H NMR Lecture

Rule of Thirteen

•When the formula of the unknown is not given, you are in trouble…..unless

you know the Rule of Thirteen

•Molecular weight of the compound is M

M/13 = n + r/13

where n is the integer derived from dividing M/13, and r is the remander

Note: do not use the calculator for this rule

•Base formula (not necesseraly the formula of your given compound): CnHn+r

•Parent formula: CnH2n+2

IHD: [2n+2-(n+r)]/2 =(n-r+2)/2

•Before you go on with the IHD calculations, check on the presence of

heteroatoms:

For example, if there is an oxygen present, then O=16

Substitute it in your base formula by subtracting oxygen and adding a

molecular fragment with the same weight (C=12 and 4H=4)

Examples

O

The molecular weight of the unknown aldehyde is 134g/molPropose several structures

Subtract oxygen134-16=118

Rule of 13: 118/13= 9+1/13Base formula is C9H10OParent formula is C9H20IHD = 20-10/2=5benzene ring and a double bond=aldehyde

O

Nuclear Magnetic Resonance

•Atoms with nuclei containing an odd number of protons or neutrons (1H,13C, 14N, 19F, and 31P) have a magnetic moment µ induced by the nuclear

spin.

•Hydrogen atoms are more than 99% 1H. Other isotopes have special

names: 2H = deuterium, 3H = Tritium (radioactive).

•Spin quantum number I=1/2, meaning their magnetic moments can only

have two equal but opposite values +µ and -µ.

•In the applied magnetic field, they can be oriented parallel ! (m=+1/2)

or anti-parallel " (m=-1/2). The energy is slightly different.

S

A spinning nucleus with it's magnetic field alinged with the magnetic field of a magnet

N

S

N

!- spin state,

favorable,

lower energyN

S !- spin state,unfavorable,lower energy

A spinning nucleus with it's magnetic field alinged against the magnetic field of a magnet

S

N

Nuclear Magnetic Resonance and Relaxation

Applied magnetic field (field strength H0 and flow density B0)

A proton experiences an ‘impact’ generated by a high frequency transmitter

A resulting movement proceeds in two directions (directional quantization):

o Direction of the external field (low energy)

o Direction opposite to the applied field (high energy)

o When the frequency of the transmitter equals to the precession

frequency, an inversion of the magnetic moment occurs (Nuclear Magnetic

Resonance).

o The process of returning to the lower energy position is called

relaxation.

S

N

E2

E1

#E = E2 – E1

high frequency

transmitter ($0)

• Number of peaks: number of non-equivalent protons in the sample

• Chemical shift: kinds of protons in the molecule

• J-coupling: protons that are near other protons.

• Integrals: the ratio of each kind of proton in our sample.

Interpreting SpectraThe Chemical Shift (%) Scale

Simple presentation:

•Choose a standard sample, an NMR of that standard

•Measure its absorbance frequency.

• We then measure the frequency of our sample and subtract its frequency from that

of the standard.

•We then divide by the frequency of the standard. This gives a number called the

“chemical shift,” %

•% does not depend on the magnetic field strength.

• Why not?

•Lets examine the same sample at 300 and 600 MHz field

•On 300 MHz instrument

•standard absorbs at 300,000,000 Hz (300 MHz)

•sample absorbs at 300,000,300 Hz.

•The difference is 300 Hz 300/300,000,000 = 1/1,000,000

•1 part per million (or 1 ppm).

•On 600 MHz instrument

•standard absorbs at 600,000,000 Hz (600 MHz)

•sample absorbs at 600,000,600 Hz. (the frequency of sample increases proportionally)

•The difference is now 600 Hz, 600,000,000 (600/600,000,000 = 1/1,000,000, = 1 ppm).

CH3

Si

H3C

H3C CH3

Chemical Shifts

•Most protons appear between 0 and 12 ppm.

! ppm

TMS

CH3CH3

RONR2

CH3OCH3

RO

HR

R R

HH

RO

Ph CH3

HR

Cl

CH3

Ph

OH

OH

R

NHR

Upfield regionof the spectrumDownfield region

of the spectrum

012345678910

CH3HO(R)

1112

OH

RO

TMS = Me Si

Me

Me

Me

Inductive electron withdrawal by neighboring electronegative groups.

