I was given a problem in which I had to identify the most reactive site for electrophilic substitution out of the positions 1,2,3,4 (shown in the figure).

The molecule

My Approach: Upon drawing the resonating structures of the molecule, negative charge appears on position 3 and 2.

In the resonating structure in which the negative charge is under 3, there is a positive charge under -NO2 group making that structure less stable. This does not happen with the structure in which the negative charge is under 2.

So I concluded that 2 would be the most reactive site. But the answer given is '4'.

What is wrong in my reasoning?

  • $\begingroup$ @AvnishKabaj I think that should happen only if bulky groups are attached to the phenyl ring which would align them in mutually perpendicular planes because of free rotation. Can you give a source which mentions this? $\endgroup$ Nov 16, 2018 at 7:12
  • $\begingroup$ @AvnishKabaj Also, phenyl ring is said to be ortho para directing because of its resonance effect, (while it deactivates the ring via inductive effect). $\endgroup$ Nov 16, 2018 at 7:15
  • $\begingroup$ consider only the deactivating effect of the ring.. 4 is the best option because that ring is comparitively less deactivated $\endgroup$
    – Ava
    Nov 16, 2018 at 10:13
  • $\begingroup$ @AvnishKabaj Look at the substitution on that specific biphenyl in the linked question... $\endgroup$
    – Zhe
    Nov 16, 2018 at 14:00
  • $\begingroup$ Waaiiiit the answer below just reiterated what I said $\endgroup$ Nov 17, 2018 at 10:08

1 Answer 1


As you know, in deciding which position undergoes aromatic electrophilic substitution, three elements must be taken in account: steric effect, mesomeric effect and inductive effect.

Let's call the ring containing atoms 1 and 2 "A", and the one containing atoms 3 and 4 "B".

Before deciding the importance of any of those effects, some considerations about the geometry of the molecule should be taken in account. It could be guessed, by thinking about the spatial arrangement of such molecule, that the high steric effect between rings A and B prevents their rotation, thus keeping them in a constantly "skewed" position in respect to the nitrated ring.

A representation of the molecule, with the Van Der Waals radius of the elements, follows:


Steric Effects: Because of the presence of a nearby phenyl (the nitrated one) to any of the ortho positions of rings A and B, and because of the presence of the other phenyl in the middle of the possible electrophile's path, all the ortho positions might be considered as non favourable.

Mesomeric effect: One might guess that the resonance structures of the nitrated ring could extend to ring B, while leaving ring A untouched. Actually, such a resonance structure requires a "double bond behaviour" for the aromatic carbons between ring B and the nitrated ring. But the molecule is extremely unlikely to behave that way, since the copresence of two aromatic rings prevents their own rotation, and locks them in a skewed position. This conformation, in other words, renders the pi system "less conjugated", and the mesomeric effect much less relevant.

Inductive effect: This is the final and, in this specific molecule, most important consideration. Simply, everything boils down to the proximity of the nitro EWG: this makes ring A more electron-poor, and ring B more electron-rich.

The answer is, hence: the meta position of ring B is attacked, or the "4" position in your image.

  • $\begingroup$ 's if rings are skewed and mesomeric effect is very very less then it stands to reason that meta substitution is not possible. During eas intermediate formed could be more stable if para attack takes place.Inductive effect from NO2 inductive effect decreases with distance and could be negligible . $\endgroup$ Nov 17, 2018 at 4:51

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