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CHEM 212 – NMR Spectroscopy

Spring 2014Spring 2014

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR1. NMR is rarely used in a vacuum to do a “forensic” analysis of an

unknown

2. NMR (all nuclei) is usually used:– To analyze the product of a chemical reaction along with IR– To elucidate the structures of natural products (like the

spice lab compounds in CHEM 213) in conjunction with mass spectrometry (which gives molecular weights and formulas), IR and UV

3. In this course, you will be given one of three pieces of data with an 1H NMR for consideration:

– A molecular formula– An IR spectrum– The first part of a chemical reaction – for example:

O

NaBH4

EtOH

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 1: Do a quick assessment of the information you are given

Molecular formula – one of the most important pieces of information

Use the index of hydrogen deficiency (HDI) to determine the possible number of rings, double and triple bonds in the molecule:

For a chemical formula: CxHyNzO (halogens count as Hs)

HDI = x – ½ Y + ½ Z + 1

Example: C4H8O -- HDI = 4 – ½ (8) + ½ (0) + 1

HDI = 1 This compound contains either one double bond or one ring

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 2: Do a quick assessment of the 1H NMR you are given– Is the molecule simple or complex?

– Is the molecule aromatic, aliphatic or both?

– What are the total number of resonances that you observe

Be careful with overly simple spectra – remember a large molecule may appear to be small and simple if it is highly symmetrical

Consider: Durene, C10H14

1H NMR spectrum consists of two singlets!

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Example: 1H NMR for C4H8 (which has an HDI of 1)

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 3: Use the integration along with the molecular formula to make sure you can find all of the 1Hs (this is 1H NMR after all!)

If you do not have a molecular formula, use the integration to attempt to tabulate the number of 1Hs in the formula (does it make sense?)

If one hydrogen appears to be “missing”, you may suspect it is acidic or exchangeable with the dueterated NMR solvent

Remember, integration gives you the least common denominator of the total number of protons of each type

Keep in mind organic molecules contain –C-H’s, -CH2- ’s, -CH3 ’s and multiples of chemically equivalent ones!

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis Continue our example:

For C4H8O we need to find 8 protons

The integral for the quartet at 2.5 measures 30 mm, the singlet at 2.2 measures 44 mm, the triplet at 1.1 measures 43 mm

The ratio: 30:43:42 is roughly 2:3:3

2 + 3 + 3 = 8We found all 8 protons

30

44

43

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 4: Classify each of the proton resonances by using the general correlation table

Reconsider the number of rings, double or triple bonds that are possible given the HDI, and reconcile this data with what the 1H chemical shifts are telling you

Some hints:If you calculate an HDI > 4, you probably have an aromatic ring, and this should show on the spectrum

If you calculate an HDI of 1 or 2 and see no protons that are part of an alkene and alkyne – suspect rings if no oxygen's are present, carbonyls if (C=O) they are

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis Continue our example:

For C4H8O we have 3 families of chemically equivalent protons in a ratio of 2:3:3 or –CH2-, -CH3, -CH3

Both the 2.2 and 2.5 resonance correspond to protons on carbons next to electron withdrawing groups

The 1.1 resonance corresponds to protons on carbons bound to other aliphatic carbons

CH3EWG

CH3RCH2R

EWG

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis Continue our example:

We found –CH2-, -CH3 and

–CH3, subtract these from our original formula:

C4H8O - CH2 = C3H6O

C3H6O – CH3 = C2H3O

C2H3O – CH3 = CO

We needed an HDI of 1 and there is no evidence for 1H-C=C on the 1H NMR, our missing HDI, and EWG is a C=O!

CH3EWG

CH3RCH2R

EWG

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 5: Analyze the spin-spin coupling multiplets to elucidate the carbon chains of the molecule

Hints:• Singlets indicate you have protons on carbons that have no

chemically non-equivalent protons on any adjoining atom

• Multiplets mean you have chemically non-equivalent protons on adjoining carbons (or atoms), use the n+1 rule in reverse to find out how many

• Spin-spin coupling or splitting is MUTUAL, if you observe a multiplet there must be another multiplet it is related to (split by)

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis

For our example, this is trivial; we have concluded the C=O is the electron withdrawing group

The –CH3 singlet at 2.2 obviously has no 1Hs on adjoining carbons, as it is next to the carbonyl

The 2.5 –CH2- quartet is next to a –CH3 (n+1 = 4, so n = 3, it is next to a –CH3)

The 1.2 –CH3 is a triplet, so it is next to a –CH2

CH3EWG

CH3RCH2R

EWG

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis – Introductory 1H NMR

Step 6: Construct the molecule and double-check consistency• Does the HDI match? Have you accounted for all atoms in

the formula?

• From your constructed molecule, pretend you are trying to verify if that spectrum matches, and quickly re-do the problem

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Spectral Analysis – 1H NMR NMR Spectroscopy

NMR Spectral Analysis

We concluded we have:

or 2-butanone

C4H8O, HDI = 1

3 proton resonances2 mutually coupled (split)

H3C CH2C CH3

O

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 2: C9H9BrO

HDI =

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 2: C9H9BrO

HDI = 9 – ½ (10) + 0 + 1 = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 2: C9H9BrO

HDI = 9 – ½ (10) + 0 + 1 = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 2: C9H9BrO

HDI = 9 – ½ (10) + 0 + 1 = 5

O

Br

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 3: C4H10O

HDI =

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 3: C4H10O

HDI = 4 – ½ (10) + 0 + 1 = 0

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 3: C9H9BrO

HDI = 0

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 3: C4H10O

HDI = 0

HO

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 4: C5H10O2

HDI =

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 4: C5H10O2

HDI = 5 – ½ (10) + 0 + 1 = 1

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 4: C5H10O2

HDI = 1

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 4: C5H10O2

HDI = 1 HO

O

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 5: C10H12O2

HDI =

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 5: C10H12O2

HDI = 10 – ½ (12) + 0 + 1 = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 5: C10H12O2

HDI = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 5: C10H12O2

HDI = 5OH

O

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 6: C6H4ClNO2

HDI =

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 6: C6H4ClNO2

HDI = 6 – ½ (5) + ½ (1) + 1 = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 6: C6H4ClNO2

HDI = 5

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Spectral Analysis – 1H NMR NMR Spectroscopy

Example 6: C6H4ClNO2

HDI = 5

NO O

Cl