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Organic Chemistry

Blazing goIITian

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29 Jan 2008 16:57:31 IST

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For the following compounds......

None

Please click here to have a look at the diagram.

For the following compounds,

The rate of electrophilic substitution in the decreasing order is

(A) I > III > II

(b) III > I > II

(C) III > II > I

(d) I > II > III

--------------------------------

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Comments (11)

Akhil

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29 Jan 2008 17:57:54 IST

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i think its A) 1>3>2

In 1 4 methyl groups are attached ...so o,p are activated.

In 2 ,no groups are activating...phenyl group is -I so deactivating

In 3, only 1 C is attached to 2 rings so lesser activation

Rate if right

nudge if wrong

Akhil

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29 Jan 2008 18:01:38 IST

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The only reason i can find it to be c is maybe theyve considered 1 as
sterically hindered...so it goes right in the end behind 3 and 2
so they get 3>2>1

sinjan jana

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29 Jan 2008 18:03:22 IST

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nitigya

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29 Jan 2008 19:16:14 IST

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can we say that 1 is strically hindered.

2 has no activating groups

3 has an activating grp but is less sterically hindered

hence ans (c)

fwks_phoenix

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29 Jan 2008 19:25:40 IST

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i think it's A

cuz electrophiles : electron deficient

in [I] :due to +I effect of alkyl grps attached to the ring, the electrophilic sub is increased.

[II]:in this too, CH3 is bonded to the 2 rings.. it's lesser than I cuz the same C is attached to the ring in this case while in I there r four CH3 grps attached to the rings (2 each)

[III]:no +I effect.. so it's d least...

GAURAV SHARMA

Hot goIITian

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29 Jan 2008 20:13:07 IST

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but inductive effect is overruled by resonance which is observed in all three.so steric factor has to be considered. since its most difficult to react with 1st for the steric reasons it is last,then in between 2nd n 3rd, 3rd will have the maximum rate because it has an additional ring attached to it,increasing its stability than the 2nd compound.hence the correct order is

3>2>1.

Neeraj Agarwal

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29 Jan 2008 21:42:22 IST

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should be (A)

1. 4 CH3 means 3 alpha hydrogens on each => more hyperconjugation => more -ve charge density

2. since there is only 1 CH2 grp. attached, there are in total only 2 alpha hydrogen=> less hyperconjugation => less -ve charge density

3. there are two rings which exert -I effect on each other and thus electrophillic sub. becomes difficult...

so (A)

Neeraj Agarwal

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29 Jan 2008 21:42:57 IST

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srijan...can u please tell me how did u create and publish that page...???

sinjan jana

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29 Jan 2008 22:11:18 IST

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www.skrbl.com

Karthik M

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29 Jan 2008 22:11:20 IST

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Do it by the resonance concept. III has amazing resonance. II has better conjugation

Prof. Sanjay Sharma

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Joined: 18 Sep 2007

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2 Feb 2008 14:46:37 IST

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Dear Sinjan.j
Among the options given (A) is the correct option. The reason is very clear just take 2 as the basic structure and think about the fact that electrophilic substitution depends upon the electron density over the rings in question. Now monosubstitution in both the rings with a group of the order of alkyl group increases the electron density in both the rings equally thus increasing the electrophilic substitution here as compared to 2nd.
Similarly if we look at compound 1, here each ring show di substitution with methyl group. This group in real sense is a ring activator thus activating the rings for further substitution to the most among all he three structures.

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