1456 (m), 1397 (m), 1117 (w), 1016 (w), 985 (w), and 828 (w) cm-1 

Choose (circle) the best structure for compound X from the possible structures shown below. Explain your reasoning. (Hint: You do

not need to utilize any absorptions in the “fingerprint” region of the IR spectrum to answer this question.)

H

O

O

O

H

 

The IR spectrum of X displays a strong C=O stretch at 1727 cm-1

, which is more consistent with the non-conjugated

C=O of the middle structure than the conjugated C=O of the first and third structures, which would be expected to appear

at ~1675ncm-1

. Also, the presence of the weak, low-frequency aldehydic C-H stretch at 2713 cm-1

 is compatible with the

middle and third structures, but not the first structure.

OCH2   H

O

Cl

OH

OH

A   B