SCH4UC Name _________________________________ Date _____________________________

Spectroscopic Identification of Organic Compounds

Degree of unsaturation or index of hydrogen deficiency (IHD) can be used to determine from a molecular formula the number of rings or multiple bonds in a molecule

To determine IHD from molecular formula (CcHhNnOoXx):IHD = (0.5)(2c + 2 – h – x + n)

Practice #1: Determine the IHD for the following.

a) C3H6 b) C6H9Cl

1) Infrared spectroscopy

Practice #2:

Practice #3: Which of the following compounds will have an infrared absorption band in the 1700-1750 cm—1 region?

a) but-2-ene b) propanalc) CH3COCH3 d) CH3CH2CH2CH2Cle) CH3COOCH3 f) CH3CH2CH(OH)CH3

g) CH3CH2COOH

2) Mass spectrometry Involves ionizing a gaseous molecule to form a molecular ion, M+

Molecular ion peak corresponds to molecular mass of compound (since it is assumed only +1 ions are produced, thus m/z = mass)

Molecular ion peak is not necessarily the biggest peak in terms of abundance, but it has the highest m/z value

Mass spectrum may also show a peak with a mass one unit higher than the molecular ion, an (M+1)+ peak caused by the presence of an atom of C-13 in some molecules

Smaller fragment ions provide info on functional groups present See data booklet Table 28 (page 24)

Example: Mass spectrum of propane, CH3CH2CH3

Practice #4: The molecular ion peaks for a series of compounds containing only carbon, hydrogen, and oxygen, occur at the values shown below. Suggest possible molecular formulas for the compounds.

a) 44 b) 60 c) 72 d) 88

3) Proton nuclear magnetic resonance (HNMR) spectroscopy Gives information about different chemical environments of hydrogen atoms (and, thus, types

of bonds with hydrogens in molecule) Chemical shift – position of the NMR signal relative to a standard (tetramethylsilane, TMS);

symbol δ, expressed in ppm Advantages of TMS as standard:

1) The 12 protons are in the same chemical environment, so just one single, strong peak assigned 0 ppm

2) It is inert3) It absorbs away from most other protons in organic compounds4) Can be easily removed from the sample after measurement because of its volatility

Chemical shifts given in data booklet (Section 27, page 23) Ratio between heights of each step in the integration trace is the same as the ratio between the

numbers of protons in each environment

Example:

Practice #5:

High-Resolution HNMR Can show further splitting of some absorptions resulting from spin-spin coupling DUE TO

HYDROGEN ATOMS ON ADJACENT CARBONS Rule 1: If a proton, Ha, has n protons as its nearest neighbours, that is n x Hb, then the peak

of Ha will be split into (n + 1) peaks Rule 2: The ratio of the intensities of the lines of the split peak can be deduced from Pascal’s

triangle

A singlet occurs for H’s which have 0 H’s on adjacent carbons A quartet occurs for H’s which have 3 H’s on adjacent carbons A triplet occurs for H’s which have 2 H’s on adjacent carbons

Example:

Practice #6: The NMR spectrum for a compound with molecular formula C4H8O is shown below. Deduce the structure of the compound.

4) X-Ray Crystallography Involves irradiating a crystal with X-rays and looking at the positions and intensities of the

diffracted beams Can give a 3-dimensional picture of molecule with bond lengths and angles

Practice #7: