Name. Department of Chemistry SUNY/Oneonta. Chem Organic Chemistry I. Examination #3 - November 11, 2002 ANSWERS

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1 Name INSTRUTINS Department of hemistry SUNY/neonta hem rganic hemistry I Examination #3 - November 11, 2002 ANSWERS This examination has two parts. The first part is in multiple choice format; the questions are in this Exam Booklet and the answers should be placed on the "Test Scoring Answer Sheet" which must be turned in and will be machine graded. The second part requires your responding to questions in this Exam Booklet by writing answers into the spaces provided. The Exam Booklet must be handed in and will be returned to you with a grade. n the Test Scoring Answer Sheet, using a soft pencil, enter the following data (in the appropriate places): your name, instructor's name, your S Student Number or Social Security number, course number ( ) and the test number (03); darken the appropriate bubbles under the entries (if you are using your student number which begins with a letter, leave the bubbles under the letter blank, but darken the bubbles under the numbers), making dark black marks which fill the bubbles. You may use a set of molecular models and the periodic table (last page) provided, but no other aids, during the exam. Answer all questions. The questions on Part I are worth 4 points each. You have 50 minutes. Good luck! Answers are shown highlighted. Explanations are shown in red type.

2 November 11, 2002 hem Exam #3 Page 2 of 8 1. Provide the IUPA name for the compound shown to the right. 3 l 2 l (a) 2,2,3,3-tetrachloro-5-heptyne, (b) 2,2,3,3-tetrachloro-5-heptene, (c) 2,2,3,3-tetrachloro-5-octyne, (d) 2,3-tetrachloro-5-heptyne, (e) 5,5,6,6-tetrachloro-2-heptyne, (f) None of the previous answers is correct. 2. Select the structure(s) of the intermediate(s) that is(are) formed when (Z)-2-butene reacts with bromine I II III 3 3 (a) I only, (b) II only, (c) III only, (d) I & II, (e) I, II & III, (f) None of the previous answers is correct. The adds stereospecifically across the double bond. This means that if the methyl groups started out cis they will be cis in the product and if they started out trans they will be trans in the product. 3. Select the major organic product of the following reaction. 3 3 NBS, 2 DMS solvent Note: The N-bromosuccinimide provides a low concentration of bromine. 3 (a) 3 omine attaches itself first in the bromohydrin formation, so this is a Markovnikov type addition. 4. Select the major organic product of the following reaction ) B 3 / T solvent 2) 2 2, K 3 (c) 3 3 (b) 3 3 (d) 3 (e) None of the previous answers is correct 3 (a) 3 3 (c) 3 3 (b) 3 3 (d) 3 (e) None of the previous answers is correct ydroboration-oxidation gives anti-markovnikov addition to form an alcohol.

3 November 11, 2002 hem Exam #3 Page 3 of 8 5. What is (are) the principal organic product(s) of the following reaction? ) g(ac) 2, 2 /T 2) NaB 3 (a) (b) R-isomer only (c) S-isomer only (d) unequal amounts of R & S isomers (e) racemic mixture of R & S isomers Markovnikov formation of an alcohol. When achiral reactants form chiral products racemic mixtures form. 6. Select the principal product of the following reaction sequence. 1. B major , K (p=8) product (a) (b) ( 3 ) 2 3 (c) (d) (e) Anti-Markovnikov addition of water to the triple bond leads initially to compound (d), but this undergoes keto-enol tautomerization to give the aldehyde. 7. Select the appropriate reagents to carry out the indicated reaction. (a) 2, Lindlar catalyst, 3 (b) 2, Pt, (c) 2, 2 S 4, gs 4, 3 3 (d) Li, N 3, 3 (e) NaN 2 in N 3 Answer (a) would produce the cis isomer. 8. Which of the alkyl chlorides below is likely to be most successful in the following reaction, where R is an alkyl group: 3 /: - Na + + R-l 6 3 /-R + Na + l - (a) 3 2 l, (b) ( 3 ) 3 l, (c) ( 3 ) 2 l (d) Bogus question, dude! Like, none of these compounds will react this way. Primary alkyl halides work best in this reaction. 9. A pair of stereoisomers that are not enantiomers are (a) constitutional isomers. (b) meso structures. (c) diastereomers. (d) conformational isomers. (e) Another bogus question, dude! Like, all stereoisomers are enantiomers. By definition.?

4 November 11, 2002 hem Exam #3 Page 4 of If a chiral substance that has just one chiral center is dextrorotatary the chiral center is (a) R, (b) S, (c) E, (d) Z, (e) Actually it is impossible to tell whether the chiral center is R or S based only on the rotational direction of the optical activity. 11. Select the statement that best describes the relationship between the two structures shown to the right. They are (a) constitutional isomers. (b) enantiomers. (c) diastereomers. (d) the same molecule and not meso structures. (e) the same molecule and are meso structures. The structures are the same because they are the same, i.e. superimposable. These two structures represent a meso molecule because there are chiral centers in the molecule and the molecule is achiral (it is superimposable on its mirror image). 12. Select the statement that best describes the relationship between the two structures shown to the right. They are (a) constitutional isomers. (b) enantiomers. (c) diastereomers. (d) the same molecule and not meso structures. (e) the same molecule and are meso structures. They are non-superimposable mirror images. 14. Select the answer that correctly indicates the R/S configuration around the chiral centers in the molecule to the right. (a) 1-R, 2-R, (b) 1-R, 2-S, (c) 1-S, 2-R, (d) 1-S, 2-S. 15. Which of the following is not true of enantiomers? N 2 3 l (a) They have the same boiling point. (b) They have the same melting point. (c) They have the same specific rotation. (d) They have the same density. (e) They have the same chemical reactivity toward achiral reagents. Specific rotation is the only physical property that differs between enantiomers. 16. Which of the following molecules is chiral? 3 3 l 3 2 (a) (b) (c) (d) (d) is the only molecule in this group that is not superimposable on its mirror image. 3 3

5 November 11, 2002 hem Exam #3 Page 5 of Which of the molecules in question #16 is a meso structure? (a) is meso. (a) has (two) chiral centers but is achiral (it is superimposable on its mirror image); therefore it is meso. 18. Which of the following is not a reactive intermediate in the free radical chlorination of methane? (a), (b) 3, (c) l, (d) Bogus question. All of them are intermediates. 19. Select the two products that form in the following reaction. NBS (N-bromosuccinimide) UV light, l 4 solvent I II III IV (a) I&II, (b) I&III, (c) I&IV, (d) II&III, (e) II&IV, (f) III&IV The allylic free radical (more stable than alternatives) that forms as an intermediate is shown to the right. It can pick up a bromine at the two sites that share the unpaired electron. 20. Which of the following alcohols will be most reactive toward in terms of converting the alcohol to an alkyl bromide? (a) (b) (c) (d) These alcohols would be essentially equally reactive. rder of reactivity: tertiary>secondary>primary. Reason: tertiary alcohol forms tertiary carbocation, etc. and tertiary carbocation is more stable than secondary, etc. Directions for Part II --- Answer the questions in the space provided. If there is insufficient space continue your answer on the back of the sheet but clearly indicate on the front of the sheet that you have done this.

6 November 11, 2002 hem Exam #3 Page 6 of 8 Mechanism. (a) Show the mechanism for the following reaction. > Be certain to clearly show the configurations around the chiral carbon(s) in the product(s). > Be certain to show all intermediates and their stereo (3-dimensional) structure(s). > Be certain to show direction(s) of approach of reacting species if it has stereochemical consequences. or example, if an approach from one direction leads to one stereoisomer and approach from a different direction leads to a different stereoisomer, show this l l (a) + l 3 + l (b) from path (a) from path (b) l l 3 3 l The carbon that is the site of positive charge in the intermediate carbocation is trigonal/planar (sp 2 ) and has a vacant 2p orbital, as shown. This orbital can join to a chloride ion from the top [path (a)] or from the bottom [path (b)]. These paths are equally accessible so the products (enantiomers) will be formed in equal amount. Now, it is true that I ve shown a conformation of the intermediate carbocation in which the methyl group on the left is pointed toward the front, not toward the top or bottom. If this methyl group is oriented up toward the top, instead of front, it will be easier for chloride to approach via path (b). owever, there will be an equal number of carbocations in which the methyl group is oriented down (mirror image of the carbocation in which it is up). or these carbocations approach of chloride via path (a) will be easier. These mirror-imaged conformations of the carbocation will have the same energy and hence will be present in equal amount, thus leading to equal amounts of the product enantiomers.

7 November 11, 2002 hem Exam #3 Page 7 of 8 (b) Is the product, or mixture of products taken as a whole, of this reaction optically active? No. A racemic mixture of enantiomers results in this reaction as explained above. owever, one need not know the mechanism of this reaction to know the answer to this question because optically inactive reactants (the case here where neither reactant is chiral) give optically inactive products. 2. Synthesis. utline syntheses which would produce each of the following compounds in good yield. [Note: In outlining a synthesis you should show explicitly what compounds you are using and any special conditions. You need not balance equations or show mechanisms; doing so correctly will gain you no additional credit, doing so incorrectly will cost you.] (a) You must start each synthesis with cyclopentene, and may use any other materials you need to carry it out. More than one step may be required. (i) 3-bromocyclopentene NBS UV light or heat 3-omocyclopentene is a chiral molecule. Will just one enantiomer form, or will unequal amounts of the two enantiomers form, or will the product be racemic? Racemic. Achiral reactants give racemic mixtures of enantiomers when chiral products are formed.

8 November 11, 2002 hem Exam #3 Page 8 of 8 (ii) 3-(3-cyclopentenyl)cyclopentene Li Li ui u from part (i) Li It should be apparent from the question that the idea here is to join two cyclopentenyl units together. urrently, you know (or should know) only two synthetic methods that will join two smaller units together to give a larger one. ne of these involves using terminal alkynes; that method will not work here. The other method is to employ a Gilman reagent. That is the method employed here. Part I (80) PartII 1. (10) 2. (15) Total(105)