Chemistry 251/253

Answer Key

Homework 10

1.What would the electron density be in an "imaginary benzene molecule" that had the same electron/proton ratio as ? Express your answer to 3 decimal places. 6*[6/[(3*6)+7+8]]=1.091 electrons per nucleus

2. Arrange the following compounds in order of increasing reactivity in electrophilic aromatic substitution reactions. C<A<B

3. Enter the number of the atom in that would be protonated first in an acidic solution. 2 The lone pair of electrons on the nitrogen is perpendicular to the pi system of the heteroaromatic ring.

4. Enter the number of the atom in where electrophilic aromatic substitution would occur preferentially. 1 While there is in reality only one pi system in this molecule, you may consider the two rings independently. The 5-membered ring is electron rich, while the 6-membered ring is electron poor. Therefore, an electrophile will react preferentially with the 5-membered ring.

5. How many of the following structures are consistent with the MO diagram ?2 The 2nd and 3rd.

6. Draw the structure of the major product of the following reaction.Cc1c(Cl)oc2c(C)cccc12

7. Arrange the following compounds in order of increasing basicity:2<3<1

8. Draw the resonance structure that makes the most important contribution to the intermediate that is formed in the following reaction:ClC1C=CC=[O+]1

9. Bromination of thiophene-3-carboxylic acid, , produces 5-bromothiophene-3-carboxylic acid and none of the 2-bromo isomer. Draw the resonance structure that indicates most clearly why attack of the Br⁺ at C-2 generates a less stable intermediate than attack at C-5.

OC(=O)[C+]1C=CSC1Br If the Br⁺ bonds to C-2, it generates a carbocationic intermediate in which the positive charge develops on C-3, adjacent to the partial positive charge on the carboxyl carbon. This is a destabilizing interaction. A comparable interaction is not possible when the Br⁺ bonds to C-5.

10. Use the principles of resonance theory to predict the number of the atom in pyrrole, , that would be protonated preferentially. 2

 

11. Bromination of pyridine, , under forcing conditions produces 3-bromopyridine in 86% yield. How many of the structures shown below represent legitimate resonance contributors for the intermediate that is involved in this reaction? 2 In the first structure C-2 is tetravalent, while the last structure is not a valid resonance structure because the lone pair of electrons on the nitrogen is orthogonal to the pi system.

12. Pyridine is classified as an electron deficient heteroaromatic compound. As such it is susceptible to reaction with nucleophiles. For example, treatment of pyridine with sodium amide produces 2-aminopyridine in 70% yield:. What is the equilibrium constant for this reaction? 10e3

13. Furans are relatively unstable when treated with aqueous acid. For example, 2,5-dimethylfuran reacts with an aqueous mixture of acetic and sulfuric acids to generate the compound that gives the attached ¹H-NMR spectrum in 90% yield. Draw the structure of this compound.CC(=O)CCC(=O)C

14. How many isomeric tertiary amines are there with the molecular formula C₅H₁₃N? 3

15. Many physiologically active compounds are derivatives of 2-phenylethanamine, C₆H₅CH₂CH₂NH₂. During a recent drug bust, police seized a large quantity of a compound that gave the attached ¹H-NMR spectrum. Draw the structure of the compound that the drug dealer was selling.COc1cc(CCN)cc(OC)c1OC

16. Select the reaction conditions you would use to affect the following transformation:.

17. 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.

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

The structure of the compound is .

18. Draw the starting material A that was used in the following synthesis:O=C1CCC=C1 This sequence begins with a 1,4-addition of HCN to the a,b-unsaturated ketone. Reduction of product B followed by acidification of the reaction mixture generates the final product.are less desireable, albeit credit-earning, alternatives because elimination of HBr or HCl would be a significant side reaction with these secondary alkyl halides.

19. Predict the product of the following reaction:OS(=O)(=O)c1cccc2cccnc12 The electrophile will react preferentially on the ring that does not contain the nitrogen atom. Drawing the appropriate resonance structures indicates that the intermediate carbocations formed by attack of the electrophile on positions 1 and 3 should be more stable than the alternatives generated by reaction at 2 or 4. H-bonding favors sulfonation at position 1 over position 3.

20. Enter the number of the carbon atom in pyrimidine, , where an electrophile would be most likely to bond. 3 When an electrophile bonds to C-1, C-2, or C-4, the resulting carbocationic intermediates all have two resonance contributors in which the positive charge is localized on one of the nitrogen atoms, e.g. . Note that in this structure there are only 6 electrons around the positively charged nitrogen. Not good. Bonding the electrophile to C-3 does not generate comparable structures.

21. The synthesis of involves three of the four compounds shown below. Which one is not involved? B

22. Arrange the following compounds in order of increasing acidity:A<B<C The acidity of an acid is determined by the stability of its conjugate base. In this case that means you have to assess the relative stabilities of . In particular, you are assessing the potential energy of the lone pair of electrons on the nitrogen atoms, which correlates with the hybridization of the nitrogen. The hybridization changes from sp³ to sp² to sp on going from the primary amine to the nitrile. Electrons in an sp hybridized orbital have lower potential energy than those in sp² hybridized orbitals, which have lower potential energy than those in sp³ hybridized orbitals.

23. Draw the structure of the major product of the following reaction sequence: Cc1ccc(N)cn1

24. Watson and Crick were arguing over whether or not was aromatic. Watson did not believe it was and demanded that Crick prove otherwise, which he did by simply drawing another form of the compound that made its aromatic character more obvious. Draw the structure that Crick drew to convince Watson that this compound is aromatic.Oc1cnc(O)nc1 The original structure is a diketone. In the dienol form, the 6 pi electron system is more obvious.

25. Draw the product of the following reaction sequence: COc1cccc(N)c1

26. When 1 mmol of CH₃CH₂CH₂CH₂NH₂ is treated with 1 mmol of CH₃CH₂Br, the contents of the flask at the end of the reaction are best described as

27. Select the solvent in which the reaction would occur the fastest.

The reaction should be an Sn2 reaction. Since there is a build-up of charge as the transition state is formed, a polar solvent will stabilize the transition state more than it does the starting materials. This results in a decrease in the activation energy and a faster reaction.

28. How many of the following compounds could be described as b-phenylethylamines? 3 Only the second structure is not a b-phenylethylamine.

29. Figure 8 of the topic Amines II presents the structures of four alkaloids that are derived from the amino acid tryptophan. How many of them may be classified as b-phenylethylamines? 3 Only quinine does not qualify.

30. Predict the product of the following reaction. Assume the reactants are mixed in a 1/1 ratio and that there has been an appropriate work-up. CCNCCO The outcome here is different than in the O-methylation reactions involving S-adenosyllmethionine. In the enzyme catalysed reactions the amino group is bound to the active site of the enzyme and is not available to act as a nucleophile. In a test tube, the inherently greater reactivity of the nitrogen atom leads to reaction at that atom.