Organic Structure Analysis

Organic Structure Analysis
Professor Marcel Jaspars

Aim This course aims to extend student’s knowledge and experience with nuclear magnetic resonance (NMR) and mass spectrometry (MS), by building on the material taught in the 3rd Year Organic Spectroscopy course, and also to develop problem solving skills in this area.

Learning Outcomes By the end of this course you should be able to:
Assign 1H and 13C NMR spectra of organic molecules. Analyse complex first order multiplets. Elucidate the structure of organic molecules using NMR and MS data. Use data from coupling constants and NOE experiments to determine relative stereochemistry. Understand and use data from 2D NMR experiments.

Synopsis A general strategy for solving structural problems using spectroscopic methods, including dereplication methods. Determination of molecular formulae using NMR & MS Analysis of multiplet patterns to determine coupling constants. Single irradiation experiments. Spectral simulation. The Karplus equation and its use in the determination of relative stereochemistry in conformationally rigid molecules. Determination of relative stereochemistry using the nuclear Overhauser effect (nOe). Rules to determine whether a nucleus can be studied by NMR & What other factors must be taken into consideration. Multinuclear NMR-commonly studied heteronuclei. Basic 2D NMR experiments and their uses in structure determination.

Books Organic Structure Analysis, Crews, Rodriguez and Jaspars, OUP, 2010 Spectroscopic Methods in Organic Chemistry, Williams and Fleming, McGraw-Hill, 2007 Organic Structures from Spectra, Field, Sternhell and Kalman, Wiley, 2008 Spectrometric Identification of Organic Compounds, Silverstein, Webster and Kiemle, Wiley, 2007 Introduction to Spectroscopy, Pavia, Lampman, Kriz and Vyvyan, Brooks/Cole 2009

Four Types of Information from NMR
Chemical shift (electronic shielding/functional groups) Integrals (ratio of H atoms) Multiplicity/coupling (connectivity) Peak width (relaxation)

1.) Complex multiplets 2.) Overlap 3.) Confusing integrals 4.) Many peaks can be assigned to similar structural features 5.) General rules R-NH2  R-ND2 CD3OD water CD3OD No NH (CD3OD) a-H

1.) Peaks different sizes
2.) Peak overlap 3.) Some peaks not present 4.) General rules CD3OD Aromatic aliphatic C=O a-C

Spectral Dispersion The average 1H spectrum is from 0-10 ppm. At 400 MHz this is 4000 Hz, at 600 MHz this is 6000 Hz. The average 13C spectrum is from ppm. At 100 MHz (400 MHz 1H) this is Hz and at 150 MHz (600 MHz 1H) this is Hz. In this case spectral dispersion is 5 times greater for 13C than for 1H.

1H NMR Chemical Shifts in Organic Compounds

13C NMR Chemical Shifts in Organic Compounds

5 Minute Problem #1 MF = C6H12O Unsaturated acyclic ether 2H t J = 7 Hz 3H t J = 7 Hz 2H sext J = 7 Hz 1H d J = 7 Hz 2H quint J = 7 Hz 1H d J = 14 Hz 1H dd J = 14, 7 Hz

Six Simple Steps for Successful Structure Solution
Get molecular formula. Use combustion analysis, mass spectrum and/or 13C NMR spectrum. Calculate double bond equivalents (DBE). Determine functional groups from IR, 1H and 13C NMR Compare 1H integrals to number of H’s in the MF. Determine coupling constants (J’s) for all multiplets. Use information from 3. and 4. to construct spin systems (substructures) Assemble substructures in all possible ways, taking account of DBE and functional groups. Make sure the integrals and coupling patterns agree with the proposed structure.

