Please help. Answer is B
Also help woth 14.Answer is A. How to kbow which Cl to replace
Help @Shwetanshu_2018 @Bhuvanitha_2020
Chlorine is ortho and para directing so it will replaced at para position by oet group as ortho elimination is not possible as no substitute present originally at ortho possible and since chlorine doesnt show meta directing so chlorine will not be replaced from meta position
Nitro group is meta directing. So at its meta position, the chlorine had a high binding tendency compared to the para one. So para chlorine will be easy to remove.
I searched in internet that for aromatic nucleophilic substitution cl shows more than bromine
Question in post#1
Rate of SnAr depends only on the stability of carbanion (i.e more is the -I , -M greater is the rate ) it has got nothing to do with leaving group tendency .
Usually students think that rate of bromo benzene will be greater than chloro benzene because Br is better leaving group .
But here we have to consider only the stability of carbanion formed and hence rate of chloro benzene is greater than bromo benzene
Oh thanks for clarifying
Q14 ) if the Cl in the parra position is substituted , then the carbanion formed in the intermediate step is in conjugation with No2 and hence gets stabilised , whereas if meta Cl is replaced then the C- formed is not in conjugation with No2 and hence is lesser stable .
That’s why , parra Cl will get substituted
@Samarth_2020 don’t you think that here rate will not be decided by the bond strength , rather it will be decided by the stability of intermediate formed ?
Your reasoning also leads to same result , but when it comes to explaining the concept , i think , this reason won’t be much appropriate.
(I’m just presenting my perspective , If you have read it somewhere , then do let me know , because that will help me to learn something new )
@Shwetanshu_2018 bhaiya you are right. I have used that concept of mine in many questions and got it right as well but as you said it is not appropriate. It is just cultivated over sometime of practice and luckily I got them right.
I didnt get thr conjugation part in your answer. @Shwetanshu_2018
Thanks a lot @Shwetanshu_2018
@Shwetanshu_2018 Our teacher has told us that it only has to do with -I effect i.e. greater the -I more the rate. Can you clarify.
What your teacher has taught is correct , i.e leaving group can only show -I effect .
But when it comes to No2 it will show -I as well as -M (like in next questions , the rate is decided by -M of No2 )
I think i should have clarified it there
@Shwetanshu_2018 Yeah, but NO2 isn't leaving group and here we're only talking about the effects shown by the leaving group.
@Abhishek_2020_5 check out question number 14
I did and it has nothing to do with my doubt as in question 14 the product is decided on the basis on the effect shown by NO2 not by leaving group i.e. Cl.
If you’ll read this sentence , then you’ll figure it out , that there’s nothing wrong in this sentence . Here i have no where mentioned that carbanion must be stable due to -M of leaving group .
All i said is rate depends on stability of carbanion.
Clearing your doubt , as i have mentioned in post#15
Only -I of leaving group will be considered , but all other effects of other substituents are to be considered
I hope , the above statement clears it all , and really sorry if any statement of mine confused you.