Chapter 13

NMR Spectroscopy

Recall that electrons have two “spin states”: spin up (1/2) and spin down (-1/2).

Similarly nuclei have spin quantum states….

Nuclei of interest.

By coincidence, each of these has two states, ½ and – ½ .

By the way, note that 14N has three states: -1, 0, 1. They each differ by 1.

The nuclear spin quantum number determines how many spin states there are

Spectroscopy involves using energy to excite a system from one state (ground state) to another of higher energy (excited state).

Nuclear spin quantum number

Quantum numbers of spin states

Number of spin states

0 0 1

1/2 - ½ , ½ 2

1 -1, 0, 1 3

3/2 -3/2, -1/2, 1/2, 3/2 4

Normally, nuclei in different spin states have the same energy. Can not do spectroscopy. We need to have a ground state and excited state.

In a magnetic field they have different energies. Now we can do spectroscopy….

We apply a magnetic field and create a ground state and a higher energy excited state (perhaps more than one).

5249 CHEM 51. TR9A KUNJAPPU 148NE

5250 CHEM 51. TR9J HOWELL 133NE

5251 CHEM 51. TR9C FANG 133NE

5252 CHEM 51. TR9E KUNJAPPU 148NE

5253 CHEM 51. TR9G HANS 1310N

5255 CHEM 51. TR9B METLITSKY 1310N

5278 CHEM 51. TR9K LUSHTAK 148NE

5279 CHEM 51. TR9D MOLLICA 1310N

Final Exam Schedule, Thursday December 18, 8AM

Apply a strong external field…..

Both orientations have same energy if no magnetic field

Figure 13.4, p.499

•Two kinds of hydrogens in methyl acetate: two peaks. (Peak at zero is tetramethyl silane to standardize the instrument. )

•Chemical shift: where on horizontal axis the signal from a nucleus occurs. Question: What causes nuclei to appear with different chemical shift??

•Answer: the sigma bonding electrons in a molecule will be set in motion to establish a magnetic field that opposes the external magnetic field. The nuclei are shielded.

•The shielded nuclei experience less of a magnetic field, closer energy states.

•The shielded nuclei require less energy to excite and their signal occurs to the right in the spectrum.

More shielded, nuclei experience lesser magnetic field.

Less energy to excite.

Example of nmr spectrum: methyl acetate.

p. 481

Doing nmr spectroscopy:

•the magnetic field creates the energy difference between the spin states of the nucleus and

•Radio waves provide the energy needed to excite the nucleus from the lower energy state to the excited state.

More Shielding

Simplifying

•The energy supplied by the radio waves has to match the energy gap created by the magnetic field.

•We can vary either the magnetic field or the frequency of the radio waves to match the exciting radiation energy with the energy needed to reach the excited state.

Hold the external magnetic field constant, vary radio frequency. Less energy needed to excite the nuclei when more shielded.

Hold excitation energy (radio waves) constant, vary magnetic field. Stronger magnetic field needed to overcome shielding.

More Shielding

More Shielded, Less energy needed from radio waves

More shielded, stronger magnetic field needed to create the right energy difference.

Since we control energy of excited state (magnetic field) and the energy being supplied by radiowaves: two ways for an nmr spectrometer to function:

or

Terminology based on this approach: downfield (lower ext field) on left; upfield on right

Remember that methyl acetate only gave two peaks in its spectrum. There were two sets of equivalent hydrogens.

Equivalent hydrogens

Hydrogens are equivalent if

• They are truly equivalent by symmetry.

-or-

• They are bonded to same atom and that carbon atom can rotate freely at room temperature to interchange the positions of the hydrogens making the equivalent to the spectrometer.

Figure 13.6, p.500

Equivalence by Symmetry

Equivalent by rotation

H

H

H

H

Cl

H

equivalent

Note that if it were not for rotation the methyl hydrogens would not be equivalent. Two are gauche to the Cl and one is anti.

p.501

Some molecules which have only one type of hydrogen - only one signal

p. 484

Figure 13.7, p.503

Signal area: proportional to the number of hydogens producing the signal

Looking at the molecular structure

# Methyl hydrogens : # tert butyl hydrogens = 3:9 = 1:3

In the spectrum we find two peaks

23 : 67 = 1 : 2.91

Conclude smaller peak due to methyl hydrogens; larger due to tert butyl hydrogens.

p.504

Now return to chemical shift and factors affecting it. Look at two isomeric esters to get some feeling for chemical shift. The electronegative oxygens play the key role here.

