UNIVERSITY OF CAMBRIDGE

DEPARTMENT OF CHEMISTRY

IA CHEMISTRY 2009/2010

The Shapes and Structures of Molecules

Part One

OO

OO

O

O

HO

HO

Me

OH

Me

Cl

OH

O

HO Me

OMe

HO

H

OMe

OO

Supervisors'

version

Lecturer: Dr Peter Wothersemail pdw12@cam.ac.uk

Representing MoleculesChemists need to be able to communicate with other chemists about the work they do. Overthe years, many different systems for drawing and naming compounds have evolved. Whilstchemists have more or less settled on how to draw structures, naming them is more complicated.As we shall see, the systematic rules are often put to one side in favour of names that are easierto handle. In any case, we need to learn the language before we can study the chemistry.

Drawing moleculesThere are many different ways to depict molecules: butane for example could be written C4H10,CH3(CH2)2CH3, CH3CH2CH2CH3 or in a pictorial manner. Which way we use depends on whatwe are trying to show. The formula C4H10 only tells us the elemental composition – it does nottell us which isomer we are referring to. The displayed formulae indicate which isomer we arereferring to, but not the shape of the molecule.

C

H H H H

H H H H

HH C C C

This structure shows the connectivity of all the atoms but is rather misleading about the 3D shape of the molecule.

Don't draw structures like this again!

The structure of a tetravalent carbon atom is based on a tetrahedron – methane could be drawn:

C

H

HH

H

these bonds are in the plane of the paper

this bond goes back into the paper

this bond comes out of the plane

HC

CC

C

H

H H

H HH H

HH

A more informative drawing of butane would be:

The drawing shows the tetrahedral structure but is rather cluttered. With larger molecules, it getseven harder to see what is important and what is not. What we need is a way of representingstructures in an informative yet clear way. Most chemists would draw butane like this:

this represents a CH3 group

this represents a CH2 group

1

Guidelines for drawing organic structures

1. Draw chains of atoms as zig-zags with 120◦angles to show their shape.

2. Miss out the capital C’s representing carbon atoms.

3. Miss out the H’s attached to carbon atoms, along with the C–H bonds.

4. Draw in all other atoms, together with their bonds and all the atoms attached to them.

5. If a carbon atom is drawn in, include all the other atoms attached to it.

Examplespropanol

(propan-1-ol)

3-hexene 3-hexyne

cyclohexane benzene

propanoic acid glycine

propionitrile(CH3CH2CN)

propylamine(CH3CH2CH2NH2)

propanol(propan-2-ol)

OH

C–O bond

no carbon in here

OH

cis trans

or

OHCOOH

Oor

H2NOH

O

N CNor

NH2

draw triple bonds linear

2

Common abbreviationsSome frequently used groups have their own abbreviations to save time writing them out fully:

Me Et

nPr

nBu

t Bu

Ac Ph

R Ar

i Pr

= methyl, -CH3 = ethyl, -CH2CH3

= normal propyl, -CH2CH2CH3

= normal butyl -(CH2)3CH3

= tertiary butyl -C(CH3)3

= acetyl CH3CO-

= any alkyl group = any aryl group(a substituted benzene ring)

= phenyl -C6H5

= iso-propyl,

O

Inorganic moleculesIn organic structures, a line joining two atoms represents a bond but in inorganic structures thisis not necessarily the case. The lines indicate that the atoms are in some way connected, butin some cases they are often no more than a guide to the eye to indicate the geometry of thestructure.

S

F

F

F F

F F B

B B

BH

HHH

H

H

HH

H

S has 6 bonds NaClwould not say each

Na & Cl has 6 bonds

?should the borons be joined?

does H have 2 bonds?

3

Naming compoundsOften the simplest way of referring to a compound is to draw its structure – in scientific papers,for example, structures are given and referred to by numbers e.g. “16 was obtained in goodoverall yield and converted smoothly to the cyclic ether 17 by treatment with N-iodosuccinimide,followed by tri-n-butylstannane reduction.”

HOMe OMe

O O

HH

MeO MeOMeOMeO

HO HOH

1. N-iodosuccinimide2. nBu3SnH

(71% yield)

16 17

Example taken from an issue of one of the chemical journals, Tetrahdron Letters.

Nonetheless, chemists still need to talk about the chemicals they use. There are books of rules(and even computer programs) which enable any compound to be given a systematic name, butthese names are rarely used in the laboratory or even in scientific papers. In the above example,N-iodosuccimide was used. This is not a systematic name but a trivial name. Trivial names arepreferred, mainly because they are often considerably shorter. The names of the very commonreagents will simply have to be learnt.

O

OH

O

O

O

O

CHCl3

O

OH

OH

O

O

N

O

O

I

acetone

acetic anhydride chloroform

succinic acid N-iodosuccinimide

acetic acid (diethyl) ether toluene

Determination of molecular structure

Chemists have an array of techniques to help them work out what compound they have madeincluding X-ray diffraction, nuclear magnetic resonance, IR spectroscopy, UV-visible spec-troscopy, microwave spectroscopy and mass spectrometry. Each technique reveals differentinformation and each has its own particular advantages and disadvantages.

4

X-ray crystallography

The spacing between atoms in most compounds is comparable to the wavelength of X-rays, bothbeing around 1 Å (100 pm). As a result, when a beam of X-rays is focused onto a single crystal,the beam is diffracted. By careful analysis of this diffraction pattern, the positions of the atomswithin the crystal can be deduced. This is certainly the most powerful technique for determiningthe structure of crystalline materials.

N(3)

C(4a) C(3a)

F(2) F(1)C(3)C(4)

C(2) C(1)

N(1)S(1)

S(1a)N(1a)F(1a)

C(8)C(5)

F(2a)

or or just

doesn't show double or triple bonds but does give bond lengths

CN C

NS

SN

F F

F F

C

NS

SN

NS

SN

atomic displacement parameters showuncertainty in position of atoms

5

Not only does the crystal structure give the bond lengths and angles within a single molecule,but it also shows how all the molecules pack together. The packing influences the behaviour ofthe solid. unit cell - shows

repeating structure

The X-rays are diffracted by interactions with the electrons in the molecule rather than with thenucleons (protons and neutrons). Hence the technique really produces electron density maps.

O(11)

C(14)

C(13)

C(12)

C(11)

N(11)

O(12)

C(15)

N(12)O N

N O

H

H

HH

H

H

contours join regions of equal electron density

note oxygens have most electron density

Notice that the hydrogen atoms do not show up in the above electron density map. This is oftenfound to be the case in X-ray structures – hydrogen atoms have little electron density associatedwith them and so are often difficult to locate.

Advantages and disadvantages of X-ray crystallography

Advantages

• The ultimate method for structural information. Gives locations of all the atoms.

Disadvantages

• Need good quality crystals

• (Sometimes hard to locate hydrogen atoms)

6

Mass spectrometry

This technique effectively weighs molecules. The sample is introduced into the instrument,vaporised and ionised. By focusing the ions onto a detector using magnetic and electric fields,it is possible to determine the masses of the ions present. The results can be very accurate andmeasure to one part per ten million.

Different techniques are used to ionise the sample: the crudest is to knock electrons out of thesample by firing high energy electrons at the vapour. This technique yields a positive molecularion, M+.

A much more effective, gentle ionisation technique is electrospray. In this technique, thesample is introduced into the instrument as charged aerosol droplets. The solvent evaporates inthe vacuum, leaving charged molecules. What is detected is not the M+ molecular ion, but themolecule with an ion stuck on. The detected ion may be positive (for example with a sodiumion attached) or negative (for example with a methoxide ion, MeO− or acetate ion, CH3CO2

−,attached).

When working out the masses of different ions in a mass spectrum, it is important to use theatomic masses for the specific isotopes present and not the usual relative atomic masses whichare the average of these isotopic masses.

79 81

E.g. naturally occurring Br is a 50:50 mixture of 79Br and 81Br. Hence relative atomic mass of Br is 80 but in the mass spectrum of Br+ no peak is seen at 80, instead see one at 79 and one at 81.

Naturally occurring carbon, hydrogen, oxygen and nitrogen all consist of largely one isotope:12C, 1H, 16O and 14N with accurate masses 12 (exactly, by definition), 1.00782, 15.99491 and14.00307 g mol−1 respectively.

By very accurately measuring the masses of ions, it is possible to determine their molecularcomposition. As an example, consider the molecules CO, N2 and C2H4 all of which have a massof 28 g mol−1. Using the accurate masses for the most common isotopes, we may calculate theaccurate masses of these different molecular ions:

12C16O 27.99491 12C21H4 28.03130 14N2 28.00614

Thus a peak at 28.031 could only be due to [12C21H4]+.

7

Example – Spirastrellolide ASpirastrellolide A is a natural product obtained from a marine sponge, and has been synthesisedin these laboratories. The structure, as worked out in May 2004 using MS and NMR, is givenon the front of the handout.

Accurate mass spec of [methyl ester + Na+ ion] = 1049.52026

Best sensible formula = C53H83O17Cl Na (calc mass = 1049.52110, error = 0.8 ppm)

Fragmentation in Mass SpectrometryOnce ionised, the sample molecule begins to fall apart and so typical mass spectra contain manypeaks. Clues may be gained about the structure of the molecule by analysing these peaks butthese days there are many better techniques for structural determination.

