Can First Year Organic Students Solve Complicated Organic Structures?

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1 an First Year rganic Students Solve omplicated rganic Structures? Tools of the trade (for problems in this discussion): 1. Mass Spec information limited data (M+, M+1, M+2, M+3, etc.) 2. I information functional groups, alkene, aromatic substitution patterns 3. NM information most useful ( 1, 13, DEPT, ET, SY, MB) ommon reasons and an approach to identify organic or biological structures. Prepare a new compound or Isolate a new compound Purify ompound hromatography (t.l.c., flash chromatography, G, PL, etc.) Distillation of liquids (simple, fractional, steam, vacuum, etc.) ecrystallisation of solids Sublimation of solids Pure ompound Elemental analysis (ombustion analysis) Empirical Formula Mass Spectrum Molecular weight Molecular Formula (ule of 13) (calculate degree of unsaturation & number of rings and/or pi bonds) Spectral properties I = functional groups and = substitution patterns 1 NM = number & types of hydrogens 13 NM = number & types of carbons UV-VIS = pi bond information If structue is simple enough. Similar to known structures? Identify structure. If a structure is too complex, chemical modification might make it simpler. Derivatives or fragments of a compound ompare to literature values.

2 Determining a Formula, degrees of unsaturation, pi bonds and rings Limited mass spec data (M+, M+1, M+2, etc.) is used. Structure problems assume this data is available, though this not always the case. N 2 S hemical Formula: 5 12 Exact Mass: M+ = (100.), M+1 = (5.5%) hemical Formula: 4 11 N Exact Mass: M+ = (100.), M+1 = (4.7%) hemical Formula: 4 10 Exact Mass: M+ = (100.), M+1 = (4.5%) hemical Formula: 4 10 S Exact Mass: M+ = (100.), M+1 = (5.1%), M+2=92.05(4.6%) l Br Br Br l l hemical Formula: 4 9 l Exact Mass: M+ = (100.), M+1 = (4.3%), M+2 = (32.), M+3 = (1.4%) hemical Formula: 4 9 Br Exact Mass: M+ = (100.), M+1 = (4.3%), M+2 = (97.3%), M+3 = (4.3%) 5 10 Br 2 Exact Mass: M+ = (100.), M+1 = (5.4%), M+2 = (194.6%), M+3 = (10.7%), M+4 = (94.6%), M+5 = (5.1%) 5 10 l 2 Exact Mass: M+ = (100.), M+1 = (5.5%), M+2 = (63.9%), M+3 = (3.5%) M+4 = (10.2%), S S Br l Br S l S 5 12 S 2 Exact Mass: M+ = (100.), M+1 = (7.1%) M+2 = (9.), 5 10 Brl Exact Mass: M+ = (100.) M+1 = (5.6%), M+2 = (129.2%), M+3 = (7.1%), M+4 = (35.), M+5 = (1.7%) 5 11 BrS Exact Mass: M+ = (100.), M+1 = (6.3%), M+2 = (101.8%), M+3 = (5.6%), M+4 = (4.4%) 5 11 ls Exact Mass: M+ = (100.), M+1 = (6.3%), M+2 = (36.5%), M+3 = (2.), M+4 = (1.5%) N S F l Br I M+ M % % % % % M+2 M % 32% 97% (l) 2 (Br) 2 (Br,l) (S,l) (S,Br) M M+1 0.7% 0.7% M+2 64% 194% 129% 36% 102% M % 31% 1.4% 4.4%

