The following homework assignment contains 26 questions valued at ¾ point/question

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1 Chemistry 262 Homework 2: Principally Chapters 17, 20, and 21 ut: 04/27/18 Due: 05/04/18 The following homework assignment contains 26 questions valued at ¾ point/question Name: KEY 1. What would be the principal product(s) of the following reaction NBS RR I II III I c. I & II d. III e. I 2. Which of the following would you use to transform toluene to benzoic acid 2, h b. 2, Fe 3 c. (1) KMn4, H -, (2) H3 + beware the benzylic oxidation d. HN 3/H 2S 4 e. S 3/H 2S 4 1

2 3. What is the final product (C) of the following reaction sequence Toluene + KMn 4, H -, A A + HN 3/H 2S 4 B B + 2, Fe 3 C C 2 H C 2 H C 2 H N 2 N 2 N 2 I II III CH 3 N 2 e. N 2 I C 2 H Look for the oxo group directly outside the benzene ring for deactivation/meta direction 4. Which free radical would be most stable I II III I e. 2

3 5. Which hydrogen atom is most susceptible to abstraction by free radicals CH 3 CH 3 CH CH CH CH CH CH 3 I II III I Super conjugated and tertiary e. 6. Which would be the best synthetic route from propene to 1,2-dibromo-3- chloropropane CH 2 CHCH 3 CH 2 CHCH 2 a. Propene 2, C 4 2, h b. c. d. e. Propene Propene Propene Propene 2, 400 o C 2, C 4 H 2, h NBS, C 4 2, C 4 2, C 4 NBS, C 4 Diatomic chlorine will homolytically dissociate into chlorine atoms at high temperature it is important to carry out this reaction first, since once the double bond is removed there is no allylic position 3

4 7. The HM of the allylic radical has how many electrons in the ground state a. 1 b. 2 c. 3 d. 4 e What product(s) would you expect from the following reaction N-bromosuccinimide 14 CH 2 CH CH 3 RR, C 4 a. 14 CH 2=CH CH 2 alone b. 14 CH2=CH CH2 and CH2=CH 14 CH2 in equal amounts c. CH 2=CH 14 CH 2 alone d. More 14 CH 2=CHCH 2 but a little CH 2=CH 14 CH 2 e. More CH 2=CH 14 CH 2 but a little 14 CH 2=CHCH 2 9. Treatment of 4-methylcyclohexene with N-bromosuccinamide in C 4 would yield mainly I II III I e. All of the above As with #8, once the bromine atom abstracts an allylic hydrogen, the allylic radical is free to resonate this gives all 4 allylic substitution product outcomes shown 4

5 10. What is the principal product of the following reaction sequence H i) P 3 ii) C 6 H 6, Al 3 iii) LiAlH 4 iv) H 2 H I II III H H I e. 11. The IR spectrum of a compound exhibits a broad absorption band at cm -1 and a sharp band at 1710 cm -1. Which of the following compounds is a possibility a. 1-Butanol b. Propyl acetate c. Butanoic acid d. Acetyl chloride e. Acetic anhydride 12. What is the IUPAC name for the following compound a. 2-Methylbutyl 2-methylbutanoate b. 2-Methylbutyl 3-methylbutanoate c. 3-Methylbutyl isovalerate d. Isopentyl isovalerate e. Isopentyl isobutyrate 5

6 13. What is the IUPAC name for the following compound a. 2,3-Dimethylbutyl acetate b. 2,3-Dimethyl-4-oxoethanal c. 2,3-Dimethylbutyl methanoate d. 2,3-Dimethylbutyl methylate e. 2,3-Dimethylbutyl formylate 14. Which of the following would be the strongest acid C 2 H C 2 H C 2 H I II III C 2 H C 2 H I e. Recall chlorine is deactivating (even though ortho/para directing) which is synonymous with electron withdrawing 15. A compound has the molecular formula C 6H The IR spectrum shows a strong absorption band near 1740 cm -1. The NMR spectrum consists of 2 singlets, one at 1.2 ppm and the other at 3.6 ppm. Which of the following is consistent with this information I II III I e. 3.6 ppm singlet is key must be attached directly to 6

