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interesting....

:CCl2 > :CBr2 > :CH2 > :CF2

WHAT HAPPENS:

This order is so because, :CF2 is relatively very stable due to back bonding.

comparing the left 3 carbenes.

the extent of back bonding is very poor, due to high energy difference in orbitals of the overlapping orbitals (:CH2 doesn't do back bonding)

therefore we come to know that Cl and Br impart unstability to the molecule, whereas H doesn't do anything to the carbene

therefore :CH2 is more stable than :CCl2 and :CBr2

so now we consider Inductive effect, which is certainly more for Cl and therefore :CCl2 is less stable than :CBr2.

hence the final order is:

(reactivity)

:CCl2 > :CBr2 > :CH2 > :CF2

Whats going on...???

i'm explainin the whole thing agian

The answer is a)

looking at the question option c) and d) can be clearly ruled out.

WHAT HAPPENS: H-X bond dissaciates and H+ attacks.

Markowonikoff addition occurs and carbocation intermediate is formed

the carbocation formed will be stabilized by hyperconjugation and inductive effect of the Me group/s in c) and d)

now we are left with a) and b)

in option b) the Carbocation is not stabilized by any means...

but in a) the carbocation is stabilized by resonance of Cl, but not to forget the inductive effect of Cl, which is much more dominating/instability providing than the stability provided by resonance.

so the story is clear.

option a) carbocation but electron withdrawing effect(on the whole)

b) carbocation, with no effect on it...

AND MORE STABLE THE REACTION INTERMEDIATE, MORE IS THE RATE OF REACTION.

so the answer is clearly a)

hope its clear now :)

@smarsh

i didn't mean to be proudy, i'm sorry if i hurt ya by any means

but ya for sure...its NOT AN ADDITION REACTION, ITS TERMED AS AN SUBSTITUION REACTION. (and thats what i said was wrong)

and Here there is no addition by any means....

and dude, i just TRIED to explain it more clearly.

@conjurer

a carbocation intermediate is formed and not a carbanion..

1st H+ will attack, and the reaction will proceed according to Markonikoff Addn.

and more stable the intermediate, more is the speed of the reaction.

@tarin
but won't angle strain be generated there...???

please can you explain the whole thing...and clarify my concepts..

edited

because the carbocation formed will be stabilized by hyperconjugation and inductive effect of the 2 Me groups

well i come here to learn new things/concepts and help others...
not to point out mistakes like you...If you don't play the game, don't decide the rules...

The answer is a)

looking at the question option c) and d) can be clearly ruled out.

now we are left with a) and b)

in option b) the Carbocation is not stabilized by any means...

but in a) the carbocation is stabilized by resonance of Cl, but not to forget the inductive effect of Cl, which is much more dominating/instability providing than the stability provided by resonance.

so the story is clear.

option a) carbocation but electron withdrawing effect(on the whole)

b) carbocation, with no effect on it...

so the answer is clearly a)

hope its clear now :)

awssy article dude...

@ankitgg

it was act as an lewis acid also because of the empty d-orbitals, it can accomdate extra electrons...

rest of your answer is correct.

hence SOCl2 can act as an Lewis Base/Acid

NBS does substitution raeaction, because of its low concentration, it diverts the path of reaction from Addition to substitution..

ANSWER:

The carbon with 2 methyl grps is '1'

going in clockwise direction i've numbered the carbons as '2' and '3' and '4' respectively.

Actually what happens is that, as the Br radical is released, it attacks on carbon '4' because its an allylic carbon, and the intermediate radical formed will be resonance stabilized.

Now if you see which Carbon will be stable with a free radical on it, the answer will be Carbon no '2'

it will be slightly more stable than free radical on '4'

because of more number of hyperconjugation and more inductive effect.

Therefore, atom no. '2' will be the free radical and now it will attack on the Br molecule and the desired product will be formed, hope its clear now :)

@akshaytalwar2
well if not IIT then surely this :P :D

@sankydreams
i too agree with you :)

@tarin
thanks 4 thinkin mah article worth to be put up there :)

and thanks to all others too...