CEM 351 Final Exam, Version A

Tuesday, December 14, 2004 12:45-2:45 p.m., Room 138, Chemistry

Name (print) Signature Student # Section Number

Grade? 1.(25 pts.) 2.(25 pts.) 3.(25 pts.) 4.(25 pts.) 5.(25 pts.) 6.(25 pts.) 7.(25 pts.) 8.(25 pts.) 9.(25 pts.) 10.(25 pts.) TOTAL (out of 200 pts.) Note: Answer any 8 questions for a total score of 200 pts. Be sure to look over all the questions first before beginning the exam, and indicate which eight questions are to be graded by checking the corresponding box. Recitation Day and Time Instructors: Room 1. T 9:10-10:00 a.m. Cory Newman 323 2. M 10:20-11:10 a.m. Jun Yan 218A 3. T 10:20-11:10 a.m. Cory Newman 220 4. Th 10:20-11:10 a.m. Vijay Gopalsamuthiram 220 5. T 10:20-11:10 a.m. Cory Newman 220 6. T 11:30-12:20 p.m. Cory Newman 218B 7. Th 11:30 - 12:20 Vijay Gopalsamuthiram 323 8. M 4:10-5:00 p.m. Jun Yan 110

1. Consider the four compounds shown below.

(a) (4 pts) Within each of the two pairs A/B and C/D, one form is cis. Which are the two cis compounds? A and C .

(b) (4 pts) Within each of the two pairs A/B and C/D, one form is predominantly

diequatorial. Which are the two diequatorial compounds? A and D .

(c) (4 pts) Among the four compounds A-D, two are chiral. Which are the two chiral compounds? A and B .

(d) (4 pts) Label stereogenic centers in the chiral compounds with the appropriate

configuration (R or S) designation. (See above)

(e) (6 pts) Make careful chair drawings of the two diequatorial forms you identified in (b). You may use "t-Bu" for the tertiary butyl group. If the compound is chiral, be sure you are drawing the correct enantiomer.

(f) (3 pts) Rank compounds A, B, C, D in order from lowest to highest energy (some may be essentially the same) and explain your reasoning. Highest B ~ C > A ~ D Lowest

A and D have both substituents equatorial, and no interactions between them, so they are essentially equal in energy. Both B and C have one methyl group axial (see right), but again, no other unfavorable interactions.

(R)

(S)

(R)

(R)

t-Bu t-Bu

A D

t-Bu t-Bu

B C

2. (25 pts) 1,3-Butadiene reacts with HCl to give two products: 3-chloro-1-butene (C) and 1-chloro-2-butene (B). The reaction involves a common carbocation intermediate (A). On the basis of the energy diagram shown below, answer the following questions:

(a) (3 pts) Assign the labels A, B, and C to the structures in the diagram below. (b) (7 pts) Write the reaction step that forms A from 1,3-butadiene and HCl.

(c) (5 pts) Which product, B or C, is the kinetic product—i.e. will form faster? Explain.

Product C has a lower barrier to formation from A, so it forms faster.

(d) (5 pts) Which product, B or C, is the thermodynamic product—i.e. is more stable? Explain. Product B is the lower energy (i.e. more stable) product as the diagram shows. It has a disubstituted double bond, wheras C’s is only monosubstituted.

(e) (5 pts) Intermediate A is a resonance hybrid. Only one of the resonance forms is shown.

Draw the other major resonance form at right.

H Cl HA1,3-Butadiene

C

B

A

A The otherresonancestructure

3. (25 pts) The reaction shown below is a reversible acid-catalyzed process, with K near unity.

(a) (5 pts) Write the expression for the equilibrium constant K.

(b) (4 pts) Assuming K = 1, and that the concentration of each starting material is 0.5 M (there’s no product yet), what are the concentrations of all the species after the reaction has reached equilibrium?

All four species would be present at 0.25 M

(c) (4 pts) Assuming your expensive ingredient is the t-butanol (CH3)3COH, suggest two cheap ways of increasing the yield of the product MTBE.

1. Increase [CH3OH] 2. Remove H2O as it’s produced.

Now consider a different process, a highly exothermic reaction where A + B —> C + D according to the equation rate = k[A][B]. What happens to the rate if…

(d) (3 pts) …the concentration of A is doubled?

Rate doubles

(e) (3 pts) …the concentration of B is halved? Rate is cut in half

(f) (3 pts) …the concentration of C is doubled (i.e. extra C is added)? No effect

(g) (3 pts) …the concentrations of A and B are both doubled?