Example – the negative ion mass spectrum of [Rh6(CO)16 +MeO−]

500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 12000

50

100

m/z

inte

nsity

%

1096.6

1040.6

1012.6

984.6

694.6

666.6

638.5

610.6

1068.6

Rh6(CO)16 + MeO-

Rh4(CO)9 + MeO--CO -CO

-CO

-CO

-CO

-CO

-CO

The complexity of the fragmentation pattern means that each one is very different for differentcompounds – in a sense, it is like a fingerprint for that compound. This makes mass spec anexcellent analysis tool for identifying compounds whose spectra have already been recorded,especially since only a small sample is needed (just a few million molecules!).

It is also possible to isolate certain ions in the instrument and then follow these individualions to see how they fragment. This technique is called MS/MS. In the following example ofnegative ion MS/MS, the spectrum has been processed to show only the ions which fragmentedto give NO−2 ions (called the daughter ion).

Example - analysis of an explosive mixture

daughter ion

46

221

226

267

281

315

272

50 75 100 125 150 175 200 225 250 275 300 325 350 m/z25

rela

tive

inte

nsity

%

MS/MS parent-ion scan spectrum of explosive mixture

0

12.5

6.2

negative ion MS/MS of mixture + formate ions + acetate ions

[PETN-H+]

[TNT-H+]

[NG-H+]

[RDX+AcO-][RDX+HCO2

-]or

(NO3-)

NO2-

8

CH3

O2N NO2

NO2

O2NO ONO2

ONO2

N

N N

NO2

O2N NO2O2NO

O2NO

ONO2

ONO2

nitroglycerine, NG (227)TNT(227)

RDX(222)

PETN(316)

all give NO2- ion

on fragmentation

make up SEMTEX

Advantages and disadvantages of mass spectrometry

Advantages

• Gives molecular formula

• Excellent for analysis of mixtures

• Tiny sample needed – much less than for any other technique

Disadvantages

• Often difficult to interpret, especially when working with unknowns

Spectroscopic techniquesYou may already be familiar with the idea that if we were to measure the energy of electronin an atom, we would find that it could only have certain energies. We say the energy levelsare quantised. To promote an electron from one energy level to another requires a certain fixedamount of energy.

Similarly, whilst molecules are constantly moving around and therefore possess transla-tional, vibrational and rotational energy, the amount of rotational and vibrational energy amolecule may possess is also restricted. We say that the vibrational and rotational energy levelsare also quantised. It is possible for a molecule to be promoted from one vibrational energy levelto another or from one rotational energy level to another by absorbing light energy. The separa-tion between the two energy levels, ∆E, is proportional to the frequency of the light absorbed,ν.

∆E = hν where h is Planck’s constant = 6.626 × 10−34 J s.

9

The exact frequencies of light absorbed depends on the particular molecule in question – dif-ferent molecules absorb different frequencies of light. This is the basis of all spectroscopictechniques.

The type of transition caused by the light depends on the frequency (and hence the energy)of the light; in other words, different regions of the electromagnetic spectrum cause differentprocesses in atoms and molecules.

Radio Microwave Infrared Visible Ultraviolet X-rays γ-Rays

λ (nm) 109 (1 m) 3 × 105 (0.3 mm) 780 380 10 0.1 0.001 ν (cm–1) 0.01 33.3 12 800 26 300 106 108 1010

ν (Hz) 300 × 106 1012 384 × 1012 789 × 1012 30 × 1015 3 × 1018 300 × 1018

E (kJ mol–1) 0.00012 0.4 150 310 1200 1.2 × 106 108

∼

red

oran

geye

llow

gree

nbl

uein

digo

viol

et

The Electromagnetic Spectrum(all regions are approximate)

wavelength:

wavenumber:

frequency:

energy:

transitions in alignment of nuclear spins (see NMR)

rotational transitions

nuclear processes

vibrational transitions

electronic transitions

In this course, we shall look at two types of spectroscopy – infrared spectroscopy, which looksat how bonds vibrate, and nuclear magnetic resonance spectroscopy, NMR.

Nuclear Magnetic Resonance, NMR

10

You may already be familiar with the idea that electrons possess a property called ‘spin’ andthat we denote electrons with opposite spin with up and down arrows, ↑ and ↓. Spin becomesimportant, when, for example, working out the lowest energy arrangement of electrons in atoms.

In an analogous manner, certain (but not all) nuclei also possess spin and as a result, thenucleus has a very weak magnetic field associated with it. When placed in a strong magneticfield there is an interaction between this nuclear spin and the applied field which gives rise toa set of nuclear spin energy levels. Radiowaves of the appropriate frequency cause transitionsbetween these energy levels and this gives rise to the NMR signal.

in magnetic field

E2

E1

"spin down"

"spin up"

radiowaves

hν

ν

gives

frequency

NMRsignal

The nuclear spin is specified by the nuclear spin quantum number, I. It may take values of 0, 12 ,

1, 32 , . . . , with the particular value depending on the nucleus concerned. A nucleus with spin I

gives rise to (2I + 1) different energy levels when placed in a magnetic field.

Nucleus % Natural abundance I Number of energy levels1H, a single proton 99.985 1

2 22H, a deuterium nucleus, D 0.015 1 3

10B 20 3 711B 80 3

2 412C 98.9 0 113C 1.1 1

2 214N 99.6 1 316O 99.8 0 119F 100 1

2 2

Note that different isotopes of the same element can have different values of I. The value of Ifor a particular nucleus depends on the number of protons and neutrons in the nucleus. It is notsimple to predict the observed value of I but there are some useful guides:

• nuclei with odd masses have half-integral spin (e.g. 11B, 13C)

• nuclei with odd numbers of protons and odd numbers of neutrons have integral spin(e.g. 2H, 10B, 14N)

• nuclei with even numbers of protons and even numbers of neutrons have zero spin(e.g. 12C, 16O)

11

The exact difference in energy between the different spin states depends on the strength of themagnetic field the nucleus is in and on the nucleus itself. The stronger the external magneticfield, the larger the energy separation between the different spin states.

energies of 1H spin states

energies of 13C spin states

strength of magnetic field strength of magnetic fieldB B

same energy in absence of mag field

1H

1H

energy separation of 1H spin states in mag field of strength B

smaller energy separation of spin states for 13C

13C

13C

As a result, the exact frequency a nucleus resonates at depends on the particular nucleus inquestion and the strength of the magnetic field it is in:

NucleusResonance frequency in

a 4.7 T magnetic field/MHz

Resonance frequency inan 18.8 T magnetic field

/MHz1H 200 8002H 30.7 122.813C 50 20031P 81 324

195Pt 43.3 173.2

(Earth’s magnetic field is approx 50 µT.)

Local magnetic environmentsNMR would be of little use if all hydrogen nuclei (simply called protons in NMR) resonated atthe exact same frequency or if all carbon-13 nuclei resonated at the exact same frequency. Butthis is not the case. The exact difference in energy between the different spin states depends onthe local magnetic field experienced by each of the nuclei. This in turn varies depending on theelectron density surrounding each nucleus. In a magnetic field, electrons move in such a way asto set up their own magnetic field which opposes the applied field.

appl

ied

field

electron 'cloud'

magnetic field set up by moving electrons opposes the applied magnetic field

12

The nucleus then experiences a weaker magnetic field than it would have done had the electronsnot been there. The weaker the magnetic field, the smaller the difference in energy between thenucleus’ spin states and hence the smaller the frequency of radio waves absorbed. Electrons aresaid to shield the nucleus – the more electrons, the more shielded the nucleus is and hence thelower the frequency that it resonates at. Conversely, anything which removes electron densityfrom around a nucleus, deshields it and shifts the resonance to a higher frequency.

Since electronegative elements withdraw electrons towards themselves, they will deshieldother atoms attached to them.

CH4CH3ClCH3OHCH3F

increasing frequency of 13C nucleus

As you might expect, the frequency shift caused by the different degrees of shielding is quitesmall, just hundreds of Hz. This is in contrast to the different frequencies that different nucleiresonate at which are typically hundreds of MHz apart.

Equivalent nuclei – number of peaksSince the exact frequency at which a nucleus resonates at depends on the electronic environmentthe nucleus is in, each ‘different’ nucleus will resonate at a different frequency. Consider the 13CNMR spectra for two isomers of butanol: in n-butyl alcohol, each carbon atom is in a slightlydifferent environment and so gives a separate peak.

0 ppm50100150200

OH

1 13

3

2

2

4

4

solventCDCl3nBuOH

In t-butyl alcohol, the symmetry of the molecule means that three of the carbon atoms are equiv-alent and so only two peaks are observed.

0 ppm50100150200

OH

1

1

2

2

2

2tBuOH

13

This simple example shows how NMR can help to distinguish between two isomers, simply bylooking for some symmetry in the molecule. When predicting what a spectrum would look likefor a given compound, the first step is to look for symmetry in the molecule to see how manyresonance signals we should expect.

