Units of UnsaturationThis is also called “Degrees of Unsaturation” or “Double BondEquivalents (DBE)”. By looking at a molecular formula, it ispossible to find the number of double bonds and/or rings in thecompound. The procedure is as follows:

1. For the purpose of this calculation, find effective number of H.a) For each halogen, add 1 H.b) For each nitrogen, subtract 1H.c) Oxygen has no effect on this calculation.

2. Using the fact that saturated compounds have the formulaCnH2n+2 find the number of H if saturated.

3. Use the formula (Saturated H – Effective H) / 2 to find thenumber of DBE.

4. If DBE is 4 or more, an aromatic ring is present.

Examples

C4H8O2 ----> C4H8 ----> C4H10 DBE = (10 – 8)/2 = 1

Effective H = 8 Saturated H = 4x2 + 2 = 10

C6H4Cl2 ----> C6H6 ----> C6H14 DBE = (14 – 6)/2 = 4

Effective H = 6 Saturated H = 6x2 + 2 = 14

C5H5N ----> C5H4 ----> C5H12 DBE = (12 – 4)/2 = 4

Effective H = 4 Saturated H = 5x2 + 2 = 12

Infrared Spectroscopy

Nice introductions to IR spectroscopy can be seen at thefollowing links:1) MSU2) General resources

IR is commonly used to divine functionality. It gives little useful information regarding exact structure. IR can be diagnostic (able to suggest the presence or absence of a particular functional group) or confirmatory (possibly able tosupport evidence of a particular functional group found by another means).We will concentrate on the diagnostic areas, since they give themost useful information.

Correlation Chart

This chart gives you an idea of where you will find functional groups.

The 3300-3500 cm-1 region

OH, NH and NH2 stretches are found in this region. They maybe distinguished by their differing shapes.

OH examples

(a)(b)

OH stretches are typically strong (comedown a long way) andbroad (wide).(a) is more typical ofaliphatic OH stretch. (b) is more typical of acarboxylic OH stretch(even though these often appear outside the usualOH stretch region). Thisone is “masking” other peaks.

(a) (b)

NH2 and NH Examples

(a) is a typical NH2 stretch, note the two peaks.(b) is a typical NH stretch. These are not usually as strong or broad as OH stretches.

The 2900 – 3100 cm-1 Region

CH stretch (alkane only).Notice how all peaks in the 2900-3100 regionare below 3000 cm-1.

CH stretch (alkane AND alkene)Notice how the peaks in the 2900-3100 region “straddle”3000 cm-1.

CH stretches are found in this region. They can be differentiated as shown below.

The 2000-2300 cm-1 Region

CC and CN stretches are found here.

CC stretchesare typicallyless intense of the two and usually occur at2100-2200 cm-1.

CN are usually more intense of the two and usually occur at 2200-2300 cm-1.

The 1700 – 1850 cm-1 Region

C=O stretches are usually found here.They are strong and fairly sharp.

Distinguishing C=C bonds.

Typically, C=C that are in chainsshow a peak inthe 1650 cm-1

region.

Aromatic C=Cstretches may beseen in the 1500-1600 cm-1

region.

Be careful, though. These peaks are confirmatory, becauseother things can turn up in this region. This is useful only if you know a double bond is present.

Distinguishing Aldehydes and Ketones

Be careful, though. These peaks are confirmatory, becauseother things can turn up in this region. This is useful only if you know an aldehyde is a possibility.

O

2-butanone(a ketone)

H

O

butanal(an aldehyde)

Aldehydes (RCHO)Typically show 2Peaks in the 2700-2900 cm-1 region.This is for the C-H stretch fromthe O=C-H group.Peaks in this regionare typically absentfor ketones. Notethat both still showthe C=O stretch.

Proton NMR Spectroscopy

For a great introduction to proton NMR spectroscopy go to:1) MSU2) Wikipedia...click on the links in blue in the wiki. Chemical Shift

< ------ downfield upfield ------ >

Chemical Shift is the relationship between peak location and thekind of hydrogen producing that peak.There are certain trends that are evident from the chart.

a) Hs on alkanes have shifts of around 1 ppm.b) Adding electronegative groups like N, O, Cl usually results in a downfield shift (2.5 ppm – 5 ppm).c) If the H is attached to a double bond or aromatic ring it can usually be found around between 5 and 8 ppm.

There is one set of peaks for each DIFFERENT H.

CH3

CH2

CH2

CH2

C

CH3

O

CH3

CH2

C

CH2

CH3

O

O

a

b

c

d e

5 peaks

a

b

2 peaks

b

a

a

bc

b

a3 peaks

a

b

c

de

5 peaksc,d,e will overlapbecause theyare similard

c

Each different H can be labelled with a unique letter.There should be as many peaks as there are letters.

Each letter has a corresponding peak. Look at the affect of chemical shift. a,b,c Hs are attached to an aromatic ring, so theseare between 7-8 ppm. e H is attached to a C-Br, so this is at about 4 ppm.

