Spectroscopic data: identifying organic compounds

Three techniques, three questions

• Each technique answers a different question about an unknown organic compound, and none of them is enough on its own.
• For this standard the compounds come from the organic families you have met: hydrocarbons (alkanes and alkenes), alcohols, haloalkanes and amines, plus the C=O families: aldehydes, ketones, carboxylic acids, esters, amides and acid chlorides.
• You are given the spectra and data tables in the assessment; the skill is reading them and linking each feature to part of the structure.

Mass spectrometry: the molecular ion and fragments

• The sample is vaporised and hit by high-energy electrons, which knock an electron off each molecule to make a positive molecular ion, M⁺.
• Ions carry a charge of 1+, so the peak furthest to the right (apart from small isotope peaks) is at the molar mass of the compound.
• Many molecular ions break apart; only the charged fragments are detected, and the gap between M⁺ and a fragment tells you what was lost (a gap of 15 means a CH₃ was lost).
• The tallest peak is the base peak, set to 100%: it is the most stable fragment, not necessarily the molecular ion.
• Chlorine is ³⁵Cl and ³⁷Cl in a 3 : 1 ratio, so a chloroalkane shows M and M + 2 peaks of heights 3 : 1; bromine (⁷⁹Br and ⁸¹Br) gives two peaks of equal height.

Infrared spectra and functional groups

• Covalent bonds stretch and bend, and each kind of bond absorbs infrared at its own range of wavenumbers (cm⁻¹); the spectrum shows transmittance, so an absorption is a dip.
• The scale runs from 4000 cm⁻¹ on the left to about 500 cm⁻¹ on the right, and the useful dips for this standard are above 1500 cm⁻¹, plus C–O.
• Below 1500 cm⁻¹ is the fingerprint region: too many overlapping dips to assign, but unique to each compound, so it can match an unknown to a known spectrum.
• Hydrogen bonding broadens O–H dips: an alcohol's O–H is a broad rounded dip, and a carboxylic acid's is very broad, running into the C–H dips.

Try it: which bond made that dip?

Carbon-13 NMR: counting carbon environments

• In a magnetic field, each ¹³C nucleus absorbs radio waves at a frequency set by the atoms around it, reported as a chemical shift δ in ppm from 0 to about 220.
• Carbons in identical surroundings give one peak, so the number of peaks is the number of carbon environments, not the number of carbons.
• Look for symmetry: in propan-2-ol, CH₃CH(OH)CH₃, the two CH₃ carbons are equivalent, so three carbons give only two peaks.
• The shift shows the neighbours: a carbon bonded to an electronegative atom such as O or Cl is pulled to a higher δ, and a C=O carbon sits highest of all.
• Peak heights in a ¹³C spectrum do not reliably show how many carbons are in each environment, so use the number and the positions of peaks, not their heights.

Try it: count the carbon environments

Combining the data

• Start with the mass spectrum: M⁺ gives the molar mass, and with the elements present it fixes the molecular formula.
• Use the IR spectrum to name the functional group, and say which groups are ruled out by missing dips.
• Draw every structural isomer with that formula and group, then use the ¹³C NMR peak count and shifts to cut the list to one.
• Check the answer against all the data: the fragments in the mass spectrum must be pieces your structure can actually make, and every NMR peak must belong to a carbon in it.

How it is assessed, and what Merit and Excellence look like

• This is an internal: you are given spectra and data tables for unknown compounds, usually under test conditions, and write your analysis.
• Achieved identifies separate features: a functional group from an IR dip, the molar mass from M⁺, the number of carbon environments.
• Merit uses the data to work out the full structure; Excellence justifies that structure by linking all three spectra and explaining why the other isomers do not fit.