Infra Red (IR) Spectroscopy | A-level Chemistry | OCR, AQA, Edexcel

[@kieranwright7323]

I feel like I'm being taught by Hermione Grainger

[@blazebangerz3122]

wingardium leviosa.

[@emilym2742]

@@blazebangerz3122 it’s wingardium leviosaa not leviosar

[@blazebangerz3122]

@@emilym2742 lmao

[@ihfazhassan1548]

@@blazebangerz3122 Go on Harry... Don't stop

[@euansimpson5010]

literally was about to say that haha

[@mayamelie20]

In the breathalyser test, they actually test for CH bonds as there are OH bonds in the breath naturally from water vapour.

[@nareshhkumar7576]

Really? I thought it depends on the functional group. OH bonds in alcohol show different reading from OH bonds in water.

[@neithonosmani9783]

@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?

[@aleenaasif4712]

EXACTLY MY QUESTION

[@thaliaissa3009]

@@aleenaasif4712 dd u get why we couldnt

[@aleenaasif4712]

​@@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?

[@beverlymatemura820]

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)

[@Ronantheaccuser04]

Secondary alcohols only form ketones when oxidised.. Primary alchohols could form aldehyde

[@ActuallyLinden]

Anyone else using this for distance learning? :/

[@ShahFahadKhan]

Amazing lecture👍👍

[@magdalenapovanhu1478]

how do you know which peak to look at??

[@eliascath6608]

The mic really do be on a mad one

[@charliestewart885]

B could also be an ester by the same logic? and peaks and troughs are different things

[@15hanjm]

Why is carbon dioxide a bent molecule here? IR absorption of carbon dioxide is to do with C=O bond stretching, not bending.

[@mohsinraza2589]

that is most likely because CO2 has a trigonal planar shape with bond angles of 120 degrees. thats just how its arranged

[@Leo_BS-ex2xz]

@@mohsinraza2589 But CO2 is linear.

[@aniketmajety3795]

Do we have to memorize the wavenumbers or will they be provided?

[@abdullahbaig8700]

They will be provided in your data booklet which will be given to you with your paper.

[@aniketmajety3795]

@@abdullahbaig8700 tq bro!!

[@abdullahbaig8700]

@@aniketmajety3795 Np bhai.

[@aniketmajety3795]

@@abdullahbaig8700 When r u writing urs?

[@abdullahbaig8700]

@@aniketmajety3795 writing? I didn't get what you just said.😅 Are you talking about my exam?

[@АнитаТен]

perfect explanation!

[@Kiyotaka_Narumi]

Man, did this help!

[@ninjadog5800]

Our data sheet doesn't have the "intensity" part, would this be this an aqa thing?

[@user-jr9tf1iq8e]

10:36 why couldn't it be an aldehyde?

[@hamidas7890]

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.

[@user-jr9tf1iq8e]

@@hamidas7890 Thanks

[@madvexing8903]

@@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?

[@hamidas7890]

@@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.

[@madvexing8903]

@@hamidas7890 My bad, I didn't read the question properly then.

[@dylanmaxim4909]

herro, fank yu for the kemistree

[@mandip863]

Speak loud 😂

[@punjabihits3028]

Tuadi angrezi smjh nhi lgdi payee 😅

[@some1ulove162]

madad chahidi ae?