Infra Red (IR) Spectroscopy | A-level Chemistry | OCR, AQA, Edexcel
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- Опубліковано 4 сер 2024
- Infra Red (IR) Spectroscopy in a Snap!
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The key points covered in this video include:
1. Vibration and IR - Global Warming
2. Making an IR Spectrum
3. Application of IR
4. The Fingerprint Region
Vibration and Infrared Radiation
A pair of atoms constantly vibrates. Infrared Radiation can be absorbed by molecules, causing them to vibrate more. This causes their covalent bonds to vibrate. These bonds can either: Stretch, Bend. The bonds vibrates at their own frequency. This is usually between 300 and 4000cm^-1. This is the IR region of the electromagnetic spectrum. The degree of vibration depends on: Bond Strength: Stronger bonds vibrate at a higher frequency. Bond Length. Mass of Atoms: Heavier atoms vibrate at a lower frequency.
Vibration and IR Radiation: Global Warming
The bonds present in Greenhouse Gases absorbs Infrared Radiation well. These gases absorbed IR that is given off by the Earth’s surface. This heat would otherwise be lost to Space. This contributes to the increasing temperature of our planet. Increasing levels of Greenhouse gases may result in further Global Warming.
Making an IR Spectrum
Process. 1. A beam of infrared radiation is passed through the sample. This beam will have a frequency in the infrared region of the electromagnetic spectrum. 2. Molecules of our sample will absorb some of these frequencies. 3. The beam that passes through is analysed. 4. We can plot a graph of transmittance against frequency. This is the infrared spectrum of the molecule. We can analyse the spectra that are formed. The peaks in the spectrum represent the vibration caused by the absorbance of IR radiation by a specific bond in the molecule.
Applications of IR Spectroscopy
Breathalysers. IR spectroscopy can be used to identify the -OH functional group that is indicative of alcohol. The level of absorption that is observed on the spectrum is related to the concentration of alcohol in the blood. England/Wales: 80mg/100ml. More sensitive methods of testing are used in addition to ensure that tests are specific. Other chemicals may produce similar spectra. Air Quality. IR spectroscopy can be used to monitor the levels of pollutants. The levels of CO and NO can be measured. Forensics. IR spectroscopy can be used to identify the chemicals that are present at crime scenes. Fuels, Accelerants, Plastics.
The Fingerprint Region
This is the region between 500cm^-1 and 1500cm^-1. The absorptions within this range are interesting. They are caused mainly from bending, Together, they are largely unique to the specific molecule. This region can be used to identify the chemical.
Summary
IR Radiation is absorbed by molecules
Covalent bonds stretch and bend
The amount of vibration depends on:
a. Bond strength
b. Bond length
c. Mass of each atom
Most bonds vibrate between 300 and 4000cm^-1
IR beam is passed through the sample and emerging beam is analysed and a graph plotted
Each different functional group has a typical range and shape of absorption
a. These can be found in the data booklet
I feel like I'm being taught by Hermione Grainger
wingardium leviosa.
@@blazebangerz3122 it’s wingardium leviosaa not leviosar
@@emilym2742 lmao
@@blazebangerz3122 Go on Harry... Don't stop
literally was about to say that haha
@10:30 how do we know its a ketone? if there's no carboxylic acid peak but there is the 1700cm-1 peak to indicate that a c=o bond is present, wouldn't it be an aldehyde too? how would we know if its an aldehyde or a ketone?
EXACTLY MY QUESTION
@@aleenaasif4712 dd u get why we couldnt
@@thaliaissa3009 yes i think, it would have to be because of the peak not actually being at 2700-2775 right coz it goes further than that so therefore no peak due to C-H bond in aldehyde?
I think it’s because the peak isn’t broad like the O-H peak should be, it’s sharp (I’m not really sure)
Secondary alcohols only form ketones when oxidised.. Primary alchohols could form aldehyde
Amazing lecture👍👍
Anyone else using this for distance learning? :/
In the breathalyser test, they actually test for CH bonds as there are OH bonds in the breath naturally from water vapour.
Really? I thought it depends on the functional group. OH bonds in alcohol show different reading from OH bonds in water.
Man, did this help!
perfect explanation!
Our data sheet doesn't have the "intensity" part, would this be this an aqa thing?
how do you know which peak to look at??
B could also be an ester by the same logic? and peaks and troughs are different things
The mic really do be on a mad one
Do we have to memorize the wavenumbers or will they be provided?
They will be provided in your data booklet which will be given to you with your paper.
@@abdullahbaig8700 tq bro!!
@@aniketmajety3795 Np bhai.
@@abdullahbaig8700 When r u writing urs?
@@aniketmajety3795 writing? I didn't get what you just said.😅 Are you talking about my exam?
Why is carbon dioxide a bent molecule here? IR absorption of carbon dioxide is to do with C=O bond stretching, not bending.
that is most likely because CO2 has a trigonal planar shape with bond angles of 120 degrees. thats just how its arranged
@@mohsinraza2589 But CO2 is linear.
herro, fank yu for the kemistree
10:36 why couldn't it be an aldehyde?
it wouldn't be an aldehyde because we're told that alcohol A has the molecular formula C5H12O and that it is branched - if you draw this out, you'll notice that it's a secondary alcohol (OH is attached to a C that's attached to 2 other C atoms)
Upon heating/reflux of a secondary alcohol, a ketone forms.
@@hamidas7890 Thanks
@@hamidas7890 Hello Hamida, I also thought it could be an aldehyde. When you draw C5H12O it could be branched, HOWEVER, there is also the non-branched chain primary alcohol version that is simply Pentan-1-ol. This is also C5H12O, and thus when oxidised would form an aldehyde.
Have I gone wrong here, if so, could you correct me?
@@madvexing8903 if the question did not specify that the structure is branched then we could assume that because it's a primary alcohol it would form pentanal or pentanoic acid (dependent on conditions) when oxidised. However, because we are explicitly told in the q that we have a branched structure we know it is not a primary alcohol so an aldehyde would not form, instead we get a ketone.
@@hamidas7890 My bad, I didn't read the question properly then.
Speak loud 😂
Tuadi angrezi smjh nhi lgdi payee 😅
madad chahidi ae?