Double Bond Equivalents
CaHbOcNdXe [(2a+2) – (b-d+e)] DBE = 2 (4+2)-3-1 C2H3O2Cl = = 1 2

The Easy Way to Calculate DBE
You are calculating the number of H2 needed to obtain a fully saturated hydrocarbon (CnH2n+2) For example – for benzene C6H6  C6H14 takes 4H2 hence DBE = 4 Each N can be replaced by a CH (three remaining valences) so for pyridine C5H5N  C6H6  C6H14 hence DBE = 4 Each halogen is replaced by an H in this method

Tabulate Data Shift (ppm) Int. Mult (J/Hz) Inference 6.48 1H dd, 14, 7
sp2 CH with 1 cis & 1 trans neighbour 4.17 d, 14 sp2 CH with 1 trans neighbour 3.97 d, 7 sp2 CH with 1 cis neighbour 3.69 2H t, 7 CH2-O next to CH2 1.65 quint, 7 CH2 next to 2 x CH2 1.42 sext, 7 CH2 next to CH2 & CH3 0.95 3H CH3 next to CH2

Solution dd d sextet t d quintet d

Shift Prediction Prediction

THE PROCESS OF STRUCTURE ELUCIDATION
Organic Structure Analysis, Crews, Rodriguez and Jaspars

Dereplication Dediscovery

Dereplication OH

Dereplication 20 hits + mass = 1 hit 3 MF 10 hits 3 hits 33 hits

Determining the Molecular Formula Using NMR and MS Data
DEPT-135 CH3 CH3 CH CH CH CH2 Organic Structure Analysis, Crews, Rodriguez and Jaspars

CH3 CH CH2 CH2 CH3 CH CH2 CH3 C C 17 = OH Total = (C)2 + (CH)2 + (CH2)3 + (CH3)3 = C10H17 = 137 154 = C10H18O

Rule of 13 (CH = 13) 154/13 = 11 remainder 11 154 is C11H22 N = CH2, O = CH4 So: Choices are: C10H20N C10H18O C9H18N2 C8H18N3

Monoisotopic masses Isotope Mass % Abundance 1H 1.008 99.99 12C 12.000 98.89 13C 13.003 1.11 14N 14.003 99.64 15N 15.000 0.36 16O 15.995 99.76 18O 17.999 0.20 32S 31.972 95.03 33S 32.971 0.76 34S 33.968 4.20 35Cl 34.969 75.77 37Cl 36.966 24.23 79Br 78.918 50.52 81Br 80.916 49.48 For many elements, there is more than one isotope, each with a different natural abundance. For the most common elements in organic molecules. MS measures monoisotopic masses and these must be used to calculate molecular masses in MS.

MS Errors Experimental accurate mass measurement (from MS) was suggesting C10H16 is the correct formula. The error between calculated and experimental mass is: = = 0.8 mmu Formula dbe Accurate mass C10H16 3 C9H12O 4 C8H8O2 5 C7H4O3 6 C9H14N 3.5 C8H12N2

Molecular Formula Calculators
James Deline MFCalc

Isotope Ratio Patterns: C100H200
For 12Cm13Cn (12C = 98.9% 13C = 1.1%) m n 1403.6 1404.6 2

ThermoFinnigan LTQ Orbitrap,
Determining molecular formulae by HR-ESI/MS ThermoFinnigan LTQ Orbitrap, Xcalibur software examples from MChem group practicals 2009 (Rainer Ebel)

[M+H]+

[M+Na]+

Analysis of isotope patterns
experimental calculated “monoisotopic peak” (mainly 12C713C1H935Cl14N16O2)

Shift Additivities and Computational Alternatives
The identity of a compound purchased is in doubt. It is either A or B. Chemical shift additivities can be used to select the correct structure quickly.

Answer 156 109 119 Base Values: 123 109 156 137 95

ChemDraw

The NMR effect

The NMR effect DE = hno

The NMR effect Larmor Equation no = gBo/2p g = gyromagnetic ratio
Bo = magnetic field strength no = resonance frequency DE = hno

Origins of NMR

Spin-spin coupling (splitting)
No coupling Jab Jab Jab Coupling da db

Spin-spin coupling (splitting)
The distance between the members of each multiplet in Hz is the coupling constant J, and can be calculated from the spectrum: J (Hz) = Dd (ppm) x spectrometer frequency in MHz Coupling constants do not change with magnetic field. The proton proton interaction is transmitted through the intervening (s) electrons making up the chemical bonds, so the magnitude of J is an indication of the number and type of bonds. %s 1JCH sp3 25 ~125 sp2 33 ~175 sp 50 ~225

