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by: Brady Spinka


Brady Spinka
GPA 3.98


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Class Notes
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This 74 page Class Notes was uploaded by Brady Spinka on Monday September 7, 2015. The Class Notes belongs to CH 310N at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/181886/ch-310n-university-of-texas-at-austin in Chemistry at University of Texas at Austin.

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Date Created: 09/07/15
Lecture 5 Chem 31 ON NMR Continued February 1St Dr Callaway Homework from last time How many equivalent hydrogens 1 C1 3 for both cis and trans Web Sites for Reference and Practice httpWwwndedusmithgrpstructureworkbo0khtml httpWWWchicacukloca1r1mr httpWWWchemunipotsdamdetools But this one is great for figuring our chemical shifts We are not quite ready for these require 13C NMR and IR to solve but you can at least take a 100k for now Our hard unkmwn fmm last timeg Let s Begin to Figure this Guy Out Solving nmr unknowns 1 Molecular Formula and unsaturation number C2H3O mass 43 17243 4 9 CgHIZO4 Cng8 our reference Cng2 our hydrocarbon formula H6 62 3 Three rings or three pi bonds or a combo of these Solving nmr unknowns 2 Number of distinct signals is 3 three different types of equivalent hydrogens 3 Integration ratio is 204060 123 123 6 126 2 The ratio of hydrogens is then 246 4 Pattern of Chemical Shifts 5 12t 42q 67 s 5 Classic triplet and quartet is likely due to ethyl groups Singlet at 67 ppm is more of a mystery Shift Table Type of Hydrogen Type of Hydrogen R alk Chemical R Ikyl Chemical Ar aryl Shift 6 Ar aryl Shi 83 Clhhsi 0 by de nition U Rug 09 RCOCEJ 37 39 RC zR 12 14 W R3C 14 17 RCOCH l 4147 R1CCRCHR2 16 26 ne zr 3133 any 2033 Rugzsr 34 36 C33 21 25 R zcl 36 33 ArC zR 2343 R 39 lquot 44 45 Rog 05 60 3 1 Rome 3440 R1CG 3 39539 RCBZUR 33 40 54quot R ZCCHR mag News cmmmn Enhanced ire wizard 1000 entries Daubase NMREa k L caum Dun s eshage p d kn s XMP vsleuer Please Egan ennrsm 5m sch The Jame Generator is how a regular Item of this servicel The Newsletter has a own page and a help page is created subscribe Newsletter unsubscribe Newsletter Since 1Feh1999 Capynght 20m 20m Steffen Thnmzs 45256 visxtm39s since 2001mm lastupdate 02m 02 x Wemel MW CC om RO Alkyl H H mm CC om ROC Aryl H H 6 I up a 21 CC om ROC COR H n Solving nmr unknowns 6 We suspect that we have an H CC fragment at 12 ppm an HCO structure at 42 ppm and after looking at the chemical shift tables perhaps a linkage of the sort H CCCO 7 Signal Splitting Patterns 12 ppm has two neighbors 42 ppm has 3 neighbors and 67 ppm has no neighbors This and total H count C8H1204 leads us to think about two oxygenlinked ethyl groups and two HCCCO fragments 8 You now have completed the analysis the final task is to synthesize the structure from the clues to put it all together This gets easier with practice Enmm iaa JF 011mm El30 7 7 7 7 7 7 r 7 7 7 7 7 7 7 1 7 7 7 7 7 7 7 7 w 7 7 7 7 7 7 7 7 w 7 7 7 7 7 7 7 7 w 7 7 7 7 7 7 7 7 x Solving nmr unknowns 8 You now have completed the analysis the nal task is to synthesize the structure from the clues to put it all together This gets easier with practice Note Our chemical shift analysis isn t good enough to tell cis from trans apart In this case there aren t any symmetry 0r splitting hints to help However we know it is not a mixture How Unsymmetrical cis vs trans vs geminal disubstituted ole ns aka alkenes can often be distinguished based on splittings As a general rule Trans disubstituted alkenes give very large splittings J 1119 Hz Cis disubstituted alkenes larger than normal splittings J 514 Hz Although these overlap those of trans alkenes in terms of range for any ole n the cis splittings are smaller than those of the corresponding trans isomer Geminal systems give small splittings J l3 Hz Note that these can be complex in the case of trisubstituted alkenes 7 11Illlll lTlIl39llTllllIllllJ rlllII1I j I l l l I I I I300 I200 noo l I IOOO 900 000 700 600 500 400 300 H J140H r J l I I I l I I l I I I I L LujlllLLlllllllllTLlllrllllllJ IIJJJ 0 60 50 40 30 20 I 0 0 ppm Cl H NMRspectrumof CClt H CN IIIIIIIIIIITITI l I I I l I l I I I l I l I 7 I I I T 1 I I I I I I I l I I I I I300 I200 00 I000 900 900 700 SW 500 400 300 200 3900 OH H J7I7Hz J I I I I I I 1 I I I I I I IIILLLIIIIIIIIIIIIIIIIIIIIIIIII I II 70 60 50 40 30 20 LO 0 ppm Cl CN NMR ml f C C S 10 p H H l V I I l I l I I I l l I r I I l l l l l l 00 I200 IIOO IOOO 900 000 700 600 500 400 300 200 I00 0 Hz I I l I I I I I l I I I I L I I I l I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I 70 60 50 40 30 20 LO ppm 8 NMR spectrum of 33 dimethyll butene CH33CCHCH2I Jai a b E CII3CH3CH32 Multiple interactions of nonequivalent neighbors Where J ab gt ch Moving On CleCCHZCH3 Splitting Analysis Chemical Shift 1HNMR Type of H 6 Type of H 8 C H34Si 0 ROH 0560 RCH3 09 RCH2 OR 3340 RCH2 R 1214 R2 NH 0550 R3 CH 1417 0 R2 CCRC HR2 1626 RHCCH3 2123 RCE CH 2030 O Arc H3 2245 RECHZ R 2226 ArC H2 R 2328 a b C CH3 CH2 C Hg N g TMS 7 6 5 Alcohol and Carboxylic Acid OH s and Amine and Amide NH s are Special Cases l V l 39 39 I 39 39 39 I a b 396 CH3 CH3 UH all m i c t FFa ms 411 33 36 14 12 10 39 IhlillIlllllllllllerlvnAlllllll llllvIvlriAIllAtlllllllllliltIIIIAII a 5 4 3 2 1 0 EH Wm Rapid exchange can average out the sign of the spin vector on H attached to O and N Chemical Exchange Hydrogens on electronegative atoms such as Oxygen and Nitrogen Undergo rapid exchange and often do give only a relatively broad singlet due to averaging Their position chemical shift can change due to hydrogen bonding and hence solvent and concentration These hydrogens also exchange equilibrate with Deuterium in D20 and disappear from the spectrum IIIIIIIIIIIIIIIIIIIIII I I I I I I I I I I I I 1300 I200 IIooIoooooo m we we Ioo m I I I I I I I I I I I I III IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 10 60 50 4 50 20 Lo 0 WM IIIIIIIIIIIIIIIIIIIIIIIIIIIII I I I I I I I I I I I I I r I300 I200 Moo Iooo no no 100 we we too no too Inc ON 41 I I I I I I I I I I I I I LLIIIIIIIIIIIIIIIIIIIIIlIIIIIIIIIIIIIII 10 an 50 a 30 20 Lo 0 ppm 3 FIGURE 1346 NMR L 39 39 39 39nu Pllull I L I lion for IJIe up spectrum is 10 M for the bouom 01 M I I I I I I l I I I I I I I I I I I I I I I I l I I I I l I I I I I I I I I I I I I I I I I I I I300 I200 IIOO I000 900 800 700 600 500 400 300 200 loo 0 Hz 44 There is no Visible OH signal Due to exchange with the D s of D20 I I I I I I I 4 I I I I I l I 1 I I l I I I I I I I I I l I I I I l I I l I I I 70 60 50 4 0 20 11 391 4 341 1 LO 0 ppm 8 FIGURE 1347 NMR spectrum of ethyl alcohol in carbon tetrachloride The solution was shaken with D20 and the layers separated before the spectrum was measured Aromatic Systems Sometimes Easy but Often Complex Because of Multiple Splittings The rule of next door neighbors being the only ones that lead to splitting breaks down Look for symmetry count number of hydrogen atoms and keep an eye out for classic socalled AB splitting pattern that arises from two adjacent ortho protons in different chemical and magnetic environments Use the handout I gave to practice Mll1 39 mu El 290 100 1F 39 ca tiufjgLian I 1400 quot quot 31m Carbonyl Chemistry 111 Lecture 14 Great to be back Y Chemistry 310N Alcohols React with Aldehydes and Ketones to From HemiAcetas and HemiKetas R THIS IS REVIEW C 0 FROM LAST TIME RI a hemiacetal FF R6C 6 H Product is called R Hemiacetal reacts further in acid to yield an aceta T nquot quot This product is R C 9R called an acetal RI ROH H Ff R6C 5H This product is n a hemiacetal R Example Cu QCH ZCH30HZOH I HCI Q39CHOCHZCH32 H20 Benzaldehyde diethyl acetal 66 In general Position of equilibrium is usually unfavorable for acetal ketal formation from ketones Important exception Cyclic ketals can be prepared from ketones Example 0 C6H5CH200H3 HOCHZCHZOH benzene ptoluenesulfonic acid H20 CH2 O O H0 78 C 2 Mechanism of Acetal Formation First stage is analogous to hydration and leads to hemiacetal acidcatalyzed nucleophilio addition of alcohol to CO Mechanism Mechanism C 26 H e 5 U H Resonance stabilized cation Mechanism R oC 39 4 4 H H Mechanism Mechanism Mechanism R O C O This is our