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 14 N has three states: -1, 0, 1. They each differ by 1.

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

Nuclear spin quantum number 0 Quantum numbers of spin states 0 Number of spin states 1 1/2 ½ , ½ 2 1 3/2 -1, 0, 1 -3/2, -1/2, 1/2, 3/2 3 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 spectroscopy….

they have different energies.

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

Final Exam Schedule, Thursday December 18, 8AM 5249 CHEM 51.

TR9A KUNJAPPU 5250 CHEM 51.

TR9J HOWELL 5251 CHEM 51.

TR9C FANG 5252 CHEM 51.

TR9E KUNJAPPU 5253 CHEM 51.

TR9G HANS 5255 CHEM 51.

TR9B METLITSKY 5278 CHEM 51.

TR9K LUSHTAK 5279 CHEM 51.

TR9D MOLLICA 148NE 133NE 133NE 148NE 1310N 1310N 148NE 1310N

Apply a strong external field…..

Both orientations have same energy if no magnetic field

Figure 13.4, p.499

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.

p. 481

More Shielding Doing nmr spectroscopy: •the

magnetic field creates the energy difference

the nucleus and between the spin states of •

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.

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

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.

Equivalence by Symmetry

Figure 13.6, p.500

Equivalent by rotation

H H H H H Cl 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

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

p.501

p. 484

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

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.

Less shielded, more deshielded, downfield Most electron density around the H atoms, most shielded,

upfield

.

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

p.504

Chemical shift table…

Figure 13.8, p.505

Relationship of chemical shift to electronegativity Less electrons density around hydrogens as ascend table.

For C-H bond as the hybridization of the carbon changes sp 3 to sp 2 to sp the electronegativity of the C increases and expect to deshield (move left) the H peak.

sp 3 sp sp 2 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.

•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

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 • 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.

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

Example

H Cl C H 3 C CH 3 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).

Attempt to anticipate the splitting patterns in each molecule.

p. 491

p. 491

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

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

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 The original single peak of H1 has been

split into two

peaks by the effect of the neighbor H2. The energy difference is J neighboring hydrogens will augment the applied magnetic field and about 50% will decrement it. Get two peaks, a double H1, being excited Here H2 augments external field, peak moved downfield.

Here H2 decreases external field, peak moved upfield.

3-pentanone Coupling constant, J, in Hz Same as gap here.

The left side of molecule unaffected by right side.

Peak identification… O

Figure 13.14, p.511

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 = h n ) 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.

anti vinyl systems gauche

Table 13.4, p.511

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

H a is being excited. H b is causing spin spin splitting

by slightly increasing or decreasing the magnetic field experienced by H a .

Spin-Spin Splitting H b is augmenting external field causing a larger energy gap.

H b decrementing external field causing a smaller energy gap.

Again H a is flipping, resonating. The two H b are causing spin spin splitting

by slightly changing the magnetic field experienced by H a .

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”.

Recall that for the two H b 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

Three neighboring H b ’s causing splitting when H a is excited.

H a being excited.

Three equivalent H b causing spin spin splitting.

All H b augment Two augment, one decrement.

One augment, two decrement.

All decrement.

Figure 13.15c, p.512

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

Figure 13.17, p.513

Three nonequivalent nuclei. H a other.

and H b split each other . Also H b and H c split each

Technique

: use a tree diagram and consider splittings sequentially.

Figure 13.19, p.513

More complicated system

Figure 13.20, p.514

Return to Vinyl Systems

Not equivalent (

R 1 is not same as R 2 ) because there is no rotation about the C=C bond.

Figure 13.21, p.514

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.

Analysis Each of these patterns is different from the others.

J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Now examine the left most signal….

J AB = 11-18 Hz J AC = 0 - 5 J BC = 5 - 10 H a being excited. Both H b and H c are coupled and causing splitting.

H b causes splitting into two peaks (big splitting, J AB ) H c causes further splitting into a total of four peaks (smallest splitting, J AC )

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.

J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 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 H a .

Analysis in greater depth - 2.

Each of these patterns is different from the others.

J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 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 H b .