Coupling Invariance

Origin of spin-spin coupling
The number of spin-spin splitting peaks of a particular proton can be calculated from: The n+1 rule: The number of neighbouring protons (n) nonequivalent to the proton in question plus 1. This is a specific case of the 2nI + 1 rule where I = ½ for 1H

Coupling in ethanol deshielded shielded 3 nearest non-equivalent
neighbours. n + 1 = 4 (quartet) 2 nearest non-equivalent neighbours. n + 1 = 3 (triplet)

Coupling is mutual

Coupling is mutual Coupling between neighbours
is mutual – the coupling constant from Ha to Hb is the same as from Hb to Ha (Jab = Jba)

Coupling in ethanol To see why the methyl peak is split into a triplet, let’s look at the methylene protons (CH2). There are two of them, and each can have one of two possible orientations- aligned with, or against the applied field This gives rise to a total of four possible states: E Hence the methyl peak is split into three, with the ratio of areas 1 : 2 : 1

Coupling in ethanol Similarly, the effect of the methyl protons on the methylene protons is such that there are 8 possible spin combinations for the three methyl protons: E The methylene peak is split into a quartet. The areas of the peaks have the ratio of 1:3:3:1.

Pascal’s triangle n relative intensity multiplet 0 1 singlet
doublet triplet quartet quintet sextet septet

Coupling patterns

First Order In CH3CH2OH we could explain coupling by the n + 1 rule, this is called 1st order coupling 1.1 ppm at 500 MHz = 550 Hz 3.6 ppm at 500 MHz = 1800 Hz J = 7 Hz Dn/J = ( )/7 = >>> 6

Non First Order Like CH3CH2OH expect 7 lines but get many more. Dn/J < 6

Common Coupled Spin Systems

Complex 1st Order Spin Systems

Iterative application of the n + 1 rule

Deriving complex 1st order systems
dA (ddd, J = 12, 4, 2 Hz) dB (dt, J = 10, 5 Hz) 10 Hz 12 Hz 5 Hz 4 Hz 5 Hz 2 Hz

5 Minute Problem #2. Given the 1H NMR chemical shifts and coupling constants for allyl alcohol, explain the observed spectrum (OH peak omitted)

A doubled quartet (dq) 5.15 ppm 17 2 2 2 2 2 2 2 2 2 2 2 2

What about this? δ 5.86 (ddt, J = 17.2, 10.4, 5.2) 5 5 5 5 10 17

Spin Simulation Real Spectrum

Pro-R and Pro-S R S 3 4 pro-R 4 3 pro-S 2 1 2 1
In a chiral environment (eg enzyme, chiral solvent), the two hydrogens are no longer equivalent and the symmetry between them is destroyed.

Homotopic, Enantiotopic, Diastereotopic
Ha = Hb da = db Jab Jab da db Jab about 15 Hz

Methyl groups

Chemical Equivalence/Magnetic Non Equivalence
B B’ A’

What is going on? JAB ≠ JAB’ JA’B’ ≠ JA’B A B A A B’ A A A’ But:
From the viewpoint of A The two B protons are non-equivalent A and A’ are chemically equivalent They cannot be distinguished by chemical reaction They have the same chemical shift But A and A’ are magnetically non-equivalent The same is true for B and B’ JAB = JA’B’ and JAB’ = JA’B

Result Expect: Quantum mechanics gives: NO J Values can be obtained!

Using Coupling Constants
Jab > Jac The size of J depends on the dihedral angle 60o = 4 Hz 180o = 11 Hz

Glucose H1, d, 3.5 Hz H4, t, 10 Hz H5 & H6’ H3, t, 9.7 Hz
H6, dd, 12.5, 4.0 Hz H2, dd, 10.3, 3.5 Hz

Glucose 3.5 10 9.7 10.3

5 Minute Problem #3 Work out which of d 2.1 and d 2.5 is equatorial and which is axial. Also work out the 3 dihedral angles for d 2.1, d 2.5, d 2.8, d 6.8. There are also peaks at: 6.80, 1H, d, J = 0.5 Hz; 1.95, 3H, s; 0.93, 9H, s. dd, 15, 9.5 dd, 15, 1.8 ddd, 9.5, 1.8, 0.5 30 Hz

