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Chemistry 303fall, 2006

FIRST EXAMINATION

7:00 PM, OCTOBER 17TH, 2006Duration: 2.0 hr

Name_______________________________KEY_____________________________

Lab TA___________________________________________________________(if you do not know his/her name, give day of lab section)

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—

We love to give partial credit.If there seems to be more than one good answer, explain your thinking.If you invoke resonance delocalization as part of your answer, draw the relevant resonance structures

WRITE SOMETHING

BUT: write legibly! We will not work too hard to decipher sloppy writing.

Write only in the space provided for each question.

Score:

p2___________/13 p3___________/12 p4___________/15 p5___________/14

p6___________/09 p7___________/06 p8___________/13 p9___________/13 p10___________/06

Total: /101

There are 10 pages in this exam; please check now to be sure you have a complete set.

Please be aware that a small number of students will be taking the exam at different times up until the morning on Tuesday. It wouldbe well not to discuss the exam until after that time.

Pledge:_________________________________________________________________________________

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I. (13 pts). Consider the molecules, A, B, C.H3C

H3C CH 3

OH3C

H3C H

OCH2

H2C

CH2

OH

tert-butyl alcohol isopropyl methyl ether[2-methyoxypropane]

n-butanol[1-hydroxybutane][2-hydroxy-2-methylpropane]

A B C

CH 3H H3C

A. (5 pts). Write here the list of molecules in order to increasing bp:

__B___ < _A ____ < __C___lowest highest

B. (4 pts). Give the single most important reason for the difference in bp for the one with highest bp and the one that issecond highest. Use words and pictures.

Both A and C are alcohols and enjoy H-bonding as the primary intermolecular attraction. However, C is a linear,straight chain alcohol while A is a branched chain. The primary effect is due to crowding (steric) in A . There is a minor

difference in van der Waals interaction, where the linear molecule would have stronger forces, but H-bonding factors aremuch larger. The effects lead to the same conclusion and either will be accepted.

The H-bonding for A will be less effective because of the steric crowding when two molecules associate. C has less stericcrowding.

H3C

H3C CH3

O

H2C

CH2H2C

OH

CH 3H3C CH3

CH3O

H H3C CH3

CH3OH

crowding

crowding

CH2

H2C C

H2OH3C

CH2

H2C

CH2

O CH3

H

H

H

C. (4 pts). Give the single most important reason for the difference in bp for the one with second highest bp and the one thatis lowest. Use words and pictures.

Cpn B cannot be an H-bond donor, and therefore self-association is simply due to dipole-dipole and van derWaals attraction. Cpn. A can H-bond as shown above, and has stronger self-association and therefore a higher bp. Thedipole-dipole and vdW forces would be similar for both.

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II. (12 pts). Consider the pair of molecules below.

A. (02 pts). Draw in all of the non-bonded electron pairs (lone pairs). B. (04 pts). Both molecules have two more-or-less basic sites.

Draw the conjugate acid of 1 and explain why the site you chose is the most basic.

C. (02 pts). Draw the conjugate acid of (2) .

NH 2N NH 2

N1 2

H + H+

NH 2N

H

NH 3N

Protonation takes place on the –NH2 group in (1). This nitrogen is much more basic due to hybridization effects. It is besp3 hybridized, and that provides on 25% “s” character to the orbital bearing the lone pairs. The other N is required to besp2 hybridized, more “s” character (33%). “s” character adds stabilization to the N lone pair and makes protonation lessfavorable.

Same story for (2), comparing lone pairs in sp (50% s) or sp3 (25% s) orbitals

D. (04 pts). Which is more basic, ( 1) or ( 2)? Give the single most important reason for the difference in basicity; explain.

(1) is more basic. The hybridization is the same for both and there is no opportunity for delocalization: noadjacent pi bond. So the primary difference is due to the electron withdrawing effects of the CN group vs theC=N- group. Three bonds to the electronegative atom in the –CN group makes is a stronger electron withdrawinggroup. It tends to remove electron density from the other N and stabilize the electron pair, reducing the basicity in(2).

