From the following, the least stable structure is :
A.textOption 1$\text{Option 1}$
Resonance Effects
B.textOption 2$\text{Option 2}$
Resonance Effects
C.textOption 3$\text{Option 3}$
Resonance Effects
D.textOption 4$\text{Option 4}$
Resonance Effects
Solution & Explanation
### Core Logic
In structure 3, there are positive formal charges on two adjacent atoms (Oxygen and the Carbon adjacent to it). Like charges on adjacent atoms cause extreme electrostatic repulsion, making the structure highly unstable.
Resonance Effects diagram for Q62 - JEE Main 2026 Morning
### Pattern Recognition
Rules of resonance stability:
1. Complete octets are more stable.
2. More covalent bonds = more stable.
3. Least charge separation is more stable.
4. Like charges on adjacent atoms create massive destabilization (Least stable scenario).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#least stable resonance structure#JEE Main 2026 Morning Q62#Organic Chemistry - Some Basic Principles and Techniques JEE Main 2026#Resonance Effects JEE Main 2026
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 3
Qjee_main_2025_03_april_eveningIUPAC Nomenclature of Multi-substituted Benzenes
What is the correct IUPAC name of the compound given below? Substituted benzene derivative with carboxyl, hydroxyl, bromo, and nitro substituents.
A. 3-Bromo-2-hydroxy-5-nitrobenzoic acid
B. 3-Bromo-4-hydroxy-1-nitrobenzoic acid
C. 2-Hydroxy-3-bromo-5-nitrobenzoic acid
D. 5-Nitro-3-bromo-2-hydroxybenzoic acid
Solution
### Related Formula
According to IUPAC rules for nomenclature of aromatic compounds:
- Principal functional group has highest priority:
-mathrmCOOH > -mathrmOH$$-\mathrm{COOH} > -\mathrm{OH}$$
- The principal functional group carbon is designated as Carbon-1, and numbering is directed to give substituents the lowest possible locants.
### Core Logic
Assign priority and number the ring:
- Carbon-1: -mathrmCOOH$-\mathrm{COOH}$ (Carboxyl carbon, parent name 'benzoic acid')
- Carbon-2: -mathrmOH$-\mathrm{OH}$ (Hydroxyl substituent)
- Carbon-3: -mathrmBr$-\mathrm{Br}$ (Bromo substituent)
- Carbon-5: -mathrmNO_2$-\mathrm{NO}_2$ (Nitro substituent)
This numbering yields substituent locants at positions 2, 3, and 5.
### Step 1: Arrange alphabetically
List the substituents alphabetically with locants:
- 3-Bromo
- 2-Hydroxy
- 5-Nitro
Combining these names:
text3-Bromo-2-hydroxy-5-nitrobenzoic acid$$\text{3-Bromo-2-hydroxy-5-nitrobenzoic acid}$$
This matches Option (1).
### Pattern Recognition
Carboxylic acid always dictates position 1 in ring numbering over alcohol. Numbering clockwise gives 2-hydroxy, 3-bromo, and 5-nitro, whereas counterclockwise numbering would yield much higher locants (2-nitro, 4-bromo, 5-hydroxy) which violates the lowest-locant rule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Qjee_main_2025_03_april_eveningStoichiometry of Nitration
Xmathrm~g$X\mathrm{~g}$ of nitrobenzene on nitration gave 4.2mathrm~g$4.2\mathrm{~g}$ of m-dinitrobenzene. The value of X$X$ is ________ mathrmg$\mathrm{g}$. (nearest integer)
[Given: molar mass (in mathrmg~mol^-1$\mathrm{g~mol}^{-1}$ ) C: 12, H: 1, O: 16, N: 14]
Numerical Answer.Answer: 3 to 3
Solution
### Related Formula
Balanced reaction for nitration of nitrobenzene:
mathrmC_6H_5NO_2 + mathrmHNO_3 rightarrow mathrmC_6H_4(NO_2)_2 + mathrmH_2O$$\mathrm{C_6H_5NO_2} + \mathrm{HNO_3} \rightarrow \mathrm{C_6H_4(NO_2)_2} + \mathrm{H_2O}$$textMoles = fractextMasstextMolar Mass$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$
### Core Logic
From the balanced stoichiometry:
- 1 mole of nitrobenzene yields 1 mole of m-dinitrobenzene.
