Haan. D id say as well.
At first glance I was like kya rangoli ka pattern banadiya hai ?
Answer given is a
Okay look I also thought A at first. Then thought D is more correct
If you use H2/Ni first, isn’t there a chance all Ph rings will be hydrogenated?
I also choose D in the exam but they give answer as a...
Galat ans pe Kabhi Kabhi Apne knowledge se Bharosa uth jata hai😅
Is there anything we can say about the middle rings stability?
You mean it’s less stable than the others?
If A is the right answer then something like that must be true.
Yeah, that’s true tho
@arush_kumar_singh solution mey kya likha hai? Is it the Fiitjee special “factual”??
In some site they gave a reaction in which naphthalene on mild hydrogenation gave benzocyclohexane. SO maybe something like that is happening here.
Ohhh. Maybe yeah. I’ll see if I find anything more.
Sol nahi diya hai
Even after revised keys were uploaded still it is A
Its be cuz some square are antiaromatic so they have to be reduced first
Ohh yes. You're right bro. Nice one
How squares are anti aromatic ...
There are 4 aromatic rings and e^- of benzene ring resides on itself .. and there isn't be any conjugation in square ring....
If squares ring are in conjugation the compound is anti aromatic but nature prefer stability so it will not occur..
Let the ans is A so why D isn't possible ??
Is it hard to reduce middle ring when extra bonds get formed ????
Bro this was a last yr ques and i remember this was reasoning it was very old ques around 1990 or something
2001 I guess
I guess central ring should be reduced first so that anti aromatic butadienes go away and the compound attain aromaticity.
Then it will be easy for bromination.
If you count the no. Of pi-electrons in peripheries then you see that the whole compound in anti-aromatic. So if middle ring is reduced, then we get three individual aromatic parts from an anti-aromatic compound which is much more stable.
Aren't there only 2 pi electrons for a square which one are the other 2?