Analysis in greater depth - 3.

Each of these patterns is different from the others.

J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 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 H c .

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.

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

Figure 13.25, p.516

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

H a will be a triplet (two H b ); Likewise for H c .

We analyze H b .

A triplet of triplets Here H a and H c have same coupling with H b (J ab = J bc ), ,, coincidental overlap: splits to 5, four equivalent neighbors.

Analyze what happens as J ab becomes equal to J bc .

First get peak heights when J ab does not equal J bc .

Recall heights in a triplet are 1 : 2 : 1

1 1 2 1 2 1 First split the H b by H a in ratio of 1:2:1.

Each component is split by H c ratio of 1:2:1.

in Result for each final peak is product of probabilities 1 x 1 1 x 2 2 x 1 2 x 2

Examine middle peak.

Let J bc become larger until it equals J ab and add overlapping peaks together.

1 2 1 2 4 2 1 2 1 Peak heights shown when J ab does not equal J bc . 1+4+1

Now adjacent peak.

1 2 1 2 4 2 1 2 1 2+2 1 4 6 4 1

Fast Exchange

H ethanol H H H H OH 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?

C(CH 3 ) 3 H 3 C H 2 C CH 3 H C(CH 3 ) 3 Seem to be equivalent until we look at most stable conformation, the most utilized conformation.

H H 3 C CH 3 H Are these two hydrogens truly equivalent? H

Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic

.

H 3 C 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.

H C(CH 3 ) 3 H CH 3 H

Homotopic H H replace one H with D H D Achiral Achiral The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions.

Enantiotopic H H replace one H with D H D Achiral Chiral, have two 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 Pro R hydrogen H D Pro S hydrogen.

This structure would be S

Diastereotopic H H H D D H H 3 C Cl H replace H with D H 3 C Cl H H 3 C Cl H produced diastereomers 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 R hydrogen H 3 C Cl H D H This structure would be S Pro S hydrogen. (Making this a D causes the structure to be S.)

Example of diastereotopic methyl groups.

c H H 3 C a H 3 C a' OH CH 3 b H H d a and a’ Diastereotopic methyl groups (not equivalent), each split into a doublet by H c

13

C NMR

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

• H atoms can spin-spin split a 13 C peak. ( 13 CH 4 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

4 peaks  4 types of carbons.

13

C chemical shift table

Hydrogen NMR: Analysis: Example 1 Fragments: (CH 3 ) 3 C-, -CH 2 -, CH 3 -(C=O) O 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 CH 2 .

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

Carbonyl group

!

6. Absence of splitting between CH 2 and CH 3 . Conclude

: they are not adjacent

.

Example 2, C

3

H

6

O

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

OH O OCH 3 HO O O

Chemical shift table… Observed peaks were 2.5 – 3.1 vinylic ethers Observed peaks were 2.5 – 3.1.

Ether!

Figure 13.8, p.505

Possible structures

OH O OCH 3 HO O O

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.

Now look at chemical shifts 1. Formula told us that there are two pi bonds/rings in the compound.

2. From chemical shift conclude geminal CH 2 =CR 2 . Thus one pi/ring left.

3. Conclude there are no single C=C

H

- vinyl hydrogens. Have CH 2 =C-R 2 . This rules out a second pi bond as it would have to be fully substituted, CH 2 =C(CH 3 )C(CH 3 )=C(CH 3 ) 2 , to avoid additional vinyl hydrogens which is

C 8 H 14

.

X

In CH 2 =CR 2 are there allylic hydrogens: CH 2 =C(C H 2 -) 2 ?

Do the R groups have

allylic

hydrogens, C=C-CH?

1. Four allylic hydrogens. Unsplit. Equivalent!

2. Conclude CH 2 =C(CH 2 -) 2 3. Subtract known structure from formula of unknown… C 7 H 12 - CH 2 =C(CH 2 -) 2 ----------------------------------------- C 3 H 6 left to identify Remaining hydrogens produced the 6H singlet.

Likely structure of this fragment is –C(CH 3 ) 2 -.

But note text book identified the compound as