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@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.OH- kicks out Br- I guess
aldehyde pe rxn nahi hoga?
Uske baad fir I guess nucleophilic attack.....ahmmm
Kuch significant hota dikh nahi raha
NaOH concentrated hai kya?
Like is anything mentioned about that?
nahi
should i send complete ques
Ek baar bhej do
I'll see
Multi correct
i hope the Q i mentioned is major
on second thoughts it might not be
Major ortho vala hoga
But that doesn't ring a bell
š„
Usse aage ki reaction ka nahi samajh aaya kuch ki kya hoga
yeah fr
Yaar I'll see abhi thoda busy hun
Isme ether nahi banega?
(as I told you)
NaOH takes phenol ka hydrogen
Then does nucleophilic atk
That O won't be nucleophilic anymore
Resonance
Haan right
In resonance
Ahmm I'll see yaar I'm kinda into something rn š
alr no problem
Chcl3 naoh forms :ccl2
First step is peroxo wala compound?
first step forms phenol
and then reimer tiemann
Last step would be snar rxn?
not sure
Sorry sn2 reaction hoga
Williamson ether
Phch2 o - ph cho
Cho is an edg right
It withdraws sir:)
Sorry
Haha
Withdrawing yes
Both inductively and by resonance
Electronegative atom next to C
Yup yup
Similar to no2
Yeah that's an even stronger one
But for bhi phenoxide is a good nu: na
I feel like SN2 shouldn't be the way to go
Naaah sir it isn't
O ka charge is very delocalised
And O pe negative stable hai
I mean it could but doesn't seem favourable imo
I'm out here eating and kinda somewhere outside (I'm on a trip rn)
So dekhta hun jab mile thoda leisure š
Since you have a benzylic halide it would react na?
Generally phenoxide is not that great agreed
But with allylic ya benzylic halide it may react?
Enjoy bro
That seems like the only thing possible
Won't be in very great yields is I'm sure
Yeah with a family full of motion sickness vaale log at a trip to Shimla on a bus......
Definitely
š
Ouch.
I mean still would try to enjoy tomorrow lol
Aaj bohot thak gaye hain
Reached very late
is it still crowded asf
Yeah it is
Too crowded and shit is way more expensive than when we came last time
~popular spots
Like it was too long back so fair enough
Fr
its better to go higher up in less known places and eat normal dhabbe ka khana in peace
I think the question is looking for an SN2 yaar
Yeah I guess
But yahan lesser known places dekhne me hi eww hain š¤¢
Like you wouldn't even wanna go inside
so hard to travel š
i dont wanna spoil the ques by giving the ans
Yeah don't I'll see later dw
alr alr
Ortho major hoga
And O- will kick out Br
Aldehyde pe reaction nai hoga since agle step pe phch2br dediya hai
Yeah
Oh- takes acidic h oh phenol wala oh and forms o- right ?
didnt understand your reasoning in 3rd line
ether formed tho right?
Yeah
Aldehyde won't react because final step mein to Ph Ch2 Br hai
NaOH must be limited
This is hinting Sn2
how is this hinting sn2
Or maybe it becomes carboxylic acid due to self cannizaro
It shouldve mentioned NaOH excess
Wo 2-4 DNP wala option bhi dikkat karega
Last step mein phch2Br
how do u know its correct
conc naoh for that
Nah jus NaOH(excess) would've worked
Or agar Williamson ether karwana tha to NaOH(1 eq)
doesnt canizzaro require conc naoh
Advanced wale dedete h dw
Mains wale na bhi de
It's applied
Mains mein bas NaOH bhi dete h bohot baar
ah
Fir aldol cannizaro karwate rehte hai
You gotta understand the situation
so unless dil noah ho canizzaro hi karu
What the question wants you to do
Agar Cannizaro nahi hoga then it's correct
for aldol and canizaro ik alpha H condition
Aldehydes give 2-4 dnp test
but how do i know sn2 hogi and not canizzaro
why does phch2br hint sn2
Dono hogi agar NaOH excess dediya to
Lekin agar 1 eq hai to acid base reactions are much much faster than nucleophilic reactions
Isiliye bas Sn2 hogi
NaOH bas acid base karega
Agar NaOH 1eq hai
Br good leaving group and ph Ch2 mein easily Sn2 hojaaegi
And OH group with base dikh raha hai so acid base bhi hogi
O- banega and will attack
Just note that agar 1eq reagent hai to humesha acid base pehle hogi
Then aldol cannizaro wgera
ohhh
1 eq hi lagraha h isme I guess
agar naoh excess hota tab pehla kya hota
Pehle to acid base hi hota
Fir cannizaro hojaata baad mein
O- banta
Yeah
with another compound
right?
Another molecule*
Compound to ek hi hai jisme aldehyde h
Simple oxidation reduction hojaata
yeahh
Isme Sn2 hi krwana chah rha hai I guess
But ideally 1eq ya excess Dena chahiye
Advanced wale dete h humesha
Mains thoda dikkat krta hai
Kabhi kabhi
naoh excess hota toh aldol nahi hoti
Cannizaro hi hoga
Aldol keliye Alpha H chahiye
uhhh no alpha H
Cannizaro No Alpha H pe hi hota h
š
so oh- first attacks on c=o
right?
han tbhi toh canizzaro ho rha idhar
Yea
Fir hydride transfer
yeah
got it
thanxx
Np dude
@Enamine solved mark kardu?
Dono ko dede
š„ŗ
can u look a the current one if u got some spare time
Physics wala doubt hogaya solve?
Haan wahi dekhne jaraha tha
Lmao
woh toh ofc dunga hi
i just wantd to know if he got inputs/doubts
alr W
I guess I see the problem here
It is the fact that NaOH and PhCH2Br are shown to react in the same step
They should be subsequent
And then the simple williamson synthesis could be a conclusion
(although bad yield but nothing else is possible then)
And cannizzaro ofc agar NaOH daala hai kaafi saara to ho to sakti hai
Although it's not really favoured much unless the solution is very concentrated and stuff is heated up a bit
Ab ye mains kya bolta hai advanced kya bolta hai idk but this question could've been made less ambiguous
Just by marking the reagents in the last step as 1)NaOH 2)PhCH2Br
I've said all I had to
ambiguous ques but cant help it
ans is ABC tho
Satisfies the product we made right ?
yeahh
(I didn't see the options)
solved it is then
Alright then good
+solved @Enamine @Dexter
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