11/6/2009 1

Fri 6 Nov 09

More IR

Mass spectroscopy

Good web site for IR, Mass, NMR spectra:

http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.cgi?lang=eng

Hour exam 3 Fri 11-13Covers Chaps 9-12Wednesday: Review

11/6/2009 2

H3C

O

CH3

H3C

O

CH3

H3C

O

CH3

IR is good for identifying functional groups, but not good for Mol Wt.

© 2009 John W. Keller

11/6/2009 3

O

IR is good for distinguishing some isomeric structures e.g. C4H8O

(NIST webbook)

H2C OH

H3C

O

CH3

C-O

C-O

C=O

O-HC=C-H

C=C

C-O© 2009 John W. Keller

11/6/2009 4

IR frequencies Table 12.1. Do not memorize – handed out in exams.

© 2009 John W. Keller

11/6/2009 5

IR frequencies Table 12.1. Do not memorize – handed out in exams.

© 2009 John W. Keller

11/6/2009 6

The compound from the following list that exhibits the infrared spectrum below is...

a b c d e

20

7

0

21

1. a

2. b

3. c

4. d

5. e

1804

© 2009 John W. Keller

11/6/2009 7

no

no

no

no

no

no

© 2009 John W. Keller

11/6/2009 8

1-hexanol

5-hexen-1-ol

CO--H C-C C-O

C=C

C--H

H3C OH

OH

H

H

H

=C--H

11/6/2009 9

The compound from the following list that best

exhibits the infrared spectrum below is...

a b c d e

0 0

11

0

1. a

2. b

3. c

4. d

5. e

33112962

2120

1432

630

11/6/2009 10

33112962

2120

1432

630

The compound from the following list that best exhibits the infrared spectrum below is...

The answer is... -C≡C-H bend

(alkyne bending: not in table)

11/6/2009 11

This compound must be a terminal alkyne, in spite of the fact that the C≡C stretch falls slightly outside the nominal range given in the OWL table.

1) No other bond types absorb in the entire 1810-2400 cm-1 region except C≡N, which is not among the choices. 2) The terminal C≡C stretch at 2110 cm-1 does show “medium”

intensity, meaning roughly 50% transmission.3) Can’t be an alcohol, there is no “strong” C-O stretch at 1100 cm-1.4) Can’t be an amine or amide, since none of compounds contain N.

5) Can’t be a carboxylic acid, since there is no C=O.6) Can’t be aromatic: no medium peaks at either 3100 or 1600 cm-1.

11/6/2009 12

Mass spectrometry

Butane C4H10 � C4H10+ + e-

C4H10+ � C3H7

+ + CH3

C4H10+

C3H7+

Vac pump removes neutrals

Path of the

heavier ion is

bent less by

mag field.

11/6/2009 13

Molecular ion M+.

Fragment ions

"Base peak" = highest

CH3CH2CH2CH2OH C4H10O mw=74

x-axis in "m/z" = "mass-to-charge ratio" (usually z = +1, so effectively its m).

11/6/2009 14

Molecular ion M+.

M+1 ion

Natural carbon contains 1% 13C

isotope, so if there are 5 C’s in a molecule, then 5% of the molecules will

have one 13C.

If there are 20 C’s, the M+1 peak will be ~20% and the M+2 peak (two 13Cs) will

be about 2%.

CH3H3C

C5H10

70.13

(The exact % M+1 and

M+2 ions are also

influenced by other isotopes such as 2H, 18O, etc)

11/6/2009 15

Cl and Br are common elements that

occur as mixed isotopes with high

percentages.

35Cl 75.5%37Cl 24.5%

79Br 50.5%81Br 49.5%

M 12C21H5

79Br

M+1 13C12C1H579Br

M+2 12C21H5

81Br

M+3 13C12C1H581Br

C

H

H

H

C

H

H

Br

11/6/2009 16

See isotopic abundance web applet at

http://www.chem.uoa.gr/applets/AppletMS/Appl_Ms2.html

11/6/2009 17

• Routine mass spectra are resolved to about ±0.1 amu.

High resolution mass spectra may be resolved to ±0.00001 amu

C8H14

110.19

110.109550

O

C7H10O

110.15

110.073165

OH

OH

C6H6O2

110.11

110.036779

NH2

NH2

C6H10N2

110.15

110.084398

Molecular weights using most common isotopes (1H/12C/14N/16O ) are significantly different at high resolution.

11/6/2009 18

Which electron is lost?

• Non-bonding electron if O (or other heteroatom)

• Sigma bond for alkanes

The “electron hole” is actually delocalized by orbital overlap.

Therefore different bonds are susceptible to breakage.

These electron losses cause fragmentation. We will discuss fragmentation patterns of three types of molecules (and not amines):

Alcohols…………………...…….alpha cleavage and dehydrationAlkanes……………………………C-C bond breakage Ketones (& aldehydes)…..alpha cleavage and McLafferty rearrangment

11/6/2009 19

1-butanol

CH3CH2CH2CH2OH

MW 74

M+.

M - 31

M - 43

31

43

A. alpha cleavage

H3CO

HH

H

HH

1. Alcohols

11/6/2009 20

H3CO

HH

H

HH H3C

HH

O

HH

H

31

43

H3CO

HH

H

HH

H3CH

HO

HH

H

43

31

How does the fragmentation occur?

Recall that only cations – M+. or others – are accelerated by the ms

source. So each fragment peak must be due to a cation (+).

This is a “homolysis” that forms a neutral propyl radical fragment (it is

lost), and a protonated formaldehyde.

This is a “heterolysis” that forms a neutral hydroxymethyl radical

fragment (it is lost), and a propyl carbocation.

11/6/2009 21

H3CO

HH

H

HH

H3CO

HH

H

HH

H3CH

H

O

HH

H

An alternative, and equivalent, way of looking at alpha

cleavage process. This considers the "electron hole" to be in

the alpha C-C bond. The fragmentation occurs when that

single electron moves to one carbon or the other.

H3C

H H

H H

OH

H3C

H H

H H

OH

11/6/2009 22

1-butanolCH3CH2CH2CH2OH

MW 74

M+.

M - 18

56

B. dehydration

1856

1. Alcohols

H3C

H H

H H

OH

H3C

H H

H H

OH

H3C

H H

H H

OH

H3C

H

H H

OH

H

H3C

H

HH

OH

H

H3C

H

HH

H3C

H

H

H

The initial alcohol cation-radical

The O atom removes a H

atom from the β-carbon, then

splits off as water molecule...

...leaving behind the dehydrated

cation, which can be thought of as

an alkene cation-radical.