Q.Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

Resonance Structures with Curved-Arrow Notation

Concept: Resonance structures show delocalisation of π-electrons and lone pairs through conjugated systems. Curved arrows track electron movement from source (lone pair or π-bond) to destination.

(a) C₆H₅OH (Phenol)

The lone pair on oxygen delocalises into the benzene ring:

     OH              OH⁺             OH⁺             OH⁺
      |               |               |               |
   ⟨benzene⟩  ↔   ⟨o-negative⟩ ↔ ⟨o'-negative⟩ ↔ ⟨p-negative⟩

Three ring contributors place the negative charge at the two ortho positions and the para position (never meta).

(b) C₆H₅NO₂ (Nitrobenzene)

The NO₂ group withdraws electrons; ring π-electrons shift toward the nitro group:

Positive charge appears at ortho/para positions; the nitro group itself has $\ce{N^+=O^- ↔ N-O}$ resonance.

(c) CH₃CH=CHCHO (Crotonaldehyde)

Conjugated π-system allows electron flow from C=C to C=O:

Two main forms: neutral and charge-separated (positive on β-carbon, negative on oxygen).

(d) C₆H₅—CHO (Benzaldehyde)

Ring π-electrons delocalise into the carbonyl:

Similar to nitrobenzene: positive charge at ortho/para, negative on oxygen.

(e) C₆H₅—CH₂⁺ (Benzyl cation)

The positive charge delocalises into the ring via π-donation: …

Resonance is not about molecules flipping between forms—it's our way of depicting electron delocalisation that a single Lewis structure cannot capture. When π-electrons, lone pairs, or charges sit adjacent to double bonds or aromatic rings, they spread out over multiple atoms. The curved arrow is the language: it shows where an electron pair moves from (the tail) and where it goes (the head). A double-barbed arrow means two electrons; we never break octets on second-row elements or move single electrons in these neutral/cationic systems.

The key insight: electrons flow toward electronegative atoms in neutral molecules, but positive charges pull electron density toward themselves. Aromatic rings are particularly good at delocalising both.

(a) $\ce{C6H5OH}$ (Phenol)

The oxygen lone pairs are adjacent to the benzene π-system. One lone pair on oxygen can donate into the ring.

Step-by-step:

1. Identify the source: Oxygen has two lone pairs; one is in a p-orbital that overlaps with the aromatic π-cloud.

2. First resonance structure: Move the oxygen lone pair into the C–O π-bond, simultaneously pushing the π-electrons of the ring around. The oxygen becomes positively charged (it "gave away" electrons), and the ortho carbon gains electron density.

3. Continue around the ring: The π-electrons shift in a cyclic fashion. Each movement places negative charge (or a new π-bond) at ortho and para positions.

Curved arrows and structures:

• Arrow from O lone pair to the C–O bond.
• Arrow from the ortho C=C bond toward the adjacent carbon.
• This propagates around the ring.

The resonance structures show negative charge density at the ortho and para positions:

(b) $\ce{C6H5NO2}$ (Nitrobenzene)

The nitro group $\ce{-NO2}$ is electron-withdrawing. The nitrogen is bonded to the ring and has a formal positive charge in the conventional structure $\ce{C6H5-N^{+}(=O)-O^{-}}$.

Step-by-step:

1. Nitro group itself: The $\ce{N=O}$ bonds have resonance. Move a lone pair from $\ce{O^{-}}$ to form a double bond with N, while the existing $\ce{N=O}$ π-bond moves to the other oxygen. This swaps which oxygen carries the negative charge.

2. Ring conjugation: The π-electrons of the benzene ring can shift toward the electron-deficient nitrogen. Move ring π-electrons toward the C–N bond, creating negative charge on the ring (ortho/para) and further delocalising the positive charge on nitrogen.

Curved arrows:

• Within $\ce{NO2}$: Arrow from $\ce{O^{-}}$ lone pair to N, arrow from $\ce{N=O}$ π to the other O.
• Ring to nitro: Arrow from ring C=C π-bond toward the ipso carbon, then toward nitrogen.

The major effect: electron density is pulled from the ring into the nitro group, placing partial positive character at ortho and para positions (the opposite of phenol).

(c) $\ce{CH3CH=CHCHO}$ (Crotonaldehyde)

A conjugated enone system: the C=C double bond is conjugated with the C=O of the aldehyde.

Step-by-step:

1. Identify conjugation: Four p-orbitals in a row (the two carbons of C=C, the carbonyl carbon, and oxygen).

2. Push π-electrons from C=C toward oxygen: Move the C=C π-bond toward the carbonyl carbon, forming a C–C single bond and a new C=O π-bond. Simultaneously, the existing C=O π-bond moves onto oxygen as a lone pair.

Curved arrows:

• Arrow from C=C π-bond to the middle C–C bond.
• Arrow from C=O π-bond to oxygen.

The resonance structure has negative charge on oxygen and a positive charge (or electron deficiency) on the β-carbon (the $\ce{CH}$ next to methyl). This is the classic α,β-unsaturated carbonyl pattern—nucleophiles can attack either the carbonyl carbon or the β-carbon (Michael addition).

(d) $\ce{C6H5CHO}$ (Benzaldehyde)

The aldehyde carbonyl is directly attached to the benzene ring.

Step-by-step:

1. Ring to carbonyl: The benzene π-electrons can donate into the electron-deficient carbonyl carbon. Move a ring π-bond toward the C–CHO bond, then the C=O π-electrons onto oxygen.

2. Propagate around the ring: This places positive charge on the ring (ortho/para) and negative charge on oxygen.

Curved arrows:

• Arrow from ring C=C π to the ipso C–C bond.
• Arrow from C=O π to oxygen.

Resonance structures show the carbonyl carbon is even more electrophilic due to conjugation, and the ring ortho/para positions carry partial positive charge (though less pronounced than in nitrobenzene because the carbonyl is less strongly withdrawing than $\ce{NO2}$).

(e) $\ce{C6H5CH2^{+}}$ (Benzyl cation)

A carbocation adjacent to benzene—classic benzylic stabilisation.

Step-by-step:

1. Ring donates to the cation: The positive charge on $\ce{CH2^{+}}$ is an empty p-orbital. The benzene π-electrons can flow into it. …