3 ur approach to obtaining a molecular formula, degrees of unsaturation, pi bonds and rings will be as follows. 1. Determine if 35 l, 79 Br or 32 S are present (and how many) using the mass spec information listed above (M+2 peaks, etc.). 2. A proton count can be determined by using the proton integration from the NM. This will be given until NM is covered. 3. A carbon count can be estimated by dividing the M+1 mass spec peak by 1.1, since the 13 isotope is 1.1% of the 12 isotope. This can be confirmed by counting the number of carbon peaks in the proton decoupled 13 NM (taking into account any symmetry features). This will be given until 13 is covered. A minor complication is that 33 S is 0.8% of 32 S and two 15 N is about 0.8% of 14 N. Both of these possibilities can appear as an extra carbon atom in a formula unless subtracted from the M+1 percent. 4. The total masses of any 35 l, 79 Br, 32 S (use the exact masses of the lower atomic mass isotopes), plus the total mass of hydrogen ( 1 ) and carbon ( 12 ) is subtracted from the exact mass, the M+ peak. This will leave a residual mass most likely composed of 14 N and/or 16. Usually, how many of each element becomes clear by inspection, looking for some combination of 14 and 16 to equal the residual mass. 5. nce all of the atoms are determined, one uses the formula to calculate the total degrees of unsaturation, which equals the total pi bonds and rings. 6. Pi bonds are estimated from the chemical shifts in the 13 NM, looking for alkene, alkyne (possibly seen in the I), aromatic, nitrile (easily seen in the I) and = functionality. ertain types of pi bonds will not show up in the 13, such as N= and N=N pi bonds (nitro, nitroso, azido, etc.). Most of these are rare for us, but the nitro (-N 2 ) is common enough and can be seen in the I spectrum (1550 cm -1 and 1350 cm -1 ). The number of pi bonds is subtracted from the total degrees of unsaturation to find out how many rings are present. The procedure described here is a good first step towards finding an unknown structure. Two examples are provided below. aving an I and/or NM data will give many additional clues to help decide what a structure is. We ll look at those techniques after the formula examples below.

4 Example 1 finding a molecular formula and the number of pi bonds and rings Mass spec data: Exact Mass: M+ = (100.), M+1 = (17.7%), M+2 = (99.6%), M+3 = (17.2%), An odd mass indicates an odd number of nitrogen atoms. Example 2 1. M+ M+2 so bromine is present (exact mass = 79). 2. Proton count from 1 NM = 20 (given for now, until a NM is provided) 3. arbon count from MS data = (17.7)/(1.1) = 16 arbon count from 13 NM is 16 (given for now, until a NM is provided) 4. Total mass of 79 Br = esidual mass = = 78 = = 4 x + 1 x N Molecular formula = BrN 4 maximum single bonding positions = 2(16) = 35 degrees of unsaturation = [(35) - (21)] / 2 = 14 / 2 = 7 degrees NM shows 6 x (=) in alkene/aromatic region = 3 x (=) and 1 x (=) for a total of 4 pi bonds. owever, there are two absorptions in the I at 1550 and 1350 cm -1, indicat that a nitro group might be present. That brings the total pi bonds to 5. The number of rings = (total unsaturation) - (pi bonds) = 7-5 = 2 rings Mass spec data: Exact Mass: M+ = (100.), M+1 = (22.9%), M+2 = (36.5%), M+3 = (8.3%), M+4 = (1.5%), An even mass indicates no nitrogen atoms or an even number of nitrogen atoms. 1. M+2 = (M+) x (36.5%) so l (32%) and S (4.5%) are likely present (exact mass = = 67). Also M+4 = 1.5%, supports this conclusion. This also adds 0.8% to M+1 peak. 2. Proton count from 1 NM = 27 (given for now, until an NM is provided) 3. arbon count from MS data = ( ) / (1.1) = 20 (subtract 0.8 for each sulfur atom present) arbon count from 13 NM is 20 (given for now, until a NM is provided) 4. Total mass of 35 l + 32 S = esidual mass = = 80 = 5 x 16 = 5 x Molecular formula = l 5 S maximum single bonding positions = 2(20) + 2 = 42 degrees of unsaturation = [(42) - (28)] / 2 = 14 / 2 = 7 degrees NM shows 6 x (=) in alkene/aromatic region = 3 x (=) and 2 x (=) for a total of 5 pi bonds. The number of rings = (total unsaturation) - (pi bonds) = 7-5 = 2 rings

