Q.Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.

The key idea is ortho/para directing — halogens are deactivating but still ortho/para-directing because they can donate electron density through resonance, even though they withdraw inductively.

Reasoning:

1. The lone pairs on the halogen (X) can delocalise into the ring, creating resonance structures where the negative charge appears at the ortho and para positions. …

Resonance structures of halobenzene ($\mathrm{C_6H_5{-}\ddot{X}\!:}$). In addition to the Kekulé forms, a lone pair on X can be donated into the $\pi$-system, giving three charge-separated contributors in which X carries a positive charge and a negative charge appears on the ring:

1. $X^{+}$ with the negative charge on one ortho carbon,
2. $X^{+}$ with the negative charge on the para carbon,
3. $X^{+}$ with the negative charge on the other ortho carbon.

In every contributor the negative charge appears only at the ortho and para carbons — never at a meta carbon, because conjugation cannot deliver charge to the meta position. …

Concept: Resonance Effects in Halobenzenes — Directing Nature of Halogens

Method: Resonance Structure Analysis for Directing Group Determination

Step 1: Identify the functional group and its electron effects

• The molecule is halobenzene ($\mathrm{C_6H_5-X}$), where X is a halogen (F, Cl, Br, or I).
• Halogens have three lone pairs and are electronegative — they exert two opposing effects:
  • -I effect (inductive withdrawal) — pulls electron density away from the ring
  • +R effect (resonance donation) — pushes electron density into the ring via lone pair delocalisation

Step 2: Draw resonance structures showing lone pair delocalisation

• The curved arrow in the given structure shows a lone pair from X moving into the ring, forming a new pi-bond toward the ortho position.
• Tracking formal charge on X: before donating, X has 3 lone pairs (6 e⁻) + 1 bond (1 e⁻ owned) = 7 electrons owned, matching its 7 valence electrons -- neutral. After donating one lone pair into a new pi-bond, X has 2 lone pairs (4 e⁻) + 2 bonds (2 e⁻ owned) = 6 electrons owned -- X now carries a +1 formal charge.
• The extra electron pair X pushed into the ring displaces the ring's own pi-electrons around the ring in a cascade, ending up as an extra lone pair (negative formal charge) on the ortho or para ring carbon -- exactly the same pattern as the standard phenol/aniline resonance picture (O or N donates a lone pair into the ring; the heteroatom becomes positive, the ortho/para ring carbons become negative).
• This generates three charge-separated resonance structures in addition to the original (one for each ortho position, one for para; none for meta), exactly as drawn below:

• Structure I (original): $\mathrm{C_6H_5-\ddot{X}\!:}$ — lone pairs on X, neutral molecule.
• Structure II (ortho): $\mathrm{X^{+}}$ double-bonded to the ring, negative charge on one ortho carbon.
• Structure III (para): $\mathrm{X^{+}}$, negative charge on the para carbon.
• Structure IV (other ortho): $\mathrm{X^{+}}$, negative charge on the other ortho carbon.

Step 3: Identify where negative charge appears

• In the resonance forms, the negative charge (the extra electron density) appears only at:
  • ortho positions (two equivalent positions)
  • para position (one position)
• No resonance structure places extra negative charge at the meta position. …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Drawing resonance structures that break the octet rule for the halogen

• The error: Students often draw a structure where the halogen (X) forms a second bond to the ring but also keeps all three lone pairs, giving it 10 electrons in its valence shell.
• Why it's wrong: Halogens (F, Cl, Br, I) are in Group 17 and have only 7 valence electrons. In the original structure, they have 3 lone pairs (6 electrons) plus 1 bond (1 electron owned) = 7 electrons. When they donate a lone pair into the ring, they form a second bond and are left with 2 lone pairs (4 electrons) + 2 bonds (2 electrons owned) = 6 electrons owned -- a positive formal charge on X, not a negative one, and definitely not 10 electrons.
• How to avoid: Always count electrons. After drawing the arrow from a lone pair on X into the ring, remove one lone pair from X. Then check the formal charge: X now has 2 bonds and 2 lone pairs → formal charge = 7 − (4 + 2) = +1.

Mistake 2: Placing the positive charge on the ring carbon instead of the halogen

• The error: Students conclude that the ring carbon (ortho or para) ends up with the positive charge, and the halogen ends up negative or neutral.
• Why it's wrong: It's the opposite. The halogen is the atom that GAVE UP a lone pair, so it loses electron ownership and becomes positively charged (+1, worked out in Mistake 1). The ring carbon, on the other hand, RECEIVES the extra electron density that gets pushed around the ring's pi system -- it ends up with an extra lone pair and a negative formal charge, not positive. This is the same pattern as phenol's or aniline's resonance structures (O/N lone pair donation → negative charge on ortho/para ring carbons, positive charge on the donating atom) -- halobenzene behaves the same way, with the halogen playing the donor role.
• How to avoid: Follow electron ownership, not intuition. The atom that DONATES a lone pair loses an electron and becomes more positive; the atom/position that RECEIVES the extra electron pair becomes more negative. Halogen donates → halogen is positive. Ring carbon receives → ring carbon is negative.

Mistake 3: Claiming both the halogen and a ring carbon are simultaneously positive

• The error: Some explanations state the halogen carries +1 AND a ring carbon also carries +1 in the same resonance structure.
• Why it's wrong: The molecule started neutral overall, and resonance structures must conserve total charge. If the halogen is +1, there must be a compensating −1 somewhere -- on the ortho or para ring carbon, exactly where the extra electron pair ended up. Two positive charges with nothing negative to balance them violates charge conservation.
• How to avoid: After assigning formal charges in a resonance structure, always add them up and confirm they sum to the molecule's actual overall charge (here, zero).

Mistake 4: Forgetting that the halogen is an ortho/para director despite being deactivating

• The error: Students see that halogens are deactivating (they withdraw electron density inductively overall) and conclude they must be meta directors.
• Why it's wrong: Halogens are a unique case: they are deactivating overall (from the strong inductive electron withdrawal, since they're highly electronegative) but ortho/para directing (from resonance donation of a lone pair into the ring specifically at those positions). The resonance structures should show negative charge concentrating at ortho and para -- never meta -- which is exactly why an incoming electrophile prefers those positions even though the ring as a whole is less reactive than benzene.
• How to avoid: Memorise the rule: Most deactivating groups are meta directors -- except halogens, which are deactivating but ortho/para directing (because their DIRECTING effect comes from resonance, not from their overall inductive DEACTIVATING effect -- two separate effects, pointing in different directions on which aspect they control).

Mistake 5: Drawing only one resonance structure and stopping

• The error: The question asks for "other resonance structures" (plural). Students draw just one and think they are done.
• Why it's wrong: The lone pair on the halogen can delocalise toward either of the two ortho positions or the para position -- three distinct resonance contributors, each with the negative charge landing on a different ring carbon. …