Homework 6 - Answers - Homework Problem Set 6 Iverson...

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Unformatted text preview: Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 NAME (Print): _____________________________ SIGNATURE: _____________________________ Chemistry 310N Dr. Brent Iverson 6th Homework February 27, 2008 Please print the first three letters of your last name in the three boxes Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 Score: _________ Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 1. (10 pts) On the left is drawn the Lewis structure of a simple amide. Draw the two next most important contributing structures in the spaces provided. Be sure to show all lone pairs and formal charges. H H C H C O H N H H H C H C O H N H H H C H C O N H H 2. (6 pts) Because of the resonance you described with the above structures, several atoms of an amide bond are in the same plane. On the amide below, circle all the atoms that are in the same plane. Think carefully about this one!! Note that one of the H atoms on the methyl groups will also be in the same plane at times, depending on the exact conformation. The atoms circled are ones that are always in the same plane. O H C H H H C N H H C H 3. (4 pts) In the boxes provided, write the hybridization state of the given atoms. sp2 H H C H O C N H H H C H sp3 sp3 sp2 4. (6 pts) In the spaces given, answer the question as "yes" or "no". H C O C H C H H N H H Does this bond rotate freely? Yes H Does this bond rotate freely? Does this bond rotate freely? No! Yes Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 5. (8 pts) Carboxylic acids exist in solution as characteristic "dimers". Draw the structure of the dimer formed by butanoic acid, CH3 CH2 CH2 CO2 H. H O O H O O A Hydrogen-bonded dimer 6. (4 pts.) Rank the following in terms of reaction with a strong nucleophile such as HO-, with a 1 under the molecule that is least reactive, and a 4 under the molecule that is most reactive. O O O O Cl O O O NH2 3 4 2 1 7. (6 pts) List two attributes of amide bonds that lead to stabilization of the folded structures of proteins. 1) The amide carbonyl oxygen and hydrogen on nitrogen can make strong hydrogen bonds 2) The C-N bond cannot rotate freely, so the backbone of proteins is relatively rigid (less flexible) and easier to stabilize in a folded conformation. 8. (3 pts each) In the space provided give the IUPAC name for the given molecule. O O Ethyl 5-methylhexanoate N O N,N-Dimethylpropanamide Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 9. (12 pts.) The molecule drawn below undergoes a conformational change to give a new conformation that reacts to give two new products when heated. Draw this reactive conformation inthe box provided. Now, on the molecule you just drew, draw arrows to indicate the flow of electrons and then draw the stuctures of the two products initially produced. Note that one molecule undergoes a further transformation to give the final stable product. Draw the final stable product in the space provided. Make sure to draw all lone pairs and formal charges. O O H O O H O O Reactive Conformation O H + O C O 2nd product Enol Initial product O Keto Final product Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 (3 or 5 pts each) Fill in the box with the product or products that are missing from the following chemical reaction equations. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES. O OH HO H2CrO 4 O OH H2 / Pd 1) LiAlH 4 2) H2 O O OH HO HO OH O OH SOCl 2 O Cl OH H2 SO 4 (catalytic amount) OH O O Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 10. (7 pts) Using any reagents turn the starting material into the indicated product. All carbon atoms must come from the starting material. Draw all molecules synthesized along the way. When it doubt, draw the molecule! Hint: this should look familiar as a homework problem. All of the carbons of the produce must come from the given starting material. OH ? O H2 SO4 OH O H2 CrO 4 OH (Fischer esterification) O (Alternative approach) H2 CrO 4 OH OH O SOCl2 O Cl Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 10. (17 pts) Using any reagents turn the starting material into the indicated product. All carbon atoms must come from the starting material. Draw all molecules synthesized along the way. When it doubt, draw the molecule! Hint: this should look familiar as a homework problem. All of the carbons of the produce must come from the given starting material. O O OH ? Racemic O OH O SOCl2 Cl O H 1) MgBr O 2) HCl / H2O HO HO O H2CrO 4 PBr3 Br PCC Mg ether Racemic Recognize that the product is actually an ester, made from the corresponding alcohol and an acid chloride. Note that you would not want to carry out a Fischer esterification on this secondary alcohol because in the presence of H2 SO4 it would probably dehydrate rather than form an ester. Recognize that the acid chloride can be derived from the starting alcohol via H2 CrO4 oxidation followed by SOCl2 . Recognize the complex alcohol used in the last step is the product of a Grignard reaction between the corresponding aldehyde and Grignard reagent, which in turn come from the starting alchohol via PCC and PBr3 followed by Mg in ether, respectively. Homework Problem Set 6 Iverson CH310N Due Wednesday, March 5 10. (13 pts) Using any reagents turn the starting material into the indicated product. All carbon atoms must come from the starting material. Draw all molecules synthesized along the way. When it doubt, draw the molecule! Hint: this should look familiar as a homework problem. All of the carbons of the produce must come from the given starting material. OH + PBr3 Br Mg ether MgBr 1) CO2 CO 2 ? O O O O Racemic H2SO4 (catalytic amount) (Fischer Esterification) OH OH H2SO4 H2O OH O O OH OH O Racemic 2) HCl / H2O Recognize that the product is a lactone, that is a cyclic ester. It is formed by a Fischer esterification of the corresponding 5-hydroxyhexanoic acid. Counting carbons in the starting material and product indicates this must come from adding CO2 to a Grignard reagent. Recognize the hard part of this problem is realizing that a Grignard reagent is too basic to be made in the presence of an -OH group in the same molecule, so the Grignard reaction used to add the extra carbon atom must take place before the -OH group is created, ...
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This homework help was uploaded on 04/21/2008 for the course CH 310n taught by Professor Iverson during the Spring '08 term at University of Texas at Austin.

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