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Carbon-13 NMR

Only 1.1% of all carbon atoms are carbon-13 isotope, the only isotope of carbon that displays the ability to create the magnetic field when its nucleus spins. Taking a carbon NMR of 1-pentanol using the same number of scans as a proton NMR:

Which are real and which peaks are “noise”? Now, taken with multiple extra scans (can sometimes take HOURS):

Out of 100 carbon atoms only around 1 is a carbon-13 isotope. So – how is it possible that we can study carbon-13 NMR? Consider a solution of 1-butanol molecules:

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CH3-CH2-CH2-CH2-OHCH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH CH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH CH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH

CH3-CH2-CH2-CH2-OH

When averaged over an entire solution, probability and statistics will show that every carbon position will have a carbon-13 in some molecule. This generally requires more scans (more time) and a larger sample (more molecules!) than a proton NMR, in order to get enough data acquired to identify true peaks away from “noise” on the spectrum. Calibration Peak at 0.0000 ppm – all four carbons of tetramethylsilane are equivalent and given a value of 0.0000 ppm.

• Can also calibrate on the triple set of peaks caused by CHCl3, the center peak of which appears at 77.0000 ppm.

Symmetry: Every signal on the carbon NMR represents a different carbon atom in the molecule. Symmetrical carbons produce twice the data points during acquisition of a spectrum so those peaks generally appear larger as a result. How many different types of carbons? CH3CH2CH2CH3

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0510152025PPM

(CH3)3CH

0510152025PPM

(CH3)2CHCH2Br

051015202530354045PPM

Aromatic Compounds – Symmetry: Monosubstituted:

Br

4

020406080100120140PPM

Disubstituted: Para:

Br

Br

Br

Cl 1-hydroxy-4-methoxybenzene:

020406080100120140160PPM

Meta:

Br Br Br Cl

1-bromo-3-methylbenzene:

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020406080100120140PPM

Ortho:

Br Cl

Br Br 1-bromo-2-methylbenzene

020406080100120140PPM

Same sort of analysis with trisubstituted aromatic rings: 1,2,3-substituted:

Br ClBr

ClBr

Br

Which one is it?

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020406080100120140PPM

1,2,4-substituted:

CH3

Cl

OH

020406080100120140160PPM

1,3,5-substituted:

ClBr

CH3

BrBr

Cl

BrBr

Br Which one is it?

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020406080100120140PPM

and this time?

020406080100120140160PPM

Chemical Shift – X-axis – 0-220 ppm 0-50 ppm alkyl carbons also benzylic C’s 45-90 ppm C-O (N, X) 110-160 ppm aromatic carbons (also alkene) 115-130 ppm nitrile carbons 170-220 ppm carbonyl carbons Simple Problem: Consider each of the following and analyze for what you EXPECT to find on the Carbon NMR?

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Now: Match the Carbon NMRS a.

020406080100120140PPM

b.

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Integration: Peak areas are not typically measured on carbon-13 NMR spectra so integration is not performed. Instead, one can view the general peaks heights to determine where on the spectrum signals containing symmetrical carbons are found. 2-Butanone has four carbons, all of which are different. Peak heights make no difference.

The are two factors that affect the height of the peak: 1. The number of carbons producing the signal: the more carbons (symmetry!), the stronger the signal, the taller the peak.

2. The presence of hydrogen atoms attached to carbon increase the strength of the signal on the spectrum. Those carbons without a hydrogen attached tend to have smaller, weaker signals.

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For your Qual lab, you must: 1. Identify the number of carbons on your 13C spectrum and determine if you have any symmetry 2. Identify what each type of carbon is, based on the chemical shift value. Now: Puzzle Solving Using 13C NMR DEPT-13C NMR Spectroscopy: -“Distortionless Enhancement by Polarization Transfer” -Distinguishes between CH3, CH2, CH and quarternary C’s Ex: 6-methyl-5-penten-2-ol: Symmetry?

OH

A DEPT experiment is done in three stages: Stage 1: run ordinary spectrum to find chemical shifts of all peaks Peaks: 18, 23, 24, 25, 40, 64, 124, 132 ppm

Stage 2: Run a DEPT-90 spectrum – only CH peaks appear on the spectrum -

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Peaks: 64, 124 ppm Stage 3: Run a DEPT-135 spectrum – CH3 and CH peaks appear as positive peaks (UP from baseline) and CH2 peaks appear as negative peaks (DOWN from baseline).

Positive: 18, 23, 25, 64, 124 ppm Negative: 24, 40 ppm Those not appearing on either the second or third spectrum are quaternary carbons. Peak? 132 ppm. Examples: C6H13Br

01020304050PPM

DEPT-90: no peaks DEPT-135: positive peak at 29 ppm; negative peaks at 28, 49 ppm C6H12O

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020406080100120140160180200220PPM

DEPT-90: 24 ppm DEPT-135: positive peak at 23, 24, 31 ppm; negative peaks at 55 ppm MORE Puzzle Solving: Using Splitting Patterns: In normal 13C NMR mode, splitting cannot occur between carbons because of the low natural abundance of carbon-13 (not likely to find on adjacent positions).

In spin-coupled mode, the magnetic fields of the protons are allowed to interact with the carbon atom magnetic fields and splitting can occur between the carbon-13 atoms and the protons directly attached to those carbon-13 atoms.

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N+1 Rule for Carbon-13 NMR: The signal for a carbon (or equivalent carbons) with N hydrogen atoms attached will be split into N+1 peaks in spin-coupled mode. N N+1 CH3 3 4 (quartet) CH2 2 3 (triplet) CH 1 2 (doublet) C 0 1 (singlet) See Problems online.