➢ How can you differentiate between an amine and an alkyne?

➢ Why do primary amines have two signals and secondary amines have one?

➢ How to tell apart a C=O bond and a C=C bond?

CHEM 2320: Dr. Werner  Organic Chemistry II Exam 1  Chapter 15: Infrared Spectroscopy and Mass Spectroscopy Key vocabulary to know Key numerical data to know Key concept 15.1 Introduction to Spectroscopy ➢ Frequency is directly proportional to energy; wavelength is inversely proportionaIf you want to learn more check out integrating functions using long division and completing the square
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l ➢ IR spectroscopy uses infrared light and causes molecular vibration and stretching ➢ IR spectroscopy is used to tell the functional groups and bonds in a compound ➢ Neat means a pure sample is used ➢ Only organic molecules can absorb IR radiation NOT ionic molecules; this is why salt plates can be  used  15.2 IR Spectroscopy ➢ wavenumber is the x-axis measured in inverse cm; higher wavenumber= higher energy ➢ % transmittance is the y axis telling the percent of IR radiation that does not get absorbed by the  molecule and hits the detector ➢ We are focused on diagnostic region above 1500 cm-1; region below is fingerprint region which  requires a computer to detect differences ➢ 3 key characteristics of IR spectroscopy: wavenumber, intensity, shape 15.3 Effects on wavenumber o Size and bond strength ➢ Lower mass and greater bond strength (smaller bond distance) means higher energy. Think of  a ping-pong ball being more impacted by higher frequency than a bowling ball.  ➢ Increasing energy: Single bonds < double bonds < triple bonds < bonds to H ➢ Triple bonds: 2300cm-1 ➢ Single bonds: less than 1500 cm-1 o Hybridization ➢ Triple bonds are highest in energy b/c they have shortest bond length (most s character)  ➢ Sp3 C-H bonds: 2900 cm-1 single bonds  ➢ Sp2 C-H bonds: 3100 cm-1 double bonds ➢ Sp C-H bonds: 3300 cm-1 triple bonds ➢ Always draw in C-H bonds; compounds with no sp2 C-H bonds will have sp3 C-H  bonds only and no signals above 3000 cm-1 o Resonance ➢ Resonance lengthens the single bond character, weakening the bond ➢ Conjugation means pi bonds are separated from each other by one single bond; more  conjugation= more resonance=lower wavenumber  ➢ Exception: Esters have more resonance but have a higher wavenumber (1740) than  ketones (1720)  o Symmetry ➢ Symmetrical alkenes and alkynes produce no signal in double bond or triple bond  region.  15.4 Effects on Signal Intensity o Dipole strength ➢ The greater the difference in inductive strength between vinyllic positions the greater the  dipole and the greater the intensity of the signal ➢ How to tell apart a C=O bond and a C=C bond? ▪ C=O bonds are stronger signals at 1720 cm-1and C=C are weaker signals at 1650  cm-1check for one of the other around the 1700 cm-1line15.5 Effects on Signal Shape o Hydrogen Bonding ➢ Hydrogen bonds are transient and weak thus they vary in their bond lengths; why alcohols  produce broad signals around 3400-3600 cm-1 ➢ More concentrated alcohols make broader signals due to more Hydrogen bonding; dilute  alcohols have less hydrogen bonding thus a narrow signal ➢ Carboxylic acids are even more broad than alcohols because they form hydrogen bonding  dimers.  ➢ Signal at 1720 and broad alcohol signal ~3400 is always a carboxylic acid o Primary vs. Secondary Amines ➢ Why do primary amines have two signals and secondary amines have one? o Secondary amines have 1 N-H bond which must rotate either symmetrically or  asymmetrically; primary amines have 2 N-H bonds where one rotates symmetrically  and the other asymmetrically ➢ How can you differentiate between an amine and an alkyne?  o An amine will not have a signal in the triple bond region near 2300 cm-1. If there is a  triple bond there may also be a signal at 3300cm-1for the sp C-H bond if it is a  terminal alkyne. There will not be a signal at 3300cm-1if it is an internal alkyne.  ➢ How strong are amines? o They are less polar than O-H bonds and more polar than C-H bonds. Thus, their  wavenumber is between O-H and C-H sp bonds C-H bonds have many signals  because of how numerous they are in a molecule and the various stretching  combinations that are possible.  