Examples O O

Cl

Cl

Cl

Cl

Cl

Cl

1

1 1

1

1

1

3

3

33

33

22

2

2 2

2

22

22

4

4

55

4

5 signals

3 signals 4 signals 2 signals

9 signals - some may be very close to each other

(spectra on p483)

The NMR chemical shift scaleSince the exact frequency that a nucleus resonates at depends on the magnetic field it experi-ences, the same compound would resonate a different frequencies in different strength magneticfields. To avoid the confusion this would create, we quote the resonance frequency of a lineas its shift from an agreed reference compound, expressed as a fraction to compensate for themagnetic field strength.

increasing frequency

peak of interest

peak from reference compound

The chemical shift of a resonance is calculated from

chemical shift, δ, in ppm = 106 frequency of resonance - frequency of reference

frequency of reference

difference just a few Hz

millions of Hzmultiply by 106 to get a reasonably

sized number expressed in ppm

Any differences in the external magnetic field strength affects the frequecies of all the signalsequally by the same factor. Because of the way it is defined, the shift scale remains unaffected:

chemical shift (in ppm) = 106 K x frequency of resonance - K x frequency of reference

K x frequency of reference

14

The reference compoundWhat do we want in an ideal reference compound?

• Chemically inert, since it is often added to the sample.

• Preferably just on signal – nothing too complicated.

• Preferably want the signal to be out the way of other peaks in the spectrum.

Si

H

H H

H Si

CH3

CH3

CH3H3C

Compound used is TMS - tetramethylsilane

tetramethylsilane, Si(CH3)4silane, SiH4

All the methyl groups in TMS are chemically equivalent, that is, they all have the same electrondensity around them. This means that the compound shows just one signal in its carbon NMRspectrum. All the protons are equivalent too so this compound also shows just one signal inits proton NMR. Silicon, being below carbon in the periodic table, is less electronegative thancarbon which means that rather than withdrawing electrons from the carbon, it is actually slightlyelectron donating and so the carbons are more shielded relative to most organic compoundswhich have just C–C bonds. The net result is that in the 13C NMR spectra of most organiccompounds, the TMS reference signal is over to the right, out of the way.

Compound Shift of methyl carbon / ppm

CH3Li –14

Si(CH3)4 0 by definition

CH3CH3 7

CH3Cl 26

CH3NH2 28

CH3OH 50

CH3NO2 61

CH3F 72

CH3Li is like “CH−3 Li+” – the carbon is heavily shielded

15

Dividing up the scaleCarbon NMR spectra are usually run from 0 to just over 200 ppm since most organic moleculesfall in this region. The scale can be conveniently broken down into different regions:

sp2 carbons with very electron withdrawing groups attached

sp3 carbons with very electron withdrawing groups attached

sp2 carbons sp3 carbons

200 150 100 50 0 ppm

C O C C

C C

C C

C NC O

(sp carbons)large shifts('downfield')

small shifts('upfield')

These divisions act as guide lines only – for example, given enough electron withdrawinggroups, an sp3 hybridised carbon will creep into the 100-150 region.

OMe

H OMeOMe

51 ppm114 ppm

CH3Cl 26 ppmCH2Cl2 54CHCl3 77CCl4 96

an orthoester

The carbonyl region ca. 150-200 ppmWhilst most carbonyl groups fall in the 150-200 ppm region, it is useful to sub-divide this regionfurther:

i) around 200 ppm

ii) around 160 -170 ppm

O

H

O

OH

O

OR

O

Cl

O

NH2

O

ketonestypically just over 200 ppm

aldehydestypically just under 200 ppm

acid derivatives

Note that the carbonyl carbons for the acid derivatives resonate at a lower shift than the ketonesand aldehydes, even though they have more electron withdrawing groups attached! This indi-cates that the carbon atoms in the acid derivatives are more shielded and that the heteroatoms(i.e. O, N, Cl) must be donating some electron density. This is an idea we shall return to later inthe course.

16

0 ppm 50100150200

Example spectrum - hallerone O

HO O

O

1

1

3 3

3 2

2

4 4

45

5

6

6

7

7

8

8

solvent, CDCl3 at 77 ppm

Notice that the resonance lines from the quaternary carbon atoms (i.e. ones with no hydrogenatoms attached) appear much smaller than those of the carbons that do have protons attached.This effect is a consequence of the way the spectra are recorded and need not concern us further.With care this feature can be used to help identify quaternary carbons, although weak signalscould also be due to impurities.

Coupling between nucleiCoupling between nuclei with spin 1

2

The exact frequency a nucleus resonates at depends on the magnetic field it experiences. Wehave already seen that the electron density surrounding a nucleus affects the local magnetic fieldit experiences – so too does the magnetic field exerted by nearby nuclei. Consider the 13C NMRspectra of acetic acid and fluoroacetic acid.

0 ppm 50100150200

0 ppm 50100150200

x 5 expansion

HOC

CH3

O

HOC

C

H H

F

O

note smallquaternarypeak

acetic acid

the signal from each carbon is split into a doublet

17

Apart from causing the resonance for the sp3 carbon to shift to a higher ppm, the fluorine atomhas another affect on the 13C spectrum – where before there were single lines for each carbonresonance, now each signal has split into a pair of lines. Each pair is called a doublet. The twolines which make up the doublet are placed symmetrically about the chemical shift position andtheir separation gives the value of the coupling constant, J, which is measured in Hz. The carbonand the fluorine nuclei are said to couple with each other.

in 13C spectrum in 19F spectrumwith no coupling

with coupling same splitting, J

two doublets

J J

To understand the coupling, consider the 13C nuclei. In addition to the applied magnetic field,the 13C nuclei also experience an additional field due to the nuclear spin of the fluorine atoms.Since for fluorine I = 1

2 , the fluorine nuclei could be in either of two spin states: up or down.Those fluorine nuclei that are spin up reinforce the magnetic field experienced by the carbonnuclei and shift them to a slightly higher frequency than they would otherwise resonate at.

Conversely, those fluorine nuclei that are spin down reduce the magnetic field experiencedby the carbon nuclei and shift them to a slightly lower frequency.

13C

F spin down decreases field felt by 13C and hence

a smaller shift

F spin up increases field

this is where the carbon signal would have come if unaffected by the spin of the F

13C -19F19F -13C

The effect of coupling between nuclei is often determined using a “tree” diagram in which weshow a stick spectrum without coupling and then add the coupling in the next row and show howthe lines in the second row are related to those in the first.

original shift position with no coupling

resultant doubletJ

18

Notice in the spectrum of fluoroacetic acid that the coupling to the sp3 hybridized carbon is muchlarger than that to the carbonyl carbon. This is because coupling is a through bond interactionand gets weaker the more bonds there are between the coupling nuclei. It is vary rare to see anycoupling at all between nuclei more than four bonds apart. In 13C NMR, it is rare to see anycoupling through more than one or two bonds.

Coupling constants are represented xJY–Z where x is the number of bonds through whichnuclei Y and Z are coupling. In the spectrum of fluoroacetic acid, the coupling constants aredenoted 1JC–F and 2JC–F.

Sometimes splittings are not seen• C–H coupling

No C–H coupling is usually seen because 13C NMR spectra are usually proton decoupled (seelater).

• C–C couplingNo C–C coupling is usually seen because of the low abundance of 13C in naturally occurringcarbon. Could see coupling if the sample has been artificially 13C-enriched.

HOC

CH3

O

1212

HOC

CH3

O

1312

HOC

CH3

O

1213

HOC

CH3

O

1313

most molecules are like this and not NMR active

1% have a 13C here. Gives 1 peak in spectrum at 177 ppm

another 1% have a 13C here. Gives 1 peak in spectrum at 22 ppm

only 0.01% have 2 13C atoms next to each other. Each signal would be a doublet but is lost in the noise.

• Equivalent nucleiSplittings due to couplings between equivalent nuclei are not seen! Equivalent spins dointeract with each other, but the effects of this coupling are never seen as splittings. To allintents, it appears that equivalent nuclei do not couple with each other.

13C 13C

OO

OHHO

no splittings would be seen in 13C enriched oxalic acid because the 2 carbon atoms are equivalent

19

Coupling to more than one nucleiWe can use a tree diagram to predict what the 13C NMR spectrum for 13C-enriched fluoroaceticacid would look like.

Coupling to the fluorine splits this line into a doublet with coupling constant 1JC–F

Coupling to the other 13C splits each of these lines into a doublet with coupling constant 1JC–C

Coupling to the other carbon splits this into a doublet with coupling constant 1JC–C

The longer range 2-bond coupling to the F splits each line into another doublet with coupling constant 2JC–F

The order in which we draw the splittings makes no difference!We still get the same spectrum.

HOC

C

H H

F

O

13

13

1JC–F

1JC–F

1JC–F

1JC–C

1JC–C

1JC–C

1JC–C

1JC–C

1JC–C

2JC–F

2JC–F

2JC–F

a doublet of doublets

a doublet of doublets

20

Difluoroacetic acid

0 ppm50100150200

× 5 expansion

HOC

C

F F

H

O

1JC-F 239 Hz

2JC-F 28 Hz

Now each carbon appears as a triplet. Each of the carbon nuclei couples to two equivalentfluorine nuclei. The spins of these can be arranged in four ways:

with both F spins up, the13C has the greatest shift

with one F up and one F down, no overall effect on shift of carbon

since the F nuclei are indistinguishable, these two arrangements are magnetically equivalent

with both F spins down, the carbon resonance is shifted to a smaller chemical shift

Thus we expect to see three lines in the spectrum and because it is twice as likely that the fluorinespins are one up and one down rather than both up or both down, the net result is a triplet withratio 1 : 2 : 1.

