Chemistry 251/253

Answer Key

Homework 6

Note- 03/17/01 10:00 a.m. I have regraded question 8.

1. The most common reagent for converting carboxylic acids into the corresponding acid chlorides is thionyl chloride, SOCl2. However, other reagents may be used to achieve the same result. How many of the following reagents might you reasonably expect to be suitable for effecting this change? 3 BCl3, PCl3 , and AlCl3 all contain a central atom that has an unfilled valence shell which acts as an electrophilic center.

HCl
NaCl
BCl3
PCl3
Cl2
AlCl3

2. Draw the structure of the product of the following reaction sequence:CCN(CC)C(=O)c1cccc(C)c1

3. Arrange the following compounds in order of increasing rate of reaction with NaOH.C<A<D<B This reactivity sequence parallels the relative stabilities of the leaving groups, which correlates inversely with the pK values of the conjugate acids of those leaving groups: HCl < HOAc < HOCH3 < NH2CH3.

4. Draw the structure of the product of the following reaction sequence: OCCCCC(=O)O This is a simple saponification of a cyclic ester.

5. The mechanism for the esterification of a carboxylic acid was deduced, in part, from the results of isotopic labeling studies. The results of one such study are shown below, where [A] represents a putative, electrically neutral tetrahedral intermediate. Draw a structure for intermediate A that is consistent with the information given and with mechanistic principles we have discussed on numerous occassions in this course.

C[18O]C(O)(O)c1ccccc1

Questions 6 and 7 involve the following isotopic labelling experiment:

6. Assuming that you start with a 1/1 ratio of benzoic acid and isotopically labeled water, to the nearest whole number, what percentage of the isotopic label that was originally in the water ends up in the benzoic acid at equilibrium? 67 The tetrahedral intermediate has three identical OH groups. There are three ways this intermediate can lose water to regenerate the starting material. In two of them the 18O remains in the benzoic acid, while in the third it does not. Therefore, 2/3 of the label ends up in the equilibrated benzoic acid.

7. Which of the following structures best represents the distribution of 18O in the benzoic acid at equilibrium? Note- The implication of structure C is that either of the oxygen atoms is labeled. It does not mean that both oxygen atoms are labeled at the same time. C

Questions 8 and 9 involve the polymer known as kevlar:, where the structure within the brackets is called the "repeat unit" of the polymer and n is an integer representing the number of "repeat units in the polymer.
8. Draw the structure of the carboxylic acid from which kevlar is made.OC(=O)c1ccc(cc1)C(=O)O

9. Draw the structure of the amine from which kevlar is made.Nc1ccc(N)cc1

10.represents the "repeat unit" of a polymer called nylon[6]. Draw the structure of the monomer from which nylon[6] is made.NCCCCCC(=O)O

11. Draw the structure of nylon[4,4] where the number of repeat units equals 2.One repeat unit is shown in red and one in blue. Note that both the diamine and dicarboxylic acid components of each repeat unit contain 4 carbon atoms. NCCCCNC(=O)CCC(=O)NCCCCNC(=O)CCC(=O)O

12.Glycine, H2NCH2CO2H, is the simplest of the naturally occuring amino acids. Polymerization of glycine produces polyglycine, a polymer that biochemists sometimes use as a model for the investigation of the chemistry of proteins, which are polymers composed of many different amino acids. Draw the structure of polyglycine where the number of "repeat units" equals 2.NCC(=O)NCC(=O)O Since glycine contains both an amino group and a carboxylic acid group, each glycine represents one "repeat unit". Compare this to the case in question 11 where each repeat unit consists of one molecule of a dicarboxylic acid and one of a diamine.

13. Draw the structure of the product of the following reaction:COC(=O)CCCC(C)O The reaction conditions imply a competition reaction. The borohydride ion will react preferentially with the carbonyl group that is the more electron deficient, namely that of the ketone.

14. Draw the structure of the product of the following reaction: CC(C)CC(=O)OC1CCCC1 This is an example of a Fisher esterification.

15. When Polly Ester attempted to perform the synthesis shown below, the 1H-NMR spectrum of the product she obtained indicated that things had not gone as planned. Draw the structure of the compound that Polly actually prepared.

CCC(O)(CC)c1ccccc1

The ethyl lithium reacts with the ethyl benzoate to produce propiophenone. The carbonyl group of this ketone is more reactive than the carbonyl group of the ester, so as soon as the ketone is formed, it reacts with the ethyl lithium to generate a tetrahedral intermediate that is protonated during the work-up. The NMR spectrum is interesting; the protons of each of the two methylene groups are diastereotopic and therefore non-equivalent. One of the methylene hydrogens resonates at 1.87 ppm as a doublet of quartets, while the other occurs at 2.06 ppm, also as a doublet of quartets.

