Chemistry 251/253

Answer Key

Homework 9

1. Cyclooctatetraene, COT, is a molecule that looks, at first glance, as though it should be "anti-aromatic". However, the terms aromaticity and "anti-aromaticity" apply only to planar molecules. In fact, COT is not planar. Rather, as shown below, it has a tub shape. The four double bonds are not conjugated to each other, and COT displays chemistry that is similar to that of simple alkenes.

An interesting question arises about COT derivatives where the 8-membered ring is forced to be planar; will such compounds be aromatic or "anti-aromatic" ? Two recent articles in the Journal of the American Chemical Society, 123, 1755,1768 (2001) report the results that address this question. The authors prepared the COT derivative 1. X-ray analysis demonstrated that the 8-membered ring is planar and that bond lengths in the 8-membered ring alternate in length. Select the statement that is most consistent with this data.

A. Compound 1 is aromatic.

B. Compound 1 is antiaromatic

C. Compound 1 is non-aromatic.

2.The following reaction yields a complex mixture of products. The 1H-NMR spectrum of one of them is attached. Draw the structure of this compound.Oc1ccccc1Cc2ccccc2O The first part of this transformation involves nucleophilic addition of the aromatic ring to the aldehyde to produce an alcohol. Under the acidic reaction conditions this alcohol undergoes an Sn1 reaction with a second molecule of phenol acting as the nucleophile to produce the product.

3. Draws the structure of the major product of the following reaction: CC(=O)c1ccc2CCCC(=O)c2c1 The keto group is a meta director, while the methylene group is an ortho, para director. The para position is sterically preferred.

4. From first principles, deduce the number of the carbon atom at which the electron density should be the highest. 4 The inductive effect of the substituent withdraws electron density from the ring. The magnitude of this effect decreases regularly from C-1 to C-4. The rseonance effect of the substituent increases the electron density in the ring. It does so preferentially at C-2 and C-4. The net effect of the substituent is to increase the electron density at C-4 the most.

5. Arrange the following compounds in order of increasing rate of reaction with a mixture of nitric acid and sulfuric acid.C<A<D<B

6. Draw the structure of the compound that the NMR data in Table 1 of the topic entitled Electrophilic Aromatic Substitution Reactions II indicates is the best electron donor. Nc1ccccc1, C6H5NH2

7. Draw a resonance structure that indicates how the substituent in benzaldehyde, C6H5CHO, withdraws electron density from the pi system of the aromatic ring. CC=C1C=C[CH+]C=C1 or [O-]C=C1[CH+]C=CC=C1

8. Select the statement that is most consistent with the following equation:

A. This is not the actual product.

B. The cyano group activates the meta position the most.

C. The cyano group deactivates the meta position the least.

D. Relative to a H, the CN group stabilizes the transition state.

9. Arrange the following compounds in order of increasing rate of Friedel-Crafts acylation. C<A<D<B

Consider the following Friedel-Crafts reaction:

10. Which product would be formed the fastest? B The methoxy group is strongly activating. It activates the postions ortho and para to itself. The first methyl group will enter the position para to the methoxy group for steric reasons. The second methyl group will enter ortho to the methoxy group to give B rather than meta to the methoxy group to form C.

11. Which product would be formed the slowest? A The first methyl group experiences a steric interaction with the methoxy group and the second methyl group experiences a steric interaction with the first methyl group. Both steric interactions reduce the rate of substitution.

12. Draw the structure of the product of the following reaction. Assume the Br2 is the limiting reagent. COc1ccc(cc1Br)[N+](=O) The directing effects of the two groups reenforce each other, so the incoming bromine bonds exclusively to the position ortho to the methoxy group.

13. Draw the structure of the major product of the following reaction. CC(C)(c1ccccc1)c2ccccc2 The proton adds to the double bond according to Markownikov's rule to produce a tertiary benzylic carbocation. This electrophilic species then reacts with benzene to form the cyclohexadienyl carbocation, which ultimately loses a proton to form the final product. This question is closely related to problem 2.

14. Draw the structure of the major product of the following reaction. CC(=O)Oc1ccc(Cl)cc1

15.Which of the following structures is not involved in the bromination of toluene, C6H5CH3 ? A When the Br+ bonds to the ring, the positive charge develops on the adjacent atom, not on the atom the Br bonds to.

16. In the reaction shown below, only one of the four structures shown is a reasonable representation of an intermediate that might actually be formed. Which one? B While attack at the meta position is less likely than it is at the ortho or para positions, B is the structure of the ion that is formed when attack occurs at the meta position. None of the other structures are consistent with the rules of resonance that we have studied.