The tree diagram shows that a triplet is really just a special case of a doublet of doubletswhere because of the equivalence of the two fluorines, the coupling constants are the same andhence two of the signals overlap to give twice the intensity.

original shift

coupling to one F splits the signal into a doublet

coupling to the second F splits each line into a doublet again

Since the coupling constant is the same, the 2 central signals fall on top of each other

1 : 2 : 1net result is a triplet

21

In trifluoroacetic acid, each carbon resonance appears as a quartet.

0 ppm 50100150200

HOC

CF3

O1JC-F 287 Hz2JC-F 38 Hz

1 : 3 : 3 : 1

3 ways 3 ways

3 spins up, larger shift

3 spins down, larger shift

net 1 spin up, small shift

net 1 spin down, small shift

We can construct a tree diagram for the quartet:

original shift

coupling to 1st F splits the signal into a doublet

coupling to 2nd F splits each line again

same coupling constant each time

coupling to 3rd F splits each line into 2 to give rise to the quartet

J

J

J

1 1: : :3 3

22

In general, when a nucleus couples to n equivalent nuclei with spin I , its resonance signal issplit into (2nI + 1) lines. If the nucleus couples to n equivalent nuclei with spin 1

2 , the formulasimply becomes (n + 1). Notice how for coupling to spin 1

2 nuclei, the ratio of the intensitiesmay be predicted by Pascal’s Triangle.

1 4 6 4 1 quintet

1 3 3 1 quartet

1 2 1 triplet

1 1 doublet

1 singlet

:

: :

: : :

::::

Examples – PF3 and XeF+3

FP

F

F

In 31P spectrum, the P 'sees' 3 equivalent Fs and so is split into a

quartet with intensity ratio 1 : 3 : 3 : 1

19F spin 1/2

In the structure of PF3, all the fluorine atoms are equivalent. This is not the case in the ion XeF+3which has a ‘T-shape’.

Xe

Fa

Fb

Fa

+The three fluorine atoms are no longer all equivalent; the two marked Fa

are equivalent, but Fb is different. This means that in the 19F spectrum,there would be two signals due to the fluorine atoms – one resonancedue to Fa and one due to Fb (these would then be split further by cou-pling to the 129Xe). The 129Xe NMR spectrum has been recorded forthis compound (129Xe, I = 1

2 ). The xenon couples to the two equivalentFas to give a triplet, ratio 1 : 2 : 1, but then each of these lines is furthersplit into doublets through coupling to Fb. The net result is described asa triplet of doublets. This is best shown using a tree diagram.

550650 600 ppm

129Xe NMR spectrum of XeF3+

coupling to 1st Fa gives a doublet

coupling to 2nd Fa gives a triplet

coupling to Fb splits each of these lines into doublets

JXe-Fa

JXe-Fa

JXe-Fa

JXe-Fb

JXe-Fb

1 : 1 : 2 : 2 : 1 : 1

Net result is a triplet of doublets

2380 Hz

2600 Hz

23

Natural abundance and satellitesThe low abundance of 13C (1.1%) compared to 1H (almost 100%) means that more sample isneeded to record a 13C NMR than a proton NMR. We saw earlier how this low natural abundanceof 13C means that no C–C coupling is observed in 13C NMR spectra. A further consequence isthat no coupling to carbon is observed in proton NMR.

Consider the NMR spectra for chloroform, CHCl3. When recording a 13C NMR each 13C isattached to a spin 1

2 proton and so unless we applied broadband proton decoupling (see later),coupling to the proton would mean the 13C signal would appear as a doublet. In contrast, whenrecording a 1H NMR spectrum of chloroform, most protons in the sample are attached to spin-zero 12C and so no coupling is observed. However, in 1% of the sample, the proton is attached toa 13C which means if we looked very closely at the baseline, we just might see the tiny fractionthat appears as a doublet.

13C1H

×100

proton coupled 13C

NMR of CHCl3

1H NMR

13C satellites

1JC-H1JC-H

The 13C satellites are usually lost in the background noise of the 1H spectrum, but there are someisotopes of other elements where satellites are much more prominent because they are presentat higher levels. One such case is with platinum. There are six naturally occurring isotopes ofplatinum and all but one have spin I = 0. The exception is 195Pt which is 34% abundant and hasspin I = 1

2 .

Example – CisplatinCisplatin was one of the earliest suc-cessful drugs used in chemotherapyunder such names as ‘Neoplatin’.

Pt

Cl

Cl NH3

NH3

–60–50 –70 –80 ppm

15N NMR spectrum of 15N enriched cisplatin

This is not a triplet but a doublet superimposed on a singlet

ratio 17 : 66 : 17

most 15N is attached to Pt with spin I = 0 so appears as a singlet

smaller amounts of 15N is attached to spin I = 1/2 195Pt

JN-Pt

Most of the nitrogen nuclei are attached to spin zero isotopes of platinum and hence resonate asa singlet. However, one third of the cisplatin molecules contain 195Pt which splits the nitrogenresonance into a doublet – what is observed is a doublet superimposed on the singlet.

24

Decoupling13C spectra are usually recorded in such a way so that coupling to protons does not appear –such spectra are said to be proton decoupled. All the 13C NMR spectra you will usually meetare recorded in this way. While the 13C spectrum is being recorded, the protons are irradiated insuch a way as to cause their nuclear spins to move rapidly between their “up” and “down” spinstates. If this happens fast enough the couplings to 13C average to zero and all of the splittingsdisappear. This process is called broadband proton decoupling. The “broadband” refers tothe fact that we are irradiating over a range of frequencies so that all the protons are rapidlyinterconverting between their different spin states.

Shown below are the proton-decoupled and the proton-coupled spectra of 3-methyl-4-hydroxybutan-2-one.

200 150 100 50 0

200 150 100 50 0

proton decoupledspectrum

proton coupledspectrum

1

1

1

3

32

2

2

45

5

4H3C OH

O

CH3

HH H

The proton coupled spectrum gives us more information than the decoupled spectrum in thatit is easy to see whether the carbons are CH3, CH2, CH or quaternary since these appear asquartets, triplets, doublets or singlets respectively. However, there are several disadvantages.The signal strength is now much weaker (compare the relative sizes of the solvent peaks) whichmeans that some signals could easily become lost in the background noise. In most compoundsthe multiplets will overlap and it would be very hard to say which peaks go together as part ofa multiplet and which are due to different carbon nuclei. This particular example was chosenbecause all the signals are well separated.

There is a way to get the best of both worlds – an experiment called the Attached ProtonTest, APT, is run on the NMR spectrometer. How this works need not concern us, but the resultis that the peaks from the carbon atoms with an even number of protons attached, 0 or 2, pointone way (the same way as the deuterated solvent) and the carbon atoms with an odd number ofprotons, 1 or 3, point the other way.

200 150 100 50 0

quaternary & CH2s

CH & CH3

APT

solvent (CDCl3)no H's

25

Proton NMRThe chemical shift scale in 1H NMR is much smaller than that in 13C NMR. The same refer-ence compound, TMS, provides the zero, but the protons in most organic compounds resonatebetween 0 and about 14 ppm. Remember that the actual frequencies that carbon and hydrogennuclei resonate at are significantly different – this 0-14 ppm is not the start of the carbon 0-200ppm scale!

Proton NMR spectra rapidly get very complicated due to the coupling. However, this alsomakes 1H NMR incredibly useful since so much information can be deduced from the spectra.Proton NMR will be covered in much greater depth next year – at this point we will only lookat some very simple examples.

Example: 1H NMR spectrum of ethyl acetate

5 4 3 2 1 0 ppm

H3 CCH2

O CH3

O

quartet since couple to 3

equivalent Hs on Me

triplet since couple to CH2

protons

no coupling seen between

equivalent nuclei

Example: 1H NMR spectrum of styrene

4.55.05.56.06.57.07.58.0 ppm

Ha

Ha

Hb

Hb

Hc

Hc

phenyl protons too complicated!

doublet of doublets

each couple to Ha to give a doublet. Coupling between

Hb and Hc too small to show up!

Jab

Jab

Jac

Jac

26

Coupling to spins greater than 12

Many nuclei have spins greater than 12 and in principle these should show a coupling to nuclei

such as 13C. However, in most cases, this is simply not observed. This is because the spinstates rapidly interconvert due to a process called relaxation. Such a process is common to allnuclei with spin, but it turns out to be very rapid for most nuclei with spins greater than 1

2 .The resulting rapid interconversion of the spin states causes all the couplings to that nucleus toaverage to zero, just as in the case where we deliberately irradiated the protons in broadbanddecoupling (see page 25). However, there are a few examples of nuclei with spin greater than 1

2which do relax sufficiently slowly for coupling to be seen. A common example is deuterium.

