Plz check third and solve fourth.... ble hou (3) (4) ch,ch ch, c

stability goes like i>ii>iii>iv

stability of carbocation directly proportional to +I ,+H and +R

and inversely proportional to -I and -R

ok

have you check the -I series shweta??

but in third and fourth I have used dnp rule

no

thus is right n

but in 3 two chlorine atom more unstable but dominant factor is Br which is nearer ..so fourth one is least stable

yes

got Shweta??

sir no2bhi to hai

which has more minus I

which makes carbication unstable

No2 hai lekin dominance distance factor karega shweta...Br near hai..so jyada -I effect lagega

4th one

sir no2kyin le re hai dnp main

no2 to hai he dominant ....I am considering dnp for third and fourth only

two chlorine hai 3rd carbon pe..but Br 2nd carbon pe hai..so iska jyada -I effect hoga

ye pta chal gya

jbki no2 is different

No2 ka effect jyada hota hai bt wo 3rd carbon pe hai...

acha

distance dependent..first priority dena hai

ok

samagh aa gya

👍👍

thanku

welcome 😊and do watch the whole video which i send

I have already seen the video sir

good👍

got it??

but I have done third like this

but here we will give priority to distance.

did you check my explanation??

but sir why second is more stable it's no2 gruop electron withdrawing group more minus I than cl and br here we are interestrd in stability of carbication

ii is more stable than iii because here in iii there is two Cl so ii have more -I effect than ii. but stability is reversed so ii is more stable

please accept the answer if you got it 🙏🙏

thanku sir

you welcome 🙏

okay