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Chemistry 2301 Thursday, August 4, 2011

Final Exam Answer Key

Final Exam Mean: 121 Final Exam Median: 120 Final Exam St. Dev.: 37

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Multiple-Choice Problems Please answer these problems on the bubble sheet.

(2 pts each) For ethylformamidinium cation, drawn at right:

For each atom marked “hybridization”, indicate whether the atom is hybridized(a) sp, (b) sp2, (c) sp3, or (d) none of these.

For each bond angle marked “angle”, indicate whether the angle is closest to (a) 109.5°, (b) 120°, or (c) 180°.

Although the nitrogen atom on the left end might look like it should be sp3 at first glance (with its three bonds and one lone pair), its hybridization lowers to sp2, because the nitrogen has a lone pair and is adjacent to a multiple bond.

7. (3 pts) Of the resonance structures on the right,

which contributes least to the overall electronics in ethylformamidinium cation?

All of these are valid resonance structures. Structures A and C have more bonds than structure B, and they have filled octets where B does not. Admittedly, structures A and C have a positive charge at a more electronegative atom, but Wade’s list of priorities in making resonance structures says that this isn’t as important as bonds or octets. So B is minor, and A and C are major.

H N C

H

N

H

C

H

C

H

H

H

H

H

a.

H N C

H

N

H

C

H

C

H

H

H

H

H H N C

H

N

H

C

H

C

H

H

H

H

H

c. b.

hybridization

#1 (b) sp2

H N C

H

N

H

C

H

C

H

H

H

H

H

hybridization

#2 (b) sp2

hybridization

#3 (b) sp2

hybridization

#4 (c) sp3

angle

#6 (a) 109.5°

angle

#5 (b) 120°

3

8. (3 pts) Which of the structures below represents the most stable 3-dimensional conformation of ethylformamidinium cation?

(2 pts each) Does the proton-transfer equilibrium for each acid-base pair shown below favor products, or starting materials? (Is the acid strong enough to protonate the base?) 9.

Both the starting material side and the product side have a cation--which side accommodates the charge better? Nitrogen is less electronegative than oxygen, so it tolerates the charge better, favoring the product side.

NC

N

H

H CC

H

HH

H H

Ha.

NC

N

H

H CC

H

H

HH H

Hb.

NC

N

H

H

H CC

H

HH

H H

c. d.

N

CNH

HH

CC

H

HH

H H

Does the equilibrium favor:

N

H

H HO

H

H H+

a. products, or b. starting

materials?

conjugate acid-base products

OH H

+ N

H

H H

H

4

10.

Again, both the starting material side and the product side have an ion. Both ions are stabilized by resonance, but the trifluoroacetate anion on the product side is also stabilized by induction from the electronegative fluorine atoms. So the equilibrium favors the product.

11.

Both anions are primary carbanions, but the starting material cation is stabilized by resonance. This equilibrium favors that cation.

(2 pts each) For each of the molecules drawn

on the right:

For each atom marked “chirality”, indicate whether the atom would be labeled as (a) an (R)-chiral center, (b) an (S)-chiral center, or (c) not a chiral center, according to the Cahn-Ingold-Prelog classification system.

Indicate whether the molecule would be chiral or achiral.

chirality

#12

b.(S)

CH3

H3C HBr

BrH 14. Is this molecule

a. chiral, or

b. achiral?

chirality

#13

a.(R)

This molecule has two stereocenters with opposite Cahn-Ingold-Prelog assignments, and exactly the same groups attached. That means the molecule is meso, and achiral.

CH3

O

OF3C

O

OH

+ conjugate acid-base products

a. products, or b. starting

materials?

CH3

O

OH

F3C

O

O+

H H

H H H H

H H H

H

H

H

H

+ conjugate acid-base products

a. products, or b. starting

materials?

H

H

H

H

H H

H

H H

H H

H

H

+

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17. (2 pts) Are the methyl-group hydrogens

in the structure on the right

a. enantiotopic, b. diastereotopic, or c. neither of the above?