0.8ppm 3.0ppm 5.3ppm 7.3ppm

Chemical Shifts

H C

H

H

H

H C

H

H

Cl

H C

H

Cl

Cl

H C

Cl

Cl

Cl

H C

H

H

H

I C

H

H

H

Cl C

H

H

H

HO C

H

H

H

F C

H

H

H

Br C

H

H

H

electronegativity

of X

2.1 2.5 2.8 3.1 3.5 4.0

chemical shifts, ppm0.23 2.16 2.68 3.05 3.40 4.26

0123

PPM

O

Cl

012345

PPM

O Cl

O

01234

PPM

O

Hybridization effect

•More S-character to the adjacent carbon, more electronegative, stronger is the deshielding

effect on the H’s of the methyl group

C CH3 CH3CH3

0.2ppm!! 1.6-1.9ppm 2.1-2.5ppm

H

C CH3

H

CH3

CH3

H

Wrong!!!

•In this case, your expectations for chemicals shifts of protons attached to the carbons

of sp3, sp2, and sp hybridization should increase in the same order. Right?

Magnetic Anisotropy Effect

C

C

H

H

H

applied field Ho

enhancement of applied magnetic field shielding of applied magnetic field

+

+

(-) (-) C C

+

+

(-) (-) C CH H

(-)

(-)

+ +

6.6 -8.5 ppm 4.9-6.2 ppm 2.4-3.1 ppm

Spin-spin splitting or J-coupling

•When the field created by HB reinforces the magnetic field of the NMR machine (B0 ), HA

feels a slightly stronger field.

•When the field created by HB opposes B0, HA feels a slightly weaker field.

• So, we see two signals for HA depending on the alignment of HB. The same is true for HB, it

can feel either a slightly stronger or weaker field due to HA’s presence. So, rather than see a

single line for each of these protons, we see two lines for each.

C C

HBHA

HA HB

HA is split into two lines because it feels the magnetic field of HB.

HB is split into two lines because it feels the magnetic field of HA.

For this line, HB is lined up with the magnetic field

(adds to the overall magnetic field, so the line

comes at higher frequency)

For this line, HB is lined up against the magnetic field(subtracts from the overall magnetic field, so the line

comes at lower frequency)

C C

HBHA

HA HB

HA is split into two lines because it feels the magnetic field of HB.

HB is split into two lines because it feels the magnetic field of HA.

For this line, HB is lined up with the magnetic field

(adds to the overall magnetic field, so the line

comes at higher frequency)

For this line, HB is lined up against the magnetic field(subtracts from the overall magnetic field, so the line

comes at lower frequency)

HB

HA

HB

HA

•When there is more than one proton splitting a neighboring proton, we get more lines.

More on J-coupling

C C

HBHA

HA'HA + HA' HB

HA and HA' appear at the same chemical shift because they are

in identical environments They are also split into two lines (called a doublet) because they

feel the magnetic field of HB.

HB is split into three lines because it feels the magnetic

field of HA and HA'

Note that the signal produced by HA + HA' is twice the size

of that produced by HB

C C

HBHA

HA'HA + HA' HB

HA and HA' appear at the same chemical shift because they are

in identical environments They are also split into two lines (called a doublet) because they

feel the magnetic field of HB.

HB is split into three lines because it feels the magnetic

field of HA and HA'

Note that the signal produced by HA + HA' is twice the size

of that produced by HB

HB

HA’

HB

HA

HB

HA’

HB

HA

HA

HB

HA’

HA

HB

HA’

HA

HB

HA’

HA

HB

HA’

Splitting Pattern (Spin – Spin Coupling)

•Splitting can occur via space or binding electrons.

•Protons on the same carbon are called geminal: rarely observed

•Protons on adjacent carbons are called vicinal

CC C

H

H

H

H

1

2

1

2

3

geminal splitting vicinal splitting

Coupling Constants: Alkenes

•Coupling constants in alkenes differs depending on whether the protons are cis or trans to

each other.

