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Introductory Organic Chemistry Laboratory

by: Ms. Jerrell Lind

Introductory Organic Chemistry Laboratory CHEM 344

Marketplace > University of Wisconsin - Madison > Chemistry > CHEM 344 > Introductory Organic Chemistry Laboratory
Ms. Jerrell Lind
GPA 3.55

Nicholas Hill

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Nicholas Hill
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This 18 page Class Notes was uploaded by Ms. Jerrell Lind on Thursday September 17, 2015. The Class Notes belongs to CHEM 344 at University of Wisconsin - Madison taught by Nicholas Hill in Fall. Since its upload, it has received 36 views. For similar materials see /class/205339/chem-344-university-of-wisconsin-madison in Chemistry at University of Wisconsin - Madison.


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Date Created: 09/17/15
1H NMR Spectroscopy Background Molecules are two small to be observed with the naked eye In fact it was only recently that the technology was developed to directly observe molecules by a specialized microscope The molecules that can be observed by such a technique are still quite limited or the instrumentation is prohibitively expensive As a result most characterization of organic molecules is done by Nuclear Magnetic Resonance Spectroscopy Nuclei have a quantum property called spin The 13C nuclei and 1H nuclei protons have two separate spin states 1 and 1 as their spin quantum number is 1 The 12C nuclei have only one spin state 0 as their spin quantum number is 0 The spin quantum number is a measured property related to a nucleus39s angular momentum and cannot be calculated based on number of neutrons and protons in a nucleus Normally the nuclei does not care what spin state it resides in as the both states have equal energy However when a magnetic field is applied the two spin states start to have different energies as the two spin states do not have the same inherent magnetic field If the magnetic field is substantially large enough 712 Tesla roughly 200 thousand times that of Earth39s magnetic field a noticeable energy difference between the two spin states occur The slight energy difference is approximately equal to that of a radiowave AE Radiowave lt INCREASING MAGNETIC FIELD It should be noted that modern day spectrometers work differently than described below but the principle is still the same Since the energy difference is similar to that of a radiowave a sample can be bombarded with a wide range of radio frequencies When the energy of the gap and the radiowave match there is an absorbance and the spectrometer registers a peak Moreover depending on symmetry the nuclei in a molecule are in different environments that have various small local magnetic elds As a result a different peak is observed for each different nuclei allowing a unique spectrum for each molecule Furthermore this pattern can be interpreted to ascertain the basic structure ofa molecule Sinoe the 12C nucleus only has one spin state no separation and thus no peaks can be seen rendering 12C nuclei invisible to NMR instruments 1H NMR more commonly known as proton NMR can lead to far more insight into the structure of a molecule than 13C NMR specially when the molecules are small There are three characteristics to an NMR spectrum chemical shift integration and ooupling Chemical Shift Chemical shift works the same in proton NMR spectroscopy as it does in 13C NMR Like C NMR chemical shift is relative and is based on the distance in ppm of a pmk from the peak at 0 ppm tetramethylsilane The basic unit for chemical shift is parts per million For every signal you have a unique proton Chemical shift tables for proton NMR have been developed that list very various functional groups and approximate chemical shift values A more extensive chemical shift table can be found at the back of the lab manualMost signals in proton NMR lie between 0 and 10 ppm Chemical shift in proton NMR is very descriptive of functional groups and with a knowledge of the structure of the compound the values can be predicted and readily assigned Prulon NMR Chemical Shth Table quotCG HAF H HACl j HAEr a H H mh A LF H I l HAD Adv L H VH Since chemical shift is governed by the local magnetic eld that the nucleus experiences it is important to know the factors that increase or decrease that magnetic field Electron density has the largest impact on the localized magnetic eld as electrons themselves have a magnetic field associated with them This eld is situated to directly oppose the large external magnetic field so it partially quotshieldsquot the nucleus from the magnetic field The higher the electron density the more the nucleus is