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CHE 311 Answers in BOLD RED EXAM 3 Answer Key (Ch. 1115) Multiple Choice (60%; 2% each) Please mark the letter of the BEST answer to each question clearly on your answer sheet. Thank you! 1. The MO pi bonding model of benzene consists of ______ molecular orbitals in total, ______ of which are bonding. a. four ; two b. eight ; four c. six ; three d. three ; all 2. The common name xylene is used to refer to a benzene derivative having ______________ on the ring. a. two methyls b. a hydroxyl c. an amino (‐NH2) d. a carboxyl 3. Which of these benzylic groups will NOT oxidize to a carboxyl (‐COOH) when heated with a strong oxidant? a. –CH3 b. –CH2CH3 c. –C(CH3)3 d. –CH2OH 4. Which of these annulenes is aromatic according to Hückel’s Rule? only C (10 pi electrons) a. b. c. d. 5. The most common alkenylbenzene compound, used in making a variety of polymers, is: a. styrene b. anthracene c. butadiene d. naphthalene 6. In order to be useful for substitution reactions of aromatic rings, a reagent must be classified as a(an) a. carbocation b. nucleophile c. electrophile d. Lewis base 7. Sulfonation of benzene is unusual in two ways; which two? I. The active reagent is NOT an ion II. Yields are quite low III. It is very reversible IV. Products may explode a. I and II b. II and III c. II and IV d. I and III 8. In Friedel‐Crafts Alkylation, one of the main complications is NOT that a. alkyl groups may rearrange b. alkyl groups may come off easily c. products have increased reactivity d. alkyl halides are expensive 9. An example of an activating group which is also ortho, para directing would be a. –NO2 b. –Cl c. –COOH d. –CH3 10. The normal reagent sequence used to place an nbutyl group on a benzene ring would be a. (CH3)3CHCl + AlCl3 b. CH3CH2CH2CH2Br + AlBr3 c. CH3CH2CH2COCl + AlCl3 d. step c, then HCl + Zn(Hg) 11. If you treat 1‐bromo‐2‐nitrobenzene with Cl2 + AlCl3, where will the Cl most likely be placed? a. C‐3 and C‐4 b. C4 and C6 c. C‐4 and C‐5 d. C‐3 and C‐6 12. Which analytical method does NOT rely on the presence of energy levels of some type within a molecule? a. UV b. FTIR c. NMR d. MS 13. The two common forms of ‘hyphenated MS’ combine Mass Spectroscopy with a. extraction b. filtration c. distillation 14. The hydrogens on the second carbon atom of 1‐chlorobutane should have a H‐NMR signal which is a. a triplet b. a triplet of triplets c. a quarter d. a quintet 15. Which is a limitation of 13C‐NMR as it is usually performed? a. peaks fall over a narrow range of ppm values b. not every 13C atom will give an NMR signal c. there is a poor correlation of shifts with structure d. peak area does not correspond to number of C 16. In Infrared Spectroscopy, the peaks that are largest are produced by: a. the most frequent functional group b. nonpolar bonds c. highly polar bonds d. symmetrical bonds 17. Which of the following should have a significant UV spectrum (produce absorbances in the UV range)? a. CH2=CHCH2CH2COOH b. (CH3)2C=C=O c. Cyclopentene d. 2Phenylpropane 18. Organometallic reagents like Grignard and Lithium reagents are usually made by reacting a. metal halides + alkenes b. metals + alkanes c. alkyl halides + metals d. alcohols + metal oxides 19. When a Grignard reagent is reacted with formaldehyde (H2C=O) and then worked up normally, the product is a. a primary alcohol b. a secondary alcohol c. a tertiary alcohol d. a hydrocarbon + CO gas d. chromatography CHE 311 Exam 3 Page 2 20. If we react excess CH3CH2MgBr with the ester CH3CH2COOCH3, the major product formed after workup is a. (CH3CH2)3COH b. CH3CH2C(=O)CH2CH3 c. CH3CH2CH(OH)CH2CH3 d. C(CH3CH2)4 21. The SimmonsSmith reagent (CH2I2 + Zn(Cu)) is used to transform alkenes into a. epoxides b. vicinal di‐iodides c. cyclopropyls d. geminal di‐iodides 22. When organometallic chemists study the bonding & stability of complexes, they see if the electron count on the metal is ______, which indicates greater stability. a. 10 b. 8 c. 18 d. 32 23. Ethanol is most often made commercially (for use as a solvent, not beverage) by: a. reduction of CH3COOH b. hydrolysis of ethylene oxide c. oxidation of CH3CH3 d. hydration of CH2=CH2 24. Reduction of aldehydes and ketones by NaBH4 has several advantages. Which is NOT one of them? a. NaBH4 is a mild and safe reagent to use b. NaBH4 is an easier & safer to use reagent than H2 gas c. NaBH4 is stable in acidic environments d. NaBH4will not reduce C=C bonds like H2 would 25. To reduce carboxylic acids, esters, or their salts to alcohols the preferred reagent is usually a. NaBH4 b. H2 + Pt c. LiAlH4 d. Na in liquid NH3 26. Epoxides are useful