### Step 1: Determine molar masses
- Molar mass of Nitrobenzene (mathrmC_6H_5NO_2$\mathrm{C_6H_5NO_2}$):
M_1 = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123mathrm~g/mol$$M_1 = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123\mathrm{~g/mol}$$
- Molar mass of m-Dinitrobenzene (mathrmC_6H_4(NO_2)_2$\mathrm{C_6H_4(NO_2)_2}$):
M_2 = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168mathrm~g/mol$$M_2 = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168\mathrm{~g/mol}$$Stoichiometry of Nitration
### Step 2: Calculate moles and find X
Moles of m-dinitrobenzene produced:
$
Stoichiometry of Nitration
### Step 2: Calculate moles and find X
Moles of m-dinitrobenzene produced:
$n = frac4.2mathrm~g168mathrm~g/mol = 0.025mathrm~mol$n = \frac{4.2\mathrm{~g}}{168\mathrm{~g/mol}} = 0.025\mathrm{~mol}$
Since stoichiometry is $
Since stoichiometry is $1:1, the moles of nitrobenzene required is also $, the moles of nitrobenzene required is also $0.025\mathrm{~mol}:
$:
$textMass of nitrobenzene X = 0.025mathrm~mol times 123mathrm~g/mol = 3.075mathrm~g$\text{Mass of nitrobenzene } X = 0.025\mathrm{~mol} \times 123\mathrm{~g/mol} = 3.075\mathrm{~g}$
Rounding to the nearest integer gives $
Rounding to the nearest integer gives $3$.
### Pattern Recognition
Electrophilic aromatic substitution stoichiometry is straightforward: each aromatic precursor ring converts to exactly one product ring. Finding moles from the heavier substituted product and converting back using the reactant's molecular weight quickly yields the answer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Amines
Qjee_main_2025_03_april_eveningIsomerism in Benzene Derivatives
The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula mathrmC_9mathrmH_12$\mathrm{C}_9\mathrm{H}_{12}$ is ________.
Numerical Answer.Answer: 8 to 8
Solution
### Related Formula
Degrees of Unsaturation (Double Bond Equivalents, DBE):
mathrmDBE = C + 1 - fracH2 + fracN2$$\mathrm{DBE} = C + 1 - \frac{H}{2} + \frac{N}{2}$$
For formula mathrmC_9H_12$\mathrm{C_9H_{12}}$:
mathrmDBE = 9 + 1 - frac122 = 4$$\mathrm{DBE} = 9 + 1 - \frac{12}{2} = 4$$
These 4 degrees of unsaturation match a benzene ring exactly (one ring + three double bonds).
### Core Logic
Since the question specifies 'substituted benzene derivatives', we must keep the benzene core (mathrmC_6H_5-$\mathrm{C_6H_5-}$ or similar) intact. This leaves 3 carbon atoms to be distributed as alkyl substituents.
### Step 1: Categorize by substitution patterns
1. **Mono-substituted benzene** (one propyl group containing 3 carbons):
- n-Propylbenzene: mathrmC_6H_5-CH_2-CH_2-CH_3$\mathrm{C_6H_5-CH_2-CH_2-CH_3}$ (Isomer 1)
- Isopropylbenzene (Cumene): mathrmC_6H_5-CH(CH_3)_2$\mathrm{C_6H_5-CH(CH_3)_2}$ (Isomer 2)
2. **Di-substituted benzene** (one ethyl group and one methyl group):
- 1-Ethyl-2-methylbenzene (ortho-ethylmethylbenzene) (Isomer 3)
- 1-Ethyl-3-methylbenzene (meta-ethylmethylbenzene) (Isomer 4)
- 1-Ethyl-4-methylbenzene (para-ethylmethylbenzene) (Isomer 5)
### Step 2: Tri-substituted benzenes
3. **Tri-substituted benzene** (three methyl groups):
- 1,2,3-Trimethylbenzene (Hemimellitene) (Isomer 6)
- 1,2,4-Trimethylbenzene (Pseudocumene) (Isomer 7)
- 1,3,5-Trimethylbenzene (Mesitylene) (Isomer 8)
### Step 3: Total Count
Summing all options:
textTotal structural isomers = 2 + 3 + 3 = 8$$\text{Total structural isomers} = 2 + 3 + 3 = 8$$
### Pattern Recognition
For alkyl benzenes with N$N$ extra carbons, systematically group them as single chain substituents down to multiple methyl substituents. This hierarchical sorting prevents duplicates or missing patterns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Hydrocarbons
Q36jee_main_2025_03_april_eveningDumas' Method for Nitrogen Estimation
In Dumas' method for estimation of nitrogen 0.4mathrm~g$0.4\mathrm{~g}$ of an organic compound gave 60mathrm~mL$60\mathrm{~mL}$ of nitrogen collected at 300mathrm~K$300\mathrm{~K}$ temperature and 715mathrm~mm~Hg$715\mathrm{~mm~Hg}$ pressure. The percentage composition of nitrogen in the compound is :
(Given: Aqueous tension at 300mathrm~K = 15mathrm~mm~Hg$300\mathrm{~K} = 15\mathrm{~mm~Hg}$)
A. 15.71%
B. 20.95%
C. 17.46%
D. 7.85%
Solution
### Related Formula
Pressure of dry nitrogen gas:
P_mathrmN_2 = P_texttotal - textAqueous tension$$P_{\mathrm{N}_2} = P_{\text{total}} - \text{Aqueous tension}$$
Using Ideal Gas Law:
n_mathrmN_2 = fracP_mathrmN_2 VR T$$n_{\mathrm{N}_2} = \frac{P_{\mathrm{N}_2} V}{R T}$$\%mathrmN = fractextMass of nitrogentextMass of organic compound times 100$$\%\mathrm{N} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Given parameters:
- Mass of compound m = 0.4mathrm~g$m = 0.4\mathrm{~g}$
- Volume of nitrogen V = 60mathrm~mL = 0.060mathrm~L$V = 60\mathrm{~mL} = 0.060\mathrm{~L}$
- Total pressure P_texttotal = 715mathrm~mm~Hg$P_{\text{total}} = 715\mathrm{~mm~Hg}$
- Temperature T = 300mathrm~K$T = 300\mathrm{~K}$
- Aqueous tension = 15mathrm~mm~Hg$= 15\mathrm{~mm~Hg}$
### Step 1: Calculate dry nitrogen pressure
P_mathrmN_2 = 715mathrm~mm~Hg - 15mathrm~mm~Hg = 700mathrm~mm~Hg$$P_{\mathrm{N}_2} = 715\mathrm{~mm~Hg} - 15\mathrm{~mm~Hg} = 700\mathrm{~mm~Hg}$$P_mathrmN_2 = frac700760mathrm~atm approx 0.921mathrm~atm$$P_{\mathrm{N}_2} = \frac{700}{760}\mathrm{~atm} \approx 0.921\mathrm{~atm}$$
### Step 2: Calculate moles of nitrogen gas
Using $
### Step 2: Calculate moles of nitrogen gas
Using $R = 0.0821\mathrm{~L\cdot atm\cdot K^{-1}\cdot mol^{-1}}:
$:
$n_mathrmN_2 = fracleft(frac700760right) times 0.0600.0821 times 300 = frac0.0552624.63 approx 2.2436 times 10^-3mathrm~mol$n_{\mathrm{N}_2} = \frac{\left(\frac{700}{760}\right) \times 0.060}{0.0821 \times 300} = \frac{0.05526}{24.63} \approx 2.2436 \times 10^{-3}\mathrm{~mol}$
Mass of $
Mass of $\mathrm{N}_2 gas:
$ gas:
$textMass = 2.2436 times 10^-3 times 28mathrm~g approx 0.06282mathrm~g$\text{Mass} = 2.2436 \times 10^{-3} \times 28\mathrm{~g} \approx 0.06282\mathrm{~g}$
### Step 3: Calculate percentage of Nitrogen
$$
### Step 3: Calculate percentage of Nitrogen
$$\%\mathrm{N} = \frac{0.06282\mathrm{~g}}{0.4\mathrm{~g}} \times 100 \approx 15.71\%$$
This matches Option (1).
### Pattern Recognition
In Dumas' method calculations, always subtract the aqueous tension to obtain the pressure of dry nitrogen gas. Do not use the raw moist gas pressure, as doing so will overestimate the nitrogen content.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q40jee_main_2025_03_april_eveningHyperconjugation and Cation Stability
Given below are two statements:
Statement I: Hyperconjugation is not a permanent effect.
Statement II: In general, greater the number of alkyl groups attached to a positively charged C-atom, greater is the hyperconjugation interaction and stabilization of the cation.
In the light of the above statements, choose the correct answer from the options given below :
A. Statement I is true but Statement II is false
B. Both Statement I and Statement II are false
C. Statement I is false but Statement II is true
D. Both Statement I and Statement II are true
Solution
### Related Formula
The number of hyperconjugation structures is directly related to the count of alpha$\alpha$-hydrogen atoms:
textNumber of hyperconjugative structures = textNumber of alphatext-hydrogens$$\text{Number of hyperconjugative structures} = \text{Number of } \alpha\text{-hydrogens}$$
### Core Logic
Statement I Analysis:
- Hyperconjugation (no-bond resonance) involves the delocalization of sigma$\sigma$ electrons of mathrmC-H$\mathrm{C-H}$ bonds of an alkyl group directly attached to an atom of unsaturated system or a positively charged carbon atom. This is a permanent ground-state electronic effect, not dependent on external reagents. Thus, Statement I is False.
### Step 1: Analyze Statement II
- Statement II states that more alkyl groups attached to a carbocation center increase hyperconjugative stabilization. Each alkyl group brings additional sigma_mathrmC-H$\sigma_{\mathrm{C-H}}$ bonds adjacent to the empty p-orbital, increasing the total count of alpha$\alpha$-hydrogens and enhancing charge delocalization. Thus, Statement II is True.
### Step 2: Conclusion
Therefore, Statement I is False but Statement II is True, matching Option (3).
### Pattern Recognition
Permanent organic effects include: Inductive, Mesomeric (Resonance), and Hyperconjugation effects. Temporary electronic effects include: Electromeric and Inductomeric effects (which require an attacking reagent to manifest).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2026_21_jan_morning
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