Chemistry: Post your doubts here!

[student2]

wht is da difference b/w H bonded alcohol nd free alcohol in IR spectroscopy? ny1 knws?

 

[Lee Qian Yi]

can explain why answer is D?

 

[abbas haider]

View attachment 61889
can explain why answer is D?

Read This






look at (ii) if if there is one alkyl group and one hydrogen at one end of the double bond a carboxylic acid will form ,now look at the question the organic compumd has one alkyl group and one hydrogen on either side of double bond this means two Carboxylic acids will form (COOH) howevver what this pic didnt tell is that if OH is already present in a compund it will always form a COOH.



Using elimination method
A) not possible because double bond has to rupture so 2 molecules have to form so this is not possible. (because its the only link between two group of atoms, THIS MAY NOT HAPPEN IN CYCLIC COMPUNDS)
B)double bond always rupture but this is nt the case since C=C is safe . so this is not possible.
C) although C is partially correct because if we have one alkyl group and one hydrogen at one end of the double bond , first aldehyde is formed , C=C breaks ,1 carbon forms COH. however the process oxidation hasnt ended . the -H in COH also gets an O forming -COOH . thus the end product is -COOH. thus c is not possible
D) Final answer and already told why.

 

[Lee Qian Yi]

oh

Read This

View attachment 61890




look at (ii) if if there is one alkyl group and one hydrogen at one end of the double bond a carboxylic acid will form ,now look at the question the organic compumd has one alkyl group and one hydrogen on either side of double bond this means two Carboxylic acids will form (COOH) howevver what this pic didnt tell is that if OH is already present in a compund it will always form a COOH.



Using elimination method
A) not possible because double bond has to rupture so 2 molecules have to form so this is not possible. (because its the only link between two group of atoms, THIS MAY NOT HAPPEN IN CYCLIC COMPUNDS)
B)double bond always rupture but this is nt the case since C=C is safe . so this is not possible.
C) although C is partially correct because if we have one alkyl group and one hydrogen at one end of the double bond , first aldehyde is formed , C=C breaks ,1 carbon forms COH. however the process oxidation hasnt ended . the -H in COH also gets an O forming -COOH . thus the end product is -COOH. thus c is not possible
D) Final answer and already told why.

oh i see , really thx very much !!

 

[abbas haider]

wht is da difference b/w H bonded alcohol nd free alcohol in IR spectroscopy? ny1 knws?

The IR sepectrum gives different peaks for A free bonded alcohol and a hydrogen bonded alcohol



A hydrogen bonded alchol has a BROADER peak then Free alcohol


check this https://webspectra.chem.ucla.edu/cgi-bin/webspectra.cgi/rp5/I/rp4/I?Zoom=141
RED = dilute t-butanol in choloroform
BLUE = Conc - butanol


An hydrogen bonded alcohol means = more molecular contact (e.g concentrated Alcohol) = more hydrogen bonding thus in a sample there are more -OH to absorb the infrared waves thus a broad peak.
a free alchol means = less molecular contact (e.g dil alchol or alchol in gas phase) thus in a sample there are less -OH to absorb the infrared waves thus a sharp peak.

 

[abbas haider]

oh


oh i see , really thx very much !!

Your Welcome

 

[AnonymousX9]

Guys pls help how do I balance this equation using oxidation numbers???

K3Fe(C2O4)3 -----> K2C2O4 + FeC2O4 + CO2

Thank you

 

[darks]

Can someone give me the complete answer to this question? it's 5 marks and from Oct/Nov 2014 paper 43. Thanks..

 

[student2]

The IR sepectrum gives different peaks for A free bonded alcohol and a hydrogen bonded alcohol

View attachment 61896

A hydrogen bonded alchol has a BROADER peak then Free alcohol


check this https://webspectra.chem.ucla.edu/cgi-bin/webspectra.cgi/rp5/I/rp4/I?Zoom=141
RED = dilute t-butanol in choloroform
BLUE = Conc - butanol


An hydrogen bonded alcohol means = more molecular contact (e.g concentrated Alcohol) = more hydrogen bonding thus in a sample there are more -OH to absorb the infrared waves thus a broad peak.
a free alchol means = less molecular contact (e.g dil alchol or alchol in gas phase) thus in a sample there are less -OH to absorb the infrared waves thus a sharp peak.

Vry much grateful. Thnkz

 

[Thought blocker]

View attachment 61897
Can someone give me the complete answer to this question? it's 5 marks and from Oct/Nov 2014 paper 43. Thanks..

- Each carbon is sp2 hybridized as 3 sigma bond with each C atom.
- As it is sp2, thus bond angle is 120 degree
- All C - C have same bond length.
- Also, sigma bond between C - H and C - C and pi bond between C=C

I guess this is enough.

All the best buddy. (Y)

 

[nadeen64]

is this question included in this year's syllabus?

 

[darks]

is this question included in this year's syllabus?

no...

 

[darks]

d)i) E for CH,, F for CH2
while d)ii E triplet and adjacent 2H,, F doublet adjacent 1H
i got d)i wrong. How is E CH and F CH2?? am i missing a concept?

 

[Holmes]

9701/12/O/N/10 (Chemistry)
October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39.

Help me please!!
Thanks in advance

 

[Holmes]

explain why the answer is B

 

[kungfuPanda]

Hi
Can you guys help me out with paper 3 ,I have a simple question.
What do i have to keep in mind while taking readings from instruments?
Like measuring masses or temperature,Is it always good to round off numbers during measurements for final readings at the start of Question 1 or 2 ?

 

[Laveeza]

Help! Can somebody plz explain me question 33

 

[Lee Qian Yi]

1) In an experiment 12.0dm3 of oxygen measured under room conditions is used to burn completely 0.10mol of propan-1-ol. What is the final volume of gas measured at room conditions? ANS: 8.40dm3
Anyone can explain?

 

[yahyub123]

does cold NaOH react with phenol

 

[darks]

1) In an experiment 12.0dm3 of oxygen measured under room conditions is used to burn completely 0.10mol of propan-1-ol. What is the final volume of gas measured at room conditions? ANS: 8.40dm3
Anyone can explain?

CH3CH2CH2OH + 4.5O2------>3CO2 + 4H2O
so 0.1 mol propanol reacts with 0.45 mol of the o.5 mol oxygen provided (12dm3 oxygen is 0.5 mol)
0.05 mol of oxygen is left over as gas.
0.3 mol of CO2 is formed which is the gaseous product as H2O will be liquid at r.t.p.
0.35mol of total gases.
0.35x24=8.4dm3 gases in the final mixture under ROOM CONDITIONS.