15.6 -15.7Analyzing an IR Spectrum o Always do 3 steps:  ➢ Check 1700cm-1 mark for C=O or C=C bond ➢ 2. Check 3000cm-1 mark for what type of hybridized C-H bonds exist ➢ 3. Check for broad alcohol peak near 3400 cm 15.8 Mass Spectrometry o used to determine molecular weight and formula o high energy electrons are expelled generating a high energy radical cation that gets passed  through a magnetic field and fragments (Most likely: nonbonding electrons>pi bonds>sigma  bonds) o Only charged ions get detected by magnetic field; smaller compounds that are more easily moved  generate higher energy o Mass spectrum:  ➢ X axis is mass per charge; Y is relative abundance ➢ Base peak is the tallest and most abundant; gets assigned 100%; everything else is relative ➢ Base peak is most stable fragment; molecular ion peak corresponds to molecular weight of  molecule and is the tallest peak that found in the far-right cluster of signals 15.9 Analyzing Molecular Ion Peak (M)+* o First step is to find molecular ion peak from the graph or using molecular weight if structure is  given o Nitrogen rule: if the molecular weight is odd then there is an odd number of nitrogens in the  compound. If the weight is even then there are likely zero (or two) 15.10 Analyzing (M+1)+* Peak o Most carbons are C-12 isotopes; 1.1% are C-13. Thus, C-13 will be 1.1% as tall as the molecular  ion peak.  o Can mathematically determine number of carbons in a molecule if (M+1)+* and (M)+* peak heights  are known. Divide the (M+1)+*: (M)+* abundance ratio by 1.1 to get the number of carbons 15.11 Analyzing (M+2)*+ Peak  o Molecules with chlorine will have an (M+2)+* peak that is 1/3 the size of the (M)+* peak. o Molecules with bromine will have an (M+2)+* peak that is equal in size to the (M+2)+* peako If the (M)+* is smaller than the weight of bromine or chlorine it is possible that they have  fragmented 15.12 Analyzing Fragments o Alkanes ➢ M-15= methyl fragmentation ➢ M-29= ethyl fragmentation ➢ M-43= propyl fragmentation  ➢ Add 15 for each methyl group. Remember highest peak indicates most stable fragment. If  M-29 is the highest an ethyl group likely fragmented.  ➢ Fragmentation will always favor the most stable (tertiary) carbocation  o Alcohols and Amines ➢ Alpha cleavage (alcohols and amines) ▪ Bond on alpha position of alcohol cleaves to form resonance-stabilized cation  and radical  ➢ Dehydration (alcohols only) ▪ Loss of a water molecule; peak at (M-18) ➢ McLafferty rearrangement (ketones and aldehydes only) ▪ Hydrogen atom at the gamma position (alpha is next to double bond) joins the  oxygen and forms a neutral alkene and resonance-stabilized radical fragment.  ▪ Double bond formed at Beta-gamma position 15.13 High-Resolution Mass Spectrometry o Used to predict molecular weight to a high level of accuracy. o Useful for structures with the same molecular weight but different formula; Not effective for  compounds with the same molecular formula  15.14-15.15 Gas Chromatography and Mass Spectrometry of Large Biomolecules o Gas chromatography-mass spectrometry is used to separate compounds by their boiling points and  retention time; used in drug screenings and urine samples o Electrospray ionization is used for proteins and nucleic acids because it is a gentler method of  separation for molecules that don’t fragment easily 15.16 Hydrogen Deficiency Index and Degrees of Unsaturation o To be fully saturated, # of Hydrogens= (# of Carbons x 2) + 2 o Every 2 hydrogens missing is one degree of unsaturation o When halogens are in the formula add one hydrogen to the formula for each one o When oxygens are in the formula they have no effect on HDI because they bring a hydrogen with  them to the compound o When there is a nitrogen in the formula subtract one hydrogen from the formula because each  nitrogen takes the place of one hydrogen but brings two in instead o Every double bond is 1 degree of unsaturation; every ring is one degree of unsaturation; one  double bond and one ring= 2 degrees of unsaturationChapter 16: Nuclear Magnetic Resonance Spectroscopy 16.1Introduction to NMR Spectroscopy o Study of interaction between electromagnetic radiation and atomic nucleio of protons o Tells how the carbon and hydrogen atoms are connected in a molecule o a magnetic force from radio waves ccreates spinning magnetic moment in nuclei of protons  causing them to align in either an alpha (aligned with the field) or beta (aligned against the field)  spin state o when the nucleus absorbs enough energy to flip from the alpha to the beta energy state it is in  “resonance”  o More electron density around the nucleus the less magnetic force the nucleus experiences (called  diamagmetism); this shielding effect decreases the energy gap between the two spin states. o Stronger magnet also means bigger energy gap 16.2-16.3 Characteristics of an 1H NMR Spectrum  o NMR spectrometers commony use chloroform-d because they lack protons and have an even mass  number, causing them to be invisible to the spectrometer o Machine kept at very low temperatures to preserve liquid helium and nitrogen, which are excellent  superconductors o 4 characteristics: number of signals, signal location (shift), signal area (integration), and  signal shape (splitting pattern). 