shielded Peaks that are shielded are on the right hand side of the NMR spectrum and are described as quotupfieldquot Inductive effects When electronegative elements are present such as oxygen or nitrogen the signals are shifted to the left of the spectrum and are regarded as quotdown eldquot The electronegative elements pull electrons away from the protons hydrogen nuclei through single bonds thus deshielding the protons This effect is called the inductive effect and is a very short range effect It is felt typically through one or two single bonds Diethyl ether 1butanol 4 bonds 2 bonds 2 bonds away from O 139 ppm 363 ppm 347 ppm H 3 bonds away from O 5 bonds H H A 121 ppm 094 ppm H H N o O 1 bond H 224 ppm 3 bonds 157 ppm As can be seen in the examples of diethyl ether and butanol the oxygen39s effect greatly diminishes when it is further than two bonds away It is interesting to note that in the case of 1butanol that the proton directly attached to the oxygen is less affected that the proton two bonds away Protons directly attached to heteroatoms O N S non carbon atoms often have broad and sometimes unpredictable chemical shift ranges This is mainly due to hydrogen bonding which can have a significant impact on the environment a proton resides in The amount of water present in the NMR sample can cause different NMR of these protons Protons attached to carbons rarely partake in hydrogen bonding and their chemical shifts are consequently more predictable Resonance effects Inductive effects are weak effects with a limited range and are transmitted through the sigma bond network When an oxygen or nitrogen is attached to a sp2 or sp hybridized carbon a much more powerful effect takes precedence due to resonance This resonance effect sometimes referred to as a mesomeric effect can drastically increase or decrease the electron density about a carbon and subsequently protons attached to that carbon Consider the cases of p xylene and pdimethoxybenzene Even though there are ten protons present only 2 signals are observed for each molecule due to symmetry pxylene pdimethoxybenzene 705 ppm 705 m pp 683 ppm 683 ppm HD HD H5 H5 2 30 HA ppm HA 375 ppm Hc O O Hc HB HB HD HD As expected the presence of the oxygen has a deshielding effect on Hc However the same oxygen has a shielding effect on HD This is due to resonance The oxygen is a strong electron donating group By donating electrons into the aromatic ring it effectively adds negative charge on groups ortho and para to itself The increase in negative charge means an increase in electron density and thus a strong shielding effect for the protons attached to the carbons with the negative charge Hc is directly attached to an sp3 carbon that has no p orbitals and consequently does not feel the effect of the resonance Electron withdrawing groups have the opposite effect pxylene pacetylbenzene 705 PPm 705 ppm 803 PM 803 ppm H H H H 230 ppm B B 230 ppm 264 PM D D 264 ppm HA HA HC HC 0 0 HB HB HD HD Hc feels a deshielding effect due to the carbonyl inductively drawing electrons away The carbonyl has a much weaker inductive effect than the oxygen in p dimethoxybenzene as oxygen has a higher inherent electronegativity than the carbonyl group The carbonyl does have a substantial resonance effect on HD as it is a strong electron withdrawing group when it comes to resonance HD HD HD HD Hc Hc Hc Hc J 0 yo 0 a 89 HD I HD HD HD HD HD HD HD Hc HD HD HD HD Instead of placing negative charge into the ring like the oxygen the carbonyl group pulls electrons out of the ring creating positive charge within the ring As a result the carbons that bear that partial positive have a reduced electron density and therefore the protons attached to those carbons are deshielded shifted to the left of the spectrum acetophenone 79 ppm 73 PPm Deshielded Hc 259 ppm HA HD 76 ppm 0 Deshielded Hc HB anisole 688 ppm 726 PPm Shielded HG HF 375 ppm HE HH 0 692 ppm Shielded The resonance structures using electron withdrawing and electron donating groups also remove or donate electrons to and from the ortho and para positions Both positions are affected but in proton NMR these groups have more effect in the ortho position then they do in the para position This is seen in the examples of acetophenone and anisole above Common Electron Donating Groups 10 a lkoxy CLN amino Common Electron Withdrawing Groups 90 0 Ii1 0 nitro e carbonyl N nitrile Integration Unlike the rare 13C nucleus the 1H nucleus represents 99 ofall hydrogen nuclei As a result of this abundance proton NMR is not as sensitive allowing for faster accumulation times and more importantly integration In proton NMRthe area underneath is directly proportional to the number of protons that signal