Rate is quadrupled

K = [(CH3)3COCH3][H2O][CH3OH][(CH3)3COH]

4. (25 pts) For each of the following pairs of compounds, identify the relationship between molecules (5 pts each). Use the following letter designations and circle your choice at right:

A. Resonance structures B. Different conformations of the same compound C. Enantiomers D. Diastereomers E. Identical structures F. Constitutional (framework) isomers G. None of the above

5. (25 pts) Quick true/false questions (student written; read carefully here; your classmate(s) are slightly tricky): a. (5 pts) The “n” in the Hückel 4n+2 rule for aromaticity represents the number of p

orbitals in the pi system.

T F the “n” refers to the number of electrons in the π system

b. (5 pts) The stereoisomers of 2,4-dibromo-3-chloropentane include two different meso

compounds.

T F

c. (5 pts) Conformations are constantly equilibrating with each other.

T F

d. (5 pts) Diels-Alder reactions can’t form chiral products from achiral starting compounds

(Hint: draw an example).

T F

e. (5 pts) An allyl radical is more stable than a t-butyl radical.

T F

Meaningless comparison—can’t directly compare “stabilities” of species of different formulas

2

Achiral (planar) Chiral

6. (25 pts) The two synthetic antioxidants BHT and BHA are closely related. Their names stand for “butylated hydroxytoluene” and “butylated hydroxyanisole” respectively. In class we discussed the alkylation chemistry that leads to BHT. But it turns out that BHA is a mixture of two isomers. These compounds are made via Friedel-Crafts-type electrophilic t-butylation of their “hydroxytoluene” (4-methylphenol) and “hydroxyanisole” (4-methoxyphenol) cores. Here are the structures:

(a) (15 pts) Write mechanisms for the alkylation reactions to put in the last t-butyl groups, as

indicated by the arrows above. Be sure to emphasize how the directing effects work.

(b) (5 pts) Based on your mechanistic work in (a) you should be able to explain why BHT’s t-butyl groups end up in a well-defined (ortho) relationship to the –OH “business end” of this antioxidant, but the analogous synthesis of BHA gets a mixture.

Hydroxy and methoxy groups are strong ortho and para directing goups since they are strongly electron donating. Para positions are blocked in both of above cases. In the 4-methylphenyl derivative the ortho position of the hydroxy group is the most reactive carbon, and therefore the reaction happens selectively on that carbon. In the case of the second substrate, however, carbon centers ortho to both methoxy and hydroxy groups are reactive towards electrophilic substitution, and hence a mixture of products is expected in this case.

(c) (5 pts) A third major antioxidant, again made in similar ways, is tBHQ (“2,5-di-t-butyl hydroquinone”). The hydroquinone core is simply 1,4-dihydroxybenzene. Please draw it based on the name. Is the structure consistent with your findings in analyzing BHA’s alkylation chemistry? Yes—both ends similarly reactive.

7. (25 pts) (Student-written question) For each of the following reactions, select the necessary reagents from the “lab bench” below to convert 1-Deuteriopropyne into the given product. You may simply fill in the boxes with the following letter labels.

(A) O3, then H2O2 [g] (B) Hg(OAc)2/H3O+/H2O [e] (C) NaI in acetone (D) Na/NH3 [f] (E) H2/Pt (F) Igor’s magic wand

(G) HBr [d: 1 eq] (H) BH3•THF [c:reagent 1] (I) Br2 [b: 2 eq] (J) H2/Lindlar Catalyst [a] (K)H2O2/NaOH(c:reagent2) (L) tBuO-OtBu/HBr

(25 pts) Here are two phenyl-substituted 5-membered ring compounds, A and B, together with their immediate synthetic precursors, C and D. For both compounds, the precursors are alcohols, which undergo acid catalyzed dehydration, as shown.

(a) (7 pts) Why is formation of A so difficult and B so easy? Explain in terms of the structures of A and B. B is aromatic, while A is antiaromatic.

(b) (5 pts) Write a mechanism for the acid-catalyzed dehydration of D to form B. Be sure to keep track of resonance structures along the way.

(c) (5 pts) Although it's much easier to make, B is terrible at Diels-Alder reactions compared to A. Why? The Diels-Alder reaction of B goes through a transition state in which the aromatic ring of the furan is disrupted. Since aromaticity disruption takes stability from the furan, its reaction is slower. In the case of B, however, the reaction relieves the antiaromatic instability, therefore its reactions is faster than the former.

(d) (8 pts) Draw the structure of the Diels-Alder adduct E formed in the above reaction. Then draw an arrow-pushing mechanism and explain why it's so easy for E to lose CO to yield hexaphenylbenzene F. The aromatic gain upon CO loss is the driving force for the reaction.