In most of the 13C spectra we have seen so far there have been a set of peaks at 77 ppm dueto the solvent, CDCl3. The carbon resonance appears as a triplet because of the coupling to thedeuterium. The nuclear spin for deuterium is I = 1, which means that there are 3 different spinstates (2I + 1). Hence the carbon resonance is split into a triplet with ratio 1 : 1 : 1. The treediagram is now different from the spin 1

2 case since each line splits directly into three.

original shift

1 : 1 : 1

J Jcoupling to spin I = 1 splits the line into 3 with equal intensities

Notice that no coupling is observed to the chlorine nucleus in the spectra, even though both thenaturally occurring isotopes, chlorine-35 and chlorine-37, have spin I = 3

2 . Chlorine nuclearspin states relax too quickly for any coupling to be seen.

original shift

split into a triplet by the 1st D

each line is then split into another triplet by the 2nd D

1 : 2 : 3 : 2 : 1

Deuterated dichloromethane appears as a quintet, ratio 1 : 2 : 3 : 2 : 1; it comes at 54 ppm.

The number of lines into which a signal is split by coupling to n equivalent nuclei of spin I isgiven by the formula (2nI + 1). We can no longer use a simple Pascal’s triangle to work out theintensities of the lines – these are best found by constructing a tree diagram.

27

Infrared spectroscopy

IR spectroscopy looks at transitions between different vibrational energy levels in molecules.Whereas NMR gives us information about the different environments of nuclei in a sample, IRspectroscopy gives us information about the types of bonds present. A typical IR spectrum isshown below.

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By long standing tradition, IR spectra are plotted as percentage transmission versus frequency.When the sample does not absorb at all, the transmission is 100% and so the trace appears atthe top of the plot. When the sample does absorb light of a specific frequency, the amount oflight being transmitted through the sample drops and the trace moves down the plot. Absorptionpeaks thus correspond to the downward pointing features. Notice on this spectrum how the scalefor the frequency on the x-axis changes at 2000 cm−1.

The frequency is plotted in wavenumbers (cm−1). These are the number of whole wavesin one cm, or one over the wavelength expressed in centimeters. Wavenumbers are directlyproportional to frequency and may be treated as a unit of frequency.

wavenumber, ν̃, =1λ

(λ in cm)

as ν =cλ, (where c is the speed of light)

ν (in Hz) = c (in cm s−1) × ν̃ (in cm−1) i.e. ν̃ ∝ ν

28

Since ∆E = hν (see page 9), the frequency of the absorption is proportional to the energy for thevibrational transition. Hence peaks on the left of the spectrum at high wavenumbers correspondto vibrations which need more energy.

There are obviously many peaks in the IR spectrum and we need to try to make some senseout of them. There are many different ways in which a molecules can vibrate. The most usefulvibrations in an IR spectra are the stretches and these generally require most energy. Almost allthe peaks over 1500 cm−1 correspond to bonds stretching.

Modelling a vibrationWe may consider a simple diatomic molecule as two masses connected with a spring represent-ing the bond. As the bond is stretched or compressed, the energy of the system rises and there isa restoring force trying to return the bond to its equilibrium length.

r

ener

gy

0m1 m2

re

re

equilibrium bond length

extension

compression

A useful model to describe such a system, at least for small oscillations, is the harmonic oscil-lator. Such an oscillator is typified in the mechanical world by a weight hanging from a spring.When the weight is pulled down, i.e. displaced from its equilibrium position, the spring exertsa force to return it to the original position. The result is that the weight oscillates about theequilibrium position.

m

m

m

equilibriumposition

restoringforce

restoringforce

The frequency of the oscillations depends on two things, how heavy the weight is and how stiffthe spring is: the fastest oscillations occur with a stiff spring and a light mass. The relationshipbetween these parameters is given by Hooke’s law:

ν ∝!

k f

mwhere ν is the frequency of oscillation, k f is the force constant (which corresponds to how hardit is to stretch the spring and is measured in N m−1) and m is the mass.

We can adapt this model to apply to a vibrating diatomic. It would be reasonable to expectstronger bonds to be harder to stretch and so have larger force constants. Although how hard it isto stretch a bond is not the same thing as how hard it is to break a bond, there is a fair correlationbetween the bond strength and the force constant, as shown in the table below.

29

diatomicbond strength/ kJ mol−1

force constant/ N m−1

Br–Br 193 246H–H 436 575O=O 498 1177C≡O 1077 1902

The second thing we need to take into account when modifying the model, is that unlike in thecase of a mass hanging from the ceiling, when a diatomic molecule vibrates, both atoms aremoving. We can take both of the masses into account by using the reduced mass for the system,µ.

µ =m1m2

m1 + m2(units of mass)

Note how the expression for the reduced mass simplifies if one mass is much greater than theother:

when m1 ≫ m2 µ ≈ m1 m2

m1≈ m2

i.e. when m1 ≫ m2 µ ≈ m2 (lighter mass)

For a vibrating diatomic, the frequency of vibration, ν̃, in cm−1, is given by:

ν̃ =1

2πc

"k f

µ

with k f in N m−1, µ in kg (per molecule) and the speed of light, c, in cm s−1.

Because of the inverse relationship between the reduced mass, µ , and the frequency, the smallerµ becomes, the higher the frequency of oscillation. As we have seen, the reduced mass will besmall when we have a light atom attached to a heavy atom which in practice means when wehave an H atom attached to just about any other atom. This explains the first region in the IRspectrum: 2500-4000 cm−1; X-H stretches. The other regions in the spectrum correspond to thedifferent strengths of bonds (which in turn reflect changes in k f ): triple > double > single.

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X-H single bonds double bonds

triplebonds

N-H

C-H

'fingerprint region'

X-Y stretches and other vibrational

modes, e.g. bendingC C

C N

C=O

C=C

30

Group vibrations

The actual way in which a molecule vibrates may be rather complicated and involve all the atomssimultaneously moving. However, it is possible to break this complex motion down into certainso-called normal modes. The normal modes for ethyne are given below:

C C HH

C C HH C C HH

C C HH

C C HH

3374 cm-1

'C-H symmetric stretch'

1974 cm-1

'C-C stretch'

612 cm-1

'trans bend'729 cm-1

'cis bend'

3287 cm-1

'C-H antisymmetric stretch'

As we can see, it is not true to say that there will be one absorption in the IR spectrum for eachbond. Very often a single absorption corresponds to many bonds all vibrating at once. However,it is often the case that a given vibrational mode largely corresponds to one particular part ofthe molecule vibrating. For example, in the example above, the vibrations around 3300 cm−1

are mainly due to the C–H bonds stretching. It is this that makes IR so useful for identifyingfunctional groups in organic chemistry – all ketone carbonyls typically vibrate at roughly thesame frequency, slightly different from the frequency at which ester carbonyls typically vibrateand so on.

Strength of absorption

Whilst the frequency at which a bond absorbs depends on its strength, exactly how big the peakis depends on the dipole moment of the bond. Light is an oscillating electromagnetic field andthis can interact with a vibrating dipole – the larger the dipole moment in the bond, the strongerthe absorption. If there is no change in the dipole moment when the bond stretches, then it willnot absorb IR at all.

C O C C C N N N

strong dipole moment, strong

absorption

strong dipole moment, strong

absorption

no dipole moment, no absorption

of IR

If completely symmetrical,

then no dipole & no C=C absorption

(+)(+) (-) (-)

31

Just because a molecule like N2 cannot absorb IR, it does not mean it is not vibrating. Frequen-cies for such symmetrical vibrations with no change in the dipole moment may be determinedusing a technique complementary to IR spectroscopy called Raman spectroscopy. Whereas nor-mal IR spectroscopy looks at the frequencies of light absorbed by a sample, Raman spectroscopylooks at the frequencies of light scattered by a sample.

A given vibrational mode may be IR active only, Raman active only or both IR and Ra-man active. Deciding which is the case is rather involved and will be studied next year. Fordiatomic molecules however, homonuclear diatomics such as O2 and N2 will only be Ramanactive whereas heteronuclear diatomics, such as HCl and CO will be both Raman and IR active.

Example spectrum – cyanoacetamide

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N-HC-H

C N

C=Oamide bend

fingerprint

CNH2

ON

We shall now look at each of the main regions of the IR spectrum in a little more detail.

X–H regionC–H stretches usually 2900 – 3200 cm−1 – typically just less than 3000 cm−1.

exception: C≡C–H unusually strong, sharp absorption ∼ 3300 cm−1.

C–H stretches are not usually mentioned since all organic compounds contain them and thesample is often prepared as a nujol mull. Nujol is a paraffin oil whose spectrum only containsC–H (and C–C) vibrations.

32

N–H sharp absorption at around 3300 cm−1.

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NH CH3

N-H

Ph

NH2 group often shows two absorptions due to symmetric and antisymmetric stretches.

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NH H

NH H

NH2Ph

NH H

symmetricca 3300 cm-1

antisymmetricca 3400 cm-1

The shape of an O–H stretch depends on whether or not there is hydrogen bonding in the sample.With hydrogen bonding, the absorption is very broad:

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OH

OH H

O H bonding means there are lots of slightly different strengths of OH bonds

33

Here is an unusual example where the O–H stretch is very sharp. No H-bonding is possible dueto the bulky –CMe3 groups preventing the OH groups from approaching one another.

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OH

CH3non H-bonded

OH sharp

Carboxylic acids are typically more ‘V’ shaped

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CO OH

Triple bond region 2000–2500 cm−1

C≡N, strong absorption ∼ 2250 cm−1

C≡C, weak absorption ∼ 2100-2250 cm−1

Double bond region 1500–2000 cm−1

C=C, 1635-1690 cm−1, generally weak

ornumber of peaks in

range 1625-1450 cm-1 medium or weak intensity

34

–NO2 group. Two stretches – one at 1530 cm−1and one at 1350 cm−1.