•In a terminal alkene the cis and trans protons are NOT Equivalent!!!

• One is on the same side as the substituent, the other is on the opposite side.

•The coupling of trans protons to each other is typically ~ 16 Hz.

•The coupling of cis protons is a little smaller~ 12 Hz.

HA

Now, let's "turn on" HA - HX coupling. This splits the single line into two lines that are 16 Hz appart

If uncoupled, HA would appear as a singlet where the dashed line indicates

Now, let's "turn on" HA - HM coupling. This splits each of the two new lines into two lines that are 12 Hz appart for a total of four lines

12 Hz

16 Hz

12 Hz

HA

HM

HX

12Hz coupling

16 Hz coupling

HO

H

H

H

HO

H

H

H

HA

Now, let's "turn on" HA - HX coupling. This splits the single line into two lines that are 16 Hz appart

If uncoupled, HA would appear as a singlet where the dashed line indicates

Now, let's "turn on" HA - HM coupling. This splits each of the two new lines into two lines that are 12 Hz appart for a total of four lines

12 Hz

16 Hz

12 Hz

HA

HM

HX

12Hz coupling

16 Hz coupling

Common Splitting Patterns for Aromatic Compounds

Chemical Shifts 7.00 –8.3ppm

• Monosubstituted aromatic rings

012345678

PPM

Cl

Ha

Hb

Ha

Hb

Hc

dd

dd

t

a b

c

Disubstituted aromatic rings (X=Y)

012345678

PPM

Cl

Cl

Ha

Ha

Hb

Hb

d

d

a b

012345678

PPM

Cl

Ha

Cl

Hb

Hc

Hb

s

d

t

a

b c

012345678

PPM

Cl

Ha

Ha

Ha

Ha

Cl

a

s

Disubstituted aromatic rings (X,Y)

OH

Ha

Hb

NO2

Hb

Had

d

a b

0246810

PPM

0246810

PPM

OH

NO2

Ha

Hd

Hc

Hb

dd

pt

pt

dda bd c

0246810

PPM

OH

Ha

NO2

Hb

Hc

Hd

ddmight look like singlet or pt

ddd

pt

dddab dc

RO

012345

PPM

01234

PPM

X (no hydrogens)

Spinning bands

•Occur in very concentrated samples due to the coupling to 13C.

•Appear as very short and very symmetrical peaks on both sides of a peak.

•Will disappear under dilution.

Integrals

•Represent the ratio of each kind of proton.

•The heights of lines are proportional to the intensity of the signal.

3H'S

3H'S

2 H'S

O

O H H

O CH3

O

H3C O

O

Good websites for NMR tutorials:

http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/problems_fs_start.html

http://www.nd.edu/~smithgrp/structure/workbook.html

Summary:

•Information that can be obtained from H-NMR:

•Number of peaks: number of magnetically non-equivalent H’s

•Chemical shifts: chemical and magnetic environment of the given hydrogen

•Window: 0-15 ppm, signals to the left are downfield or deshielded hydrogens

•Signals to the right are upfield or shielded hydrogens

•The more electron density is on the H, the more shielded it is.

•Splitting patterns: the number of H’s on adjacent carbons

•Determined by N+1 rule for freely rotating systems, where N is the number of

hydrogens on adjacent carbons

•Most common patterns:

•Isolated ethyl group: a quartet and a triplet

•Isolated isopropyl group: a septet and a tall doublet

•Para-substituted aromatic ring: two doublets at 6.5-8.5 ppm

•Meta-substituted aromatic ring: a distinct singlet in aromatic region

•Acidic protons are usually unresolved wide signals and will change their position

with concentration and in the presence of water

•Integrations: area under the peak, represents the relative number of hydrogens

under the peak.

•If the area under the peak is not a whole number, multiply all the integrations by

an integer to obtain a whole number for all peaks.

Solving an NMR problem: C10H12O

Solving an NMR problem: C8H10BrN Solving an NMR problem: C

7H12O3

Solving an NMR problem: C10H12O3


Solving an NMR problem: C8H11N Solving an NMR problem: ExampleC

6H5N

H NMR Lecture

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