7 16. What is the initial compound being transformed in the following reaction sequence i. Mg/ether ii. C 2 iii. H + H a. HC 2CH 2C 6H 5 b. C 6H 5CH 2CH c. C 6H 5CH 2 d. C 6H 5CHCH e. =C(CH 2C 6H 5) 2 assic means of tacking on a carboxylic acid 17. What is product A i. NaCN ii. 70% H 2 S 4, reflux A + NH 4 + a. HC 2CH 2C 6H 5 b. C 6H 5CH 2CH Hydrolysis to amide and then on to carboxylic acid c. C 6H 5CH 2S 3H d. C 6H 5CHCH e. =C(CH 2C 6H 5) What is product F C 2 H i. LAH, Et 2 ii. H 2 P 3 i. Mg, Et 2 ii. C 2 iii. H 3 + F C 2 H C 2 H C 2 H I II III I Quite a bit of work to extend the carbon chain by 1 C e. Interesting in that the first 3 steps convert C at +3 to C at -3 oxidation state 7

8 19. What is the product of the following reaction 1. LAH 2. H 3 + H H H I H II III H H I e. 2 equivalents of hydride add to esters with LAH 20. Which of the following would serve as a synthetic route to (CH 3) 3CC 2H a. b. c. d. (a) & (b) e. (a) & (c) i. 2 /H (excess) ii. H 3 + i. CN ii. H 3 + (heat) i. Mg, Et 2 ii. C 2 iii. H 3 + Sorry about 20 and 21, since we really haven t talked about -C chemistry much yet. As a neat variation which tests for methyl ketones, see the iodoform reaction 8

9 21. Which of the following strategies will not generate benzoic acid a. C 6H 5CH 2H + KMn 4/H - /H 2, heat; then H 3 + b. C 6H 5CH 3 + KMn 4/H - /H 2, heat; then H 3 + c. C 6H 6 + C 2, high pressure d. C 6H 5CCH 3 + 2/H - /H 2; then H 3 + e. C 6H 5C + H - /H 2; then H Reasoning by analogy, what is the predicted product after sec-butylmagnesium bromide greets carbon disulfide and a proton source is made available S SH SH S SH I I S II S SH III e. 23. Which of the following is the mystery reactant i. NaCN ii. 70% H 2 S 4, reflux CH + NH 4 + a. CH 2C 6H 5 b. C 6H 5CH 2CH c. C 6H 5CH 2H d. C 6H 5CH 2CH 2CH e. =C(CH 2C 6H 5) What would be the principal product of the following reaction C 6 H 5 NaCN i. excess LAH, Et 2 ii. H 2 a. C 6H 5CH 2CH 2C 2H b. C 6H 5CH 2CH 2NH 2 h cyano, is there anything you can t do c. C 6H 5CH 2CH(CH 3)CN d. C 6H 5CH 2CH=NH e. C 6H 5CH 2NH 2 9

10 25. What is the principal product of the following reaction sequence i. CH 3 CH 2 H ii. dilute H, cold M H CH 2 CH 3 H H H 3 CH 2 C I II III CH 2 CH 3 H 3 CH 2 C CH 2 CH 3 H 3 CH 2 C I Ride the anhydride to esters aplenty e. Interesting point here is succinic anhydride gives mixed ester/acid notice making the diacid would require heating the alcohol in acid to get the 2 nd carboxyl group to react 26. Please clearly 1 show the mechanism for the DCC/DMAP promoted esterification between benzoic acid and benzyl alcohol K, I admit, I included this (1) because I took the time to detail the mechanism previously, and (2) It s pretty freakin cool. Extremely efficient way to get reluctant acids and alcohols to get together under mild conditions Steglich Esterification while similar to the DCC amidation mechanism (very useful for the synthesis of proteins from amino acids) differs in that the DMAP ends up attacking the initial DCC carboxylic acid adduct, generating an activated acid with a good leaving group before the alcohol attacks the carboxylic acid carbonyl (as opposed to loss of the stable dicyclohexyl urea directly upon attack of the alcohol). Here s how it looks: 1 Which implies neatly 10

11 DMAP = Dimethylaminopyidine DCC = N,N'-Dicyclohexylcarbodiimide ery good leaving group A-H A Now benzyl alcohol attacks Ureas are very stable 11