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

Showing the 12 most recent of 14 on this concept.

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. …
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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 $\{b, d\} > \{a, c\}$ . …
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KCET 2025Set D-41 markMCQ

Q. Which of the following is not an aromatic compound (A) (B) (C) (D)

›Reveal solution

Count $\pi$ electrons and apply Hückel's $4n+2$ rule: the cyclopentadienyl cation has $4\pi e^-$ ( $4n$ ) and is anti-aromatic, not aromatic.

Step 1 — The criteria for aromaticity.

A species is aromatic if it is (i) cyclic, (ii) planar, (iii) fully conjugated (an unbroken ring of p-orbitals) and (iv) contains $(4n+2)\ \pi$ electrons — Hückel's rule ( $n = 0,1,2,\dots$ , i.e. 2, 6, 10, 14 …). A cyclic, planar, conjugated system with $4n$ $\pi$ electrons (4, 8, 12 …) is anti-aromatic — actively destabilised.

Step 2 — Count the $\pi$ electrons in each option.

Option	Species	$\pi$ electrons	Verdict
(A)	Cyclopentadienyl cation (5-ring, 2 C=C, $\oplus$ )	$2\times 2 = 4$ (the positive carbon is an empty p-orbital)	$4n$ ( $n=1$ ) → anti-aromatic ✗
(B)	Cycloheptatrienyl / tropylium cation (7-ring, 3 C=C, $\oplus$ )	$3\times 2 = 6$	$4n+2$ ( $n=1$ ) → aromatic ✓
(C)	Phenanthrene (3 fused benzene rings, angular)	7 C=C $\Rightarrow 14$	$4n+2$ ( $n=3$ ) → aromatic ✓
(D)	Cyclopentadienyl anion (5-ring, 2 C=C, $\ominus$ )	$2\times2 + 2$ (the lone pair on the carbanion enters the ring) $=6$	$4n+2$ ( $n=1$ ) → aromatic ✓

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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. …
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MHT-CET 2024Set pcm-2024-05-10-M1 markMCQ
$\mathrm{-NHR}$
›Reveal solutionSolution

-NHR donates lone pair, +R

$-NHR$ shows +R (electron donation by resonance); CN, NO2, COOR are -R groups. …
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$\begin{array}{cccc} \text{(structure)} & \xrightarrow{\text{Cr}_2\text{O}_3 \\ 773\text{K}, 10-20\,\text{atm}} & (A) \xrightarrow{\begin{array}{c} \text{(i) KMnO}_4/\text{OH}^- \\ \text{(ii) H}_3\text{O}^+ \end{array}} & (B) \xrightarrow{\text{Conc HNO}_3 + \text{H}_2\text{SO}_4} (C) \end{array}$
›Reveal solutionSolution
The reaction sequence starts with toluene, which is oxidised to benzoic acid, then nitrated to give 3‑nitrobenzoic acid; the substituents D and E are –COOH and –NO₂, so the correct option is (C).

Concept & Intuition

This is a classic organic synthesis puzzle. The first step uses chromia (Cr₂O₃) at high temperature and pressure—a typical condition for the dehydrogenation of an alkylbenzene to an aromatic aldehyde or acid. But here the product (A) is then treated with alkaline KMnO₄ followed by acid, which is a strong oxidation that converts any alkyl side‑chain (or aldehyde) into a carboxylic acid. So (B) must be a benzoic acid derivative. Finally, nitration with conc. HNO₃/H₂SO₄ introduces a nitro group. The key is to identify the starting material from the given options: the final compound (C) is aromatic with two substituents D and E. Working backwards, the only combination that fits the oxidation and nitration pattern is –COOH and –NO₂.

Step‑by‑Step Reasoning

Identify the starting material

The first arrow shows a structure (not drawn here, but typical in such problems) being passed over Cr₂O₃ at 773 K and 10–20 atm. This is the dehydrogenation of an alkylbenzene (e.g., toluene) to benzaldehyde or benzoic acid. In fact, Cr₂O₃ at high temperature often gives the aldehyde, but the exact product (A) is not yet fully oxidised.

Oxidation to (B)

Step (i) KMnO₄/OH⁻ followed by (ii) H₃O⁺ is a vigorous oxidation that converts any alkyl group (–CH₃) or aldehyde (–CHO) directly to a carboxylic acid (–COOH). So (B) must be benzoic acid (or a substituted benzoic acid if the starting material already had a substituent). Since the starting material is a simple aromatic hydrocarbon (likely toluene), (B) is unsubstituted benzoic acid.