5 ommon organic functional groups and I peaks carboxylic acids ' l anhydrides esters acid chlorides N 2 amides (1 o,2 o,3 o ) nitriles N S N ' aldehydes ketones amines alcohols thiols (1 o,2 o,3 o ) ethers S sulfides N nitro compounds aromatics alkynes alkenes Br halo compounds alkanes I - typical functional group absorptions (full range, mainly stretching bands) units = cm sp 3 - sp - alkyne sp 2 thiol S- - N aldehyde - nitrile alcohol - carboxylic acid - 1 o N- 2 2 o N- acyl - phenol - alkene sp 2 - bend mono trans geminal tri cis alkoxy - aromatic sp 2 - bend = =N = alkene = aromatic N- bend nitro sp 3 - bend nitro mono ortho meta para

6 aldehydes - stretch (two weak bands) = stretch , saturated , conjugated ketones stretch , when saturated , when conjugated , when in rings esters - rule of 3 amides N 1 o 2 o N N acyl acids N , always conjugated with nitrogen lone pair, higher in small rings (4 atom = 1745) 3350 & 3180, 2 bands for 1 o amides, 3300, one band for 2 o amides, stronger than in amines, sometimes an extra overtone 3200, no bands for 3 o amides , N- bend, strong in amides with polar resonance form acid chlorides stretch (very high) , when saturated l , when conjugated -l not real useful when many peaks are present anhydrides acyl acyl - I Flowchart to Determine Functional Groups in a ompound (all values in cm -1 ) stretch , saturated , conjugated , in rings alkoxy stretch , saturated , conjugated acid very broad , has = band ( cm -1 ) very strong strong nitriles I Spectrum (all values are approximate, ) N stretch simple conjugated has a triple bond 2150 not present or weak when symmetrical sp - stretch 3300 bend 620 All I values are approximate and have a range of possibilities depending on the molecular environment in which the f unctional group resides. esonance often odifies a peak's position because of electron delocalization (= lower, acyl = higher, etc.). I peaks are not 10 reliable. Peaks tend to be stronger (more intense) when there is a large dipole associated with a vibration in the functional group and weaker in less polar bonds (to the point of disappearing in some completely symmetrical bonds). Alkene sp 2 - bending patterns monosubstituted alkene ( , ) trans disubstituted ( ) cis disubstituted ( ) gem disubstituted ( ) trisubstituted ( ) tetrasubstituted (none, no sp 2 -) Aromatic sp 2 - bending patterns monosubstituted ( , ) ortho disubstituted ( ) meta disubstituted ( , sometimes , ) para disubstituted ( ) There are also weak overtone bands between 1660 and 2000, but are not shown here. You can consult pictures of typical patterns in other reference books. If there is a strong = band, they may be partially covered up. does not have = band alkanes not useful alkenes not present or weak when symmetrical sp 3 - stretch bend sp 2 - stretch bend (see table) aromatics stretch bend (see table) stretch overtone patterns (weak) between alcohols - stretch - stretch , phenyl > 3 o > 2 o > 1 o sharp when no bonding , broad with bonding thiols S stretch 2550, weak S- stretch amines stretch 1 o 3500 & 3300, two bands for 1 o amines, N one band for 2 o amines, weaker than in amides, sometimes an extra overtone N 2 o about 3100 due to N- bend, no N- bands for 3 o amines N , N- bend, weaker in amines 800 N- bend ethers - stretch aliphatic 1120 (variable) aromatic 1250 nitro compounds (does not show in or NM, look for nitro in the I) N X = halogen X , asymmetric stretch, strong , symmetric stretch, medium variable bands, usually mixed in among other bands, not real useful(sp 3 = range) (sp 2 = range)