hemiacetal I H H I 9 Mechanism of Acetal Formation Second stage is hemiacetaltoacetal conversion involves carbocation chemistry Hemiacetal toacetal Stage H I O C Q H 9 2 R These are not separate reactions this is all one big equilibrium Hemiacetal toacetal Stage Hemiacetal toacetal Stage 0 3 Here is the water Note That it is being lost when we consider the reaction running in the forward carbonyl gt acetal direction Hemiacetal toacetal Stage R R 6 c 00 00 00 Carbocation is stabilized by delocalization of unshared electron pair of oxygen Hemiacetal toacetal Stage Hemiacetal toacetal Stage Hemiacetal toacetal Stage Hemiacetal toacetal Stage R R cog 3931 H 5R This is our acetal H Note that EVERY step is an equilibrium Therefore the reaction can be pushed forward or backward by appropriate choice of conditions The forward reaction is synthesis The backward reaction is hydrolysis Hydrolysis of Acetals 1 ii R C R39 H20 o 2RquotOH I R R ORquot mechanism reverse of acetal formation hemiacetal is intermediate application aldehydes and ketones can be quotprotectedquot as acetals DeanStark trap for removing water by azeotmpic distillation C J 2 EDII Aldehyde Alcnhnl or kelmne 1L 11 x 1 OR I a on H1 in Vapor rising toward mntlenser contains 91 I benzene and 9 water Acetal 0 l Waterjacket umdensar Condensing vapor a Cold H20 DeanS lurk trap If Upper layer contains 9994 benzene and 006 water Lower layer cuntains 00T benzene and 9993 water Acetals and Ketals as Protecting Groups Acetals and ketals act like acid sensitive ethers and are stable to Reduction Oxidation Strong Base e g Grignard conditions Typical SN2 conditions Hydrogenation Etc An Example of this Approach Q av Another Example OH Ho o W ONO BrQk lt Brlt can39t form can form Grignard Grignard Reagent Reagent 1 Mg THF CHO 2 O MgBr o OO OH H3O 2 lt A Special Kind of Acetal a THP ether Can Protect Alcohols For Example H O 0 o BrO O BrOH can39t form can YOU figure out can form Grignard mechanism Hint Grignard Reagent Remember our friend Reagent Markovnikov 1 Mg THF 2 amp HOWOH MgBr oWo 0 Info re mechanism may be found on pp 570571 Thioacetals and Thioketals These are made in analogy to oxygen series but require HgII Lewis Acid catalysis to hydrolyze Also Thioacetals may be deprotonated and used as nucleophiles to generate various kinds of thioketals The net result of this umpolung or charge reversed chemistry is the conversion of an aldehyde to a ketone with formation of a new CC bond One of many reactions that led to E J Corey s Nobel Prize Examples HSMSH m J H S S NBUU S S Stcsleaos hile Li greater polarization of sulfur permits removal of proton O Br HSCJLH 0 H20 SQ J W CH30N 0 H20 HgCI2 m S S gt CH3 WCHQ O39 Li OH Reaction of Carbonyls with Nitrogen Nucleophiles Aldehydes and ketones react condense with NH3 and 10 amines to form imines Schiff bases in an equilibrium process that often favors products A host of other NHzcontaining species react in a similar way 2 Amines react with aldehydes and ketones to form enamine The carbonnitrogen species can often be reduced to produce amines You are expected to know the contents of Table 164 the names of the starting species the names of the products and the mechanism by Which they are formed Study mech on p 574 carefully Some Examples of Many NV H2 Cat 0 HH20 H 0 H NM gt G Pd NI etc 0 2 lt imineSchi1quotf base 2 amine o H H2O HO 390 HZo 0 HN gt gt 0 enamine OH H2 o H20 N OH Pd Ni etc NH2 H2N gt gt hydroxylamine OXIme o 1 amine o R NH H20 2 R1L R2 RH R2 Table 164 Derivatives of Ammonia and Hydrazine Used for Forming lmines Reagent HgN39 R Name of Reagent Name of Derivative Formed OH Hydroxylzun i116 Au oxime NH Q P11 e nylhyd razin e A phenylhydrazone WNH N03 24 T quot 1 A J 1 OZN 0 II W NHCNH2 Semicarbazide A semicm bazone Representative Mechanism 0 H2N BI W 39 NV H20 N gt KPHZ gt imine B water amine Recommended Problems EVERYTHING IN CH 16 and 17 If you haven t done them yet start with 16191622 16241626 16301634 1637 1638 16401642 16431644 Okay Let s Quit Here For Today Next Time Chapter 17 Carboxylic Acids 39 Chemistry 310N


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