Most electron density around the H atoms, most shielded, upfield.

Less shielded, more deshielded, downfield

Most deshielded, furthest downfield. Sigma electrons pulled away by oxygen.

Figure 13.8, p.505

Chemical shift table…

Fu

rth

er le

ft,

dow

nfi

eld

,

Les

s sh

ield

ed

Relationship of chemical shift to electronegativity

Less electrons density around hydrogens as ascend table.

Expect vinylic hydrogens to be deshielded due to hybridization but acetylenic (recall acidity) should be even more deshielded and they aren’t. Some other factor is at work. Magnetic induction of pi bonds.

sp3

sp

sp2

For C-H bond as the hybridization of the carbon changes sp3 to sp2 to sp the electronegativity of the C increases and expect to deshield (move left) the H peak.

Figure 13.9, p.507

•Diamagnetic shielding

•Hydrogen on axis and shielded effectively.

•Hydrogen experiences reduced magnetic field.

•Less energy needed to excite.

•Peak moves upfield to the right.

Figure 13.11, p.508

In benzene the H atoms are on the outside and the induced magnetic field augments the external field.

Spin Spin Splitting

• If a hydrogen has n equivalent neighboring hydrogens the signal of the hydrogen is split into (n + 1) peaks.

• The spin-spin splitting hydrogens must be separated by either two or three bonds to observe the splitting. More intervening bonds will usually prevent splitting.

Example

H3C

C

CH3

H Cl

Expect the signal for this hydrogen to be split into seven by the six equivalent neighbors. Expect the peak for the

methyl hydrogens to be split into two peaks by the single neighbor.

Overall:

small peak split into seven (downfield due to the Cl).

larger peak (six times larger) split into two (further upfield).

p. 491

Attempt to anticipate the splitting patterns in each molecule.

p. 491

Spin-spin splitting. Coupling constant, J.

Split into a group of 4 Split into a group of 2

The actual distance, J, between the peaks is the same within the quartet and the doublet.

More Shielding due to electrons at nucleus being excited.

In preparation for discussion of origin of Spin-Spin recall earlier slide

Due to shielding, less of the magnetic field experienced by nucleus, Lower energy needed to excite. Peaks on right are “upfield”.

Reduced shielding, more of the magnetic field experienced, higher energy of excitation. Peaks are “downfield”.

Origin of spin-spin splittingIn the presence of a external magnetic field each nuclear spin must be aligned with or against the external field. Approximately 50% aligned each way.

Non-equivalent hydrogen nuclei separated by two or three bonds can “spin – spin split” each other. What does that mean?

Consider excitation of a hydrogen H1. Energy separation of ground and excited states depends on total magnetic field experienced by H1.

Now consider a neighbor hydrogen H2 (passive, not being excited) which can increase or decrease the magnetic field experienced by H1.

Ene

rgy

H1, being excited

Here H2 augments external field, peak moved downfield.

Here H2 decreases external field, peak moved upfield.

The original single peak of H1 has been split into two peaks by the effect of the neighbor H2. The energy difference is J

About 50% of the neighboring hydrogens will augment the applied magnetic field and about 50% will decrement it. Get two peaks, a double

Figure 13.14, p.511

3-pentanone

The left side of molecule unaffected by right side.

Coupling constant, J, in Hz

OPeak identification…

Same as gap here.

Magnitude of Coupling Constant, JThe magnitude of the coupling constant, J, can vary from 0 to about 20 Hz.

This represents an energy gap (E = h) due to the interaction of the nuclei within the molecule. It does not depend on the strength of the external field.

J is related to the dihedral angle between bonds. J largest for 0 (eclipsed) or 180 (anti), smallest for 90, intermediate for gauche.

Rapid interconversion of enol structures from 2,4-pentadione

Have mixture of keto (16%) and hydrogen bonded enol forms (84%). Spectrum shows both forms:

Keto with two types of H at 2.24 and 3.60

Enol with three types (not four) indicating rapid interconversion of the two enol structures making the methyl groups equivalent.

Table 13.4, p.511

anti gauche

vinyl systems

Now look at some simple examples. Examine the size of the peaks in the splitting.

Hb is augmenting external field causing a larger energy gap.

Hb decrementing external field causing a smaller energy gap.

Ha is being excited. Hb is causing spin-spin splitting by slightly increasing or decreasing the magnetic field experienced by Ha.