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III. (15 pts). Consider the molecule, A.

A. (06 pts).

For your benefit, on the structure shown here,draw all C and H atoms, and lone pairs.

Label each carbon with the correct hybridization:sp, sp 2, or sp 3.

B. (09 pts). 1. (06 pts) Identify the two most acidic protons in the molecule and circle them on the structure. Explain carefully why they

are more acidic than any other H in the molecule.

The two protons circled are adjacent to pi bonds. The corresponding conj. base has the opportunity for delocalization ofthe anion into the pi systems:

O

H b

H a

H b+ +

O

H H

O

H

O

H

O

H

H a+

O

The other Hs in the molecule cannot lead to resonance delocalized anions

2. (03 pts) Between the two acidic protons, one has pKa ca 18 and the other has pKa ca 22. Identify which is which andexplain the single most important reason for the difference.

Hb, on the carbon adjacent to the C=O, leads to a conj base with two good resonance structures. The other acidic H, H a ,

leads to a conj base with three significant resonance structures (above). More good structures, more stable, and the acidfrom which it is derived is more acidic.

O

A

H3C

H3C H

HH

Hsp3 sp2

sp2

sp2

sp3

sp3

sp3

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IV. (14 pts). Consider the molecule for which X is a resonance structure.A. (6 pts) Draw the two most significant additional resonance structures related to X , and label them Y and Z. Show all non-

bonding electrons and formal charges, if any.

N

B

H

HX

N

B

H

H

N

B

H

HY Z

B. (4 pts) Rank your three resonance structures, X , Y , and Z in terms of stability/significance. Explain the main reasons behind your ranking.

Y and Z are equal in energy, by symmetry. They have one more bond than X (stabilizing), although they also have twomore charges (destabilizing). By our working assumptions in 303, the extra bond dominates, and X is predicted to be lessimportant than Y/Z.

Stability: Y=Z> X

C. (4 pts) Would you expect the molecule represented by resonance structures X,Y,Z to have a greater or smaller molecular dipole than Q? Explain.

This molecule has no significant dipolar resonance structures. The structures beloware “legal” and may make a small contribution, but even those dipoles would cancel.The structure Q has zero molecular dipole due to exact canceling of bond dipoles.

N

N

H

H QN

N

H

H

more eqivalent structures

N

N

H

H Q

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V. (15 pts). When phenol is mixed with D 2O (water in which the protons are replaced with isotope2

H, deuterium) and alittle triethylamine is added, a new product is formed, D. The molecular ion region of the mass spectrum of D is shown.

Phenol has molecular weight = 94.

OHD2O

Et 3N:[D]

phenolMW 94

A. (05 pts). Draw the structure of D and explain how it is consistent with the base peak (a) inthe mass spectrum.

ODphenol has mass 94; replacing the H wth D adds one unit to the mass.So the parent ion now appears at 95.

The H on the ring are not acidic enough to exchange with the D2O under these conditions.

B. (04 pts). Explain the origin of the peak (b) at m/z 96.

For every compound with carbon, there is some probability of a molecule with C-13replacing one of the C-12 atoms, such as:

C 13

OD

The parent ion for this species appears at the M+1 position, or m/z = 96 in this case.

The intensity is proportional to the number of carbons in the molecule. About 6.6% in this case.

continued…

90m/z

m/z 95

(a)

(b)

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C. (06 pts). Write a reaction sequence (two steps) which shows how D is formed from phenol and makes clear the role of thetriethylamine. A complete answer will include the use of the arrow formalism to show the steps.

O

H

+ :NEt 3 O + HNEt 3

DO

D

O

D+ DO

NEt 3HHOD + NEt 3

Triethylamine acts as a weak base that can generate a low concentration of the phenol anion. The anion thencan return to phenol, or pick a proton off of D2O to give the deuterated analog.

