IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

man111

(1) which having highest boiling point

(a)ch₃- ch₃

(b)ch₃- ch₂-I

(c)ch₃- ch₂-Cl

(c)ch₃- ch₂-Br

explain fully .

rishabh29288

ch3-ch2-I............boiling point depends on vander waals forces of attraction and more the mol wt more the vander waals forces of attraction

rooney

Well..polar molecules have higher b.p. than non polar (strictly covalent) molecules. In that case CH3-CH3-Cl should have max b.p.

zane3004

Ch3-ch2-cl becoz of inductive effect i think (sorry kind of bad at this so not sure)
cl is more electronegative than I, Br so it will attract electrons from ch2
:. ch2 will be relatively negative
so it will have tendency to attract electrons from ch3
so bond will be stronger and hence BP must be highest

tarinbansal

I think the answer shud be CH3-CH2-Cl due to most effective hydrogen bonding as Cl is the most electronegative element among l,Br,I.
Due to better hydrogen bonding, more heat need to be given to break those temporary bonds and boil the compound.

I hope I am correct. Please corrent me if I am wrong.

sivaramakh

ch3-ch2cl has hihest boiling point has c-cl bond is stable than the other 2 so breaking it requires more heat

rishabh29288

well the answer is ch3-ch2-I...........IN DETERMINING THE BOILING POINTS, IT IS THE VANDERWAAL FORCES THAT WE CONSIDE AND NOT THE C-X BOND STRENGTH....
The large increase in number of electrons by the time you get to the iodide completely outweighs the loss of any permanent dipoles in the molecules.

harkaran

tarinbansal i think cl does not form h bonds despite high electronegativity coz of its large size do correct me if i am wrong

akhil_o

@tarinbansal

CH2Ch2Cl cannot form H-bonds!

for h-bonding,

the H needs to be directly attached to a highly electronegative atom like O,N...

in this case it is attached to only C which has electronegativity almost same as H

Answer seems to be the iodide

akhil_o

@harkaran ur assertion is right but ur reasoning is wrong

tarinbansal

Ya, I am sorry for that. I missed that point on that occasion of posting my previous solution.
It does not form H-bonds.

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