Two atoms are enantiotopic if replacing each one with a different atom—say, atom “X”—would create two new enantiomers, usually by creating a new chiral center where there wasn’t one before. That wouldn’t be true for any of the methyl group protons; replacing any of them with “X” doesn’t create a new chiral center, and so the protons aren’t enantiotopic. Two atoms are diastereotopic if replacing them creates two diastereomers instead of two enantiomers, again usually by creating a new chiral center. Once again, that isn’t what happens here. So the answer was “neither of the above”.

(4 pts each) Each of the reactions below is drawn with two possible reaction conditions. If only one of the two reaction conditions would generate the given molecule as the major product, answer with the corresponding letter. If both sets of conditions would accomplish the reaction, answer (c) “BOTH”. If neither set of reaction conditions would succeed, answer (d) “NEITHER”. 18. HBr adds to alkenes with Markovnikov orientation (with Br at the more substituted

position), unless peroxide is added, in which case the HBr adds anti-Markovnikov. Here, the product is anti-Markovnikov.

BOTH would work

NEITHER would work

HBr HBr,

benzoyl peroxide

Br

a. b.

c. d.

NH2

H3C H

O

HO

16. Is this molecule

a. chiral, or

b. achiral?

All molecules that have only one stereocenter—like this one—are chiral.

chirality

#15

b.(S)

6

19.

Both of these sets of reagents add H-OH Markovnikov to the double bond. The first set is not as clean as the second, but they both work.

20.

Both of these sets of reagents perform dihydroxylation—they add an –OH group to either side of an alkene. They differ in that mCPBA performs anti-dihydroxylation (adding the two –OH groups to opposite faces of the alkene), while KMnO4 performs syn-dihydroxylation (adding the two –OH groups to the same face of the alkene). Here, even though the product is drawn to make it look like the –OH groups are on the same side, they are on opposite sides of the starting material:

BOTH would work

NEITHER would work

H2SO4, H2O

1. Hg(OAc)2, H2O 2. NaBH4

OHa. b.

c. d.

BOTH would work

NEITHER would work

mCPBA, H2O

KMnO4

(cold, dilute in H2O/OH-) H3C

H

CH3

HH3C

CH3

OH

OH

a. b.

c. d.

H3C CH3

H H

H3C CH3

H HOH

HO

H3C

CH3H

HOH

HOequals

mCPBA

H2O

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21.

Both of these sets of reagents add H2 to alkynes. Na/NH3 adds only one molecule of H2 to yield a trans-alkene as a product. Pt/H2, on the other hand, exhaustively hydrogenates the alkyne all the way to an alkane.

22.

Although both alkylithium (R-Li) and Grignard (R-MgX) reagents will add to carbonyl groups, they are also extremely basic. In the presence of even a weak acid, such as an alcohol, either one will be protonated before it has a chance to react with the carbonyl:

(4 pts each) Each of the reactions below is drawn with two possible products, marked (a) and (b). If one of the two products predominates, answer with the letter corresponding to the correct product. If the two products are produced equally, answer (c) BOTH. If neither product would result from the reaction, answer (d) NEITHER.

BOTH would work

NEITHER would work

Na, NH3 Pt, H2

H3C C C CH3

a. b.

c. d.

BOTH would work

NEITHER would work

1. CH3Li (1 equiv) 2. H3O

+

1. CH3MgBr (1 equiv) 2. H3O

+

H3C

OHO

H3C

OHHO CH3

a. b.

c. d.

H3C LiH

O

O

CH3

CH4+ +O

O

CH3

Li

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23.

Chromate is a strong reducing agent, and will oxidize the primary alcohol all the way to the carboxylic acid.

24.

Radical bromination replaces an H with a bromine atom at the most substituted position (via the most stable radical intermediate).

25.

Combination of an acyl chloride with an alcohol to form an ester doesn’t change the stereochemistry in either starting material.

+ F

O

Cl

H3C H

OH

Na2Cr2O7 H2SO4

a. b.

H

O

OH

O

BOTH (equally)

c.

NEITHER d.

Br2

h Br

Br

a. b.

BOTH (equally)

c.

NEITHER d.

pyridine HO

HCH3

H3CO

O CH3H

F H

H3CO

O HH3C

F H

a. b.

BOTH (equally)

c.

NEITHER d.