7 eal I Spectrum Simulated I Spectrum I

8 What does NM tell us? 1 -NM - Provides information on: 1. The types of protons present ( = chemical shift and is given in parts per million, ppm, of the energy to flip a nuclear spin in a magnetic field, the usual range is = 0-12 ppm). 2. The number of such protons (integration counts the relative numbers of hydrogen atoms as a whole number ratio by summing the area under the peaks). 3. ow many neighbor protons are immediately adjacent to a specific center. a. splitting patterns = multiplicity = number of peaks (singlet = s, doublet = d, triplet = t, quartet = q,...etc.) b. J values = coupling constants (distance that peaks in a multiplet are separated in frequency units, given in z = cycles per second). These provide information about neighboring nuclei (protons and carbons). 13 -NM - Provides information on: 1. The types of carbon atoms present ( = chemical shift is given in parts per million, the usual range = ppm). arbons are more dispersed than protons (have a wider range of chemical shifts, so there is less overlap of peaks). 2. The number of distinct kinds of carbon atoms present equals the number of peaks in a proton decoupled 13 spectra. All carbon peaks appear as singlets when decoupled from the protons. 3. eveals how many protons are on each carbon. a. DEPT experiment (Distortionless Enhancement by Polarization Transfer) is a series of three 13 experiments with different mixing of proton coupling to display, 2 or 3 s as distinct patterns. arbons without hydrogen do not show up and are determined by comparison with a normal proton decoupled 13 spectrum. b. ff resonance experiment reveals coupling between protons and carbons which shows up in the multiplicity of a 13 peak as follows: a singlet (s) = (quaternary carbon), doublet (d) = (methine carbon), triplet (t) = 2 (methylene carbon) or quartet (q) = 3 (methyl carbon). This is an older method that is seldom used anymore. 2D Methods of NM that we will consider 1. SY: Proton-proton correlation spectroscopy provides proton connectivity patterns using proton spin systems based on coupling between interacting protons. 2. ET or SQ: Indicates what protons are on what carbon atoms via direct one bond coupling ( 1 J ). 3. MB: Indicates what protons are two or three bonds away from a carbon atom ( 2 J, 3 J ). It is especially helpful for connecting spin systems through quaternary carbon centers and heteroatoms.

9 Spin energies split in the presence of an external magnetic field.

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11 deshielding region, adds to B o Shielding region of the small magnetic field generated by the sigma bond electrons in opposition to B o. NM spectrum shielding side deshielding side (ppm) = chemical shift deshielding region, adds to B o B shielding region (subtracts from B o ) B o =external magnetic field B net =B o -B E=h =(h/2 )( i )(B o )(1- ) The total field, B o, is reduced by a small amount, B = shielding constant, includes a variety of factors, ( Eand are reduced too).

12 1 shiftsona300mznm These protons are least shielded because of the These protons are most shielded inductive pull of fluorine (deshielded) because of the inductive push of silicon. F 3 Si max =0-3000z hugely exaggerated 2 3 l 3 Br 3 I 3 E B 2 E B 2 B B B B E to flip a proton's spin,here = 300,000,000 z ef = TMS =0 (by definition) E=h 6 E=h 5 E=h 4 E=h 3 E=h 2 E=h TMS E 1 E 1 E 1 E 1 E 1 E 1 E 1 = 4.27 ppm 3.06 ppm 2.69 ppm 2.16 ppm 0.23 ppm 0.0 ppm z (600 Mz) 2562 z 1836 z 614 z 1296 z 138 z 0 z z (300 Mz) 1281 z 918 z 807 z 648 z 69 z 0 z z (60 Mz) 256 z 184 z 161 z 130 z 14 z 0 z more shielded larger frequency shift from the TMS reference Similar information for 13 shiftsona300mznm( 13 = 75 Mz) This carbon is least shielded. max = 0-17,000 z hugely exaggerated 2 F 2 l 2 Br 2 I 2 E to flip a carbon's spin,here = 75,000,000 z B ef = B B 3 TMS =0 Si( 3 ) 3 E=h 6 E=h 5 E=h 4 E=h 3 E=h 2 E=h TMS B B This carbon is most shielded. B (by definition) = 86 ppm 47 ppm 36 ppm 9 ppm 14 ppm 0.0 ppm z (75 Mz) 6450 z 3525 z 3375 z 675 z 1050 z 0 z