16.5 Number of Signals o Chemical equivalence ➢ Protons that occupy identical electronic environments together produce one signal ➢ Homotopic-protons have a rotational axis of symmetry. Both protons view the rest of the  molecule from the exact same perspective. When each proton is replaced with deuterium they  are identical.  ➢ Enantiotopic-protons have a reflectional axis of symmetry. The protons have stereochemistry  and there is no other stereochemistry in the molecule. When each proton is replaced with  deuterium a pair of enantiomers is formed.  ➢ Not chemically equivalent=diastereotopic. When each proton is replaced with deuterium  neither the same molecule or enantiomers are formed. The protons have stereochemistry and  there is another site of stereochemistry in the molecule.  ➢ Methyl groups will always be homotopic to each other in a symmetrical molecule because the  protons are in the same chemical environment ➢ **** think of chemical equivalence JUST like stereoisomerism. THE # of UNIQUE  CHEMICAL PERSPECTIVES= # OF SIGNALSL 1. Do the protons have stereochemistry? Yes No ???? homotopic  2. Is there already stereochemistry in the molecule? Yes ????diastereotopic No???? enantiotopic 3. Always double check with the replacement test. Ensure that each unique visual  vantage point gets its own signal. Number A-Z each unique signal. Always draw  all the protons. Hint: ASSIGN SIGNALS LIKE YOU WOULD A  STEREOCENTER. In other words, picture the entire molecule as a whole when  picturing a protons perspective. Interchangeable/ Symmetical means homotopic. o Chemical Shift ➢ Upfield: more shielded= lower chemical shift (up is more toward 0)  ➢ Downfield: less shielded= higher chemical shifto Temperature ➢ Axial and equatorial positions are not normally considered chemically equivalent; however  ring flipping occus so rapidly that axial and equatorial produce the same signal ➢ At low enough temperature the activation energy won’t be high enough for them to flip so  separate signals are observed at -100 Celsius temperatures o Inductive effects (more electronegativity correlates with less shielding and a higher chemical shift) ➢ Methyl C-H: 0-1 ➢ Allylic C-H: 2 (allylic= on carbon that is one single bond away from double bond) ➢ C-H near electron-withdrawing group (O or Cl or Br): 2-4 ➢ Vinyllic C-H: 4-6 (vinyllic=right on the carbon of a double bond)  ➢ Aryl: 7-8 ➢ Aldehyde: 10 ➢ Carboxylic Acid: 12 16.6 Integration  o The area under the peak that is relative to how many protons are giving the signal o Computer calculates integration values that are NOT absolute.  o Divide by the smallest number (which represents 1 proton) to get whole numbers (multiplying as  necessary) until the integration numbers equals the number of protons in the molecular formula.  o When there are symmetrical (homotopic) protons add all of the chemically equivalent protons  together when reporting the ratio. (what would be reported as 3:2 would be reported as 6:4 due to  homotopicity) 16.7 Multiplicity o Number of peaks in a signal; tells how many neighboring protons there are o When all of the neighboring protons are chemically equivalent: NUMBER OF PROTON  NEIGHBORS + 1= NUMBER OF PEAKS IN NMR o Why? ➢ The “home base” proton we are looking at will always spin up and fall back down but its  neighbors can either spin with or against the magnetic field. The neighboring protons affect  the magnetic field of the original proton with shielding and deshielding effects.Even if there  are no neighbors there will always be at least 1 signal o What constitutes a neighbor proton? o Protons that are viscinal; in other words in our class we are only considering protons  that are attached directly to an adjacent carbon to be a neighbor proton.  o This definition does not apply to germinal protons, meaning protons that are attached to  the same carbon cannot be neighbors and cannot split each other.  Example: Chapter 16: Nuclear Magnetic Resonance Spectroscopy 16.1Introduction to NMR Spectroscopy o Study of interaction between electromagnetic radiation and atomic nuclei of protons o Tells how the carbon and hydrogen atoms are connected in a molecule o a magnetic force from radio waves creates spinning magnetic moment in nuclei of protons  causing them to align in either an alpha (aligned with the field) or beta (aligned against the field)  spin state o when the nucleus absorbs enough energy to flip from the alpha to the beta energy state it is in  “resonance”  o More electron density around the nucleus the less magnetic force the nucleus experiences (called  diamagnetism); this shielding effect decreases the energy gap between the two spin states. o Stronger magnet also means bigger energy gap 16.2-16.3 Characteristics of an 1H NMR Spectrum  o NMR spectrometers commonly use chloroform-d because they lack protons and have an even  mass number, causing them to be invisible to the spectrometer o Machine kept at very low temperatures to preserve liquid helium and nitrogen, which are excellent  superconductors o 4 characteristics: number of signals, signal location (shift), signal area (integration), and  signal shape (splitting pattern). 