CH318N Spring 2012 Final Exam. Chemistry 318N. Spring 2012 Dr. Willson. Final Exam

Transcription

1 318N Spring 2012 Final Exam 3 hemistry 318N Spring 2012 Dr. Willson Final Exam This afternoon you will take two tests, one in chemistry and one in integrity. I want you to get A s on both of these tests but if you are to fail one, let it be the one on organic chemistry. GW Name (Print as it appears on the lass Roster) Signature Stockroom Br I KN

2 318N Spring 2012 Final Exam 1. (10 Pts) ircle the best answer for each question below. Substitutes meta An acetal Not used to shift the equilibrium to favor product in acetal synthesis F 3 P( 3 ) 2 I 3 3 Dean Stark Trap 2 S 4 Excess Alcohol Strong acids have Small pka Small Ka igh p A Phosphonium Salt ( 3 ) 3 P + 3 ( 3 ) 3 P + - ( 3 ) 4 P + l - Takes primary alcohols to acids Tollen's reagent Jones reagent Wittig reagent Reacts fastest with MgBr 3 in TF 3 l Intermediate in decarboxylation of 3 () as no non-bonding molecular orbitals False statement All meta directors are deactivators All deactivators are meta directors All activators direct ortho and para

3 318N Spring 2012 Final Exam as the smallest number of p orbital electrons B onjugate base of the weakest acid F 2 - l 2 - Br 2 - Not an equilibrium reaction Acid hydrolysis of hemiacetal Acid hydrolysis of lactone Acid hydrolysis of Nitrile Thermodynamic Anion _ An anhydride as the lowest field resonance in 1 -nmr l l l l l l Decarboxylates upon hydrolysis in hot acid 2 N 2 N 2 N An enamine N N as the longest carbon carbon bond N Did not win a Nobel Prize in hemistry Wallace arothers Paul Flory ermann Staudinger

4 318N Spring 2012 Final Exam 2. (10 pts) Please put a T in front of each true statement and an F in front of each false statement. Put an X in front of questions that you do not want to be graded. Your score will be the number of correct answers minus the number of wrong answers or zero, whichever is the largest. Please read the statements very carefully! a) The right-hand end of the scale on a typical NMR spectrum is up field. b) Frequency increases from left to right on the typical infrared spectrum c) The Diels-Alder Reaction creates six membered rings. d) Nitrile arbons resonate at higher field than carbonyls in 13 nmr. e) In mass spectra, the M+2 peak is always smaller than the M+1 peak. f) The carbon in formic acid is more highly oxidized than that in formaldehyde. g) The signature of the laisen condensation is, -unsaturated esters. h) 1,4-dimethylbenzene has 3 signals in its 13 -nmr spectrum. i) The carbon-carbon bonds in benzene are longer than those in ethene. j) Good leaving groups are the conjugate bases of strong acids. k) The higher the pka of an acid, the higher the p of its 0.1 molar solution. l) Nitriles undergo acid catalyzed hydrolysis to give ketones and ammonia. m) Amides hydrolyze faster than esters in aqueous base. n) Anhydrides can be made from acid chlorides by nucleophilic acyl substitution o) The empirical formula for dimethyl pimelate is p) Primary alcohols can be oxidized to aldehydes by PP q) Nylon is a polyamide. r) The Mn of a polymer is always greater than or equal to Mw. s) ydrogenolysis of benzyl alcohol gives toluene. t) Living polymerizations produce polymers with low polydispersity.

5 318N Spring 2012 Final Exam 3. (10 pts) omplete the reactions below by supplying missing reagents, reactants or products. + N 2 2 NN eat l Br 2 / FeBr 3 3 Monosubstitution product eat 3 3 1) 2) dil. 3 + MgBr, TF

6 318N Spring 2012 Final Exam P + + N (10 pts) We have purchased a sample of a new polyester that has unusual properties. The structure of the polymer is shown below. We analyzed the sample and learned that it was a mixture, as all synthetic polymers are. The sample is comprised of 200 grams of material with degree of polymerization 10, 200 grams of degree of polymerization 20 and 600 grams of degree of polymerization 30. Please calculate the weight average and number average molecular weight of the sample and its polydispersity. Disregard any end groups. Show your work. You do not need a calculator to do this problem! Mw = Mn = Dispersity = * * n 3 3 polyester

7 318N Spring 2012 Final Exam 5. ( 5 pts) Do you predict that nitrosobenzene will produce ortho and para isomers or meta isomers when subjected to electrophilic aromatic substitution? Please support your answer with resonance theory arguments. N Nitrosobenzene 6. (3 Pts) We have received a sample of the polyester listed below. This sample has a polydispersity of 1.5. The weight average molecular weight of the sample is 300,000 Daltons. What is the degree of polymerization? * * n 3 3 polyester

8 318N Spring 2012 Final Exam 7. (15 pts) Predict the products that you would isolate upon treatment of the compound below with hot aqueous acid. Use the curved arrow convention to show the mechanism that leads to these products. 3 + eat Predicted Products.