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32. (29 pts) For the reactions shown below, draw a mechanism that explains how the product is

generated from the starting material. In your answer, make sure that you:

Draw each step of the mechanism separately; Use “electron pushing” to show where the electrons in each step go; Use only the molecules that you are given; do not invoke reactants or solvents that

aren’t in the problem.

Resonance is not a mechanistic step per se—it’s just two ways of drawing the same molecule. As a result, you did not need to push electrons to show resonance, or even draw multiple resonance structures explicitly. You can integrate resonance into a mechanistic step; for instance, you could have combined it with the departure of H2O:

HBr CH3

OHCH3Br

Mechanism:

O

CH3

H

H BrO

CH3

H

H CH3

Br

resonance

CH3Br

CH3Br

4 4

4

2

O

CH3

H

HCH3

4 2 +

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As long as resonance is in your answer somewhere—either integrated into a step, or drawn explicitly—you get 2 points.

-OH is a terrible leaving group. If you drew the first step as -OH leaving, you only got 4 points for that step (out of 8 for the combined first two steps).

Rubric: (14 points total this part) Overall notes:

Things that have left (e.g., Br-) and spectators may be omitted until they are needed.

-2 points, for each arrow in each step, for errors in drawing arrows. Arrow must start at an electron pair, and end at nucleus where electrons will newly interact. Can only lose points if you get them.

-2 points for each error in charge, valency, structure, base, etc.; if error propagates, points are taken off only for initial error.

-2 points for each step combined with another, EXCEPT resonance. Combining resonance with another step is just fine.

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HOOH

O

Cl S

O

O

CH3

N

My mechanism would be:

HOO

H

Cl S

O

O

CH3

3

S

O

O

CH3

HOO

H

Cl

N

S

O

O

CH3

HOO

Cl

3

3

S

O

O

CH3

HOO

3

O H

S

O

O

CH3O

N O NH

3

Both proton transfer and ejection of Cl are extremely fast—hard to say which comes first. You could order steps 2 & 3 either way.

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The mechanism I’ve drawn invokes a five-coordinate sulfur atom as an intermediate in the initial nucleophilic addition to TsCl. This is analogous to the tetrahedral carbon intermediate that is observed when a nucleophile adds to a carbonyl C=O. Researchers have observed the tetrahedral intermediate in C=O additions directly, but not yet the S=O adduct; nevertheless, chemists usually infer the five-coordinate adduct based on C=O chemistry.

But not Wade. Wade shows a different mechanism for tosylation of an alcohol on page 470 of the text that does not involve a five-coordinate sulfur at the center--in other words, that has the initial oxygen nucleophile directly displace chloride.

Wade’s mechanism would be:

HOO

H

Cl S

O

O

CH3

6

S

O

O

CH3

HOO

H

N

S

O

O

CH3

HOO

O H

S

O

O

CH3O

N

O NH

4

3

3

13

I think Wade is probably wrong, but he’s not definitively wrong, and so we accepted either mechanism here.

Rubric: (15 points total this part) Overall notes:

Things that have left (e.g., Cl-) and spectators may be omitted until they are needed.

-2 points, for each arrow in each step, for errors in drawing arrows. Arrow must start at an electron pair, and end at nucleus where electrons will newly interact. Can only lose points if you get them.

-2 points for each error in charge, valency, structure, base, etc.; if error propagates, points are taken off only for initial error.

-2 points for each step combined with another. 6 points total for substitution of chloride in TsCl by alcohol oxygen.

This must occur before deprotonation by pyridine; pyridine is not a strong enough base to deprotonate an alcohol to any significant extent. However, we did not take off these 6 points if you deprotonated first.

Substitution can be two-step (with addition to sulfonyl, followed by ejection of Cl) or concerted.

3 points for first deprotonation by pyridine. This must occur after attack on TsCl. (See note above about order of events.) No

partial credit for this step if out of order. 2 points partial for deprotonation with base other than pyridine (e.g. TsO-, Cl-).

3 points for intramolecular attack by other alcohol on tosylate. Again, this must occur before deprotonation by pyridine, but you still get these

points if you put things out of order. (It’s the next ones you lose.) 3 points for second deprotonation by pyridine.

