Lecture 5 Still More nmr

Transcription

1 Lecture 5 Still More nmr three spins February 5, 2019

2 Supplemental Problem Chlorocyclobutane

3 Origins of Signal Splitting B 0 H b Magnetic field of H b subtracts from the applied field; H b signal appears at a higher applied field H a H b Magnetic field of H b adds to the applied field; H a signal appears at a lower applied field Remember it is the NET field that counts

4 The signal of Ha is split into two peaks of equal area (a doublet) J ab J ab = coupling constant

5 Origins of signal splitting no neighbors one spin two spins three spins

6 Relative Intensity of Peaks singlet double triplet quartet quintete sextete Pascal s triangle The binomial coefficients

7 The N+1 Rule The 1 H-NMR signal of a hydrogen or set of equivalent hydrogens is split into (N + 1) peaks by a set of N equivalent neighboring hydrogens All neighboring hydrogens in the analysis must have the same chemical shift (magnetically equivalent) If this condition is not met, a graphical tree or second order analysis must be used to predict the splitting pattern. We will explore this condition later

8 Splitting of Signals Nonequivalent protons split each other if they are on adjacent carbons, i.e. separated by 3 sigma bonds

9 Splitting of Signals Equivalent protons never split each other

10 Who needs DNA??

11 Corn Oil and Tristearin

12 Coloring Margarine

13 Index of Hydrogen Deficiency I knew this as the unsaturation number Valuable characteristic of structure Provides you with number of double bonds or rings in a compound Easy with CHO formulas only be careful with N Simplest, saturated normal alkanes have: H (CH 2 ) n H = C n H 2n+2

14 Index of H Deficiency IHD: the sum of the number of rings and pi bonds in a molecule. Compare the number of hydrogens in an unknown compound with the number in the reference compound C n H 2n+2

15 Index of H Deficiency IDH = (H reference - H molecule ) 2 C 6 H 12 C 5 H 10 C 9 H 14 - C n H 2n+2 C 6 H 14 C 6 H 12 - C n H 2n+2 C 9 H 20 C 9 H 14 6/2=3 2/2=1 => 1 ring or 1 double bond

16 Index of H Deficiency IDH = (H reference - H molecule ) 2 No correction is necessary for the addition of atoms of O Subtract one H for each atom of halogen added (halogens like Cl replace H one for one) Add one hydrogen For each atom of N added to the reference hydrocarbon

17 Unsaturation Number or HDI CH 3 N Add one H for N and subtract one for Cl Should be C n H 2n = C 8 H 18 but it is C 8 H 14 Cl H C 8 H 14 ClN Missing H 4 So index is. 4/2=2

18 Calculate the Unsaturation Number H C 8 H 9 N N Add one H for N H H Should be C n H 2n+2 +1 = C 8 H 19 but it is C 8 H 9 Missing H 10 So number is. 10/2=5!

19 Tricks for solving unknowns Review. Empirical formula is lowest common denominator ratio of atomic composition From Homework: unknown has an empirical formula of C 4 H 9.a single high field peak in the 1 H nmr and a molecular ion at M/Z = 114..propose a structure..and predict the M/Z of the most intense fragment you expect to see in the mass spectrum

20 C 5 H 10 O

21 Chemical Shift - 1 H-NMR Type of H (C H 3 ) 4 Si RCH 3 d Type of H ROH RCH 2 OR d RCH 2 R R 3 CH R 2 NH O R 2 C=CRC HR 2 RC CH ArC H RCCH 3 O RCCH 2 R

22 Chemical Shift - 1 H-NMR Type of H d Type of H d O RCOC H R 2 C=C H O R 2 C=C HR RCOC H 2 R ArH RCH 2 I O RCH 2 Br RCH 2 Cl RCH 2 F RCH O RCOH

23 Sample spectra 15 C H 3 CH 3 O 6 10 CH3 CH 3 5 C 7 H 14 O

24 More Practice-HW C H 14 O

25 CH 3 CH 2 CH 2 NO 2

26 Multiple interactions of non-equivalent neighbors here J ab > J bc

27 ClCH 2 CH 2 CH 2 Cl

28 Cl 2 HCCH 2 CH

29 An analysis