9 318N Spring 2012 Final Exam 8. ( 3 pts ) The reaction shown below takes place in good yield. Please propose a reasonable mechanism for the reaction. Show the step-by-step sequence of transformations and the corresponding movement of electrons using the curved arrow convention Na (4 pts) Dangerous Dan is struggling in his organic lab class. e was given an unknown to solve and has been working on it for 3 hours. e has used a number of very sophisticated analytical techniques and now knows that his unknown is either cyclohexane or cyclopentanone ( 5 8 ). The bell is about to ring so he quickly runs a high resolution mass spectrum and discovers that M + = What is his unknown? Please show your work.

10 318N Spring 2012 Final Exam 10. (15 pts) An incredibly toxic substance with 14 carbon atoms was isolated from the purple sea slug. This stuff even kills water bears! The infrared spectrum of the material shows strong peaks at 720, 1118, 1692, and 2900 cm -1 and there are medium intensity peaks at 2232 and 3027 cm -1. The substance has a strong molecular ion at M/Z=231. The 1 and 13 nmr spectra are reproduced below. n the following page please list and tabulate what you can deduce from the IR, mass spectrum, 13 -NMR, and 1 -NMR; then write the structure of the unknown. The 1 -NMR shows resonances for several different chemical shifts. The peaks are numbered 1 through 4 below. Please use these numbers to assign the resonances to specific protons in your proposed structure nmr nmr

11 318N Spring 2012 Final Exam Problem 10 continued. Infrared spectrum: 1 -nmr: 13 -NMR: Mass Spectrum: Structure and assignment

12 318N Spring 2012 Final Exam 11. (15 pts) Please provide a synthetic pathway to the substances below. You may use any reagents or compounds you choose, but all of the carbon in your final product must come from the stock room, an inventory of which is on the cover page of the exam. Please be neat!! 3 2 N 3

13 318N Spring 2012 Final Exam Exact masses of common elements and isotopes Approximate 13 hemical Shifts

14 318N Spring 2012 Final Exam EMISTRY 318N ARATERISTI PRTN EMIAL SIFTS Type of Proton Structure hemical Shift, ppm yclopropane Primary R Secondary R Tertiary R Vinylic = Acetylenic Aromatic Ar Benzylic Ar Allylic = Fluorides --F hlorides --l 3-4 Bromides --Br Iodides --I 2-4 Alcohols Ethers --R Esters R Esters --R Acids arbonyl ompounds --= Aldehydic R-(-)= 9-10 ydroxylic R Phenolic Ar Enolic = arboxylic R Amino RN 2 1-5

15 318N Spring 2012 Final Exam Table of IR Absorptions Functional Group haracteristic Absorption(s) (cm -1 ) Notes Alkyl - Stretch (m or s) Alkane - bonds are fairly ubiquitous and therefore usually less useful in determining structure. Alkenyl - Stretch Alkenyl = Stretch Alkynyl - Stretch Alkynyl = Stretch Aromatic - Stretch Aromatic - Bending Aromatic = Bending (m) (v) ~3300 (s) (v) ~3030 (v) (s) (m,m) Absorption peaks above 3000 cm -1 are frequently diagnostic of unsaturation Alcohol - Stretch (broad, s) Very strong broad peak arboxylic Acid - Stretch (broad, v) Amine N- Stretch (m) Primary amines produce two N- stretch absorptions, secondary amides only one, and tetriary none. Nitrile N Stretch (m) Medium sharp band Ketone = Stretch Ester = Stretch arboxylic Acid = Stretch Amide = Stretch (s) (s) (s) (s) The carbonyl stretching absorption is one of the strongest IR absorptions, and is very useful in structure determination as one can determine both the number of carbonyl groups (assuming peaks do not overlap) but also an estimation of which types. Ester - Stretch (s) Generally a strong peak in esters All figures are for the typical case only -- signal positions and intensities may vary depending on the particular bond environment.