13 Pi bond anisotropy: certain orientations of the molecule (relative to B o ) add to the external magnetic field and others cancel with a net contribution of what is shown below in the following figures. shielding cone allylic protons shielding cone 2 benzylic protons deshielding region, higher values 2 deshielding region, higher values deshielding region, higher values B shield shielding cone B shield shielding cone typical values = B o alkene protons = 4-7 ppm typical values = B o aromatic protons = 6-9 ppm Protons to the side of an alkene pi bond are deshielded and shifted to a larger chemical shift =. Allylic protons are shifted in a similar direction but by a smaller amount because they are farther away from the pi bond. ing current in aromatics has a larger effect than a single pi bond in a typical alkene. Aromatic protons usually have a larger chemical shift than alkene protons. Benzylic protons are shifted in a similar direction but by a smaller amount because they are farther away from the pi bonds. B pi = induced magnetic field due to pi bond electrons a. B shield opposes B o in shielding cone; B o is effectively made smaller and a smaller is the result b. B deshielded adds to B o in the deshielding region; B o is effectively made larger and a larger is the result. = -1.8 ppm, highly shielded region inside of the ring = 8.9 ppm, highly deshielded region outside of the ring [18]-annulene (aromatic)

14 Typical 1 and 13 NM chemical shift values deshielding side shielding side less electron rich more electron rich (inductive & resonance) (inductive & resonance) Typical proton chemical shifts amine N- arbon and/or heteroatoms without hydrogen do not appear here, but the influence on any nearby protons alcohol 2 1 may be seen in the chemical shifts of the protons. 5 amide N S thiols, sulfides N amines carboxylic acid aldehyde aromatic - Typical carbon-13 chemical shifts ketones no aldehydes with X carboxylic acids anhydrides esters amides acid chlorides no PPM N no X X = F > l Br I allylic alkene - benzylic - thiol carbonyl alpha - S epoxide alcohols ethers esters halogen 3 3 with & without alcohols, ethers, esters with & without 2.5 simple sp 3 - > 2 > N 2 F l Br I amines, amides with & without epoxides with & without with & without S 80 + simple sp 50 3 carbon > > 2 > 3 with & without with & without thiols, sulfides with & without PPM

15 Neighbor protons cause splitting of peaks (multiple peaks) Splitting patterns in NM (up or down) is like flipping coins (heads or tails) Two Neighbors Example 1. Protons with two nearest neighbor protons split into four populations (like two coin flips). The N+1 rule works when J values are equivalent (J 1a = J 1b ). 1 two neighbor protons a increasing, N + 1 rule (N = # neighbors) The observed proton type ( 3, 2, ), 1, sees every neighbor proton as 5 up and 5 down, like flipping a coin. 1 B o b Two neighbor protons are like two small magnets that can be arraonged four possible ways (similar to flipping a coin twice). E to flip proton E 1 J (z) Two equal energy populations here. (ppm) J (z) # peaks = N + 1 = = 3 peaks Three peaks in a 1:2:1 ratio are called a triplet = t. The ratio of these four populations is 1:2:1. If all J values are equal, the middle peaks will fall on top of one another and the N+1 rule works. J is the tiny energy differnce between these states in z. For common vicinal coupling 3 J 7 z. Two nearest neighbor protons really produces a doublet of doublets = dd. (when J 1a = J 1b = triplet) 1 a two neighbor protons splitting tree 1 The N+1 rule works when J 1a J 1b. # peaks = N + 1 = = 3 peaks The observed proton 1, b J 1b J 1a J 1b The ratio of these four populations is about 1:2:1. J (z) J (z) B o (ppm) When the coupling constants are different (J 1a J 1b ), all four peaks will be observed as two sets of doublets called a doublet of doublets (dd), which are approximately equivalent in size. splitting tree J 1a - measure peak 1 to peak 3 J 1b - measure peak 1 to peak 2 doublet of doublets = dd 1 When the J values are not equal all four peaks are observed. The ratio of these four populations 1 J 1a is 1:1:1:1. Draw splitting trees 1 with the largest coupling at the top and the smallest coupling at the bottom. J 1b J 1b J 1a J 1b (ppm)