16.5 Number of Signals o Chemical equivalence ➢ Protons that occupy identical electronic environments together produce one signal ➢ Homotopic-protons have a rotational axis of symmetry. Both protons view the rest of the  molecule from the exact same perspective. When each proton is replaced with deuterium they  are identical.  ➢ Enantiotopic-protons have a reflectional axis of symmetry. The protons have stereochemistry  and there is no other stereochemistry in the molecule. When each proton is replaced with  deuterium a pair of enantiomers is formed.  ➢ Not chemically equivalent=diastereotopic. When each proton is replaced with deuterium  neither the same molecule or enantiomers are formed. The protons have stereochemistry and  there is another site of stereochemistry in the molecule.  ➢ Methyl groups will always be homotopic to each other in a symmetrical molecule because the  protons are in the same chemical environment ➢ **** think of chemical equivalence JUST like stereoisomerism. THE # of UNIQUE  CHEMICAL PERSPECTIVES= # OF SIGNALSL 1. Do the protons have stereochemistry? Yes No ???? homotopic  2. Is there already stereochemistry in the molecule? Yes ????diastereotopic No???? enantiotopic 3. Always double check with the replacement test. Ensure that each unique visual  vantage point gets its own signal. Number A-Z each unique signal. Always draw  all the protons. Hint: ASSIGN SIGNALS LIKE YOU WOULD A  STEREOCENTER. In other words, picture the entire molecule as a whole when  picturing a protons perspective. Interchangeable/ Symmetrical means homotopic. o Chemical Shift ➢ Upfield: more shielded= lower chemical shift (up is more toward 0)  ➢ Downfield: less shielded= higher chemical shifto Temperature ➢ Axial and equatorial positions are not normally considered chemically equivalent; however, ring flipping occurs so rapidly that axial and equatorial produce the same signal ➢ At low enough temperature, the activation energy won’t be high enough for them to flip so  separate signals are observed at -100 Celsius temperatures o Inductive effects (more electronegativity correlates with less shielding and a higher chemical shift) ➢ Methyl C-H: 0-1 ➢ Allylic C-H: 2 (allylic= on carbon that is one single bond away from double bond) ➢ C-H near electron-withdrawing group (O or Cl or Br): 2-4 ➢ Vinyllic C-H: 4-6 (vinyllic=right on the carbon of a double bond)  ➢ Aryl: 7-8 ➢ Aldehyde: 10 ➢ Carboxylic Acid: 12 16.6 Integration  o The area under the peak that is relative to how many protons are giving the signal o Computer calculates integration values that are NOT absolute.  o Divide by the smallest number (which represents 1 proton) to get whole numbers (multiplying as  necessary) until the integration numbers equals the number of protons in the molecular formula.  o When there are symmetrical (homotopic) protons add all of the chemically equivalent protons  together when reporting the ratio. (what would be reported as 3:2 would be reported as 6:4 due to  homotopicity) 16.7 Multiplicity o Number of peaks in a signal; tells how many neighboring protons there are o When all of the neighboring protons are chemically equivalent: NUMBER OF PROTON  NEIGHBORS + 1= NUMBER OF PEAKS IN NMR o Why? ➢ The “home base” proton we are looking at will always spin up and fall back down but its  neighbors can either spin with or against the magnetic field. The neighboring protons affect  the magnetic field of the original proton with shielding and deshielding effects. Even if there  are no neighbors there will always be at least 1 signal o What constitutes a neighbor proton? o Protons that are vicinal; in other words, in our class we are only considering protons  that are attached directly to an adjacent carbon to be a neighbor proton.  o This definition does not apply to germinal protons, meaning protons that are attached to  the same carbon cannot be neighbors and cannot split each other.  