represents This can make assignments of signals to protons a trivial process 100 MH LH VIR n A m l 008 o MeJkOASIZMe I Me e Tnmemylslylmemyl acme 204 9 375 2 3 4 G 3 5 3 D 2 5 2 D 1 5 l U U 5 U D In the case of trimethylsilylacetate the assignments could be made based on chemical shift the numbers to the left of the signal or they can be assigned by using integration the numberto the right of the signal In this case the signal at 6 376 ppm has an integration of 2 and corresponds to the CH2 The signal at 6 204 ppm has an integration of 3 and corresponds to the single methyl group The 3 methyl groups attached to the silicon all rapidly interconvert with each other sure to rotation about CSi and CC sigma bonds causing them to all overlap and appear as a single peak with an integration of 9 008 ppm On the NMR spectrum we would simply draw the molecule out and write a b and c next to the protons and match them to the corresponding signals 300 M39Hl H NMR In K pm a O b M MeJLOASi e c I Me M e Tnmemyisaiyimnyi acetate H Hz 5 Hh TMS N l 4 u 3 5 3 o 2 5 20 l 5 1 u u 5 o u The NMR spectra that you will interpret will be spectra of the compounds that you make in lab As a result they may not be as clean as some of the spectra that you see here The format will be slightly different as well The chemical shift value will not be above the peak Instead you will have to write it out based on where the signal falls on the X axis Also the integration will not be written out on the sheet to the right of the signal It will be located under the signal itself It is also very likely the integration will not be whole numbers When dealing with integration the relative ratio is important The integration in the example above 239 could have been 4618 or even 142921436429 as the NMR instrument frequently calculates integration as a ratio totaling 100 Integration is a very useful tool not only for identifying compounds and assigning peaks but also can be used to calculate the purity of a sample should the identity of the impurities be known The proton NMR spectrum of a mixture of pXylene and methanol is shown on the next page This spectrum was taken using the instrument that will process all of your samples and the integral values are located beneath the peaks Notice that they add to 100 We can assign the signals at 71 ppm and 23 ppm to pxylene and the signals at 34 ppm and 30 ppm to methanol using a chemical shift table The integration also confirms this For pgtltylene the peak ratio should be 115 46 and is 1164 on the spectrum which is fairly close For methanol the ratio should be 13 and is actually 128 The numbers are not exact as there are some errors having to do with the how the instrumentation acquires and how the software processes the data Mixture of p xylene and methanol 1989 3510 1233 3269 With the peaks correctly assigned the ratio of methanol to pxylene can be calculated by comparing the integration values First you must choose which values to compare For the best results you should compare peaks that are close to one another but are still far enough away for other peaks so that the integration is as accurate as possible In this case let39s use the signals for the two sets of methyl groups 34 ppm for methanol and 23 ppm for pxylene Before you can compare them you must first normalize the peaks because the pxylene methyl signal consists of 6 protons and the methanol methyl signal consists of only 3 protons p xylene 549 Ratio of pxylene to methanol N 121 methanol 3510 117 3 That is just the molar ratio To get a mass ratio or volume ratio you would need to incorporate the corresponding molecular weights and densities Coupling Coupling sometimes referred to as splitting is the probably the most useful characteristic of an NMR spectrum and is at the same time the hardest to understand When protons that do not have the same chemical shift ie inequivalent protons are within a certain range 24 bonds they can influence the shape of the signals The biggest influence occurs when they are three bonds away For example isopropyl ether has two types of protons the CH methyne and the CH3 methyl The four methyl groups are related by a symmetrical mirror plane and the two methynes are related by a mirror plane as well isopropyl ether Methyne 364 ppm CH3 H Methyl HX 112 ppm H H3C o 2 3 The methyl protons are located three bonds away from the methyne proton As a result coupling occurs and the signals representing the methynes and the methyl groups no longer consist of single peaks but multiple peaks 300 MH H NMR Ll rmil Isupmpyl ether i AA L i 71 4 5 U 2 5 2 D 5 H 05 D 0 From the integration it is easy to assign which peak is which The methyl groups have integration of 12 and the methynes