Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. Since both structural statements are valid, both Statement I and Statement II are true. Always remember that 1^, 2^, and 3^ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the ligh

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and terti

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. Since both structural statements are valid, both Statement I and Statement II are true. Always remember that 1^, 2^, and 3^ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the ligh

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and terti

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement…

Year	2026	2025	2024	Total
Questions	22	49	30	101

Q46 jee_main_2025_07_april_morning Quantitative Elemental Analysis

An organic compound weighing

500 mg

, produced

220 mg

of

CO₂

on complete combustion. The percentage composition of carbon in the compound is ______ %. (nearest integer) (Given molar mass in

g mol⁻¹

of

C: 12

,

O: 16

)

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

% C = (12)/(44) \times Mass of CO₂ producedMass of organic compound taken \times 100

Core Logic

Given:

Mass of organic compound taken $= 500 mg = 500 \times 10⁻³ g$
Mass of $CO₂$ $\mathrm{CO}_2$ produced $= 220 mg = 220 \times 10⁻³ g$

Using the formula:

% C = (12)/(44) \times 220 \times 10⁻³500 \times 10⁻³ \times 100

% C = (12)/(44) \times (220)/(500) \times 100

% C = (12)/(44) \times 44 = 12 %

Thus, the percentage of carbon is $12$ $12$.

Pattern Recognition

Carbon dioxide has exactly $12/44 \approx 27.27%$ $12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of $CO₂$ $\mathrm{CO}_2$ ( $220/500 = 0.44$ $220/500 = 0.44$) by $12/44$ $12/44$ to directly get $0.12$ $0.12$ or $12%$ $12\%$.

Evaluation Rubric / Model Answer

A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise $12$ $12$ percent carbon composition.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q jee_main_2025_08_april_evening IUPAC Nomenclature

What is the correct IUPAC name of the following organic compound?

Cyclic substituted alkene organic molecule structure for Q35 — The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.

A. 4-Ethyl-1-hydroxycyclopent-2-ene
B. 1-Ethyl-3-hydroxycyclopent-2-ene
C. 1-Ethylcyclopent-2-en-3-ol
D. 4-Ethylcyclopent-2-en-1-ol

Solution

Core Logic

Let us apply official IUPAC priority indexing rules:

Principal Functional Group: The hydroxyl group ( $-OH$ $-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the $-OH$ $-\text{OH}$ group is assigned position C-1.
Numbering Direction: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates C-2 and C-3.
Locating Side Chains: Proceeding with this direction puts the ethyl group at position C-4.
The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.

Assembling the structural parts alphabetically:

Substituent: `4-Ethyl`
Parent root: `cyclopent-2-en`
Suffix: `1-ol`

Combined IUPAC format: 4-Ethylcyclopent-2-en-1-ol.

Pattern Recognition

Principal suffix priority hierarchy: $-OH > Double bond > Alkyl side-chain$ . Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q27 jee_main_2025_08_april_evening Reactive Intermediates and Reagents

Match the LIST-I with LIST-II:

LIST-I	LIST-II
A. Carbocation	I. Species that can supply a pair of electrons.
B. C-Free radical	II. Species that can receive a pair of electrons.
C. Nucleophile	III. $sp²$ $sp^2$ hybridized carbon with empty p-orbital.
D. Electrophile	IV. $sp²/sp³$ $sp^2/sp^3$ hybridized carbon with one unpaired electron.

Choose the correct answer from the options given below:

A. $A-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-IV, C-II, D-I$ $\text{A-III, B-IV, C-II, D-I}$
D. $A-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$

Solution

Core Logic

Let us analyze each term carefully:

A. Carbocation: Features a positively charged trivalent carbon atom. It represents an $sp²$ $sp^2$ hybridized carbon with an empty unhybridized p-orbital.
Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
B. Carbon Free Radical: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits $sp²$ $sp^2$ or $sp³$ $sp^3$ hybridization depending on structural environments.
Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
C. Nucleophile: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons.
D. Electrophile: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons.