Example: ❖ CHEMICAL SHIFT VALUES TO KNOW (straight from Werner’s office  ☺) o 0-2= alkyl C-H (1-2 is more likely to be ethyl groups and 0-1 is methyl  but don’t need to know that)  o 2= allylic  o 2-4: near electron withdrawing group  ▪ allylic to oxygen (or on carbon next to oxygen) is on low end/  closer to 2 (aldehyde O=-H) ▪ vinyllic or (on carbon connected to oxygen) is high end/closer  to 4 (-O- ether) o 4-5: vinyllic H o 7-8: aryl o 10: aldehyde  o Coupling constant (J value) is the space between the peaks measured in Hertz; it is the  degree to which 2 magnetic fields affect each other in terms of shielding. It will stay  constant no matter the strength of the magnetic field o When protons split each other they therefore have an equal coupling constant o Higher strength magnets do change the resolution (how clear the spaces look) and are less  likely to have overlapping, unclear-looking signals o Patterns ⮚ An isolated ethyl group will have a 2-H triplet and a 3-H quartet with  equal peak spaces ⮚ 9-H singlet = tert-butyl group ⮚ 1-H septet next to 6-H doublet= isolated isopropyl group ⮚ Ethanol: 2-H quartet, 1 H singlet and 3 H triplets o Complex splitting= proton splitting when neighbors are NOT chemically equivalent protons (homotopic or enantiotopic ) (assymetric; will form different signals); for  example complex splitting would form a doublet of doublets instead of a triplet (triplets  can only split each other if they are all chemically equivalent o You would need to know that something makes, say, a quartet of triplets or a  triplet of quartets but not know which is which o Aryls do crazy complex splitting; don’t read too much into it from the graph.  Mostly you need to be able to draw complex splitting not necessarily always be  able to see it in a graph because the graph has other factors influencing it that we  haven’t learned about. Aryls are just crazy. If they aren’t crazy then their  constituents are likely on opposite sides of the ring from each other o ALCOHOLS DO NOT SPLIT! Why? ▪ They form acidic protons which will exchange with the solvent and  become deuterated, which means they will have an even mass number  and not be seen  16.9 Using 1H NMR Spectroscopy to Distinguish between Compounds  o Can be difficult to distinguish between molecules that have the same functional groups  and are structural isomers. H NMR can use the number of signals to distinguish them 16.10 Analyzing a 1H NMR Spectrum o 1. Always calculate HDI first when given a formula ▪ If there is an HDI of 4 and ppm is between 7 and 8 it is an aromatic ring. 1  degree of unsaturation means either one double bond or one ring o 2. Look at the number of signals and integration values (3H, etc.) to tell symmetry ▪ If the fully saturated formula is the same as the number of hydrogens you are  given in the data there is no symmetry. Symmetry happens when there are fewer  signals in the NMR than you would expect ▪ if there is an integration value of 4 it suggests symmetrical CH2 groups o 3. Analyze each signal and write down each piece so you can put it together ▪ For example “3H next to 2” etc. for each signal. {Integration number} next to {#  of peaks – 1}  ▪ Analyze expected chemical shift values. If different identify if they appear to be  close to an electronegative atom which will raise the ppm value ▪ Hints: 3 next to 2 and 2 next to 3 means an ethyl group (CH3-CH2); 3 next to 0 is  often an ether (CH3-O-); watch out for tert-butyl pattern, meaning if there is a  really high integration number chances are the connectivity will involve a  tertbutyl situation) o 4. Assemble the fragments into a complete structure through trial and error and ensure it  makes sense o Drawing the chemical shift numbers next to the molecule helps to check it 16.11-16.12 Acquiring a 13C NMR Spectrum o 1H is the most abundant but 13C is the least abundant isotope; thus only see singlets  because splitting isn’t likely (due to low abundance of 13C isotopes)  o Each chemical environment is unique; if there is any signal overlap that means there is  symmetry o 13C has a much wider NMR signal (0-220 compared to 0-20) so you would have to  change the bandwidth on an NMR o In 13C only chemical shift is relevant; it is influenced by shielding and deshielding just  like 1H NMR only on a wider ppm scale ▪ sp3 Carbons (0-50) < sp Carbons (alkynes) and C next to  electronegative atom (50-100)< sp2 carbons (alkenes) (100-150)  <carbonyls (150-220) o Splitting does occur between 13C and 1H but it is so complex; instead we just view the  carbon and say that 13C is decoupled  16.13 DEPT 13C NMR Spectroscopy  o Use broadband decoupled to determine number of protons attached to each  carbon in the compound o Broadband decoupled gives all signals (CH3, CH2, CH, and C)  o DEPT 90: only gives CH signals o DEPT 135: CH3 and CH are positive but CH2 are the only negative signals you  will see out of all three scopes o CH groups give positive signals in all spectra  o Number of signals equals number of carbons unless there is symmetry