which method will give more yield???

i also think its the second path bcoz of less deactivated ring.

and rohit 11 halogens are weakly deactivating groups...not weakly activating

but Rohit in the book it is given that second method will give more yield!!

1 is better since it is easier for OH^- to attack the more ∂+.....

I think it should be method 2 because: Cl shows -I effect , while OH shows -I and +R effects, upcoming group in II is a nitro group(electrophile), so it willl prefer to accept +R , -I effect of OH other than -I effect of Cl in I . Please comment

second only

what is the answer given in the book from which the question is taken?

i spot a problem here.....how are simply using NaOH and HCl to convert a aromatic halide group to a phenolic OH group??

The second method gives more yield because the NO2 +(electrophile) prefers to attack the more electron rich species at the more electron rich position..the OH gropu makes the ortho position more electron rich than the Cl group !

i think method 1 has more yield.... because of excessive steric hinderence Cl comes outta the plane of the benzene ring and can easily be replaced by OH

ok make it NaOH + HCl at 368 K thats all . You dont need any extreme conditions because of the two NO₂ groups attached to the ring.........if it would have been simply clorobenzene then needed extreme conditions like 625 k temp and 300 atm....hope it helps u

aryl halide cannot be directly converted to phenol.extreme conditions are must.read ncert dude

i mean i dont fin a mistake in this.......see any book and aryl halide is converted to phenol in presence of NaOH and H⁺ only

what are u saying mr psycho....how should i correct the conversion? there are multiple ways to do so!!! i am expecting a reply from the originator of this thread! hope he responds soon!