have integration of two The number of the peaks within each signal is important and is due to coupling The methyl signal consists of two peaks while the methyne signal consists upwards of five peaks Both patterns are the result of the same effect Let39s take a look at the methyl group first The methyl protons are three bonds away from the single methyne proton The methyne proton can be spinning one way or another with or against the external field One spin state shields the other spin state desshields Since it is a near equal chance either way two different energy transitions appear equal in magnitude and each offset from the original frequency by the same amount Two different energy transitions mean two different peaks The peaks are separated by the coupling constant abbreviated J J is measured in hertz Hz and is typically 7 Hz in the case of linear alkyl groups A way to show this is called a coupling tree also known as a splitting diagram All of the original integration of the methyl signal is divided almost equally into two peaks Each peak offset from the original Hmethyl signal by 35 Hz half of the coupling constant Therefore if the integration of the original signal is 12 then each of the two 35 H 35 H f peaks would have integrals of 6 leethyne Hmethyne l The methyne pattern though is a bit more complicated gt 5 peaks but arises through 1 the same mechanism Except instead of J N 7 Hertz having one neighboring proton three bonds 11 doublet away it has six Each of them has a spin signal by 35 Hz The result is actually a seven line pattern called a septet with each peak 7 Hz from the next I state that shields or deshields the original methyne Unlike the case of the doublet these peaks do not have equal intensity Instead the pea at t e center have the highest intensity because the center pea s consist of a series of overlapping peaks Ift e peak heights were carefully measured the ratio would be approximately 1615201561 Peak Inbensity 1615201551 7 3 6 When dealing with coupling constants of similar magnitude an important rule to remember is the n1 rule where n equals the number of protons three bonds away or protons on neighboring atoms for short In the case of the methyl group it has one neighboring proton so 112 a doublet In the case of the methyne it has six neighboring protons so 617 a septet The intensities of these multiplets can even be predicted by the mathematical construct called 9 l r Pascal39s tn39a le Pascal39s Triangle 1 SinglEt 1 1 Doublet 1 2 1 Triplet 3 1 Quartet 1 4 5 4 1 Quintet 1 5 10 10 S 1 Sextet 1 6 15 20 15 5 1 Septet It should be noted that the septet octet and nonets can be tricky to identify Most of these have low integration values to begin with as they are invariably connected to two or three carbons each bearing multiple protons Most of that integration is in the central peaks and very little of it is on the outer edges One outer peak of the septet has 164 of the total integration Such a peak is very small and may blend into the baseline Therefore when you count ve peaks there may actually be seven or even nine present When you count six peaks there may be eight present You can tell the even number patterns from the odd number patterns as the odd number pattern have a single tall central peak The even number pattern has two central peaks of approximately the same intensity Let39s take a look at nitroethane nitroethane The methyl group is attached to a methylene CH2 so its signal should be a triplet 213 The methylene is attached 0 to a methyl group and a nitro group which does not have a proton so it will be a quartet 314 The triplet will have peaks with intensity ratio of 121 and the quartet will have a 0 peak intensity ratio of 1331 Since the methylene is directly attached to the nitro group it is the most deshielded so it will be on the left The quartet will be on the right This is indeed what the spectrum looks like 300 MHZ H NIVJR mm 158 442 Me No2 Nitroeihane 3 2 so 55 50 45 4o 35 30 25 20 15 ll 05 on Another way to look at these multiplets is in term of the spin of the neighbors The methyl group has two neighboring protons each with the possibility of an up spin or a down spin Let39s say the down spin causes shielding and the up spin causes deshielding So the possibilities are ll ll ll one two one deshielded quotneutralquot shielded combination combinations combination 121 triplet One the other hand the methylene has three neighboring protons This gives the following Situation M l H T w ill ill lll ill lll one strongly three three one strongly deshielded deshielded shielded shielded combination combinations combinations combination 1331 quartet The n1 rule does have a couple of caveats It works when the