@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?
@@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?
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 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.
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
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.
@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
Anyone else using this for distance learning? :/
Amazing lecture👍👍
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.
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
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?
The mic really do be on a mad one
perfect explanation!
Man, did this help!
Our data sheet doesn't have the "intensity" part, would this be this an aqa thing?
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?