Solution 9.5Hz 9.5 Hz = 28o or 152o 152o 1.8 Hz = 59-72o or 110-118o

Removing Couplings Changing Solvents
Jac Jab db≠ dc Each coupled to Ha Jab ≠ Jac CDCl3 Ha, dd Jab ≠ Jac db dc da C6D6 db = dc Coupled to Ha Jab = Jac Ha, t Jab = Jac

Removing Couplings Spin decoupling
Coupling due to Ha is removed See Hb, Hc at db, dc With mutual Jbc CDCl3 irradiated Signal due to Ha disappears db dc da CDCl3 Irradiate at 4.11 ppm

Spin Decoupling Two spins, A (nA), X (nX) with JAX
Irradiate nX with RF power, A loses coupling due to X OFF JAX JAX nX nA A↓ X ↓ Bo A↓ X↑ A↑ X↓ A↑ X↑

Spin Decoupling Two spins, A (nA), X (nX) with JAX
Irradiate nX with RF power, A loses coupling due to X Coupling lost ON nX nA A↓ X ↑ ↓ A↑ X↑ ↓ Average of X ↑ and X ↓

Nuclear Overhauser Effect
Two protons that are not J-coupled but which are in close proximity (<5 Å) can mutually relax each other. If one nucleus is irradiated then increasing the intensity of the peaks observed. Classify NOE qualitatively Strong 1.8 – 2.5 Å Medium 2.5 – 3.5 Å Weak 3.5 – 5.0 Å Two ways to obtain NOE spectra Difference spectrum (subtract normal spectrum from spectrum with irradiated nucleus) which only shows peaks which have increased intensity. Limited sensitivity (ca 5% NOE) Selective irradiation of nucleus shows peaks with NOEs. Very sensitive (~1% NOE).

Size of NOE NOE  1/r6 Dependence of the 1H-1H nuclear Overhauser enhancement (h) on the product of the Larmor frequency (w0) and rotational correlation time (tc). Small Molecules Large Molecules

Effect of NOE on 13C NMR Irradiating the 1H increases the
intensity of the 13C peaks in the order CH3 > CH2 > CH > C Broadband irradiation 10/5/9 3 8 2 6 4 1 7

13C – 1H NOE at equilibrium (small molecule)
H transitions are 4x C transition intensity This reflects the relative gyromagnetic ratios gH = 4gC C ● C↑H ↓ H H ●●●● C↓H ↑ C ●●●●● C↑H ↑

13C – 1H NOE irradiation on H
●● C↓H↓ Equalise H populations Saturate H transitions C transitions still OK C ●●● C↑H ↓ Hsat Hsat ●● C↓H ↑ C ●●● C↑H ↑

13C – 1H NOE irradiation on H left ON
●● C↓H↓ C Dipole-dipole relaxation ●●● C↑H ↓ Hsat Natural C relaxation Tries to restore Boltzmann distribution Hsat ●● C↓H ↑ C ●●● C↑H ↑

13C – 1H NOE result ● H transitions still saturated C↓H↓
3x increase in C transitions 3x improvement in signal-to-noise Maximum enhancement =0.5(gI/gS) I is saturated spin S is observed spin So for C and H Max = 0.5x(4gC/gC) = 2 (enhancement so 3x increase) C↓H↓ C ●●●● C↑H ↓ Hsat Hsat ● C ●●●● C↑H ↑

1H-1H NOE example NOE opposite phase to irradiated peak H1 H2O H4 H3
Not real NOE (antiphase) irradiate

1H – 1H NOE at equilibrium (small molecule)
S↓I↓ Two 1H spins I, S Relaxing each other via dipole-dipole mechanism but not J-coupled W1S W1I ●● ●● S↑I ↓ S↓I ↑ W1I W1S ●●●● S↑I ↑ nI nS Peaks same intensity Both W1S and W1I are ●●

1H – 1H NOE irradiation on S
● S↓I↓ Saturates W1S transition Equalises populations as shown W1S (sat) W1I ●●● ● S↑I ↓ S↓I ↑ W1I W1S (sat) ●●● S↑I ↑ nI nS W1S saturated W1I still ●●

1H – 1H NOE irradiation on S left ON
● Zero quantum transition W0 decreases I intensity Double quantum transition W2 Increases I intensity Both try to re-establish Boltzmann distribution In small molecules W2 dominates S↓I↓ W1S (sat) W1I W2 ●●● ● S↑I ↓ W0 W1I W1S (sat) ●●● S↑I ↑