NOO

NO O

NO O

NO O

NO O

+ +

- -or

symmetric 1350 cm-1 antisymmetric 1530 cm-1

Carbonyl groupsIR spectroscopy is particularly useful in identifying carbonyl groups. The C=O stretching modeshows a strong absorption due to the large dipole moment and the exact frequency of this stretchgives a good indication of what sort of carbonyl functional group is present. You will cover theseagain in the Reactions course where you will see how the IR stretching frequency also gives aguide to the chemical reactivity of the group.

O take as our starting reference a ketone which typically absorbs at ~1715 cm-1

ketone , 1715 cm-1

Anything that strengthens the carbonyl bond will give rise to an absorption at higher frequencyAnything that weakens the carbonyl bond will give rise to an absorption at lower frequency

The carbonyl is strengthened by anything that withdraws electrons and weakened by anythingthat donates electrons. We will see later on in the course exactly how the electrons are donatedand withdrawn and why this affects the C=O bond strength.

O

Cl

chlorine withdraws electrons and strengthens the carbonyl

nitrogen donates electrons and weakens the carbonyl

acid chloride ~1750-1820 cm-1

O

NH2

amide ~1640-1690 cm-1

Oxygen lies in between nitrogen and chlorine in its ability to donate or withdraw electrons. Thisis reflected in the frequency at which the C=O in an ester or a carboxylic acid absorbs – it liesin between the two extremes of an amide and acid chloride.

O

OH

carboxylic acid~1730 cm-1

ester~1745 cm-1

O

O

35

Aldehydes absorb slightly higher than ketones. Since alkyl groups can weakly donate electronsinto the carbonyl system, when the alkyl group is replaced by a hydrogen (for which no electrondonation is possible), the C=O stretching frequency is observed to increase.

O

H~1730 cm-1

aldehydes have an H here instead of the weakly electron donating

alkyl group in the ketone

Acid anhydrides show two stretches due to the symmetric and antisymmetric stretching modesof the two carbonyl groups.

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O O

O

O O

O

~1820 cm-1 symmetric stretch

~1750 cm-1 antisymmetric stretch

18271755

O O

O

As we shall see later in the course, the carbonyl bond is also weakened by conjugation. Thecarbonyl is said to be conjugated with a C=C double bond if the two double bonds are separatedby just one single bond. Conjugation lowers the base frequency by 20-30 cm−1.

OO OO

non-conjugated ketones1715 cm-1 C=O stretch

conjugated ketones~ 1690 cm-1 C=O stretch

2 single bonds between double

36

The final factor which influences the stretching frequency of a C=O concerns compounds wherethe carbonyl group forms part of a ring. The smaller the size of the ring, the higher the stretchingfrequency.

OO O

O

~ 1715 cm-1 ~ 1745 cm-1 ~ 1780 cm-1 ~ 1815 cm-1

6

+30 +35 +35

5 4 3

It turns out that the size of the ring does not alter the strength of the C=O significantly. The cor-rect origin of the effect can be understood by considering what happens during a C=O vibration.Both the oxygen and the carbon atoms move, but the carbon atom feels some ‘resistance’ dueto the repulsion from the other carbon atoms bonded to it. The smaller the angle in the ring, themore the C–C bonds need to be compressed during a vibration. This requires more energy andthis is reflected in the higher frequency of the absorption.

CC O O

C

C

C

C

C

C O O

small CCC angle requires considerable C-C compression

as C=O stretches harder

larger CCC angle requires less C-C compression

as C=O stretches easier

Predicting carbonyl stretching frequenciesThe numbers given for the stretching frequencies of carbonyl groups are only approximate, butnevertheless can be used to predict reasonably well the frequencies that carbonyls absorb at indifferent molecules.

O

O

O

O

ester : 1745in 5 membered ring : +30 predict : 1775 cm-1

ester : 1745in 5 membered ring : +30 conjugated : -30 predict : 1745 cm-1

37

Example spectrum – aspartame

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HO

HN

O

O

ONH2

O

1748

3319N-H

broad O-H

C-H Ph1737 1668

1737carboxylic acid

C=O stretch

amide1668

ester1748

38

Problems

1. (a) Write framework structures (as on page 2 of the handout) for the followingmolecules: propanol, propanoic acid, propanal, propan-2-ol, cyclohexanol, phenol, aceticanhydride, CH3CH=CHCOOH (two isomers), CH2=C(CH3)COOEt, PhCHO, C(OMe)2Ph2,CH3CH2CCCH2CH3.

(b) Draw structures for the following compounds and name each one:

AcOH, Ac2O, Et2O, tBuOH, nBuOH, PhMe, MeCOMe

2. Find the molecular formula (i.e. CaHb ...) corresponding to the following framework structures:

HN

N

OO

OH

HO

HO

OH

OH

HN

OHH

NO

O OMe

O

Ph

N OH

S

OEt

N

HNN

N

N

NnPr

O O

O

tetrodotoxin - the deadly poison that must be removed from Japanese Flunitrazepam - better

known as Rohypnol

cocaine

anthracenepyridine menthol

Sildenafil, the active ingredient in Viagra

N N

F

O

Me

NO2

3. The following nuclei have spin of either 0, 12 , or 1. What is the spin for each nucleus?

6Li, 14N, 15N, 76Se, 77Se, 78Se, 110Cd, 111Cd

4. Predict how many lines you would expect to see in the 13C NMR spectra of the followingmolecules; explain briefly your reasoning in each case:

(a) benzene, anthracene, 1,4-dinitrobenzene, 1-chloro-4-bromobenzene, 1-chloro-2-nitrobenzene,Et2O, EtOMe, CH3(CH2)4COEt.

39

(b)

Cl

CN

CN

the active ingredient in CS gas

5. Sketch the 13C NMR spectra that you expect from the following molecules. You should indicateon you spectrum which carbon is responsible for which resonance. (You cannot give precisevalues of the shifts, but you can predict the region of the spectrum in which each carbon willgive a resonance and for similar carbons what order the shifts will come in).

Br OMe

O

MeO

MeO

OMe

NH2

mescaline - an hallucinogenic and intoxicating drug first isolated from 'magic mushrooms'

6. Describe what you would expect for the 13C NMR spectrum of ethanol recorded (a) with and(b) without broadband proton decoupling. Explain the form of any multiplets you predict.

It is possible to synthesise ethanol in which all of the carbon is 13C. Predict the form of thebroadband proton decoupled 13C NMR spectrum of a sample of this material.

7. (a) The 19F NMR spectrum of IF5 consists of a doublet (that is two lines with intensity ratio 1:1)and a quintet (that is five lines with intensity ratios 1 : 4 : 6 : 4 : 1). Suggest a structure for IF5

that is consistent with this spectrum, explaining how you arrive at your result.

(b) The following are three possible structures for the molecule SF4. Predict the form of the 19FNMR spectrum for each of these possible structures. Structures 1 and 2 are based on a trigonalbipyramid; structure 3 is based on a square based pyramid. (I for 32S = 0; I for 19F = 1

2)

S

F

F

F

FS

F

F

FF S

F

F

F

F

1 2 3

8. Explain why the broadband proton decoupled 13C spectrum of CH2Cl2 consists of a single line,whereas that of CD2Cl2 consists of a 1 : 2 : 3 : 2 : 1 quintet. What would the spectrum ford6-DMSO (CD3SOCD3) look like?

40

9. 11B has spin 32 and relaxes sufficiently slowly that couplings to it can be seen. Explain what you

would expect to see in the proton NMR spectrum of diborane, B2H6, whose structure is shownbelow. (You may assume that all the B atoms are 11B and that any couplings between the protonscan be ignored.)

BH

HB

H H

H H

10. Explain why the 129Xe NMR spectrum of XeF+ is a doublet with J = 7600 Hz but the 19Fspectrum appears to be a triplet with J = 3800 Hz.

(19F has spin 12 and 100% natural abundance; 129Xe has spin 1

2 and 26% natural abundance).

11. Using the force constant given in the table on page 30, calculate the stretching frequencies for1H2 and 79Br2. What experimental technique(s) would you use to determine these frequencies?

[speed of light, c = 3.00× 1010 cm s−1; relative atomic masses: 1H = 1.0078; 79Br = 78.918;NA = 6.02 × 1023]

12. Given that the stretching frequency for HCl (i.e. 1H35Cl) is 2991 cm−1, estimate the stretchingfrequency for DCl (2H35Cl). What assumption have you made for this calculation and is itreasonable? What experimental technique(s) could be used to measure the stretching frequenciesfor HCl and DCl?

13. How can an APT NMR experiment distinguish between the following hydrocarbons?

14. How could the following six isomers be distinguished using 13C NMR?

A B

C D

FE

41

15. The carboxylate anion, RCOO−, shows two absorptions in the C=X region, one around 1575and one around 1430 cm−1. The C=O vibrational frequency for a typical carboxylic acid,RCOOH, is 1725 cm−1; the vibrational frequency of a C–O single bond is around 1100 cm−1.Discuss.

16. Shown below is a portion of the 1H NMR spectrum of bromoethane as measured on a 400 MHzspectrometer where 1 ppm ≡ 400 Hz. In order to magnify the baseline, the spectrum has beenvertically expanded and cropped. Explain the appearance of the spectrum as fully as possibleand determine (in Hz) the 1JC–H coupling constants and the 3JH–H coupling constant.

