1but wat will be the relation between the first and the third
arrange the following with respect to their reactivity towards Sn1
(CH3)3C-I & (Ph)2-CHBr
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19 Answers
1
1diphenyl>t-butyl>phenyl
sory abhi din saw ur ans but i can giv the reason
1. double resonance is ofcoarse is the strongest
2. for t-butyl & phenyl t-butyl is ahead as
a) 9 hyperconjugetion for t-butyl
b) some amount of -I effect goes against phenyl
1but my teacher told that first one is more stable , it is experimentaly found...............
1between the first n the third also i think it will be third one.. (resonance..)
1the second compound will be more reactive towards Sn1 as the carbocation will be resonance stabilised over two rings
whereas in the first case there will be no resonance
1the answer for Sn1
(CH3)3C-I & (Ph)2-CHBr
is (Ph)2-CHBr>(CH3)CI.
1arrey first toh wahi hai... stability of carbocation...
and dat is the rate determining step... so dat is the only imp step in sn1...
aur kuchh nucleophile toh ana hi hai.. nahi toh phir toh reaction hi nahi hoga!
1jaha tak mujhe yaad hai (:P) ... sn1 has stability of carbocation concept...
toh yaha pe nucleophile kaha se aya?
next thing,,,, wen thr will be a positive charge on C in case of 2nd ocmpound, there are TWO benzene groups to stabilize that positive charge... so my ans is 2nd one...
1ii guess (ph)ch2br
cozx here nucleophile will attck so positive charge shud b more thus ........wen 2 ph groups r ther positive chrage decreses due to -! effct thus
i guess that's d asnwer ie phch2br ........
1I THINK (Ph)2-CHBr... coz two benzene groups attached... so more e-density than in case of the third one...
(can be wrong)
1and what will be the answer if
(CH3)3C-I , (Ph)2-CHBr & (Ph)-CH2Br