4.2.4 (a,b,c,d,e) Infrared Spectroscopy

Syllabus

(a) Infrared (IR) radiation causes covalent bonds to vibrate more and absorb energy

(b) absorption of Infrared radiation by atmospheric gases containing C=O, O–H and C–H bonds (e.g. H₂O, CO₂ and CH₄), the suspected link to global warming and resulting changes to energy usage

(i) an Alcohol from an absorption peak of the O–H bond

(ii) an Aldehyde or Ketone from an absorption peak of the C=O bond

(iii) a Carboxylic acid from an absorption peak of the C=O bond and a broad absorption peak of the O–H bond

{Learners should be aware that most organic compounds produce a peak around 3000 cm^–1 due to absorption by C–H bonds.}

(d) interpretation & prediction of an infrared spectrum of familiar or unfamiliar substances using supplied data

{Restricted to functional groups studied in this specification}

(e) use of infrared spectroscopy to monitor gases causing air pollution (e.g. CO and NO from car emissions) and in modern breathalysers to measure Ethanol in the breath.

What does this mean?

Those of you who do Physics (suckers!) probably recall the equation: E = hf

Everyone else just needs to know that the energy of a wave is proportional to its frequency.

So high frequency waves have lots of energy (but a low wavelength).

Infrared radiation isn't the highest energy radiation but IR waves have frequencies that are just the right energy to make bonds stretch or bend.

You need to know that bending and stretching is what happens, and you may need to be able to draw what you mean by this.

But you don't need to name the actions above.

Obviously, different bonds have different strengths and involve atoms with different masses.

Common sense suggests that light atoms vibrate more easily.

It is less obvious that strong bonds vibrate faster than weak ones.

Either way, different bonds absorb different frequencies to bend and stretch.

Which is why some gases in the air are particularly good at absorbing IR trying to escape the planet, causing it to warm when they re-emit the energy - someof it back to Earth.

Water and Carbon Dioxide are not the best at doing this but exist in the highest concentrations.

Knowing which bonds absorb which frequencies also allows us to use IR to look at what bonds are present in an unknown compound.

The spectrometer

You don't need to know how the machine works and won't be asked about it.

But a simple explanation may help understand the rest.

Simplistically, the IR source produces all frequencies of IR.

It is split into two beams.

One goes through your sample.

The other goes through a reference which contains the same solvent as the sample.

THis way the detector can compare the frequencies absorbed by the solvent and "add them back".

The Spectra

IR spectra look like this...

Notice the labels on the axes.

Since frequencies are difficult numbers we use wavenumber in cm^-1.

You should know this, and should know that it is proportional to frequency.

So a high frequency is a high wavenumber but the wavenumber is more convenient to write down.

On the y-axis we have transmittance - how much of the original Ir makes it through your sample.

This will never be 100% even if none was absorbed because some will always be scattered in odd directions.

But where there is a deep trough there is a significant amount of absorbtion.

And this suggests a particular type of bond.

Fortunately, you will be given this list of absorbances in the data-sheet

You won't need all of these at AS but if there is a significant trough within the stated range then you can assume it corresponds to the type of bond above.

Some of them overlap.

This is less of a problem than you might think.

An examiner is unlikely to base a question on you identifying a C-Cl or C-Br bond since it will be mixed in with the C-C bonds which almost all Organic Substances have.

You'll spend most of your time looking for O-H bonds - notice that there are different ranges for Acids and Alcohols.

You can identify them by simply looking at the data-sheet but you'll probably come to be able to do so by shape.

O-H absorbances are unusually wide.

For an alcohol the wavenumber is high enough to separate it from the inevitable C-H absorbance at around 3000.

For an Acid the wavenumber is lower and so overlaps with the C-H making the absorbance even bigger.

In addition, a Carboxylic Acid must have a C=O absorbance around 1750.

And this is another bond students often learn.

Both would have a C-O absorbance but this is in the fingerprint region below 1500 where there are often too many absorbances to be sure which is which.

So you won't normally have to identify it.

Fingerprint Region

The fingerprint region is often a confusing mess of absorbances that it would be tempting to ignore.

And this is what we will mostly do.

However, you should know why it is important.

Remember that many substances may contain the same types of bonds and so will absorb in much the same places.

For instance, all Alcohols have OH, CO and CH bonds and almost all of them have C-C bonds too.

And how would you distinguish between isomers?

You and I would not.

You just have to be able to say that both the above represent Alcohols.

But a computer will compare the fingerprint regions to a data-base and identify which is which.

Practical uses of IR spectroscopy.

The two uses to remember for the exams are Breathalysing drink-drive suspects and analysing car-exhausts for MOT testing.

Breath analysis is looking for the concentration of Ethanol in breath.

Looking for OH bonds only risks confusion with other naturally occurring substances that would naturally be in breath - notably water vapour.

So, the breathyliser also detects the C-O, C-H and C-C bonds and compares the concentrations toi what would be expected in normal breath.

However, you wouldn't be charged until a blood-test had confirmed the amount of Alcohol specifically.

Exhaust analysis is looking for Carbon Monoxide, unburned Hydrocarbons, Nitrogen Oxides etc.

These all have very specific bonds that you would not expect in what is supposed to be in exhaust fumes when a catalytic converter is working well - mostly water, Carbon Dioxide and Nitrogen.

So absorbances for C=O, N-O, N=O, C-H, C-C etc allow a computer to calculate the amounts of each compound in the exhaust.

Videos

Exam-style Questions

1. Compounds C and D, shown below, are isomers of C₅H₁₀O

C D

(a) Name compound C.

.....................…………………………………………………………………………...................................................................................(1)

(b) Use the Data Sheet to help you to answer this question.

(i) Suggest the wavenumber of an absorption which is present in the infra-red spectrum of C but not in that of D.

.............................................................................................................………………………………………………………………………….

(ii) Suggest the wavenumber of an absorption which is present in the infra-red spectrum of D but not in that of C.

...........…………………………………………………………………………...............................................................................................(2)

2 (a) A student used the infrared spectra of water vapour and of carbon dioxide to try to find a link between infrared radiation and global warming. 6 (a) (i) Use information from the infrared spectra to deduce one reason why the student concluded that water vapour is a more effective greenhouse gas than carbon dioxide. ............................................................................................................................................ ............................................................................................................................................ (1 mark)

2 (a) (ii) Use your knowledge of the bonds in CO2 to state why the infrared spectrum of carbon dioxide is not as might be predicted from the data provided in Table 1 on the Data Sheet. ............................................................................................................................................ ............................................................................................................................................ ............................................................................................................................................ ............................................................................................................................................ (2 marks)

Answers

1. (a) Pentan-2-one 1

(b) (i) 1680 – 1750 (cm–1) 1

(ii) 3230 – 3550 or 1000 – 1300 (cm–1) 1

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