29 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.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
is this from aakash module?
yep
haha yeah i have solved it
let me look
wait
alrr
can you tell the que no?
assignment ques 6
set 1?
huh
its in aakash invictus module
not the normal one
just solve the ques bruh
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
typing is hard bro
2 bromo but yes
why 2 bromo?
because cl will be removed
arsn
but why is naoh + chcl3 reacting with -oh
naoh strong base
so it removes h
and its ch3cl
so cl gets removed and the ch3 attaches to o
in the second rxn , benzyne mechanism kyu nahi hua
your process makes sense to me why do we have to proceed that way
???
Shine some light on your statement
try to read up more from ncert you will understand this much better on how the strong base substitution works
Why can't this work
i think u should first read up on ncert
then it make sense more
ohh ig acid base is prefered over benzyne
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.
due to steric reasons?
Active vs. inactive hydrogen. Phenate has resonance stability, but lp on the benzene carbon isn't involved in resonance.
so its all about finding the most stable anion
when doing reactions with naoh
Pretty much always, really, assuming you aren't looking for a kinetic product of a reaction with multiple pathways
alr got it
thanx
+solved @Opt @Sephrina
Post locked and archived successfully!
Archived by
<@741159941934415883> (741159941934415883)
Time
<t:1747640656:R>
Solved by
<@763645886500175892> (763645886500175892), <@888280831863451688> (888280831863451688)