Spin-Spin Splitting

Figure 13.15b, p.512

Two neighboring atoms assist external field. More energy needed to excite. Peak is “downfield”.

One neighbor assists, one hinders. No effect.

Both neighbors oppose. Less energy needed to excite, “upfield”.

Again Ha is flipping, resonating. The two Hb are causing spin-spin splitting by slightly changing the magnetic field experienced by Ha.

Recall that for the two Hb atoms the two states (helping and hindering the external field) are almost equally likely. This give us the 1 : 2 : 1 ratio.

Figure 13.15c, p.512

All Hb augment

Two augment, one decrement.

One augment, two decrement.

All decrement.

Ha being excited.

Three equivalent Hb causing spin spin splitting.

Three neighboring Hb’s causing splitting when Ha is excited.

Figure 13.17, p.513

Naturally if there are two non-equivalent nuclei they split each other.

Figure 13.19, p.513

Three nonequivalent nuclei. Ha and Hb split each other. Also Hb and Hc split each other.

Technique: use a tree diagram and consider splittings sequentially.

Figure 13.20, p.514

More complicated system

Figure 13.21, p.514

Not equivalent (R1 is not same as R2) because there is no rotation about the C=C bond.

Return to Vinyl Systems

Example of alkenyl system

We will perform analysis of the vinyl system and ignore the ethyl group.

Three different kinds of H in the vinyl group. We can anticipate the magnitude of the coupling constants.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Each of these patterns is different from the others.

Analysis

Now examine the left most signal….

JAB = 11-18 Hz

JAC = 0 - 5

JBC = 5 - 10

Ha being excited. Both Hb and Hc are coupled and causing splitting.

Hb causes splitting into two peaks (big splitting, JAB)

Hc causes further splitting into a total of four peaks (smallest splitting, JAC)

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Each of these patterns is different from the others.

Look at it this way... This signal appears to have big (caused by trans H-C=C-H) and small (caused by geminal HHC=) splittings. The H being excited must have both a trans and geminal H. The H must be Ha.

Analysis in greater depth based on knowing the relative magnitude of the splitting constants. Aim is to associate each signal with a particular vinyl hydrogen.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Each of these patterns is different from the others.

And the middle signal.

This signal appears to have big (caused by trans H-C=C-H) and middle (cis H-C=C-H) splittings. The H being excited must have both a trans and cis H. The H must be Hb.

Analysis in greater depth - 2.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Each of these patterns is different from the others.

And the right signal.

This signal appears to have small (caused by geminal HHC=) and middle (cis H-C=C-H) splittings. The H being excited must have both a geminal and cis H. The H must be Hc.

Analysis in greater depth - 3.

As with pi bonds, cyclic structures also prevent rotation about bonds

Approximately the same vinyl system as before.

No spin spin splitting of these hydrogens. Nothing close enough

Non equivalent geminal hydrogens. Analyze this.

Figure 13.25, p.516

Note the “roof effect”. For similar hydrogens the inner peaks can be larger.

A triplet of triplets

Ha will be a triplet (two Hb); Likewise for Hc. We analyze Hb.

Here Ha and Hc have same coupling with Hb (Jab = Jbc), ,,

coincidental overlap: splits to 5, four equivalent neighbors.

Coincidental Overlap: Non-equivalent nuclei have same coupling constant.

Recall heights in a triplet are 1 : 2 : 1

Analyze what happens as Jab becomes equal to Jbc.

First get peak heights when Jab does not equal Jbc.

12

1

11

2

First split the Hb by Ha in ratio of 1:2:1.

Each component is split by Hc in ratio of 1:2:1.

Result for each final peak is product of probabilities

1 x 1 1 x 2 2 x 1

2 x 2

Peak heights shown when Jab does not equal Jbc.

1 2 1 2 4 2 1 2 1

Examine middle peak. Let Jbc become larger until it equals Jab and add overlapping peaks together.

1+4+1

1 2 1 2 4 2 1 2 1Now adjacent peak.

2+2

1 4 6 4 1

Fast Exchange

OH

H H

H

HH

ethanol

Expect coupling between these hydrogens. Three bond separation.

There is no coupling observed especially in acid or base.

Reason: exchange of weakly acidic hydrogen with solvent.

The spectrometer sees an “averaged hydrogen”. No coupling and broad peak.

Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not?

Are these two hydrogens truly equivalent?

Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic.

How to tell: replace one of the hydrogens with a D.