Questions 17-20 involve multi-step syntheses. The starting material and the product are shown in each case. The reagents needed to effect each step are available from the table of reagents shown below. For each step, enter the letter of the reagent that you would use. The number of arrows shown in each question indicates the number of reagents you need to select. You must enter the letters in the appropriate order. You do not have to specify the regents required for any work-up.

17.

18.

19.

20.

21. is one of the MOs of. Which one?
A. y1
B. y2
C. y3
D.y4
The MO picture contains two nodes. Therefore, it must be y3.

22. Arrange the following compounds in order of increasing electron density at the meta position: You can figure out the answer to this question from first principles, or you can look at the NMR data in Table 1 of the topic Electrophilic Aromatic Substitution Reactions II. The substituents in ccompounds A and B are both electron donating, while they are withdrawing for C and D. According to the data in Table 1, the chemical shifts of the meta protons for compounds A, C, and D are 7.18, 7.36, and 7.34, respectively. So C has the lowest electron density and D the second lowest. A has the highest of the three compounds for which data is given. Since the carbonyl group in B makes that substituent less electon donating than the methoxy group in A, the correct order has to be C<D<B<A.

23. Select the structure that is most consistent with the partial 1H-NMR spectrum provided. A The key here is the chemical shifts of the two signals. The doublet at ppm indicates a strongly electron withdrawing group on the ring, while the doublet at ppm suggests a strongly electron donating group.Only structures A and D have one group of each type. But the methyl group in D is not strongly electron donating as indicated by he NMR data in Table 1 of the topic Electrophilic Aromatic Substitution Reactions II.

24. Sulfanilamide was the first sulfa drug. A proposed synthesis of sulfanilamide is shown below.

However, this approach does not yield the desired compound. Draw the structure of the product that this sequence of reactions actually produces. Nc1cccc(c1)S(=O)(=O)N In the initial step the amino group is protonated under the strongly acidic conditions. The resulting ammonium group is a meta director, so the final product is the meta isomer of the desired sulfanilamide.

25. Consider the attached 1H-NMR spectrum of a monosubstituted benzene derivative. Then select the statement that best describes the outcome of nitration of this material.

A. The compound will react faster than benzene to produce a mixture of ortho and para isomers.

B. The compound will react faster than benzene to form predominantly the para isomer.

C. The compound will react slower than benzene to produce a mixture of ortho and para isomers.

The compound will react slower than benzene to produce predominantly the meta isomer.

The NMR spectrum indicates that the substituent on the aromatic ring is electron donating since the chemical shifts of all the aromatic protons are less than 7.27, the reference value for benzene. The integration tells you that the signal around 1.5 ppm contains 9 protons, i.e. it is a t-butyl group. This bulky group will direct the electrophile to the para position.

26. Suppose you wanted to carry out the following transformation:. Which set of reaction conditions would be most likely to succeed? D In order to get this reaction to go, it is necessary to create an electron deficient oxygen atom to act aas the electrophile. You do this in two ways: 1. You attach an electronegative atom to the oxygen. 2. You add an electrophilic catalyst. Compare the two reactions shown below, where Nu: represents an aromatic ring. The proton is the ultimate electrophilic catalyst.

 

 27. Indole,, is known as a heterocyclic aromatic compound. It is a 10 pi electron system. It reacts readily with electrophilic reagents, including D+. If you were to treat indole with a large excess of D+, what would the molecular formula of the product you obtained be? Hint-This is a tricky question (It is not a trick question, however.). To answer this question you have to draw resonance structures to identify those atoms where the D+ will bond. C8D7N. D will exchange for H on all those atoms where there is an H and where you can draw a resonance structure with a lone pair of electrons. In other words, all of the Hs will exchange. The H attached to the nitrogen is also an exchangeable hydrogen as we have seen previously.

28. Nitration of one of the following compounds produces mostly the ortho nitro derivative. Which one? C The OH group forms a hydrogen bond with one of the oxygen atoms of the nitro group:A comparable interaction doesn't happen with aniline because the amino group is protonated under the strongly acidic reaction conditions, and the major product is the meta nitro compound.

29. Draw the structure of the major product of the following reaction:Cc1ccc(Oc2ccccc2)c(Br)c1 The disubstituted ring is more reactive than the monosubstituted ring because both substitutents are activating. The oxygen is more activating than the methyl group, so the Br bonds to the position ortho to the oxygen.

30. Use the principles of resonance theory to predict the major product of the following reaction: Brc1cccc2ccccc12 I hope you got this one right.

You get the same number of resonance structures, 5, whether the electrophile bonds to C-1 or C-2. However, two of contributors contain an in-tact aromatic ring when it bonds to C-1, while only one of the five has an in-tact ring when it bonds to C-2.