This must occur after attack by oxygen. See note above about order of events. No partial credit for this step if out of order.

2 points partial for deprotonation with base other than pyridine (e.g. TsO-, Cl-).

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33. (25 pts) Draw the missing reactant or product in the empty boxes. For products, give the predominant, most favored product. Illustrate stereochemistry in your answer where appropriate. For reactions that yield multiple enantiomers, draw only one enantiomer in the box, and include the note “+ enantiomer”.

The acetylide anion adds to the epoxide at the less hindered carbon. Rubric: 5 points.

Full credit for omitting Na+, or for protonating oxygen. (No proton source was provided in the problem, but no points were taken off if you assumed one.)

-2 points for each trivial structural mistake (omitting a carbon, charge, etc.) -2 points for omitting or incorrectly drawing stereochemistry in (any) product. 3 points partial for adduct at more substituted carbon of epoxide.

O

H3C

H

H3C C C Na O

H CH3

CC

CH3

Na

5

CH3CC

O

H3C

H

Br2 H2O

CH3

5

+ enantiomer

OH

CH3

HBr

15

Here, formation of a bromonium ion is followed by addition of water to the more substituted carbon:

The other product enantiomer

would be generated if the bromonium ion formed on the underside of the starting material (instead of the topside, shown above).

Rubric for this part: 5 points.

4 points partial for not writing “+ enantiomer”. 3 points partial for incorrect stereochemistry (syn adduct), indeterminate

stereochemistry (no wedges/dashes), or incorrect regiochemistry (Br at more substituted carbon).

2 points partial for dibromide (any stereochemistry). -2 points for each trivial structural mistake (omitting a carbon, charge, etc.)

CH3

Br Br

CH3

Br

H

+

+

The more substituted carbon accommodates more partial positive charge, and is thus preferentially attacked by H2O.

HO

H

(from behind, SN2)

O

CH3

HBr

H H

-Br

OH

CH3

HBr

NaBH4 CH3OH

O

OO

5

O

OHO

+ enantiomer

16

NaBH4 is a mild reducing agent. It is strong enough to reduce ketones to alcohols, but not esters. As a result, only the ketone is reduced. Rubric for this part: 5 points.

No need to show stereochemistry in your answer, or to write “+ enantiomer”. (Product is racemic, true, but drawing bond as a line would imply this.)

3 points partial for reducing ester (to anything) in addition to or instead of ketone. -2 points for each trivial structural mistake (omitting a carbon, charge, etc.)

In general, sodium methoxide is a strong base, and a better base than it is a

nucleophile, so it will react with the bromide via E2 if it can. There are two H’s on carbons adjacent to ( to) the leaving group—the methyl hydrogen, and the single H next to the carbonyl (C=O bond). The -carbonyl H would give the more substituted (and resonance-stabilized) alkene product, so we’ll focus on it.

E2 requires anti-coplanar arrangement of the H being taken and the leaving group.

The way the molecule is drawn, the H and Br are not on opposite sides of the C-C bond that connects them, so that bond will need to rotate to put them in position:

O

CH3

HH3C

HBr

CH3

same as prev. page, turned on its side

rotate this bond

CH3O

H and Br are anti-coplanar

H

Br

CH3

H

H3C

O CH3

H3C H

CH3OCH3

keeping all groups that don’t change in the same place…

H3C

O

H3C

CH3CH3O Na

H3C

O

CH3Br

CH3

5

17

Rubric for this part: 5 points for correct structure. 3 points partial for bromine elimination product

with wrong or unclear stereochemistry or wrong regiochemistry.

2 points partial for any bromine substitution product.

-2 points for each clearly trivial structure mistake. The answer that you intend needs to be clear for you to get these points.