16. Draw the structure of the product of the following reaction: COP(=O)(OC)OC Esterification of phosphoric acid and its derivatives is analogous to esterification of carboxylic acids. Phosphate esters are important biological intermediates.

17. Draw the structure of the product of the following reaction: CS(=O)(=O)Cl This reaction is analogous to the formation of a carboxylic acid chloride.

18. Draw the structure of the product of the following reaction:CCI This is an example of an Sn2 reaction in which the tosylate group (p-toluenesulfonate) serves as the leaving group. It is analogous to the reaction of CH3CH2Br with NaI.

19. Draw the enolate ion that would be formed in the following reaction:CCOC(=CC)[O-]

20.Draw the structure of the product of the following reaction: CCOC(=O)c1ccccc1 This process is called transesterification. It is similar to a Fisher esterification.

21. Proteins are also called polypeptides. They are polymeric amides which are formed by the polymerization of amino acids. The digestion of proteins involves the acid catalysed hydrolysis of the peptide bonds that connect one amino acid to the next. Draw the structure of the amino acid that would be produced by digestion of the tripeptide

CC(C)C(N)C(=O)O

22. How many of the following compounds would undergo a Claisen condensation when treated with a solution of NaOEt in EtOH? 2 The 1st, 3rd.

23. Draw the structure of the compound that would be formed by a Claisen condensation with CCCC(=O)C(CC)C(=O)OC(C)C

24. Draw the structure of an ester that is isomeric with ethyl acetate, but that will not undergo a Claisen condensation.CCCOC=O or CC(C)OC=O The compound can't have any a-hydrogens.

25. When Ethyl Ester ran the reaction shown below, she did not obtain the product shown. Draw the structure of the product that she did obtain.

OC(=O)Cc1ccccc1

Ethyl chose the wrong base. She should have used KOCH3 rather than KOH. With KOH the principal reaction that occurs is saponification.

26. Draw the structure of the ester that acted as the nucleophile in the crossed Claisen condensation that produced .

CCCC(=O)OCC

27. Draw the structure of compound Q:

COC(=O)CCCCC(=O)OC

Both ester groups in Compound Q must be a methyl ester because compound Q would produce a mixture of products if one of the two ester groups were something other than a methyl ester. See question 30.

28.Al Kane ran the reaction shown below. After performing the work-up, he analysed the product by GC. How many peaks did he see?4 The top two products arise from Claisen condensations, while the other two arise from crossed Claisen condensations.

29. Al Keene attempted the reaction shown below. The 13C-NMR spectrum of the product Al isolated contained peaks at 30.5, 49.1, 50.4, 176.0, and 211.3 ppm, which he thought were consistent with the structure of his target molecule. The 1H-NMR spectrum also seemed to confirm the structure. Elemental analysis indicated that the compound contained 52.6% C, 5.3% H, and 42.1% O, which was also consistent with the proposed product. But Al was, if nothing else, thorough, and he decided that he should record the mass spectrum of his product. When he did, he discovered that its molecular weight was 228.2. When Al left the lab that night, he was puzzeled and depressed. But when he returned the next morning he was merely depressed. His puzzlement had disappeared during the night when the structure of the compound he had actually made came to him in a dream. Draw Al's dream.

COC(=O)C1CC(=O)C(CC1=O)C(=O)OC

 

Instead of undergoing an intramolecular Claisen condensation as Al had hoped, the diester reacted intermolecularly to produce intermediate I. This intermediate then underwent a Dieckman condensation to produce the final product. Note that this product contains a plane of symmetry so that C-1 is identical to C-1', etc. However, the two protons on C-3 are not identical. They are diastereotopic. They have different chemical shifts and each proton is split into a doublet. Further splitting of each of the doublets by spin-spin coupling with the hydrogen on C-2 generates two doublets of doublets.

30. When Polly Ester ran the reaction shown below, GC analysis indicated that the product contained two compounds, A and B, in a nearly 1/1 ratio. The 1H-NMR spectrum of one compound was consistent with structure A. The 1H-NMR spectrum of compound B is attached. Draw the structure of compound B.

COC(=O)C1CCCCC1=O

Since the diester has two different ester groups, there are two alternative enolate ions that may be formed. Deprotonation of Ca leads to compound A, while the enolate ion formed by deprotonation of Ca' ultimately yields compound B. The tip off in the NMR spectrum is the singlet for the methoxy group at 3.7 ppm.