If produce an achiral molecule then hydrogens are homotopic,

if enantiomers then hydrogens are enantiotopic,

if diastereomers then diastereotopic.

We look at each of these cases.

HH

CH3

H3C H

C(CH3)3

Seem to be equivalent until we look at most stable conformation, the most utilized conformation.

H3C

H2C C(CH3)3

CH3

H

H3C

C(CH3)3

H

H

CH3

H

Homotopic

HH replace one H with D DH

Achiral Achiral

The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions.

Enantiotopic

HH replace one H with D DH

Achiral Chiral, havetwo enantiomers.

The hydrogens are enantiotopic and equivalent in the NMR unless the molecule is placed in a chiral environment such as a chiral solvent..

The hydrogens are designated as Pro R or Pro S

DH

This structure would be S

Pro S hydrogen.Pro R hydrogen

DH

H3C

HCl

Diastereotopic

If diastereormers are produced from the substitution then the hydrogens are not equivalent in the NMR. Diastereotopic hydrogens.

The hydrogens are designated as Pro R or Pro S

This structure would be S

Pro S hydrogen. (Making this a D causes the structure to be S.)

Pro R hydrogen

DH

H3C

HCl

HH

H3C

HCl

replace H with D

HD

H3C

HCl

produced diastereomers

H

H3C

OH

H3C

CH3

H

H

a

a'

cd

b

Diastereotopic methyl groups (not equivalent), each split into a doublet by Hc

a and a’

Example of diastereotopic methyl groups.

13C NMR

• 13C has spin states similar to H. • Natural occurrence is 1.1% making 13C-13C spin spin

splitting very rare.• H atoms can spin-spin split a 13C peak. (13CH4 would yield

a quintet). This would yield complicated spectra.• H splitting eliminated by irradiating with an additional

frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero.

• A decoupled spectrum consists of a single peak for each kind of carbon present.

• The magnitude of the peak is not important.

13C NMR spectrum

4 peaks 4 types of carbons.

13C chemical shift table

Hydrogen NMR: Analysis: Example 1

1. Molecular formula given. Conclude: One pi bond or ring.

2. Number of hydrogens given for each peak, integration curve not needed. Verify that they add to 14!

3. Three kinds of hydrogens. No spin-spin splitting. Conclude: Do not have non-equivalent H on adjacent carbons.

4. The 9 equivalent hydrogens likely to be tert butyl group (no spin-spin splitting). The 3 equivalent hydrogens likely to be methyl group. The two hydrogens a CH2.

Fragments: (CH3)3C-, -CH2-, CH3-

5. Have accounted for all atoms but one C and one O. Conclude: Carbonyl group!

-(C=O)-

6. Absence of splitting between CH2 and CH3. Conclude: they are not adjacent.

O

Example 2, C3H6O

1. Molecular formula One pi bond or ring

2. Four different kinds of hydrogen: 1,1,1,3 (probably have a methyl group).

3. Components of the 1H signals are about equal height, not triplets or quartets

4. Consider possible structures.

Possible structures

OCH3

HO

O

OH

O

O

Figure 13.8, p.505

Chemical shift table… Observed peaks were 2.5 – 3.1

vinylic

ethers

Observed peaks were 2.5 – 3.1. Ether!

Possible structures

OCH3

HO

O

OH

O

O

NMR example

Formula tells us two pi bonds/rings

Three kinds of hydrogens with no spin/spin splitting.

What can we tell by preliminary inspection….

Now look at chemical shifts

2. From chemical shift conclude geminal CH2=CR2. Thus one pi/ring left.

3. Conclude there are no single

C=CH- vinyl hydrogens. Have CH2=C-R2.

This rules out a second pi bond as it would have to be fully substituted, CH2=C(CH3)C(CH3)=C(CH3)2 , to avoid additional vinyl hydrogens which is C8H14.

X

In CH2=CR2 are there allylic hydrogens: CH2=C(CH2-)2?

1. Formula told us that there are two pi bonds/rings in the compound.

Do the R groups have allylic hydrogens, C=C-CH?

1. Four allylic hydrogens. Unsplit. Equivalent!

2. Conclude CH2=C(CH2-)2

3. Subtract known structure from formula of unknown…

C7H12

- CH2=C(CH2-)2

------------------------------------------

C3H6 left to identifyRemaining hydrogens produced the 6H singlet.

Likely structure of this fragment is –C(CH3)2-.

But note text book identified the compound as