phenol doubt
In 6th what will be the use of fecl2 and how will the rxn proceed through mech
And in 7th I did the process in the image pls correct me anybody
22 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.Andhyd fecl2 with cl2 splits it into cl+ and cl-
Electrophilic sub reaction follows with benzene
Also keto enol tautomerism you get that the compound in the second q is just 4 hydroxy phenol.
Aka hydroquinone. Does this help?
Sir are u talking about the reactant, because I wasn't able to get 4 hydroxy phenol
6th wala hogya , thanks 🙏
Still can't figure out the last one
I think 7th me to acidic medium ke karan tautomerisation favour hona chahiye
'cause that leads to aromaticity
I'll send a picture drawing everything once I get hold of pen and paper
in the second case, the h comes from the c that has a couple of them, no?
its okay, this is on of those things that once you know you remember, dw about it
And yes, what varun said
usually hydroquinone form me rehta hai cos of aromatisation
https://en.wikipedia.org/wiki/Enol
See under the heading phenols
Enol
In organic chemistry, alkenols (shortened to enols) are a type of reactive structure or intermediate in organic chemistry that is represented as an alkene (olefin) with a hydroxyl group attached to one end of the alkene double bond (C=C−OH). The terms enol and alkenol are portmanteaus deriving from "-ene"/"alkene" and the "-ol" suffix indicating...
Thanks a lot sir
Got it I forgot the other carbon
most welcome. shall we close this one out?
So just normal nucleophilic addition for the ring right?
My ans is just a but the book says b, cl at meta
well it is ortho to one OH :)
Yes sir u can close this out i have got my ans, this the first time in my 2 years someone has actually cleared my doubts, thanks a lot
a exists with just tautomerism
Ok so it wasn't meta directing
oh most welcome dude. come and ask as many as you like here!
na no way. OH is ring activating right. but since there are 2 oh groups, all 4 positions are equivalent
Ohh
so more than meta, ortho, try and see which group causes what effect
Yes sir thanks
+fsolved
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