Nitration to (C) …
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TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$Q. Assertion (A) : pKa of phenol is 4.19 and that of benzoic acid is 10 Reason (R) : Phenoxide ion is stabilised by non-equivalent resonance structures whereas benzoate ion by two equivalent resonance structures (A) A and R are true. R is the correct explanation of A (B) A and R are true, but R is not the correct explanation for A (C) A is true but R is false (D) A is false but R is true$
›Reveal solutionSolution
The assertion is false because the pKa values are swapped (phenol ~10, benzoic acid ~4.2), but the reason about resonance stabilisation is true. So the correct choice is (D).

The key here is to understand what pKa tells us about acidity. A lower pKa means a stronger acid — the molecule more readily donates its proton. The reason given compares the stability of the conjugate bases (phenoxide vs. benzoate) via resonance. Let’s check both statements carefully.

Check the Assertion (A):

The problem states: pKa of phenol is 4.19 and that of benzoic acid is 10.

In reality, the pKa of phenol is about 10, and the pKa of benzoic acid is about 4.2.

So the assertion has the numbers reversed — phenol is the weaker acid, benzoic acid is the stronger one.

Therefore, Assertion (A) is false.

Check the Reason (R):

The reason says: Phenoxide ion is stabilised by non-equivalent resonance structures whereas benzoate ion by two equivalent resonance structures.

This is chemically correct:

In the phenoxide ion, the negative charge can be delocalised into the ring, but the resonance structures are not all equivalent (some place the charge on carbon, which is less stable than on oxygen).

In the benzoate ion, the two major resonance structures place the negative charge equally on the two oxygen atoms — they are equivalent, giving extra stability.

Greater stabilisation of the conjugate base means a stronger acid. Benzoate is more stabilised than phenoxide, so benzoic acid (pKa ~4.2) is stronger than phenol (pKa ~10). …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$\begin{array}{c} \mathrm{O_2N} \hspace{0.2cm} \mathrm{C_6H_4} \hspace{0.2cm} \mathrm{NHCOCHCl_2} \\ \hspace{1.5cm} \mathrm{CH} - \mathrm{CH} - \mathrm{CH_2OH} \\ \hspace{2.5cm} \mathrm{OH} \end{array}$
›Reveal solutionSolution
The molecule is chloramphenicol, a broad-spectrum antibiotic, so the correct answer is (A).

The key to this question is recognizing the structural features of the molecule. The given compound contains a nitrobenzene ring, a dichloroacetamide group (NHCOCHCl₂), and a chain with two hydroxyl groups and a primary alcohol. This exact arrangement is the hallmark of chloramphenicol, a well-known antibiotic.

Identify the functional groups: The molecule has:

A para-nitrophenyl group (O₂N–C₆H₄–).

An amide linkage (–NHCO–) attached to a dichloromethyl group (–CHCl₂).

A three-carbon chain with two hydroxyl groups (–OH) and a terminal –CH₂OH.

Recall the known drug structure: Chloramphenicol is a natural antibiotic (originally from Streptomyces venezuelae) with the systematic name: 2,2-dichloro-N-[(1R,2R)-1,3-dihydroxy-1-(4-nitrophenyl)propan-2-yl]acetamide. Its structure matches exactly: a p-nitrophenyl ring, a dichloroacetamide, and a dihydroxypropyl side chain.

Eliminate other options:

(B) Antiseptic: Antiseptics (e.g., phenol, iodine) are simpler, non-specific germicides; chloramphenicol is a specific systemic antibiotic.

(C) Analgesic: Pain relievers like aspirin or paracetamol lack the nitro and dichloroacetamide groups.

(D) Tranquilizer: Sedatives (e.g., diazepam) have different ring systems (benzodiazepines), not this structure. …
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TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
$\text{C}_6\text{H}_5\text{CHO} + \text{Conc. NaOH} \longrightarrow \text{A} + \text{C}_6\text{H}_5\text{COONa}$
›Reveal solutionSolution
Benzaldehyde has no $\alpha$ -hydrogen, so with concentrated NaOH it undergoes the Cannizzaro disproportionation: one molecule is oxidised to sodium benzoate and the other reduced to benzyl alcohol. Product 'A' is $\text{C}_6\text{H}_5\text{CH}_2\text{OH}$ — option (B).

The concept first

When you meet an aldehyde and a strong base, the first question is always: does it have an $\alpha$ -hydrogen?

