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Chemistry 251/253

Homework 9 Answer Key

 

1. CC(=CC1CCCCC1)C(=O)C2CCCCC2

2.* BR[C@@H]([C@@H](BR)C1CCCCC1)C2CCCCC2

3.* CC(=O)/C(=C/C1CCCCC1)/C2CCCCC2

The chemistry in the following reaction scheme parallels that described in the topic Heterocyclic Chemistry I . The letters shown in red are for compounds for which H-NMR data is available.

4) CC(=CC(=O)C1CCCCC1)C2CCCCC2 This is analogous to Eq. 1 in the topic Heterocyclic I.

5) CC(BR)(C(BR)C(=O)C1CCCCC1)C2CCCCC2 This is analogous to Eq. 2 in the topic Heterocyclic I.

6) CC(=C(BR)C(=O)C1CCCCC1)C2CCCCC2 This is analogous to Eq. 3 in the topic Heterocyclic I.

7) CC(N)(C(BR)C(=O)C1CCCCC1)C2CCCCC2 This is analogous to Eq. 4 in the topic Heterocyclic I.

8) CC1(NC1C(=O)C2CCCCC2)C3CCCCC3 This is analogous to Eq. 5 in the topic Heterocyclic I.

9) CC1(C2N1C(C)(N=C2C3CCCCC3)C4CCCCC4)C5CCCCC5 This is analogous to Eq. 7 in the topic Heterocyclic I.

10. Select the weakest baseC The lone pair on the nitrogen is part of the aromatic pi system, so it is not shared as readily with other nuclei.

11. Draw the structure of the simplest non-cyclic, tertiary, aliphatic amine that would give a 13C-NMR spectrum containing two signals

.CCN(CC)CC

12. Draw the structure of the simplest cyclic, tertiary, aliphatic amine that would give a 13C-NMR spectrum containing two signals.

C1C2CN12

13. Treatment of N-benzoyl-2,3-diphenylaziridine with NaI resulted in the formation of a new compound, G:

O1C(C(N=C1C2CCCCC2)C3CCCCC3)C4CCCCC4 This is a rearrangement that is catalyzed by the iodide ion, which is both a good nucleophile and a good leaving group.

14. The chemical shifts in the NMR suggest that the compound is aromatic. The reaction involves a double dehydration. CC1CCCN1C

15.CCC1NCC(C)[NH]1

16. How many of the alkaloids shown below contain a phenylethylamine fragment? (Enter a numerical value in the text field.) 3

17. One investigation of the biosynthesis of the alkaloid lophocerine from the amino acid tyrosine involved the following labelling experiment where * represents an isotopically labeled carbon atom: complete the transformation.

Enter the number of the carbon atom in lophocerine that contains the isotopic label. C3 The initial reaction is the formation of an imine by nucleophilic addition of the amino group to the aldehyde. This is the most likely possibility. Ring closure, which is actually an example of an electrophilic aromatic substitution reaction, then follows. Dehydration and decarboxylation

18. Draw the structure of the nitrogen-containing product of the following reaction:

CN1CCCC1

19.CC1CC(C=NC(C(=O)O)C2CCCCC2)CCN1

20. CC(CO)CC=C(C)C

21. What is the oxidation level of the alpha carbon in amino acid tyrosine? 0

22. Enter the number of the carbon that would lose a hydrogen atom most readily in the following reaction: 1 The hydrogen atom on this carbon is a tertiary, benzylic hydrogen.

23. Enter the number of the carbon that would lose a hydrogen least readily in the reaction shown in Question 22. 7 This is an aromatic H. This C-H bond is the strongest of the various possibilities.

24. What is the dihedral angle between the OH group and the CH3 group in ? Enter just the numerical value in the text field. 60 or 180 depending on which conformation you chose.

25. The protons labeled Ha and He in Question 24 are 1. identical 2. enantiotopic 3. diastereotopic Enter the number that corresponds to your choice.

26. Methylation of tyramine with S-adenosylmethionine is an enzyme catalyzed Sn2 reaction:

Would this reaction proceed faster in a hydrophilic region of the enzyme or in a hydrophobic region. (Enter the word hydrophilic or hydrophobic. Don't make any spelling errors!) Hydrophobic. Charge is dispersed as the transition state is formed. Thus a polar solvent will stabilize the reactants more than it will the transition state. This results in a larger activation energy and a slower reaction. Conversely, the reaction will proceed faster in a less polar environment.

27. Polly Ester proposed to perform the following reaction:

OC1CCNC(O)N1 or O=C1CC[NH]C(=O)[NH]1 Polly isolated starting material; the methyl lithium deprotonates the two nitrogen atoms and the aqueous acid simply puts the protons back on. Either tautomeric form of the starting material was acceptable.

28. Shortly we will study the chemistry of carbohydrates. We have already had a preview of some of that chemistry during our discussion of nucleophilic addition reactions of alcohols to aldehydes and ketones. Bearing that discussion in mind, what functional group besides an alcohol is present in D-glucose, the most abundant carbohydrate in the world. hemiacetal

29. How many 1,3-diaxial interactions are there in the other conformation of D-glucose? 4

30. As we will see shortly, D-glucose exists in several different forms. Two of those forms are b-D-glucopyranose and a-D-glucopyranose:These two forms are best described as 1. conformational isomers 2. enantiomers 3. diastereomers 4. constitutional isomers (Enter the number that corresponds to your answer.)