starting points for a variety of products; treatment with ____________ gives diols while their reaction with Grignard reagents or lithium reagents will give _______________. a. H3O+ ; alcohols b. HCl ; alkenes c. H2O2 ; aldehydes d. OsO4 ; ethers 27. Simple symmetrical ethers can be produced by a. heating 2° alcohols with strong acids c. warming 1° alcohols with a trace of acid 28. Esters are made via Fischer esterification by reacting carboxylic acids… a. with alkenes and strong acids b. with alkyl halides and AgNO3 c. with warm alcohols with a trace of acid d. with epoxides in dilute aqueous acid 29. To convert a primary alcohol into an aldehyde, the best of these reagents to use would be a. H2CrO4 b. KMnO4 c. OsO4 d. PCC or PDC 30. The biochemical reagents that are most often used to oxidize alcohols to aldehydes are called a. oxidases b. aldases c. dehydrogenases d. lipases b. treating epoxides with alkyl halides and base d. treating epoxides with dilute aqueous acid Problems (40%; 15 & 25% respectively) I. Bromination of nitrobenzene produces only one major product. Begin by drawing the proper structure for nitrobenzene. Then show how bromination occurs, including how the reactive form of bromine is made. Draw cyclohexadienyl cation intermediates (& resonance forms) and explain why only one product is formed. :Br—Br: + AlBr3 Br+ + AlBr4– generates the positive Br ion (or at least polarizes the Br2 like it) O N+ O– O– O N O N etc., puts partial + at the ortho and para O– positions, making them less reactive. [can omit this structural detail but need to at least explain meta] Br only product –O N+ – H+ O O N+ O– O Br+ resonance contributors keep + from being on C of NO2 group O O– O O– O– + + + N N N Br Br Br H H H CHE 311 Exam 3 Page 3 II. A clumsy stockroom worker (totally unlike ours at TU) has forgotten to label three bottles of similar organic substances. One is 2‐phenylethanol. Another is acetophenone (C6H5C(=O)CH3) and the third is phenylacetic acid (C6H5CH2COOH). Smell could help with the identity, but the worker has a cold, so samples of each of the three bottles are taken and H‐NMR, C‐NMR, and FTIR spectra of each bottle are made. Based on the spectra and the structures, which substance is in each bottle (A, B, and C). You must explain your answer with at least TWO spectroscopic features from each sample (each of the two coming from a different method – for example you could use the IR and H‐NMR or samples A & B and perhaps the H‐NMR and C‐NMR of C to determine the correct identity of each, which ever two are most helpful). Identities: For purposes of the key, spectra are shrunk and put with the relevant answer Substance A is____Acetophenone__________ I know this because: There are two key features in the spectra that make this ID very reasonable: 1. The CNMR has a peak at 198 ppm, indicating a carbonyl 2. The HNMR has no peak below 8 ppm, so it cannot have a carboxyl group (which would occur at 1013 ppm). 3. There is no sign of an –OH in the HNMR or in the FTIR, which is almost featureless at the area that should occur. 4. Finally, the FTIR shows a very strong carbonyl signal at 1686 WN !"#$%&&$#'()$%$ *+,-.(/01$2$ ! ! ! CHE 311 Exam 3 Page 4 Substance B is ___Phenylacetic acid____ I know this because: 1. The FTIR has BOTH a carbonyl at c. 1720 WN and a broad –OH signal at c. 3100 WN, confirming the carboxylic acid group. 2. The HNMR has the intense methylene signal at 3.7 ppm, not found in the other substance still unidentified (phenylethanol); more importantly, the carboxyl H at c. 12 ppm is a strong proof. 3. The CNMR is not very helpful in this case. !"#$%&&$#'()$%$ *+,-.(/01$2$ !"#$%#&$%'(!)*%+,%#&$%,-.//$01%233$0%3/"#%4,5./$%.#%#"36%7&+/$%#&$%8(!)*%+,%#&$%/"7$0%3/"#1% 9"#&%7+#&+:%#&$%,.-$%9";<%=&$%,$3.0.#$1%9"##"-%,3$5#02-%+,%#&$%>=?*%"@%A29,#.:5$%B<! CDCl3 QE-300 240 220 200 180 160 140 120 100 80 60 40 20 0 12 11 10 9 8 7 6 5 4 3 2 1 0 CHE 311 Exam 3 Page 5 Substance C is ___2Phenylethanol____ I know this because: 1. The FTIR has a broad –OH signal at 3348 WN, confirming the alcohol group, but no strong carbonyl signal (the weak one at c. 1722 may be a trace of oxidized material). 2. The HNMR is a very strong confirmation because it has the proper shifts and splitting for the compound in question (note the alcohol OH is at the unusual shift of 2.0 ppm). The aromatic H are at c. 7.2; the –CH2O is at 3.8 while the other –CH2– is found at 2.8 ppm. These methylenes couple and split each other into the triplets that are observed. 3. The CNMR is most helpful in a negative way: it has no peak in the carbonyl region (170210). !"#$%&&$#'()$%$ *+,-.(/01$!$ ! ! !