Draw the possible resonance structures for CH3—O—CH2^+ (the oxygen carries two lone pairs and the terminal CH2

NCERT Exemplar · Q35

Q. Draw the possible resonance structures for CH3—O—CH2^+ (the oxygen carries two lone pairs and the terminal CH2 carbon bears a positive charge) and predict which of the structures is more stable. Give reason for your answer.

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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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The positive charge on the oxygen-stabilised carbocation can be delocalised onto oxygen via a π-bond, giving a more stable oxonium ion structure. The resonance hybrid is dominated by the structure with a C=O double bond and a neutral oxygen.

Why resonance matters here

The given species is $\mathrm{CH_3-O-CH_2^+}$ . The oxygen atom has two lone pairs, and the terminal $\mathrm{CH_2}$ carbon carries a positive charge. That positive carbon is directly attached to an oxygen that is rich in lone pairs — a classic setup for resonance stabilisation. The oxygen can "donate" one of its lone pairs to form a π-bond with the electron-deficient carbon, shifting the positive charge onto the oxygen itself. This delocalisation spreads the charge, making the ion more stable than a simple localised carbocation.

Drawing the resonance structures

Structure I (the given one)

$\mathrm{CH_3-\overset{\huge\bullet\bullet}{\underset{\huge\bullet\bullet}{O}}-CH_2^+}$

Oxygen has two lone pairs and is neutral. The terminal carbon bears a full positive charge. This is a primary carbocation — highly unstable on its own.
Structure II (the delocalised form)

$\mathrm{CH_3-\overset{+}{O}=CH_2}$

One lone pair from oxygen moves to form a π-bond between O and the $\mathrm{CH_2}$ carbon. Oxygen now has three bonds and a positive charge (oxonium ion). The $\mathrm{CH_2}$ carbon becomes neutral and has a complete octet. The methyl group remains unchanged.
No other significant resonance contributors

The methyl group’s C–H bonds are not in conjugation with the π-system, so hyperconjugation from the methyl group is a separate (weaker) effect, not a resonance structure. Only these two major structures matter.

Tip

Always check: does the movement of electrons create a new π-bond without exceeding the octet rule? Here, oxygen starts with two lone pairs (octet satisfied) and ends with one lone pair and a π-bond (still an octet). The carbon goes from a sextet to an octet. Both atoms obey the octet rule in both structures.

Stability comparison

Which structure is more stable? Apply the standard rules:

Octet rule: Structure II has every atom (C, O, H) with a complete octet or duet. Structure I has a carbon with only six electrons — a major destabilising factor.
Charge location: In Structure I, the positive charge is on a primary carbon (least stable carbocation type). In Structure II, the positive charge is on oxygen, which is more electronegative and can better accommodate a positive charge (though oxygen doesn't "like" a positive charge, it is still better than a carbon with a sextet). …

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Q.Draw the possible resonance structures for CH3—O—CH2^+ (the oxygen carries two lone pairs and the terminal CH2 carbon bears a positive charge) and predict which of the structures is more stable. Give reason for your answer.

Why resonance matters here

Drawing the resonance structures

Stability comparison

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