With an $\alpha$ -H (e.g. ethanal), the base pulls it off to make an enolate → aldol condensation.

Without an $\alpha$ -H, no enolate is possible. With concentrated alkali the aldehyde has only one escape route: it oxidises one of its own molecules and reduces another. That self-oxidation–reduction is the Cannizzaro reaction.

In benzaldehyde the carbon attached to $-$ CHO is an aromatic ring carbon carrying no hydrogen on an $sp^3$ centre, so there is no $\alpha$ -hydrogen — Cannizzaro is guaranteed.

Step-by-step mechanism

Step 1 — Hydroxide attacks the carbonyl.
$\text{C}_6\text{H}_5\text{CHO} + \overset{-}{\text{O}}\text{H} \longrightarrow \text{C}_6\text{H}_5\text{CH(O}^-)\text{OH}$
A tetrahedral alkoxide intermediate forms.

Step 2 — Hydride transfer. This intermediate (or its doubly deprotonated dianion, which is the better hydride donor) collapses: the C $-$ H bond breaks and the hydrogen leaves with its bonding pair as $\text{H}^-$ , attacking the carbonyl carbon of a second benzaldehyde molecule.

Step 3 — Two different fates.

The molecule that gave away the hydride becomes benzoic acid, immediately deprotonated by the alkali to $\text{C}_6\text{H}_5\text{COO}^-\text{Na}^+$ — the oxidation product (printed in the question). …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$Q. The total number of overlapping p-orbitals present in cycloheptatrienyl cation is (A) 4 (B) 5 (C) 6 (D) 7$
›Reveal solutionSolution
The cycloheptatrienyl cation is an aromatic species where all 7 carbon atoms in the ring are $sp^2$ hybridized, each contributing a p-orbital to a continuous, delocalized $\pi$ -system. Therefore, there are 7 overlapping p-orbitals.

Concept and Intuition

In organic chemistry, the concept of "overlapping p-orbitals" is central to understanding the stability and reactivity of molecules, particularly those with double bonds and cyclic structures. When we talk about overlapping p-orbitals in a cyclic system, we are referring to the formation of a continuous $\pi$ -electron cloud above and below the plane of the ring. This continuous overlap is a prerequisite for aromaticity, a special stability found in certain cyclic, planar, fully conjugated systems.

For a cyclic system to have continuous overlap of p-orbitals, two main conditions must be met:

Each atom in the ring must be $sp^2$ or $sp$ hybridized. This ensures that each atom has at least one unhybridized p-orbital available. In most aromatic systems, the atoms are $sp^2$ hybridized.

These p-orbitals must be aligned parallel to each other. This allows for effective side-by-side overlap, forming a delocalized $\pi$ -system.

The number of overlapping p-orbitals is simply the count of atoms in the ring that contribute an unhybridized p-orbital to this continuous $\pi$ -system. In the case of a carbocation, if the carbon bearing the positive charge is part of a conjugated system, it will be $sp^2$ hybridized and contribute an empty p-orbital to the overall delocalization.

Step-by-Step Solution

Identify the structure of cycloheptatrienyl cation: The name "cycloheptatrienyl cation" indicates a 7-membered carbon ring ("cyclohept-") containing three double bonds ("-triene") and a positive charge ("-yl cation"). The structure can be drawn as a 7-membered ring with three double bonds and one carbon atom bearing a positive charge.

$\begin{array}{c} \text{C}_1=\text{C}_2 \\ / \quad \quad \backslash \\ \text{C}_7^+ \quad \quad \text{C}_3 \\ \backslash \quad \quad / \\ \text{C}_6=\text{C}_5 \\ \quad \quad \text{C}_4 \end{array}$
(This is a simplified representation; imagine a heptagon with alternating double bonds and a positive charge on one carbon.)

2. Determine the hybridization of each carbon atom in the ring:

* Carbons involved in double bonds (e.g., C1, C2, C3, C4, C5, C6) are $sp^2$ hybridized. Each $sp^2$ carbon has one unhybridized p-orbital.

* The carbon bearing the positive charge (C7) is a carbocation. Carbocations are typically $sp^2$ hybridized, with the positive charge residing in an empty p-orbital. This empty p-orbital is crucial for conjugation and delocalization.

Since all 7 carbon atoms in the ring are either part of a double bond or bear a positive charge, they are all $sp^2$ hybridized.

3. Count the number of p-orbitals involved in the cyclic overlap: …
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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. …
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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. …
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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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