1H – 1H NOE result ½ S↓I↓ W1S (sat) W1I ●●●½ ½ S↑I ↓ W1I W1S (sat)
nI nS W1S saturated W1I enhanced ●●●

NOE 3D example

NOE 3D example 32% 13% (6%) 11% 13%

K A B D

K I E C A B D

K I E C A F B D G

Events Accompanying Resonance

ONE-PULSE SEQUENCE (90o)x 1H Preparation Detection A short RF pulse
becomes broadband irradiation under Fourier transform Organic Structure Analysis, Crews, Rodriguez and Jaspars

ONE-PULSE SEQUENCE Organic Structure Analysis, Crews, Rodriguez and Jaspars

Fourier Transformation
The 90o pulse simultaneously excites all 1H frequencies. Result is data in time domain – Free Induction Decay (FID). After Fourier transformation (FT) the data is in the frequency domain. Main advantage is improved signal to noise (S/N) by acquisition of large number of FIDs and adding them together. S/N  √n where n is the number of FIDs added. FT Intensity envelope  e-t/T2 Transformed FID gives Spectrum in frequency domain FID in time domain – sum of 16 acquisitions

Relaxation and Peak Shape
T1 = longitudinal (spin-lattice) relaxation time T2 = transverse (spin-spin) relaxation time For small molecules T1  T2 The faster a nucleus relaxes (small T) the broader the peak. Normal 1H peak is Hz wide.

Rotational Correlation Time tc
wo = 2pno Small Molecules wotc << 1 Fast tumbling limit ↓ Slow Relaxation Narrow Lines Large Molecules wotc  1 Slow tumbling limit ↓ Fast Relaxation Broad Lines Small Large Molecule Molecule

Nuclear spin Example Atomic mass Atomic number Spin, I Odd Odd or Even
13C, 1H, 17O, 15N, 3H Odd Odd or Even 1/2, 3/2, 5/2 etc 12C, 16O Even 2H, 14N 1, 2, 3 etc 6 1 8 7 1 6 8 1 7

Receptivity Gyromagnetic ratio g = m/P m = magnetic moment P = spin angular momentum Larger g = easier observation Resonance frequency = no = gBo/2p Receptivity = |g3C| where C = natural isotopic abundance Nucleus C Relative g Receptivity Relative receptivity 29Si 4.7% -5.32 707 3.7 x 10-4 13C 1.1% 6.73 335 1.8 x 10-4 1H 100% 26.75 1.9 x 106 1

Multinuclear NMR

15N NMR Shifts

31P NMR Shifts

Coupling

Effect of 31P on 1H NMR 1JPH = 692 Hz

Effect of 31P on 1H NMR x x o o o x 3JHH = 7 Hz x o 3JPH = 9.1 Hz

Effect of 31P on 13C NMR 5 4 3 1JPC(3) = 126.5 Hz 3JPC(5) = 5.5 Hz 1
+ d at 210 ppm (2JPC(2) = 6.1 Hz)

The 2nd Dimension For complex molecules a great deal of overlap occurs in the 1H, and sometimes 13C NMR spectra. Many 1H-1H coupling constants will be the same or similar making it difficult to determine connectivity. Need alternative approach that gives connectivities without analysing coupling constants. Can do this by introducing a second NMR dimension correlating 1H with 1H or 1H with 13C.

BASIC LAYOUT OF A 2D NMR EXPERIMENT
Preparation Evolution t1 Mixing t Detection t2 Added evolution and mixing period t1 is increased by a small amount between each set of FIDs t is the same throughout the experiment Organic Structure Analysis, Crews, Rodriguez and Jaspars

How a 2D NMR experiment works
Contour plot n is the number of increments Organic Structure Analysis, Crews, Rodriguez and Jaspars

TYPES OF 2D NMR EXPERIMENTS
AUTOCORRELATED Homonuclear J resolved 1H-1H COSY TOCSY NOESY ROESY INADEQUATE CROSS-CORRELATED Heteronuclear J resolved 1H-13C COSY HMQC HSQC HMBC HSQC-TOCSY Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY
& & HMBC data is ambiguous (2 or 3 bond correlations – impossible to tell which) Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY

STRATEGY BASED ON C-H CONNECTIVITY – DEPT DATA
Label from left to right in capitals A B C D E F Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
A B C D E F dC f e d’ diastereotopic protons d c b a label protons in lower case letters Organic Structure Analysis, Crews, Rodriguez and Jaspars

ATOM dC (ppm) DEPT dH (ppm) A 131 CH 5.5 B 124 5.2 C 68 4.0 D 42 CH2 3.0 2.5 E 23 CH3 1.5 F 17 1.2 Organic Structure Analysis, Crews, Rodriguez and Jaspars

diastereotopic protons Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
b-f c-e a-d/d’ a-b And many more… Organic Structure Analysis, Crews, Rodriguez and Jaspars

ATOM dC (ppm) DEPT dH (ppm) COSY (HH) A 131 CH 5.5 b, c, d/d’, f B 124 5.2 a, d/d’, f C 68 4.0 a, d/d’, e D 42 CH2 3.0 2.5 a, b, c, d, e E 23 CH3 1.5 c, d/d’ F 17 1.2 a, b Organic Structure Analysis, Crews, Rodriguez and Jaspars

diagonal Organic Structure Analysis, Crews, Rodriguez and Jaspars

HSQC suggests diastereotopic protons: 3.08/2.44 ppm 1.86/2.07 ppm ? OK Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
D-c A-b B-a C-a And many more… Organic Structure Analysis, Crews, Rodriguez and Jaspars

ATOM dC (ppm) DEPT dH (ppm) COSY (HH) HMBC (CH) A 131 CH 5.5 b, c, d/d’, f b, c, d, f B 124 5.2 a, d/d’, f a, d, f C 68 4.0 a, d/d’, e a, d, e D 42 CH2 3.0 2.5 a, b, c, d, e a, b, c, e E 23 CH3 1.5 c, d/d’ c, d F 17 1.2 a, b Organic Structure Analysis, Crews, Rodriguez and Jaspars

Combinatorial explosion
STRATEGY BASED ON C-H CONNECTIVITY RETROSPECTIVE CHECKING Combinatorial explosion Pieces: Possibilities: Organic Structure Analysis, Crews, Rodriguez and Jaspars

STRATEGY BASED ON C-H CONNECTIVITY RETROSPECTIVE CHECKING
HMBC data (C-C-H & C-C-C-H) from C  H And similarly for COSY data Organic Structure Analysis, Crews, Rodriguez and Jaspars

PROSPECTIVE CHECKING Pieces:

2D EXERCISE 1. For a simple organic compound the mass spectrum shows a
molecular ion at m/z 98. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given. Atom dC (ppm) dH (ppm) 1H - 1H COSY (3 bond only) 1H  13C Long range (2 - 3 bonds) A 218 C - A-b, A-c, A-d, A-e B 47 CH2 1.8 dd b-d B-c, B-d, B-e, B-f C 38 CH2 2.3 m c-e C-b, C-d, C-e D 32 CH 1.5 m d-b, d-e, d-f D-b, D-c, D-e, D-f E 31 CH2 2.2 m e-c, e-d E-b, E-c, E-d, E-f F 20 CH3 1.1 d f-d F-b, F-d, F-e Organic Structure Analysis, Crews, Rodriguez and Jaspars

An additional peak is present in the 1H NMR at 11.6 ppm (bs). Atom
2D EXERCISE 2. For a simple organic compound the mass spectrum shows a molecular ion at m/z 114. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given. An additional peak is present in the 1H NMR at 11.6 ppm (bs). Atom dC (ppm) dH (ppm) 1H - 1H COSY (3 bond only) 1H  13C Long range (2 - 3 bonds) A 178 C - A-d, A-b B 136 CH 5.7 m b-c, b-d B-d, B-c, B-e C 118 CH 5.5 m c-b, c-e C-b, C-d, C-e, C-f D CH2 3.0 d d-b D-b, D-c E CH2 2.1 m e-c, e-f E-b, E-c, E-f F 13 CH3 1.0 t f-e F-c, F-e Organic Structure Analysis, Crews, Rodriguez and Jaspars

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