coupling constants are the same Typically when the carbon atoms are sp3 hybridized and are linear not locked in a ring the coupling constant is 7 Hz Protons that are equivalent do not split each other Protons can be equivalent based on symmetry or rapid rotation about a single bond The CH2 groups in 12 dimethoxyethane do not couple with each other even though they are on adjacent carbons Protons on heteroatoms may or may not couple depending on the conditions in the NMR sample Generally they do not couple but instead appear as a broad peak When peaks broaden they shorten The width of carboxylic acid signals can span over an entire ppm and as a result appear as baseline 300 MHZ H NMR lnl no SXCH3 CH2 NH2 4 If 25 20 15 10 D 5 00 When the molecule is in a ring or there is a pi bond present the rotation about some of the bonds is restricted As a result different coupling constants can occur First case to look at is 4isopropylaniline 300 Alliz H NMJI 1quot ix l3 Me Me augmmiamime ll 7 7 5 5 4 3 2 l D The NHZ and the isopropyl group is straightfonNard to assign as the NHZ is a broad peak 35 ppm the CH is a multiplet 28 ppm and the two methyl groups form the doublet at 12 ppm The isopropyl coupling constant is about 7 Hz as t e carbons are sp3 hybridized The ring protons are a different situation The para substituted benzene ring givs a very distinct pattern of two pairs doublets in the aromatic region 300 MHz 39H NMR 1 am 7uz ppm J U J10 HZ J10 H2 A A I 75a 725 7 an 575 550 5 25 son The protons attached to the ring are rigidly held in one geometry and do not freely rotate As a result the coupling oonstant is different from the freely rotating 7 Hz coupling constant Instead it is the much larger ortho coupling oonstant J0me can range from 612 Hz but typically lies around 10 Hz If you examine the patterns closely you will see that the peaks are not really doublets Instad mch peak oonsists of six lines These lines represent part ofa very oomplex pattem referred to as a seoond order pattern For the purposs of this course thse peaks can be thought of as pseudodoublets To actually assign the peaks to the appropriate protons resonance structures can be drawn The protons ortho to the NH2 group bear a partial negative charge and are thus more shielded It becomes more oomplicated when a meta substitution pattern is present When two protons are meta to each other they can split each other even though they are four bonds distant This ooupling constant Jmeba is smallert an t e ortho and ranges from one to three ertz This can lead to some interesting patterns Consider the aromatic region of 3nitroanisole 30H MHZ H NAIR m cncu 743 up u 771 H N01 3 779 l 723 1 l lr1 I 1 c a n ifygm a fl W 77 i 79 78 77 7E 75 74 73 72 71 The aromatic region has 4 signals Two of which appear to be triplets though one is wider than the other The other two pattems have eight lines apieoe To assign this spectra it will be neoessary to make a table of the coupling constants HA HB Hc Ho Jortho 0 1 2 1 Jmem 2 2 0 2 With the table in hand we can generate a series ofsplitting diagran39s to predict the shapes of the signals There are two possibilities for HA If the two meta coupling constants are not equal then HA is a doublet ofdoublets 4 lines If they are equal then HA is a triplet 3 lines A triplet is just a doublet of doublet will have similar possibilities as HA except l39lc will be wider After all the width of the signal is just the sum of the coupling constants and the Jam is greater than Jmeta HA HA I I JAD I IJAD I I I I JAB Iquot If JAB I four line pattern dd three line pattern t doublet of doublets triplet 1111 ratio 121 ratio When you compare the predicted patterns to the observed patterns it becomes clear that the triplet at 771 ppm is HA and the triplet at 748 ppm is Hc HB and HD give rise to a more complicated HB pattern as both have one large Jam and 2 small Jmeta If all three coupling constants are different an eight line pattern called a doublet of doublets of doublets ddd is formed Indeed this is the case and the multiplets at 779 ppm and 723 ppm are best described as ddd Since both give identical patterns the only way to assign the two structures is by JBC looking closely at the structure of the molecule itself and relying on chemical shift HE is ortho to an electron withdrawing I I I I JBD group therefore it will be strongly deshielded Conversely HD on the other I I I I I I I I hand is ortho to an electron donating group thus it will be strongly shielded With this I information the ddd at 779 ppm can be assigned to H5 and the ddd at 723 ppm belongs to HD eight line pattern ddd doublet of doublets of doublets 11111111 ratio


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