8. (25 pts) We have looked at electronic effects of substituents on phenyl ring reactivity toward electrophilic aromatic substitution. We understood them mostly in terms of resonance structures of the intermediates from electrophilic attack. Now let us look at their effects on other characteristics of aromatic systems. (a) (6 pts) Trimethylamine, like

ammonia, is a respectable base (remember that many natural amines are called alkaloids because they behave as alkalis--bases). The pKa of its protonated form is 10.8, but when one of the methyl groups is replaced by a phenyl ring, the basicity drops, so that protonated dimethylaniline’s pKa = 5.2, more than 5 pKa units less basic/more acidic than trimethyamine. Using resonance arguments and precise drawings, rationalize this shift in pKa. (Hint: remember basicity has to do with the availability and localization of the N lone pair; how freely it can be shared with a proton).

Via resonance delocalization (see below), the phenyl group lowers the nitrogen lone electron pair’s energy, and hence its basicity.

(b) (6 pts) Pyridine, with the N in the ring, is also a base, but is substantially weaker than trimethylamine. In fact, protonated pyridine happens (serendipitously) to have the same pKa as protonated dimethylaniline. Noting the differences in bonding environments between the pyridine and the trimethylamine N’s, please explain their pKa difference.

In pyridine, the electron pair that plays the base role is in an sp2 hybrid orbital, whereas that of trimethyl amine is in an sp3 one. With their higher s character, sp2 electrons are held more tightly than sp3 ones, so they are less available as Lewis bases.

(c) (7 pts) Keeping in mind the electronic effects exerted by the phenyl ring on the basicity of the dimethylamino group in dimethylaniline, now let’s look at how putting the same group onto pyridine to form DMAP affects the pyridine pKa. First of all, protonation doesn’t occur at all on the dimethylamino group—why? (Hint: If it was pKa 5.2 with phenyl, how would its pKa have changed with a pyridine ring in place of phenyl?).

In DMAP, protonation on the pyridine ring nitrogen forms a structure that can gain stabilization through resonance. However, no stabilizing resonance form is possible in the product of protonation on the sp3 nitrogen of dimethylamine.

(d) (6 pts) Finally, when it does protonate on the ring N, why is the pKa of protonated DMAP so much higher (4 pKa units) than that for unsubstituted pyridine. (Hint: Look back at your reasoning in (a) and (c) to see how this is a natural result of the parts interacting.)

Through resonance, the nonbonding electron pair of the 4-dimethylamino group in protonated DMAP helps stabilize the positive charge, making DMAP a stronger base than pyridine itself.

9. (25 pts.) Bromine adds in an anti fashion across the double bond in cyclohexene to form trans 1,2-dibromocyclohexanes. (a) (5 pts) Please draw the two products of this addition and describe their relationship.

They are enantiomers (b) (6 pts) Now draw the three products of two-fold bromine addition to 1,4-cyclohexadiene.

Again, label their three stereochemical relationships (enantiomers, diastereomers, constitutional isomers, etc.)

(c) (4 pts) Recalling that substituents on cyclohexanes usually prefer the equatorial positions,

select (circle) among your three products the one that should be lowest in energy. Now redraw it in a chair representation in the box provided. (Hint: you may actually find it easier to draw the chair first, and then figure out which of your products it is.)

(d) (4 pts) Label all stereogenic sites in your structure from (c). Is it chiral or achiral? (Hint:

look for symmetries within the molecule). Chiral Achiral (e) (3 pts) As part of a mechanistic study you decide to start with the isotopically labeled

compound cis-3,6-dideuterio-1,4-cyclohexadiene in the above story. If you made this stuff via Diels-Alder from deuterated 1,3-butadiene and acetylene (ethyne) where should the deuteriums start on the butadiene? Please add them to the drawing.

(f) (4 pts) Imagine complete bromination of your stereochemically labeled cyclohexadiene

now to make deuterium labeled versions of the all-equatorial 1,2,4,5-tetrabromocyclo-hexane you drew in (c). How many isomers of this all-equatorial tetrabromide form? Draw careful chair structures below.

One structure (it’s meso)

10. If the partners below react in SN2 fashion, what will be the likely products? (a) (5 pts) Make a good 3D drawing. Be sure your equation is balanced (i.e. show all

reactants and products). Remember that D is just another symbol for 2H (deuterium), a heavy isotope of hydrogen. It allows us to keep track of stereochemistry in the present case.

(b) (2 pts.) What is the nucleophile in the reaction in part (a)? The methoxide ion

(c) (2 pts) What is the electrophile? The deuterated ethyl iodide

(d) (2 pts) What is the leaving group? The iodide anion

(e) (5 pts.) If the similar partners below react in E2 fashion, what will be the likely products? Be sure your equation is balanced.

(f) (5 pts.) If the following reactants react in H2O, what will be the likely products? What sort of reaction occurs? Be careful – this is a trick question; think about pKa’s.

(g) (4 pts.) Assign (R) or (S) configurations to starting material and products in (a) and (f).

(R)

(R)