4.0 3.5 3.0 2.5 2.0 1.5 1.0

17. How would you distinguish spectroscopically between the following pairs of isomers (i) using13C NMR spectroscopy and (ii) using IR spectroscopy?

O

OO

O

CNNH2

OH

OHO O

O OO

CO2H

O

and and

andand

and

18. For organic molecules, “reduction” involves adding H2 across double bonds. Reduction of thefollowing molecule gave two products, both with molecular formula C6H12O.

O

From the following spectroscopic data, identify these two products and explain how the data isconsistent with your proposals for the structures of A and B:

Product A: 13C NMR: 137, 115, 85, 36 and 17 ppm; IR: 3400 and 1650 cm−1.Product B: 13C NMR: 213, 38, 31, 17 and 8 ppm; IR: 1710 cm−1.

42

19. Compound A has the formula C7H7NO3 and may be reduced to give compound B with theformula C7H9NO. The 13C NMR and the IR spectra for A and B are given below. The signalsin the aromatic region of the 1H NMR spectrum of A appear as two triplets and two doublets.Deduce the structures of A and B and, as far as possible, assign the spectra.

200210 180 160 140 120 100 80 60 40 20 0 ppm

200210 180 160 140 120 100 80 60 40 20 0 ppm

62.5

125.

012

8.5

129.

913

4.1

136.

8

147.

6

64.3

116.

011

8.2

124.

812

9.2

129.

4

146.

0

Compound A

Compound A

Compound B

Compound B

5001000150020002500300035004000

100%

40%

60%

80%

20%

0%

trans

mis

sion

frequency / cm–1

3386

1589

5001000150020002500300035004000

100%

40%

60%

80%

20%

0%

trans

mis

sion

frequency / cm–1

1512

1574

1612

1612

1334

43

20. (i) The relative molecular mass of a compound, X, has been determined by high resolutionmass spectrometry to be 68.0261. Using the accurate relative atomic masses below, determinethe formula of compound X.

1H : 1.0078 12C : 12.0000 16O : 15.9949

(ii) The IR spectrum for compound X is shown below. What is its structure? (Hint: you will needto measure carefully the carbonyl frequency to distinguish between different possible structures.)

5001000150020002500300035004000

100%

40%

60%

80%

20%

0%

trans

mis

sion

frequency / cm–1

21. Shown below are six derivatives of the steroid cholestane:

O

O O

CH3COO

O

CH3COO

O

OEtO

O

OEt

CH3COO

A B C

D FE

Shown in the table are data from six IR spectra, each of one of the cholestane derivatives (onlypeaks in the C=X region are listed). In as far as you can, match up the chemical structures withthe IR data; explain your reasoning. Strong and weak absorptions are marked s and w.

spectrum ν / cm−1 spectrum ν / cm−1

1 1715(s) 4 1695(s), 1686(s), 1608(w)

2 1724(s), 1712(s) 5 1653(s), 1626, 1592

3 1730(s), 1695(s), 1642(w) 6 1730(s), 1658(w), 1626

44

22.

OO

NanoBalletDancer

2

1

3

4

56

7

8

910

11

12 13

14

15

16

17

1823 19

21

36

2425

26 27

2829

3031

3233

3435 39 3738

40 41

22 20

In June 2003, a research paper was published announcing the synthesis of the smallest represen-tations of the human form: 2 nm tall anthropomorphic molecules, nicknamed ‘NanoPutians’ bytheir creators. The molecules synthesised included ‘NanoKid’, ‘NanoBaker’ and ‘NanoAthlete’.The compound shown above was called ‘NanoBalletDancer’ and has the formula C41H50O2.

(i) How many signals do you expect to see in the 13C NMR spectrum of NanoBalletDancer?Indicate which carbon atoms are equivalent.

(ii) There are 13 different environments of hydrogens in NanoBalletDancer; their signals arelabelled A-M in the 1H NMR spectrum below. The numbers of hydrogen atoms in each uniqueenvironment is given under the label. Using the numbers of hydrogens in a given environment,the multiplicities of the signals, and to some extent the chemical shifts of the signals, assign (asfar as possible) which signals correspond to which hydrogen atoms in the compound. As a start,signal B is due to the hydrogens on carbon atoms 19 and 23.

1.01.52.02.53.03.54.04.55.05.56.06.57.07.5 ppm

A1H

C2H

D1H

E2H

F2H G

2H

H4H

I4H

J3H

K18H

L6H

M3H

B2H

solvent

The 1H NMR spectrum of NanoBalletDancer

45

Suggested answers to problems

1a. OH OH O O

O H

i) ii) iii)

iv) v) vi)

x) xi)

vii) ix)

or

or

vii)

or

O

OH OH OHO O

OH

O

OH

O

O

OO

H

OO

Ph

Ph

OMeOMe

other variations are possible!

1b.

OH OO

OH OH

O O O

O

acetic acid

t-butyl alcohol n-butyl alcohol toluene acetone

acetic anhydride (diethyl) ether

2. pyridine C5H5N anthracene C14H10 menthol C10H20O cocaine C17H21NO4

viagra C22H30N6O4S tetrodotoxin C11H17N3O8 rohypnol C16H12FN3O3

3. 6Li I = 1; 14N I = 1; 15N I = 12 ; 76Se I = 0; 77Se I = 1

2 ; 78Se I = 0; 110Cd I = 0;111Cd I = 1

2

46

4a.NO2

NO2

Cl

Br

Cl

NO2

O O

O

1 signal 4 signals

2 signals 4 signals

6 signals2 signals 3 signals 8 signals

4b.

Cl

CN

CN

1 1

1 42

43 3

2 1

22

11

1 1

23

4 6

7

1

1

1

1

2

2

3

1

1

1

1

1

1

2

2

1

1

1

1

1

12

2

2

2

5

1 1

2 23

44

4 signals

10 signals(all different)

7 signals 3 signals 2 signals2 signals

(make a model!)

2 signals 6 signals 7 signals 4 signals

5.

200 150 100 50 0 ppm 200 150 100 50 0 ppm

200 150 100 50 0 ppm 200 150 100 50 0 ppm

Br

OMe

O

MeO

MeO

OMe

NH2

1

12

3

4

5

51

1

3 4 6

51

28 7

2

2

34

4

5

5

6

7

8

34

2

2

3

1 2 3 24

3

21

3 1 4

47

6.

100 50 0 ppm

HO1

1

2

22

with broadband 1H decoupling

100 50 0 ppm

1

21

2

100 50 0 ppm

with broadband 1H decouplingfor 13C enriched ethanol

without broadband 1H decoupling coupling constants are not the same

coupling constants are

the same

7a. Two signals, therefore two different fluorine environments. From the splitting patterns, there isone fluorine in one environment and four in the other.

I

Fa

Fb Fb

Fb Fb

unique Fa couples to 4 equivalent Fbs to give a quintet

four fluorines all equivalent, couple to Fa to give a doublet

7b. • Structure 1 (the observed structure): two fluorine environments (axial and equatorial withtwo in each). Two signals – each signal will be a triplet.

• Structure 2: again two environments so two signals. Now only one axial and three equa-torial. The axial fluorine will be split into a 1 : 3 : 3 : 1 quartet by the three equatorialfluorines; the equatorial fluorines will be split into a 1 : 1 doublet by the axial fluroine.

• Structure 3: One environment – all equivalent so no coupling observed. One singlet only.

8. CH2H2 proton decoupled so no coupling to hydrogens. Chlorine (both 35Cl and 37Cl) have spin 32

and relax too quickly for coupling to be seen.

original shift

split into a triplet by the 1st D

each line is then split into another triplet by the 2nd D

1 : 2 : 3 : 2 : 1

CD2Cl2spin for D, I = 1

48

8. (Continued)

1 type of carbon coupled to 3 equivalent Ds.

(2nI+1) predicts 7 linesIn practice, often looks more like a quintet since outer lines are lost in the noise of the spectrum.1 : 3 : 6 : 7 : 6 : 3 : 1

d6DMSO

OS

CD3

CD3

9.

BH

HB

H H

H Hterminal Hs couple to one B spin 3/2. Gives (2 x 1 x 3/2) + 1 = 4 peaks ratio 1:1:1:1.

bridging Hs couple to two B spin 3/2. Gives (2 x 2 x 3/2) + 1 = 7 peaks ratio 1:2:3:4:3:2:1.

1 2 3 4 3 2 1: : : : : :

10. The coupling constants between the xenon and the fluorine must be the same. Since thecoupling in the doublet in the xenon spectrum is 7600 Hz, this must be the coupling constantin the 19F spectrum. Therefore the 19F spectrum must be of a doublet (with J = 7600 Hz)superimposed on a singlet. Most of the fluorine is attached to spin zero xenon – only 26% isattached to spin 1

2129Xe to give the satellites. Ratio of observed peaks will be 13 : 74 : 13.

