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.)

Andhra Pradesh BieapShort· 3mImportance★★★★★est

77% · 100/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 $3^\circ > 2^\circ > 1^\circ$ ). …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$+R$
›Reveal solutionSolution
A substituent shows $+R$ effect if the atom attached to the ring has a lone pair to donate into it; of the six given groups, $-Cl$ , $-NHC_2H_5$ , $-OCH_3$ , $-NHCOCH_3$ qualify (4 total), while $-COCH_3$ and $-COOCH_3$ are $-R$ (carbonyl withdraws electron density).

Concept and Intuition

Resonance ( $R$ /mesomeric) effect on a benzene ring depends on whether the directly-attached atom can donate a lone pair into the ring ( $+R$ , activating by resonance) or whether the ring's $\pi$ electrons are pulled into the substituent through a multiple bond (like $C=O$ ), which is $-R$ (deactivating by resonance). Groups with an available lone pair on the ipso atom (halogens, $-OR$ , $-NR_2$ , $-NHCOR$ ) are $+R$ ; groups where a $\pi$ -bonded electronegative atom (as in $C=O$ , $C\equiv N$ ) sits right next to the ring are $-R$ .

Step-by-Step Solution

$-Cl$ : chlorine's lone pair conjugates into the ring ( $+R$ ), even though its strong $-I$ effect makes it net deactivating — it is still a $+R$ group.

$-COCH_3$ (acetyl): the carbonyl carbon is attached directly to the ring; ring electrons delocalise into the $C=O$ , pulling electron density away from the ring — this is $-R$ , not $+R$ .

$-NHC_2H_5$ : nitrogen's lone pair conjugates strongly into the ring, a classic strong $+R$ donor (like $-NH_2$ ).

$-OCH_3$ : oxygen's lone pair conjugates into the ring — a classic $+R$ donor. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$Q. Consider the following carbocations [FIGURE] (four labelled carbocation skeletal structures:$
(a) a secondary carbocation $\mathrm{CH_3-CH_2-\overset{+}{C}H-CH(CH_3)-CH_3}$ ;
(b) an ether-oxygen-stabilised cation $\mathrm{CH_3-\overset{+}{C}H-O-CH(CH_3)-CH_2CH_3}$ ;
(c) a primary carbocation $\mathrm{\overset{+}{C}H_2-CH_2-CH_2-CH(CH_3)-CH_3}$ ;
(d) an ether-oxygen-stabilised cation $\mathrm{CH_3CH_2-O-\overset{+}{C}H-CH_2CH_3}$ ) The correct stability order for the above carbocations is (A)
(b) >
(a) >
(d) >
(c) (B)
(b) >
(d) >
(c) >
(a) (C)
(d) >
(b) >
(c) >
(a) (D)
(d) >
(b) >
(a) >
(c)
›Reveal solutionSolution
Resonance (mesomeric) donation from an adjacent ether oxygen stabilises a carbocation far more than simple alkyl hyperconjugation does, so both oxygen-adjacent cations rank above both plain alkyl cations; within the plain alkyl pair, the secondary beats the primary. Overall order: (d) > (b) > (a) > (c).

Concept and Intuition

Carbocation stability is governed, in decreasing order of strength, by: resonance/mesomeric ( $+M$ ) donation > hyperconjugation/inductive ( $+I$ ) donation from alkyl groups. An ether oxygen directly bonded to the electron-deficient carbon can donate a lone pair into the empty p-orbital, generating an oxocarbenium-type resonance structure ( $R$ – $O^+$ = $CR'_2$ ) that delocalises the positive charge onto the (more electronegative but resonance-tolerant) oxygen. This resonance stabilisation is substantially stronger than the modest stabilisation a carbocation gets merely from being flanked by one extra alkyl group (hyperconjugation). Consequently, even a comparatively less-substituted carbocation that is directly bonded to oxygen outranks an ordinary alkyl carbocation that lacks such resonance support.

Step-by-Step Solution

Classify each cation: (a) is a plain secondary alkyl carbocation (no heteroatom assistance); (c) is a plain primary alkyl carbocation (no heteroatom assistance); (b) and (d) both have the cationic carbon directly bonded to an ether oxygen.