Acetic acid is a poor nucleophile, and the added heat tips us off that this should be an SN1 or E1 reaction. There are no protons to the Cl leaving group, so it cannot be E1; this must be SN1. SN1 passes through an achiral carbocation intermediate, which reacts to generate a racemic mixture of products:

3

2

3

(with any stereochemistry

at –OCH3)

H3C

O

CH3

CH3H3C

O

CH2

CH3

H3C

O

CH3H3CO

CH3

HCl

H3C

O

OH

H

O

CH3

O

+ enantiomer

5

HCl H

H

O

CH3

O

H

O

CH3

O

+

(when acetic acid adds to top

face of cation)

(when acetic acid adds to bottom face of cation)

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Rubric for this part: 5 points for correct structure and either (a) including “+ enantiomer” or (b) not

illustrating stereochemistry (which infers racemate). 4 points partial for omitting “+ enantiomer” from correct answer that shows

correct enantiomer only. 3 points partial for inversion of stereochemistry only (as if reaction were SN2). -2 points for each clearly trivial structure mistake.

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34. (32 pts) (a)

(or, could draw mirror image of this one; any two of these three receives full credit)

(again, could draw mirror image of this one; any two of these three receives full credit)

(b) Of the four Newman projections you drew above, which would react the fastest with OH-

in an SN2 reaction? Circle one Newman projection.

Sure, the anti-conformer is the most stable. But, the conformation with the least amount of steric hindrance opposite the leaving group—with the most room for OH- to approach—will react the fastest. So:

Br

CH3H3C

Br

Br

CH3H3C

Br

H3C CH3

H2C

HH

Br

Br

H3C CH2

CH3

HH

Br Br

H H

H2C

HH

Br

Br

CH3H3C

H CH2

H

HH

Br BrH3C

CH3

3 3

3 3 2

H H

H2C

HH

Br

Br

CH3H3C

H CH2

H

HH

Br BrH3C

CH3

OH-

OH-

steric hindrance from alkyl group

prevents approach

no steric hindrance from H

20

(c)

Br

CH3H3C

Br

HO

CH3H3C

Br Br

CH3H3C

OH

E

+ Br- + Br-

+ -OH

(starting material)

(product) (product)

reaction coordinate

H CH2

H

HH

Br

Br

H3C CH3

HO

-

-

H3C CH2

CH3

HH

BrBr

HO

-

- steric

hindrance

1

3 3

Ea Ea

1 1

2

2

for drawing one step (hump) in each direction

for right hump lower than left hump

OR

Br

CH3

CH3

BrOH

H

H

-

-

Br

CH3

H3C

H

HBr

OH

H

H

-

-

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(d) In fact, both of these SN2 products are minor products of the reaction between one equivalent of OH- and the starting material, because OH- is a stronger base than it is a nucleophile. What molecule would be the preferred product of this reaction, and by what mechanism would it be produced?

Rubric: for parts (a/b): (14 points) 3 points for each staggered Newman projection.

I have drawn one anti and one gauche, but could also draw two gauche. Any two will do.

-1 point for each trivial structural mistake. 1 point partial on each rotamer if you omitted rest of chain (& just drew atoms on

the rotated bond) 2 points for circling any conformer where –Br is anti to an -H. for part (c): (13 points) 3 points for each transition state structure: Any conformation is acceptable.

1 point for backside attack by OH- on Br; 1 point for partial bonds; 1 point for partial charges. -2 points if transition state doesn’t correspond to product (say, if TS for left

reaction was also drawn for right reaction), or if there is no clear difference between the two TS structures. Only lose these points if you get them.

6 points for potential energy diagram: 2 points for drawing each process as a single (SN2) step; 2 points for drawing unhindered transition state as lower in energy than

neopentyl TS; 1 point for labeling Ea on each curve. (Curve does not have to be correct, and

relative heights can be wrong.) 1 point for circling right-hand product. for part (d): (5 points) 3 points for correct product.

1 point partial for any alkene. 2 points for “E2”.

Br

CH3H3C

Br

NaOH (1 equiv)

by what mechanism?

(not SN2)

E2

2

3

Br

CH3H3C

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9. (14 pts) For the starting materials and product shown below, propose a multistep synthesis.

In addition to the molecules shown, you can use any reagents and reactions we’ve learned about in class. You might discover multiple answers to this problem; draw only your best (one) synthetic route. Feel free to draw an incomplete route—we will give you partial credit where we can.