11.ν̃ =

12πc

"k f

µ

µH2 =1.0078 × 1.00781.0078 + 1.0078

=1.0078

2g mol−1

≡ 1.00782 × 1000 × NA

= 8.37 × 10−28 kg (molecule−1)

ν̃H2 =1

2 × π × 3.00 × 1010

!575 × 2 × 1000 × 6.02 × 1023

1.0078= 4397 cm−1

µBr2 =78.918

2 × 1000 × NA= 6.55 × 10−26 kg (molecule−1)

ν̃Br2 =1

2 × π × 3.00 × 1010

!246 × 2 × 1000 × 6.02 × 1023

78.918= 325 cm−1

Note that the Br–Br stretching frequency (at 325 cm−1) is much lower than the H–H stretchingfrequency (at 4397 cm−1). ν̃H2 is so high due to both the small reduced mass and the strongsingle bond. These stretching frequencies can be determined using Raman spectroscopy. Neitherabsorbs IR since there is no dipole moment during the stretch.

49

12. ν̃HCl = 2991 cm−1 ∝"

k f

µHCl

ν̃DCl ∝"

k f

µDCl

Assuming the force constant is the same for both HCl and DCl:

ν̃DCl

ν̃HCl=

!µHCl

µDCl

ν̃DCl = ν̃HCl ×!µHCl

µDCl

µHCL =1 × 351 + 35

=3536

g mol−1

µDCL =2 × 352 + 35

=7037

g mol−1

ν̃DCl = 2991 cm−1 ×!

3536× 37

70= 2144 cm−1

[If we had used the approximations µHCL ≈ 1 and µDCL ≈ 2, get the answer 2115 cm−1.]

We assume that the force constants are the same for both HCl and DCl. This is reasonablesince the degree of bonding is related to the electronic arrangement, which is identical for both– the extra neutron in DCl has little effect.

Both normal IR spectroscopy and Raman spectroscopy could be used to determine thesestretching frequencies.

13. All of the spectra would show five peaks in the 0-50 region in the 13C NMR spectrum, hencewe should use the APT to distinguish. Even peaks point in the same direction as the solvent;odd in the opposite direction.

5 peaks4 even1 odd

5 peaks1 even4 odd

5 peaks3 even2 odd

5 peaks5 even0 odd

5 peaks2 even3 odd

5 peaks0 even5 odd

50

14. The number of signals can be used to a certain extent to distinguish between the isomers –A and D both have 8; B and C have 5; E and F have 4. In order to distinguish between thesepairs, we need to use the APT paying careful attention in which regions the odd / even peaks arefound. Even peaks point in the same direction as the solvent; odd in the opposite direction.

A B

C D

FE

8 peaks; 4 odd 4 evenalkene carbons are odd

5 peaks; 1 odd 4 even(alkene carbons are even)

5 peaks; 2 odd 3 even(alkene carbons are even)

8 peaks; 4 odd 4 evenalkene carbons are even

4 peaks; 2 odd 2 evenalkene carbons are odd

4 peaks; 2 odd 2 evenalkene carbons are even

1o

1o

1o

2e

3e

4e

4o 4o

4o

4o

5e

1o

1o 1o

1o

1o

1o

2e

2e2e

2e

2e

3e

3e

3e 3e

3e

4e 5o

4e 5o

3e

2e

1o

2e

2e

3o

3e

4o

4e

5e

5e

6e

6o

7e

7o

8o

8o

15.

R O

O

R O

O

-

(-)

(-)

better represented as

both C-O bonds the same strength and in between a double and single bond

see two stretches: antisymmetric ~ 1575 cm-1

symmetric ~ 1430 cm-1

both in between stretching frequencies of C=O 1725 cm-1 and C-O 1100 cm-1

16. At the slightly larger shift of 3.4 ppm is the signal from the two protons attached to the carbonwith the bromine. This signal is split into a 1 : 3 : 3 : 1 quartet by the three protons on the othercarbon. (The ratio cannot be seen since the spectrum has been cropped.)

The signal at 1.65 ppm is due to the methyl protons, split into a 1 : 2 : 1 triplet by the–CH2Br protons.

Either side of each of these signals are the 13C satellites. Note the coupling between the twosets of protons and the carbons they are attached to are not the same!

4.0 3.5 3.0 2.5 2.0 1.5 1.0

1ppm 400 Hz 40 mm1 mm 10 Hz

15 mm 150 Hz = 1JC-H

13 mm 130 Hz = 1JC-H

~2.1 mm 21 Hz = 3 x 3JH-H i.e. 3JH-H = ~ 7 Hz

CH2Br- -CH3

51

17.

O

OO

O

CN

NH2

OH

OHO O

O OO

CO2H

O

and and

andand

and

NMRketone ~ 2002 peaks in 50-1002 peaks in 0-50

NMRester (lactone) ~ 1751 peak in 50-1003 peaks in 0-50

NMR3 peaks in 100-150 (-CN ~ 120)4 peaks in 0-50

NMR4 peaks in 100-1501 peak in 0-50

IRketone + ring~ 1715 + 30= 1745 cm-1

IR~1800 & 1760 cm-1 symm & antisymm C=O stretches

IR~1715 - 30 = 1685 conjugated ketone C=O ~1725 - 30 = 1695 conjugated acid C=O + broad O-H stretch

IRester + ring~ 1745 + 30= 1775 cm-1

IR-CN ~2200 cm-1

IR2 peaks ~ 3300 cm-1 for the NH2

NMR2 peaks in total

NMR3 peaks in total

NMR5 peaks in total

NMR4 peaks in total

NMR6 peaks in total

NMR4 peaks in total

IRnot much use to distinguish

IRC=O in 4-mem ring ~1775 cm-1

IRC=O in 6-mem ring ~1715 cm-1

18.

O OOH A B

1

2

3 4

5

5 1

2

34

5

5

1 115 ppm; 2 137 ppm3 85 ppm; 4 36 ppm5 17 ppm

1 8 ppm; 2 31 ppm3 213 ppm; 4 38 ppm5 17 ppm

3400 cm-1 is the O-H stretch1650 cm-1 is the C=C stretch

1710 cm-1 is the ketone C=O stretch

52

19. OH

NO2

OH

NH2

Compound A Compound B

13C NMR - 6 signals between 100-150 indicate benzene ring with at least two substituents.Signal at 62.5 ppm suggests sp3 carbon with O attached.Substituents must be either ortho or para.

IR peaks at 1512 and 1334 cm-1 suggest NO2 group. Peaks due to benzene ring around 1600 cm-1. Broad peak around 3300 cm-1 suggests -OH group

1H NMR two doublets and two triplets in aromatic region tell us there two substituents which are ortho to each other.

13C NMR - significant changes in aromatic carbons. Suggests the reduced group is directly attached to the benzene ring.Signal at 64.3 ppm unchanged sp3 carbon with O attached.

IR peaks at 1512 and 1334 cm-1 gone. Broad peak around 3300 cm-1 still present but now joined by sharp N-H stretch at 3386 cm-1.

20.

H

OC C

Formula C4H4O (3 double bond equivalents)

2095 cm-1

3260 cm-1 C-H stretch of terminal alkyne

1685 cm-1 must be a conjugated C=OX

21. B, C & F all esters so look for the highest frequencies

left with A, D & E

must be 2, 3 and 6

must be 1, 4 and 5

2 absorptions only1724 ester1712 ketonemust be B

only ketone (1715)must be E

1695(s) conjugated ketone1642(w) C=C must be C

1695(s) & 1686(s) conjugated C=Osmust be A

1 conjugated C=O and 2 C=C must be D

1658(w) and 16262 x C=C must be F

A B C D E F4 2 3 5 1 6

53

22. (i) There are 23 signals in total in the 13C NMR spectrum. If a model is made of this structure,at at least in one conformation (rotating about the single bonds 18-17 and 17-10) there is a planeof symmetry making the left and right hand sides of the molecule equivalent.

1 ≡ 16, 2 ≡ 15, 3 ≡ 14, 4 ≡ 13, 5 ≡ 12, 6 ≡ 8, 9 ≡ 11, 19 ≡ 23, 20 ≡ 22, 24 ≡ 30, 25 ≡ 31, 26 ≡32, 37 ≡ 39.

27, 28, 29, 33, 34, and 35 are all equivalent.

7, 10, 17, 18, 21, 36 38, 40, and 41 all lie in the plane of symmetry and are unique.

Most students are caught out by carbons 40 and 41 being different. One is on the same side ofthe ‘head’ ring as the body – the other is on the opposite side of the head.

(ii)

Peak description due to hydrogens on carbon(s)

A 1H singlet 7

B 2H singlet 19 and 23

C 2H singlet 9 and 11

D 1H singlet 36

E 2H singlet 17

F 2H doublet 37 and 39

G 2H doublet 37 and 39

H 4H triplet 3 and 14

I 4H sextuplet 2 and 15

J 3H singlet 40 or 41

K 18H singlet 27, 28, 29, 33, 34, and 35

L 6H triplet 1 and 16

M 3H singlet 41 or 40

Notes: Signal (C) appears shorter than B due to the very weak coupling to the proton on carbon7; this has not split the signal, just broadened it slightly.

The ‘upper’ hydrogens on carbons 37 and 39 (on the opposite side of the ‘face’ ring as thebody) are equivalent and give rise to either signal F or G. Similarly, the ‘lower’ hydrogens oncarbons 37 and 39 (on the same side of the ring as the body) are equivalent. These ‘upper’ and‘lower’ hydrogens are not the same and couple with each other, thus giving the doublets. Nasty,but there we are!

54