If you have an aromatic ring that requires substitution of some molecule, how do you know where this molecule is supposed to substitute on the ring?
In (i) and (ii) how do we select the ring and position that will receive the molecule (HSO3 in (i) and Cl in (ii))? In (iii), how do we know where to attach the I?
The position of substitution in EAS is determined by the electron-donating/-withdrawing effects of existing substituents. The effects can be inductive (the electronegativity of heteroatoms polarize the bonds), or they can be resonating (free electron pairs are donated to a double bond in the transition state).
1. rule; resonance effects overrule inductive effects.
If a ring has both a resonating and an inductive substituent, or one substituent with both effects, the resonating effect will control the reaction.
2. rule; donating effects activate the ring (more electrons for the electrophile), withdrawing effects deactivate the ring (less electrons for the electrophile).
Strong donating groups: -OH, -OR, -NH2, -NHR, -NR2, -SH, -SR, -F, -Cl, -Br, -I, -CH=CH-X (X is any donating group,
Neutral/weak donating groups: -CH3, -(CH2)R, -CHR2, -CR3 (R is any hydrocarbon), Ph.
Strong withdrawing groups: -(C=O)R, -NO2, -SO3,
Weak withdrawing groups: -(CH2)X (X is any atom more electronegative than carbon),
3. rule; activated rings favor para and ortho positions, deactivated rings favor meta position (para and ortho are deactivated).
Deactivated rings can still react, but much higher temperatures are required, and the reaction is much slower.
4. rule; para is favored over ortho.
Because the existing substituent is "in the way" of the attacking electrophile in ortho position.
In electrophilic aromatic substitution the aromatic ring has a positive charge in the transition state (because an electrophile (E+) has "borrowed" to two electrons). The electrophile can, and will, attack all three possible positions, but this first step is reversible, and any charge on a carbon is highly unfavorable. Therefore, if the charge cannot be placed on in the bottom path in the example below, the electrophile will be ejected quickly, while the other two paths are stabilized long enough for a base (B-) to extract the proton, leaving the electrons used to bond with the proton to the ring.
Question (i), the -(C=O)R is a withdrawing group on the left ring, while the RO- group has two electron pairs that it can share with the right ring. ==> para position on the right ring.
Question (ii), the -NO2 on the left ring is deactivating for that ring, while -CH2R is weakly activating for both ring. ==> para position on the right ring.
Question (iii), the ring has a strong and a weak withdrawing group, and is thus highly deactivated. Also, as ssavec pointed out, NaI is no electrophile.
Still let us assume that you had some strong electrophile; Both substituents deactivate their ortho positions, while having little effect on their meta positions. Since -NO2 is a stronger withdrawing group, it rules, and the reaction would happen meta to the -NO2 group (ortho to the -CH2Br).
In (i), it is marked correctly, in (ii) Cl will go only to the "right" benzene, not to the nitro- substituted.
In (iii), I do not see any reasonable reaction.