One possible route:

OH C CH

H H

HO

OH O

Swern oxidation (or Na2Cr2O7/H2SO4,

or PCC)

CCH

1. NaNH2 (to generate alkynyl anion)

2. (from above)

3. H3O+

O

OH

C C

H2, Lindlar’s catalyst

(Pd/BaSO4, quinoline)

H H

HO

4

3

3

for combining fragments;

for generating alcohol

4

for preparing to receive alkynyl

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Rubric: The synthesis requires four tasks: (a) getting cyclopenatnol ready to react with the

alkyne; (b) making the alkyne reactive (probably as an alkynyl anion), and combining it with the cyclopentyl fragment; (c) creating the alcohol group in the product (which could happen as a result of a and b, or could be done separately; and (d) hydrogenating the alkyne to generate an alkene.

-1 point for each minor error in structures or reagents; if error propagates, points are taken off only for initial error.

-2 points if step reagents are incorrect, but reaction could otherwise be accomplished with correct reagents. (Especially relevant to hydrogenation step.)

-2 points if reagents are correct, but product is wrong. If this happened, and you were led down an incorrect synthetic path by your mistake, you can also lose later points.

Making cyclopentanol reactive towards alkyne (a): 4 points.

Lots of potential ways to do this. Some end up becoming alcohol in product, some do not. Examples that would be worth full credit: Oxidation to ketone w/ Swern/chromate/PCC; Tosylation (TsCl/pyridine); Conversion to halide (SOCl2, PBr3, P/I2; 3 points partial for HX).

Deprotonating alkyne, and combining alkynylsodium with electrophile (b): 3 points.

1 points partial for deprotonation alone. There is no need to show alkynyl anion itself; just “NaNH2” is enough.

Other 2 points awarded if alkynyl anion is combined with something—anything that works—to form a C-C bond. So, combining alkynyl with ketone, tosylate, halide, or epoxide—no matter whether points were awarded in part (a) or not—receive credit here.

1 point partial of 3 above if combination would not work, but could if it were engineered correctly.

H3O+ workup can be omitted (though, in real life, it couldn’t be).

Full credit for drawing parts 1-3 of this recipe as explicit steps, or combining them onto the same arrow with numbers as I have. However, if you combine them on the same arrow, you must have numbers; you can’t add all the reagents at the same time.

Generating alcohol group (c): 3 points.

If answer used alkynyl + ketone, alcohol is generated automatically. 3 points for this. If answer used alkynyl + cyclopentyl-LG, allylic halogenation (with NBS) followed by

SN2 with hydroxide is worth 1 point. The SN2 would not be clean, lots of E2. Hydrogenation to cis-alkene (d): 4 points.

2 points partial for alternative hydrogenation protocol (Na/NH3, or Pt/H2). 2 points partial if you do this too early (before there are two substituents to be cis).

24

The spectra on this page correspond to a pure molecule, isolated from a chemical reaction. High-resolution mass spectrometry determined an exact mass of 102.0681 amu for the highest-mass (parent, M+) peak in the MS spectrum, which corresponds to a molecular formula of C5H10O2.

4000 3000 2000

Tra

nsm

itta

nce

1500

1716

2896

1000 500

(wavenumbers, cm )-1

80706030 5020 4010

m z/

Rel

ativ

e In

ten

sity

102 ( )M+

87

71

4543

3115

90 100 110

012

2

2

3

3

5 36 47

3.6

8

6.6 Hz

TMS

δ (ppm)

3.7

6.6 Hz

2.62.72.8

integral(w/ relativesegmentheights)

EI-Mass Spectrum

IR Spectrum

1H NMR Spectrum (300 MHz)

25

(1 pt each) Based on the features in the IR spectrum, which of the following functional groups would you expect the unknown molecule to have? Would the molecule contain a

The IR spectrum really has only two features above 1500 cm-1—a peak at 1716 cm-1, and a C-H peak below 3000 cm-1. The C-H peak has to be C(sp3)-H only; C(sp2)-H and C(sp)-H would both be found above 3000 cm-1. The IR resonance chart says that the peak at 1716 cm-1 is a C=O bond. (Sadly, problem #31 was supposed to ask you whether a C=O was present, but I copied and pasted the list from a different test, and didn’t replace the last item. Needless to say, a molecule that has no nitrogen atoms in it cannot have a C≡N group in it.) 36. (26 pts)

(a) The mass spectrum shows a parent mass peak at m/z = 102, and two fairly high-mass fragment ions at m/z = 87 and m/z = 71. For these fragment ions to be observed, the parent molecule must have ejected neutral (invisible) fragments with mass (102 - 87) = 15 and (102 - 71) = 31 atomic mass units (amu). What do you think are the structures of these neutral fragments?

