Which of the following compounds will not exist as resonance hybrid. Give reason for your answer: (i) CH3OH (i

NCERT Exemplar · Q45

Q. Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:

(i) CH3OH

(ii) R—CONH2

(iii) CH3CH=CHCH2NH2

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Concept understanding — Resonance Structures Drawing

Resonance Structures: What They Are and How to Draw Them

Let's start with a simple question. When you draw a molecule like ozone ( $O_3$ ), you might put a double bond between the central oxygen and one of the end oxygens, and a single bond to the other. But experiments show both O–O bonds are identical — same length, same strength. So which drawing is correct?

Neither single drawing is correct. The real molecule is a hybrid of both possibilities. That's the core idea of resonance.

The Intuition: A Musical Analogy

Think of a chord played on a piano. A C major chord is made of three notes: C, E, G. No single note is the chord — the chord is the blend of all three. Similarly, a resonance hybrid is the blend of all valid Lewis structures (called resonance contributors or canonical forms) for a molecule. The real molecule is not flipping between these forms; it exists as a single, stable average.

Note

Resonance structures are not in equilibrium. The molecule does not switch from one form to another. It is a single structure that is the weighted average of all contributors.

The Precise Definition

Resonance structures are two or more Lewis structures that differ only in the placement of electrons (pi bonds and lone pairs), never in the positions of atoms. The real molecule is described by a resonance hybrid — a superposition of all contributors.

Rules for Valid Resonance Structures

Atoms never move. Only electrons (pi bonds, lone pairs, and sometimes sigma bonds in special cases) change positions.
The total number of electrons stays the same. You are just redistributing them.
Each structure must obey the octet rule (for second-period elements) and have valid formal charges.
All structures must have the same net charge and the same number of unpaired electrons (if any).

How to Draw Resonance Structures: A Step-by-Step Method

Let's use the carbonate ion ( $CO_3^{2-}$ ) as our example.

Step 1: Draw the best Lewis structure

Start with the skeleton: carbon in the center, three oxygens around it. Count valence electrons: C has 4, each O has 6, plus 2 for the charge = $4 + 18 + 2 = 24$ electrons. Place bonds and lone pairs to satisfy octets. You'll get one structure with a C=O double bond and two C–O single bonds, each single-bonded oxygen carrying a negative charge.

Step 2: Identify movable electrons

Look for pi bonds (double or triple bonds) and lone pairs that are adjacent to pi bonds or to atoms with an empty p orbital. In carbonate, the C=O pi bond and the lone pairs on the negatively charged oxygens are the movable parts.

Step 3: Push electrons using curved arrows

An arrow starts at the electron source (a pi bond or lone pair) and points to where the electrons go (to form a new pi bond or to become a lone pair). In carbonate:

Take the pi bond from C=O and push it to become a lone pair on that oxygen.
Simultaneously, take a lone pair from a negatively charged oxygen and push it to form a new C=O pi bond.

Step 4: Draw the new structure

After pushing, you get a second structure where the double bond is on a different oxygen. Repeat to get the third structure (all three oxygens take turns being double-bonded).

Tip

Always check that the total number of electrons and the net charge remain unchanged after each arrow push. A common mistake is to accidentally add or remove electrons.

Common Patterns to Recognize

Pattern	Example	What moves
Allylic system	$CH_2=CH-CH_2^+$	Pi bond shifts, positive charge moves
Conjugated diene	$CH_2=CH-CH=CH_2$	Pi bonds shift (less common in neutral molecules)
Carbonyl group	$R_2C=O$	Lone pair from O forms pi bond, pi bond becomes lone pair
Benzene ring	$C_6H_6$	Alternating double bonds shift around the ring

The Most Common Mistake Beginners Make

Breaking sigma bonds. Remember: sigma bonds (single bonds between atoms) never break in resonance. Only pi bonds and lone pairs move. If you find yourself moving an atom or breaking a single bond, you are drawing a different molecule (a constitutional isomer), not a resonance structure. …

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A resonance hybrid requires multiple valid Lewis structures with electron delocalisation. CH₃OH cannot exist as a resonance hybrid because it has no π-electrons or lone pairs positioned for delocalisation, while the other two compounds do.

Resonance structures arise when electrons—particularly π-electrons or lone pairs—can be redistributed across adjacent atoms without moving the nuclear framework. The key requirement is conjugation: alternating single and multiple bonds, or a lone pair adjacent to a π-system, that allows electron density to spread over several atoms.

A molecule exists as a resonance hybrid when we can draw two or more valid Lewis structures that differ only in electron placement. The actual structure is a weighted average of these forms, with electron density distributed across the conjugated system. If no such delocalisation is possible, the molecule has a single, fixed structure.

Let me examine each compound:

1. CH₃OH (Methanol)

The structure is $\ce{H3C-OH}$ . The oxygen atom carries two lone pairs, but they sit on a terminal atom bonded only to a saturated carbon via a single σ-bond. There is no adjacent π-system—no double bond, no empty orbital—into which these lone pairs can delocalise.

The carbon is $sp^3$ hybridized with no vacant p-orbital, and all bonds are single bonds. Electron density remains localised exactly where the Lewis structure shows it.

CH₃OH does not exist as a resonance hybrid.

2. R—CONH₂ (Amide)

The functional group is:

\ce{R-C(=O)-NH2}

The nitrogen atom has a lone pair in a p-orbital, and it sits directly adjacent to the $\ce{C=O}$ π-bond. This creates a three-atom conjugated system: $\ce{N-C=O}$ .

We can draw two resonance structures:

Structure A: $\ce{R-C(=O)-NH2}$ (lone pair on N, double bond between C and O)
Structure B: $\ce{R-C(-O^-)-NH2^+}$ (lone pair moves into a π-bond with C, creating $\ce{C=N}$ and leaving O with a negative charge)

The actual structure is a hybrid: the C–N bond has partial double-bond character (restricted rotation), the C=O bond is longer than a pure double bond, and electron density is delocalised over all three atoms.

R—CONH₂ exists as a resonance hybrid.

Tip

Amides are classic resonance hybrids. The partial double-bond character of the C–N bond is why amides are planar and why rotation around that bond is restricted.

3. CH₃CH=CHCH₂NH₂ (Allylic amine)

The structure is:

\ce{CH3-CH=CH-CH2-NH2}

The nitrogen's lone pair sits on the $\ce{-NH2}$ group, which is attached to an …

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Q.Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:

1. CH₃OH (Methanol)

2. R—CONH₂ (Amide)

3. CH₃CH=CHCH₂NH₂ (Allylic amine)

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