16 Examples where the N+1 rule works. I = integration (number of protons) = group without any coupled proton(s) = observed proton N = number of nearest neighbors N = 0 N = 1 N = 2 N = s, J=none I=1 N=0 d, J=7 I=1 N=1 t, J=7 I=1 N=2 2 q, J=7 I=1 N=3 singlet doublet triplet quartet = calc or exp N = 4 = calc or exp = calc or exp = calc or exp N = qnt, J=7 I=1 N=4 = calc or exp quintet sex, J=7 I=1 N=5 = calc or exp sextet N = 6 N = sep, J=7 I=1 N=6 2 septet 2 3 oct, J=7 I=1 N=7 octet 2 ombinations of these are possible. dd = doublet of doublets; ddd = doublet of doublet of doublets; dddd = doublet of doublet of doublet of doublets; dt = doublet of triplets, td = triplet of doublets; etc. Pascal's triangle = coefficients of variable terms in binomial expansion (x + y) n, n = integer Multiplets when the N + 1 rule works (all J values are equal). s = singlet d = doublet t = triplet q = quartet qnt = quintet sex = sextet sep = septet oct = octet = calc or exp peak = 10 1 peak = 5 1 peak = 25% 1 peak = 12% 1 peak = 6% 1 1 peak = 3% peak = 1.5% peak = 0.8% T T T = spin up T = spin down (x + y) n, n = integer (x + y) 0 = 1 (x + y) 1 = 1x + 1 y (x + y) 2 = 1x 2 + 2xy + 1 y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy y 3 etc. elative sizes of the peaks in multiplets (% edge peak shown). As the multiplets get larger, it gets harder to see the edge peaks so you have to be careful interpreting large multiplets. = calc or exp T T T T T T T T T T T T

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20 The N+1 rule works for - bonds when fully or partially coupled, but there is a more modern way to see the same information (called DEPT = distortionless enhancement by polarization transfer). 0 proton neighbors carbon peak = singlet (quaternary carbons) 1 proton neighbor carbon peak = doublet (methine carbons) 2 proton neighbors carbon peak = triplet (methylene carbons) 3 proton neighbors carbon peak = quartet (methyl carbons) singlet = s doublet = d triplet = t quartet = q No J coupling N X X =, N, l, S 1 J small when off resonance decoupled 1 J small when off resonance decoupled J small when off resonance decoupled 3 DEPT = distortionless enhancement by polarization transfer p-butylbenzaldehyde diethyl acetal DEPT-135 shows and 3 carbon atoms up and 2 carbon atoms down ppm DEPT-90 shows carbon atoms up ppm 13 NM shows a peak for each type of carbon solvent ppm ppm 0 1 = = = = = = = = = = =

21 2D NM experiements work via coupling between the atoms. Three examples for us: SY (/), ET/SQ ( 1 J ) MB ( 2 J and 3 J ). SY - uses / couplings geminal coupling when protons are diastereotopic. a b 2 J ab 12 z vicinal coupling when protons are neighbors. a b 3 J ab 7 z diagonal peaks correlates each proton with itself cross peaks correlates protons with any coupled partners SY = correlation spectroscopy (proton-proton) cross peak diagonal peak 1 NM ET (SQ) - uses / couplings ne bond / coupling constants are approximately 140 z. 1 NM J 140 z correlates directly attached protons and carbon atoms NM MB - uses / couplings Two and three bond / coupling constants are approximately 10 z. 13 ET = heteronuclear correlation spectroscopy SQ - heteronuclear single quantum correlations 13 X 2 J, 3 J 7-10 z MB - heteronuclear multiple bond correlations X = nitrogen, oxygen or sulfur atom 1 NM quaternary carbons X 4 o center MB looks for correlations that answer questions about which way spin systems are connected to the quaternary centers and across heteroatoms. There can be many possibilities.

22 Data worksheet DEPT 13 structural unit ET SY MB,NE,other 1 fragments from 1 NM