There is only one fragment formula, given the C, H, and O atoms we have to work with in the parent, that adds to mass 15: CH3-. So, our molecule must have a methyl group. There are two possible fragment structures with mass 31; we

C(sp)

H26.

? a. Yes b. No

C(sp2)

H27. ?

a. Yes b. No

C(sp3)

H28. ?

a. Yes b. No

O

H29. a. Yes b. No ?

30. a. Yes b. No ? C C

31. a. Yes b. No ? C N

neutral fragment with mass 15: neutral fragment with mass 31:

4 4

CH3 OCH3

if 2 CH2OH

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ruled out the presence of an alcohol in problem 29, so the fragment must be CH3O-. (We gave partial credit for the alcohol fragment HOCH2-.)

(b) The mass spectrum shows a small peak at m/z = 103, above the mass of the parent M+

peak (102). Assuming that the material is pure (i.e., that there are no higher-mass contaminants in the sample), how is it possible that the mass of some molecules would be higher than expected? Please be brief. You could probably answer this in 10 words or less.

Rubric for part (b): 4 points for any answer that references a heavy isotope (or the words “heavy

isotope”). (c) What is the structure of the molecule? In the box below, draw your molecule’s

structure again, including all hydrogens. Then circle each set of equivalent H’s, and label each with its unique 1H NMR chemical shift.

The 1H NMR has many useful clues in it. Out of the four resonances, two are multiplets—indicating H’s with neighbors, and presumably with each other as neighbors—and two are singlets with intensity three, each indicating three

Any molecule that has one heavy isotope in it—say, where one of the carbons is a 13C instead of a 12C (probability: 1 out of every 100 carbons), or one of the hydrogens is an 2H instead of an 1H (probability: 1 out of every 10,000 hydrogens)—will show up in the mass spectrum as one a.m.u. higher than the “parent” peak. For this molecule, with 5 carbons, the intensity of the m/z = 103 peak should be ~5% that of the m/z = 102 peak.

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your molecule (C5H10O2)

C

O

CC

OCH

H H H H

H H

H

H H

= 3.7 ppm

= 2.7 ppm = 2.2 ppm

= 3.3 ppm

10

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equivalent protons with no neighbors. We can already guess that one of those singlets is from the -OCH3 group we predicted in part (b), and the other singlet is probably another -CH3 group somewhere. Each of the other two resonances represents two H’s with two neighboring H’s (because each multiplet is a triplet). So that means our molecule has the following fragments:

Rubric for part (c): 2 points for correct structure.

Structure need not be correct to receive the other 8 points in this problem. 2 points for each value. (8 points total this box.) To get full credit, circle must

include all equivalent protons for your structure, match the integration intensity of the peak, match the type of proton that would appear at that frequency, and value must be within ±0.1 ppm of value shown above.

3.7, 3.3 ppm: Must be O-C-H protons.

1 point partial for if chemical shift matches proton type, but integral and/or splitting do not match.

2.7, 2.2 ppm: Any proton adjacent to C=O.

1 point partial for if chemical shift matches proton type, but integral and/or splitting do not match.

1 point partial if integral and splitting match, but chemical shift would normally be for alkyl.

(d) Given your answer above, what is the

structure of the daughter (fragment) cation in the mass spectrum that has m/z = 87? You do not need to do electron pushing to answer this question—just draw the cation.

Rubric for part (d): 4 points (full credit) for any structure that (i) is a cation; (ii) has mass 87; and (iii)

is related by bond cleavage to the answer written in part (c). -2 points if any oxygen has an unfilled octet. Cleavage will prioritize putting

electrons on oxygen.

fragment cation with m/z = 87

H3C C

O

CH2 CH2 OCH3

OC

OCH3

4