Because O's lone pair can donate by resonance into the empty orbital on both (b) and (d), both are oxocarbenium-stabilised and this resonance effect outweighs the plain hyperconjugative stabilisation available to (a) and (c). So $\{b, d\} > \{a, c\}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$CH_2^+$
›Reveal solutionSolution
Carbocation stability is set by how much positive charge can be delocalised by resonance (allylic/benzylic conjugation) plus hyperconjugation/induction from alkyl groups. The cation with BOTH allylic and benzylic conjugation, plus a methyl substituent, is the most stable.

Concept and Intuition

A carbocation is stabilised whenever its empty p-orbital can overlap with an adjacent π-system (resonance/conjugation) or with adjacent C–H/C–C sigma bonds (hyperconjugation), and destabilised when it sits isolated with no such support. Combining two independent resonance-donating groups (here, both an adjacent ring double bond AND a phenyl ring through that double bond) gives an extended conjugated system — much more stabilising than either alone.

Step-by-Step Solution

Option (A): a cyclohexyl cation on a saturated ring, no adjacent π-bond, no aryl group — a simple secondary cation with only ordinary hyperconjugation. Least stabilised of the set.

Option (B): a cyclohexenyl cation, cationic carbon directly next to the ring's C=C — this is a genuine allylic cation, delocalised over two carbons by resonance. More stable than (A), but only single-bond-worth of delocalisation.

Option (D): the cationic carbon is an exocyclic $CH_2^+$ attached to the ring; the phenyl group sits on a different, non-adjacent ring carbon and is explicitly not conjugated with the cationic centre. So this cation behaves essentially like an isolated primary cation — very poorly stabilised despite having a phenyl group present on the molecule. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$O^-$
›Reveal solutionSolution
A valid resonance structure of p-nitrophenoxide must pair an aromatic ring with an unperturbed nitro group, OR a quinonoid ( $C{=}O$ ) ring with a charge-migrated nitro group — option (D) illegally mixes an untouched aromatic ring with an already charge-separated nitro group, which no single curved-arrow path can produce.

Concept and Intuition

In p-nitrophenoxide, the negative charge on the phenolic oxygen is stabilised by delocalising all the way to the nitro group through the ring in between ("push-pull" conjugation). Each legitimate resonance structure must differ from the next by moving exactly one pair of electrons at a time (one curved arrow, or a linked set), so the ring's bonding pattern and the nitro group's charge state must change together, in lock-step — never independently of each other.

Step-by-Step Solution

Structure (A): aromatic ring (alternating double bonds) with $O^-$ at the top and the nitro group in its ordinary neutral-looking form (one $N{=}O$ , one $N{\to}O$ dative bond). This is simply the reference/starting Lewis structure — a legitimate contributor.

Pushing the phenoxide lone pair into the ring converts it to a quinonoid form: $C{=}O$ appears at the ipso carbon, and the negative charge now sits as a carbanion on a ring carbon (ortho to the carbonyl), with the ring no longer aromatic — a legitimate intermediate contributor (structure B).

Pushing that carbanion's electrons further, through the ring, into the nitro group converts the nitro group into its charge-separated form ( $N^{\oplus}$ bonded to one $O^-$ and one $O$ ), while the ring stays quinonoid ( $C{=}O$ retained) — this is structure (C), a legitimate, fully-conjugated final contributor. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$N$
›Reveal solutionSolution
The curved arrows push π-electron density from the ring toward the nitro group, generating positive charge on ortho/para ring carbons — this is the $-R$ (electron-withdrawing resonance) effect of $-NO_2$ .

Concept and Intuition

The resonance (mesomeric) effect describes delocalisation of π-electrons through a conjugated system. A substituent can either donate electron density into the ring by resonance ( $+R$ , e.g. $-NH_2$ , $-OH$ , halogens) or withdraw electron density from the ring by resonance ( $-R$ , e.g. $-NO_2$ , $-CHO$ , $-COOH$ , $-CN$ ). The direction of electron flow in the canonical structures tells you which type of effect is operating: if the ring becomes electron-poor (positive charges appear on ring carbons) as electrons flow toward the substituent, that substituent is exerting a $-R$ effect.

