oc2

[exotic falcon]

in i. ArSn takes place and cl is replaced by oh
cant figure out ahead

[kindred tide]

<@&1227988359910260737>

Note for OP

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[exotic falcon]

i was thinking along the lines of benzyne mechanism cuz ArSn is not possible due to ERG on ortho and no EWG on para

however that doesnt give the req product

[tight jungle]

is this from aakash module?

[exotic falcon]

yep

[tight jungle]

haha yeah i have solved it

let me look

wait

[exotic falcon]

alrr

[tight jungle]

can you tell the que no?

[exotic falcon]

assignment ques 6

[tight jungle]

set 1?

[exotic falcon]

huh

its in aakash invictus module

not the normal one

just solve the ques bruh

[tight jungle]

the koh will attack on most reactive halogen

which removes br group

adding oh group

and then the naoh and ch3 cl together remove h from oh and the ch3 gets added to o giving in end 2 chloro 1 methoxy 4 nitrobenzene

[tight jungle]

typing is hard bro

[exotic falcon]

2 bromo but yes

[tight jungle]

why 2 bromo?

[exotic falcon]

because cl will be removed

arsn

but why is naoh + chcl3 reacting with -oh

[tight jungle]

naoh strong base

so it removes h

and its ch3cl

so cl gets removed and the ch3 attaches to o

[exotic falcon]

in the second rxn , benzyne mechanism kyu nahi hua

your process makes sense to me why do we have to proceed that way

[floral raft]

???

Shine some light on your statement

[tight jungle]

try to read up more from ncert you will understand this much better on how the strong base substitution works

[exotic falcon]

Why can't this work

[tight jungle]

i think u should first read up on ncert

then it make sense more

[exotic falcon]

ohh ig acid base is prefered over benzyne

[solid kite]

No, see, you're abstracting a proton in both cases, but the OH proton is much easier to pull away than one which is connected to benzene ring.

[exotic falcon]

due to steric reasons?

[solid kite]

Active vs. inactive hydrogen. Phenate has resonance stability, but lp on the benzene carbon isn't involved in resonance.

[exotic falcon]

so its all about finding the most stable anion

when doing reactions with naoh

[solid kite]

Pretty much always, really, assuming you aren't looking for a kinetic product of a reaction with multiple pathways

[exotic falcon]

alr got it

thanx

+solved @solid kite @tight jungle