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VI. (13 pts). (The preceptors think this is too tricky since it took them 5 min to figure it out; please prove them wrong)A. (10 pts). Imagine a molecule with five carbon atoms , any number of hydrogens (no other atoms) which has three (and only

three) sp hybridized carbons .

1. (06 pts) There are two isomers that fit the criteria. Draw them here in large format. 2. (03 pts) Label clearly each of the sp hybridized carbons. 3. (01 pts) Circle the carbon-carbon bond that is overall shortest, considering both structures . [one circle]

[Note: you may request from the preceptor one or both of the structures needed for this question. If you obtain one structure, themaximum score for part A will be 6 pts; if you request two structures, the maximum score for part A will be 2 pts. ]

H2C C C C CH 2 H2C C CH

C CHsp sp sp sp sp sp

The isomers above are the ones we thought of, and expected we were giving a hint by specifying that the correctstructures were isomers.

However, some of you budding chemistry geniuses, willing to think outside the box (or outside the usual bonding rules),suggested the following one. It does satisfy the premise of the question, if you ignore the suggestion that the structuresshould be isomers. It suffers some bond angle strain and is currently not a known species.

sp

sp

sp

B. (03 pts). Which of the two isomers is more acidic? Show the conjugate base for that isomer and give the single most importantreason for the difference.

H 2 C C CH

C CH H 2 C C CH

C C:

This isomer contains an sp hybridized C-H bond. That can lead to an anion with the lone pair in the sp orbital,strong "s" character and strong stabilization.

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V. (19 pts). Consider the two reactions shown here.A. (04 pts). Complete each reaction by showing the conjugate base for each.

OH

SH

H+ +

H+ +

(1)

(2)

O

S

B. (05 pts). The pKa for eq (1) in water is 16, while the pKa for eq (2) is 11. Give the primary reason for the difference in acidity.

The sulfur-centered anion is in a large diffuse volume (orbital), spreading out the charge and making it more stable. It ismore polarizable,

The oxygen has the negative charge more concentrated, less stable, higher energy.

C. (04 pts). Illustrate the difference in acidity for the two reactions by drawing an energy diagram for each reaction. Indicate the free energy difference ( ! G) for each reaction and indicate clearly which is larger. Explain any ambiguity.

The absolute energy levels for the reactants and conj bases are not relevant (or known). The important parameter is therelative energy of reactant to conj base, and a qualitative comparison of those two quantities.

FREE ENERGY DIAGRAM or REACTION COORDINATE DIAGRAM

Rxn 2Rxn 1

E

extent of reactionreactants productsproductsreactants

extent of reaction

E

reactant

conj base

! G! G

reactant

conj base

Larger G = less favorable reaction

continued….

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D. (06 pts). The pK a for eq (1) in acetonitrile is 25, while the pK a for eq (2) is 15 in acetonitrile.

1. Why is the pK a for eq (1) 9 units higher in acetonitile compared to water?

OH H+ +(1) O

This equilibrium is strongly favored by strong solvation of the conj base through H-bonding with water.The reactant is also nicely solvated by interaction with water, but the effect is less.

In acetonitrile, a polar APROTIC solvent, no H-bond donation is possible, and the conj base is lesswell solvated. Only dipole-dipole interactions are involved, much smaller magnitude than H-bonding.The equilibrium shown is less favorable. The alcohol has a higher pK a in acetonitrile

H

H

OH

OH

OH H+ +(1) O

N C CH 3H3C

C

N

! +

! +

2. The effect of the solvent change on eq (2) is much smaller, increasing the pK a in acetonitrile by only 4 units. Why is theeffect smaller for eq (2) compared to eq (1)?

SH H+ +(2) SH

H

OH

OH

H-bond stabilization to the large diffuse S anion is not as effective as it is with the oxygen anion ineq 1. Then removing the water and replacing it with acetonitrile has a less dramatic effect. Dipole-dipoleinteractions can still be somewhat effective

Overall, the effect is the same, but smaller in magnitude.

end exam

H 3 C C N

acetonitrile