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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 .

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chapter 13

Chapter 13

NMR Spectroscopy

slide2

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.

slide3

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

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

slide4

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).

slide5

Apply a strong external field…..

Both orientations have same energy if no magnetic field

slide8

Example of nmr spectrum: methyl acetate.

More shielded, nuclei experience lesser magnetic field.

Less energy to excite.

  • 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.
slide10

More Shielding

  • 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.
  • 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.
slide11

More Shielding

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

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

More Shielded, Less energy needed from radio waves

or

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

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

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

slide12
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.

slide13

Equivalence by Symmetry

Figure 13.6, p.500

equivalent by rotation
Equivalent by rotation

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.

equivalent

slide17

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.

Figure 13.7, p.503

slide18

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.

p.504

slide19

Chemical shift table…

Figure 13.8, p.505

slide20

Relationship of chemical shift to electronegativity

Further left, downfield,

Less shielded

Less electrons density around hydrogens as ascend table.

slide21

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.

sp3

sp

sp2

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.

slide22

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 13.9, p.507

slide23

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

Figure 13.11, p.508

spin spin splitting
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
Example

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).

slide28

Spin-spin splitting. Coupling constant, J.

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

Split into a group of 4

Split into a group of 2

slide29

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

More Shielding due to electrons at nucleus being excited.

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 splitting
Origin of spin-spin splitting

In 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.

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

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

Energy

H1, being excited

Here H2 augments external field, peak moved downfield.

Here H2 decreases external field, peak moved upfield.

slide31

Same as gap here.

Coupling constant, J, in Hz

3-pentanone

The left side of molecule unaffected by right side.

Peak identification…

Figure 13.14, p.511

magnitude of coupling constant j
Magnitude of Coupling Constant, J

The magnitude of the coupling constant, J, can vary from 0 to about 20 Hz.

This represents an energy gap (E = hn) 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.

slide33

gauche

anti

vinyl systems

Table 13.4, p.511

slide34

Spin-Spin Splitting

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.

slide35

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

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

One neighbor assists, one hinders. No effect.

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

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 13.15b, p.512

slide36

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

All Hb augment

Two augment, one decrement.

One augment, two decrement.

All decrement.

Ha being excited.

Three equivalent Hb causing spin spin splitting.

Figure 13.15c, p.512

slide38

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

Technique: use a tree diagram and consider splittings sequentially.

Figure 13.19, p.513

slide39

More complicated system

Figure 13.20, p.514

slide40

Return to Vinyl Systems

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

Figure 13.21, p.514

slide41

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.

slide42

Analysis

Each of these patterns is different from the others.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

Now examine the left most signal….

slide43

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

JAC = 0 - 5

JBC = 5 - 10

slide44

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.

Each of these patterns is different from the others.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

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.

slide45

Analysis in greater depth - 2.

Each of these patterns is different from the others.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

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.

slide46

Analysis in greater depth - 3.

Each of these patterns is different from the others.

JAB = 11-18 Hz, BIG

JAC = 0 – 5 Hz, SMALL

JBC = 5 – 10 Hz, MIDDLE

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.

slide47

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

Approximately the same vinyl system as before.

Non equivalent geminal hydrogens. Analyze this.

No spin spin splitting of these hydrogens. Nothing close enough

slide48

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

Figure 13.25, p.516

slide49

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

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

A triplet of triplets

Here Ha and Hc have same coupling with Hb (Jab = Jbc), ,, coincidental overlap: splits to 5, four equivalent neighbors.

slide50

Analyze what happens as Jab becomes equal to Jbc.

First get peak heights when Jab does not equal Jbc.

Recall heights in a triplet are 1 : 2 : 1

2

1

1

2

1

1

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

2 x 2

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

slide51

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

1 2 1 2 4 2 1 2 1

1+4+1

Peak heights shown when Jab does not equal Jbc.

slide52

Now adjacent peak.

1 2 1 2 4 2 1 2 1

2+2

1 4 6 4 1

fast exchange
Fast Exchange

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.

ethanol

return to question of equivalent hydrogens stereotopicity equivalent or not
Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not?

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

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.

homotopic
Homotopic

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

enantiotopic
Enantiotopic

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

Pro S hydrogen.

Pro R hydrogen

This structure would be S

diastereotopic
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

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

Pro R hydrogen

This structure would be S

slide58

Example of diastereotopic methyl groups.

a and a’

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

13 c nmr
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.

13 c nmr spectrum
13C NMR spectrum

4 peaks  4 types of carbons.

hydrogen nmr analysis example 1
Hydrogen NMR: Analysis: Example 1

-(C=O)-

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

  • 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.

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

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

example 2 c 3 h 6 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.

slide65

Chemical shift table… Observed peaks were 2.5 – 3.1

ethers

Observed peaks were 2.5 – 3.1. Ether!

vinylic

Figure 13.8, p.505

slide67

NMR example

What can we tell by preliminary inspection….

Formula tells us two pi bonds/rings

Three kinds of hydrogens with no spin/spin splitting.

slide68

Now look at chemical shifts

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

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

X

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.

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

slide69

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

  • Four allylic hydrogens. Unsplit. Equivalent!
  • Conclude CH2=C(CH2-)2
  • Subtract known structure from formula of unknown…
  • C7H12
  • - CH2=C(CH2-)2
  • ------------------------------------------
  • C3H6 left to identify

But note text book identified the compound as

Remaining hydrogens produced the 6H singlet.

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