Step-by-Step Solution

In structure 1, the nitro group is neutral; curved arrows show the ring's π-electrons beginning to shift toward the nitrogen–oxygen system.

In structures 2 and 3, this electron shift has generated a formal positive charge on the ring carbon ortho (structure 2) and para (structure 3) to the point of attachment of $-NO_2$ , while the oxygens of the nitro group pick up negative charge.

This is precisely the pattern of a group withdrawing electron density from the ring via conjugation/resonance — electrons flow away from the ring carbons and into the substituent, leaving positive charge behind on the ring. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$Q. The relatively more electron rich sites of the following aromatic compounds are ______[FIGURE] Two benzene rings with numbered ring positions 1-6:$
(i) a ring with OH at C1 and $CH_3$ at C3 (m-cresol/3-methylphenol), positions labelled 1(OH),2,3( $CH_3$ ),4,5,6;
(ii) a ring with OH at C1 and $-CO-CH_3$ at C4 (4-hydroxyacetophenone), positions labelled 1(OH),2,3,4( $COCH_3$ ),5,6 (A)
(i) 2,5 &
(ii) 2,3 (B)
(i) 4,5 &
(ii) 3,5 (C)
(i) 2,4,6 &
(ii) 2,6 (D)
(i) 2,5,6 &
(ii) 2,6
›Reveal solutionSolution
The most electron-rich sites are determined by the strongest activating substituent (OH) via resonance; for (i) the OH directs to ortho/para positions (2,4,6), and for (ii) the para-acetyl group deactivates position 4, so the OH still activates ortho/para but the para position is blocked, leaving 2 and 6 as the most electron-rich.

The key idea is that resonance effects dominate over inductive effects for strongly activating groups like –OH. The –OH group donates electron density into the ring through resonance, making the ortho and para positions (relative to itself) the most electron-rich. We must then consider whether the second substituent (CH₃ or –COCH₃) reinforces or opposes this pattern.

Step-by-step reasoning

1. Identify the activating/deactivating nature of each substituent.

–OH (hydroxyl): Strongly activating, ortho/para-directing. Its lone pair on oxygen can delocalize into the ring, creating resonance structures that place negative charge on the ortho and para carbons.

–CH₃ (methyl): Weakly activating, ortho/para-directing. It donates electron density via hyperconjugation and inductive effect, but much weaker than –OH.

–COCH₃ (acetyl): Strongly deactivating, meta-directing. The carbonyl group withdraws electron density by resonance, making the ring less electron-rich overall, especially at ortho and para positions relative to itself.

2. Analyze compound (i): 3-methylphenol (OH at C1, CH₃ at C3).

The –OH group is the dominant activator. Its resonance donation makes positions 2, 4, and 6 (ortho and para to OH) the most electron-rich.

The –CH₃ at C3 is a weak activator and also directs ortho/para, but its effect is much smaller. It does not block any of the OH-activated positions.

Therefore, the most electron-rich sites are C2, C4, and C6.

Tip

When two ortho/para-directors are present, the stronger one (OH) determines the primary pattern; the weaker one (CH₃) only slightly modifies the electron density.

3. Analyze compound (ii): 4-hydroxyacetophenone (OH at C1, –COCH₃ at C4).

The –OH group again strongly activates ortho/para positions: C2, C4, C6.

However, the –COCH₃ group at C4 is a strong deactivator. It withdraws electron density from the ring, especially from the ortho and para positions relative to itself. Since C4 is para to OH, it is also the carbon bearing the acetyl group. …
Start your 14-day free trial to see the full solution →

←Previous questionQ35 — Draw the possible resonance structures for CH3—O—CH2^+ (the oxygen car…→Next questionQ37 — The structure of triphenylmethyl cation, (C6H5)3C^+, is given below. T…

The two structures are one cation

Which contributor is more stable?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days