Integration

Tells how many H there are of a given type.For peaks < 5 ppm the following usually applies:3H = CH3 9H = 3 x CH3

2H = CH2; NH2 (NH2 single peak) 6H = 2 x CH3 OR 3 x CH2

1H = CH; NH; OH (NH, OH single peak) 4H = 2 x CH2

For peaks 7-8 ppm (aromatic) the following usually applies. (There may be two or more peaks).4H = aromatic ring with 4H attached 5H = aromatic ring with 5H attached.

See the example on the next page.

H

H

H

H

H5H = C

H

C

H

H

C

H

H

H1H = 2H = 3H =

The squiggly lines represent “join points”, places whereone group can join to another.

Multiplicity

Tells how many H are on adjacent Carbon atoms.

n peaks = n-1 adjacent H atoms on C atoms. (Hx = H in question)

1 peak = 0 adjacent H atoms on C (CHx—O, CHx—NH, CHx—C=O);

2 peaks= 1 adjacent H atoms on C (CHx—CH)

3 peaks= 2 adjacent H atoms on C (CHx—CH2)

4 peaks= 3 adjacent H atoms on C (CHx—CH3)

5 peaks= 4 adjacent H atoms on C (CH2—CHx—CH2)

or (CH—CHx—CH3)

6 peaks= 5 adjacent H atoms on C (CH3—CHx—CH2)

7 peaks= 6 adjacent H atoms on C (CH3—CHx—CH3)

Look at “e”6 peaks 5 H on adjacent Cs

Look at “f”2 peaks1 H on adjacent C

Look at “d” 2 peaks1 H on adjacent C

Solving Structures given a Molecular Formula and a Proton NMR Spectrum

1. Find DBE using molecular formula.2. Identify the pieces that can be seen by looking at the integration.3. Find what cannot be seen by subtracting the pieces from themolecular formula.4. The structure is made up of the pieces that can be seen plusthe pieces that cannot be seen.5. Draw possible structures that can be made from those pieces.6. Use chemical shift and multiplicity to figure out which possibility is correct.7. Verify by assigning each H on the spectrum.

C5H10O

Step 1

Pieces we can see based on integration: CH2,CH3,CH2,CH3Step 2

C5H10O ----> C5H10 ----> C5H12 DBE = (12-10)/2 = 1

Step 3: Find what cannot be seen by subtracting the pieces from themolecular formula.

C5H10O - { }CH2,CH3,CH2,CH3

= C, O and 1 DBE=

C=O

Step 4: The structure is made up of the pieces that can be seen plus the pieces that cannot be seen.

CH2 CH3 CH2 CH3C=O

CH3

CH2

CH2

C

CH3

O

CH3

CH2

C

CH2

CH3

O

a

b

c d

4 peaks

a

b

2 peaks

b

a

Draw possible structures that can be made from those pieces.Step 5

Step 6

Use chemical shift and multiplicity to figure out which possibility is correct.

CH3

CH2

CH2

C

CH3

O

a

b

c dBased on multiplicities: a has 3 adj H (4 peaks)

b has 5 adj H (6 peaks)c has 2 adj H (3 peaks)d has 0 adj H (1 peak)

is correct.

Step 7 Verify by assigning each H on the spectrum.

CH3

CH2

CH2

C

CH3

O

a

b

c d

13C NMR Spectroscopy

This is the correlation chart for C-13 Spectroscopy. Attaching a more electronegative element to the carbon will pushes peak moredownfield. Note also the downfield shift if the C is part of a doublebond or an aromatic ring.

< ------ downfield upfield ------ >

CH3

CH2

CH2

CH2

C

CH3

O

CH3

CH2

C

CH2

CH3

O

O

a

b

c

d

e

a

b

3 peaks

b

a

ab

c

b4 peaks

a

bc

d

e

6 peaks

d

f

c

d

6 peaks

c

e

f

One peak is seen for each different C. ONLY Carbons are seen, so only Carbons are labelled.

a

bc

d e

f

g

h

Note how all of the peaks are single peaks. This is becauseC-13 only makes up 1.1% of all C isotopes. The probability oftwo C-13 atoms being next to each other is 1:10000.No C-13 to C-13 coupling is observed. Note that the shift of carbons (e)-(h) increases as they get closer to the ring. SimilarCs can sometimes be differentiated by height. (b) and (c) are tallerthan (a) because there are two versus one. CDCl3 is a solventand should be ignored.

If the spectrum is obtained using different setup parameters,the coupling between C-13 and 1-H can be seen. This is shownby letters above the peaks.q = 4 peaks = CH3

t = 3 peaks = CH2

d = 2 peaks = CHs = 1 peak = C

The reason that letters are shown above the peaks is for clarityand to avoid overlap of multiple peaks.

OHNH2

O

1

2

3

41

2

3

123

4

4

Differentiating spectra in a matching question.

First, differentiate spectra by number of peaks. Secondly,differentiate by chemical shift and by q,t,d and s.

OH

1

2

3

4

O

123

4

NH2

1

2

3

4

Focus on the key peaks: C-O is usually around 60 ppmC-N is usually around 40 ppm. Use the correlation chart!The C attached to the electronegative element will be farthest downfield (the “1” C here).