Which of the following two carbocations is more stable? Use resonance to explain your answer. (A) A cyclohexen

NCERT Exemplar · Q36

Q. Which of the following two carbocations is more stable? Use resonance to explain your answer. (A) A cyclohexene ring (the C=C double bond lies within the ring) bearing an exocyclic —CH2(+) group on the sp2 ring carbon that carries the double bond — i.e. the positive charge is on a primary exocyclic carbon that is allylic to the ring double bond [(cyclohex-1-en-1-yl)methyl cation]. (B) A cyclohexane ring bearing an exocyclic =CH2 (methylene) group on one ring carbon, with the positive charge on the ring carbon adjacent to that =CH2 carbon — i.e. the double bond lies outside the ring and the positive charge is on a secondary ring carbon [2-methylenecyclohexyl cation]. (A and B are the two resonance contributors of the same allylic cation.)

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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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Structures (A) and (B) are the two resonance contributors of the same allylic cation, exactly as the question states. A contributor is more stable when its positive charge sits on the more substituted carbon — so (B) (charge on a secondary ring carbon) is the more stable, major contributor, while (A) (charge on a primary exocyclic carbon) is the minor one.

Both drawings describe one and the same allylic carbocation. The allyl unit is a three-carbon system: the exocyclic $\ce{CH2}$ carbon, the ring carbon it is attached to (call it C1), and the neighbouring ring carbon C2. In an allylic cation the positive charge is not fixed on one atom — it is delocalised over the two ends of this allyl unit, which we show by drawing two resonance contributors.

The two structures are one cation

(A) places the double bond inside the ring ( $\ce{C1=C2}$ ) and the positive charge on the exocyclic $\ce{CH2+}$ — a primary carbon.
Pushing the ring $\pi$ electrons toward that charge moves the double bond out onto the exocyclic carbon ( $\ce{C1=CH2}$ ) and shifts the positive charge to the other end of the allyl unit, ring carbon C2 — a secondary carbon. That is exactly structure (B).

\ce{\overset{+}{C}H2-\underset{(C1)}{C}=\underset{(C2)}{C} <=> H2C=\underset{(C1)}{C}-\underset{(C2)}{\overset{+}{C}}}

Only electrons have moved — the connectivity of the atoms is unchanged — so the question's parenthetical is correct: (A) and (B) are resonance forms of the same allylic cation.

Important

In an allylic cation the charge is shared between the two termini of the three-carbon allyl unit — here the exocyclic $\ce{CH2}$ and ring carbon C2. It never lands on the central carbon C1 (that is where the new double bond forms). So the two positive centres to compare are primary (A) and secondary (B) — there is no tertiary carbocation here.

Which contributor is more stable?

A resonance contributor is more stable, and contributes more to the real hybrid, when the positive charge sits on the more substituted carbon (carbocation stability $3^\circ > 2^\circ > 1^\circ$ ). …

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The two structures are one cation

Which contributor is more stable?

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