Step 1: Alignment Matrix

Matching each item yields:

$A arrow III$ $\text{A} \rightarrow \text{III}$
$B arrow IV$ $\text{B} \rightarrow \text{IV}$
$C arrow I$ $\text{C} \rightarrow \text{I}$
$D arrow II$ $\text{D} \rightarrow \text{II}$

This sequence aligns flawlessly with Option (4).

Pattern Recognition

Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q38 jee_main_2025_08_april_evening Quantitative Elemental Analysis

On complete combustion,

0.210 g

of an organic compound containing C, H, and O yielded

0.127 g

of

H₂O

and

0.307 g

of

CO₂

. The mass percentages of hydrogen and oxygen in the given organic compound respectively are:

A. 53.41, 39.6
B. 6.72, 53.41
C. 7.55, 43.85
D. 6.72, 39.87

Solution

Related Formula

Percentage of Hydrogen in organic analysis:

%H = (2)/(18) \times Mass of H₂OMass of Compound \times 100

Percentage of Carbon:

%C = (12)/(44) \times Mass of CO₂Mass of Compound \times 100

Percentage of Oxygen:

%O = 100 - (%C + %H)

Execution

Step 1: Compute the mass percent of Hydrogen:

%H = (2)/(18) \times (0.127)/(0.210) \times 100 = (0.254)/(3.78) \approx 6.72%

Step 2: Compute the mass percent of Carbon:

%C = (12)/(44) \times (0.307)/(0.210) \times 100 = (3.684)/(9.24) \approx 39.87%

Step 3: Deduce the remaining mass percent of Oxygen:

%O = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41%

Thus, the values of hydrogen and oxygen percentage are $6.72%$ $6.72\%$ and $53.41%$ $53.41\%$, matches with Option (2).

Pattern Recognition

Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q42 jee_main_2025_08_april_evening Qualitative Analysis of Functional Groups

Match the reagents in LIST-I with the corresponding chemical functional groups they detect in LIST-II:

LIST-I (Reagent)	LIST-II (Functional Group detected)
A. Sodium bicarbonate solution	I. double bond / unsaturation
B. Neutral ferric chloride	II. carboxylic acid
C. Ceric ammonium nitrate	III. phenolic - OH
D. Alkaline $KMnO₄$ $\text{KMnO}_4$	IV. alcoholic - OH

Choose the correct answer from the options given below:

A. $A-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-II, C-IV, D-I$ $\text{A-III, B-II, C-IV, D-I}$
D. $A-II, B-IV, C-III, D-I$ $\text{A-II, B-IV, C-III, D-I}$

Solution

Core Logic

Let us review the chemical basis for each qualitative test:

A. Sodium bicarbonate ( $NaHCO₃$ $\text{NaHCO}_3$) solution: Carboxylic acids are sufficiently acidic to decompose $NaHCO₃$ $\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, $A arrow II$ $\text{A} \rightarrow \text{II}$.
B. Neutral ferric chloride ( $FeCl₃$ $\text{FeCl}_3$): Phenols react with neutral $FeCl₃$ $\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, $B arrow III$ $\text{B} \rightarrow \text{III}$.
C. Ceric ammonium nitrate (CAN): Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, $C arrow IV$ $\text{C} \rightarrow \text{IV}$.
D. Alkaline $KMnO₄$ $\text{KMnO}_4$ (Baeyer's Reagent): Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown $MnO₂$ $\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, $D arrow I$ $\text{D} \rightarrow \text{I}$.

Step 1: Assembly

Combining the validated relationships gives:

A-II, B-III, C-IV, D-I

This maps perfectly to Option (1).

Pattern Recognition

Baeyer's test (alkaline $KMnO₄$ $\text{KMnO}_4$) always tests for alkenes/alkynes. $NaHCO₃$ $\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Alcohols, Phenols and Ethers

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Solution & Explanation

Related Formula

Core Logic

Step 1: Conclusion Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8

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Related Formula

Core Logic

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Evaluation Rubric / Model Answer

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Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Alignment Matrix

Pattern Recognition

Chapter Mix

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Assembly

Pattern Recognition

Chapter Mix

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