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how can u solve itYes.
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how can u solve itYes.
solve what?how can u solve it
View attachment 61889
can explain why answer is D?
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.
wht is da difference b/w H bonded alcohol nd free alcohol in IR spectroscopy? ny1 knws?
oh
oh i see , really thx very much !!
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.
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..
no...is this question included in this year's syllabus?
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