In which pairs, the first ion is more stable than the second?

Q: In which pairs, the first ion is more stable than the second?

### Related Formula ### Core Logic Evaluating each specific structural pair: - **Pair (A):** The first carbocation is stabilized by strong +M back-bonding from the methoxy (-OMe) oxygen lone pair, making it significantly more stable than the second. - **Pair (B):** The first carbanion is strongly stabilized by the -M and -I electronic effects of the nitro (-NO2) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable. - *

Q58 jee_main_2026_21_jan_morning Quantitative Analysis of Elements

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass

32 g mol⁻¹

). Molar mass of barium sulphate is

233 g mol⁻¹

.

A. $4.55%$ $4.55\%$
B. $10.30%$ $10.30\%$
C. $21.97%$ $21.97\%$
D. $16.48%$ $16.48\%$

Solution

Related Formula

Percentage of Sulphur = Mass of S in BaSO₄Molar mass of BaSO₄ \times Mass of BaSO₄ formedMass of organic compound \times 100

Core Logic

Molar mass of $BaSO₄$ $BaSO_4$ = 233 g/mol. Mass of Sulfur (S) in 1 mole of $BaSO₄$ $BaSO_4$ = 32 g. Mass of $BaSO₄$ $BaSO_4$ formed = 1.2 g. Mass of organic compound = 0.75 g.

% S = (32)/(233) \times (1.2)/(0.75) \times 100

% S = (32 \times 1.2 \times 100)/(233 \times 0.75) = (3840)/(174.75) \approx 21.97%

Chapter Mix

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

Q62 jee_main_2026_21_jan_morning Resonance Effects

From the following, the least stable structure is :

A. $Option 1$ $\text{Option 1}$
B. $Option 2$ $\text{Option 2}$
C. $Option 3$ $\text{Option 3}$
D. $Option 4$ $\text{Option 4}$

Solution

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:

Complete octets are more stable.
More covalent bonds = more stable.
Least charge separation is more stable.
Like charges on adjacent atoms create massive destabilization (Least stable scenario).

Chapter Mix

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

Q66 jee_main_2026_21_jan_morning Isomerism

Identify correct statement from the following: A. Propanal and propanone are functional isomers. B. Ethoxyethane and methoxypropane are metamers. C. But-2-ene shows optical isomerism. D. But-1-ene and but-2-ene are functional isomers. E. Pentane and 2, 2-dimethyl propane are chain isomers. Choose the correct answer from the options given below:

A. $B, C and D only$ $\text{B, C and D only}$
B. $A, B and C only$ $\text{A, B and C only}$
C. $A, B and E only$ $\text{A, B and E only}$
D. $C, D and E only$ $\text{C, D and E only}$

Solution

Core Logic

A. Propanal ( $CH₃CH₂CHO$ $CH_3CH_2CHO$) and propanone ( $CH₃COCH₃$ $CH_3COCH_3$) have different functional groups (aldehyde vs ketone) but the same molecular formula. They are functional isomers. (Correct)

B. Ethoxyethane ( $C₂H₅-O-C₂H₅$ $C_2H_5-O-C_2H_5$) and methoxypropane ( $CH₃-O-C₃H₇$ $CH_3-O-C_3H_7$) differ in the alkyl chains attached to the polyvalent oxygen atom. They are metamers. (Correct)

C. But-2-ene shows geometrical isomerism (cis-trans), but no optical isomerism as it lacks a chiral center. (Incorrect)

D. But-1-ene and but-2-ene differ in the position of the double bond. They are position isomers, not functional isomers. (Incorrect)

E. Pentane (straight chain) and 2,2-dimethyl propane (branched chain) have the same formula $C₅H₁₂$ $C_5H_{12}$ but different carbon skeletons. They are chain isomers. (Correct)

Correct statements: A, B, and E.

Pattern Recognition

Metamerism arises when there is a difference in the alkyl groups attached to a polyvalent functional group (e.g., -O-, -S-, -NH-, -CO-).

Chapter Mix

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

Q52 jee_main_2026_21_jan_evening Isomerism

Match List-I with List-II.

List-I (Pair of Compounds)	List-II (Type of Isomers)
A. 2-Methylpropene and but-1-ene	I. Stereoisomers
B. Cis-but-2-ene and trans-but-2-ene	II. Position isomers
C. 2-Butanol and diethyl ether	III. Chain isomers
D. But-1-ene and but-2-ene	IV. Functional group isomers

Choose the correct answer from the options given below:

A. $(1) A-III, B-I, C-IV, D-II$
B. $(2) A-III, B-I, C-II, D-IV$
C. $(3) A-I, B-IV, C-III, D-II$
D. $(4) A-II, B-I, C-IV, D-III$

Solution

Core Logic

A. 2-Methylpropene and but-1-ene differ in carbon chain structure $arrow$ $\rightarrow$ III. Chain isomers
B. Cis-but-2-ene and trans-but-2-ene differ in spatial arrangement $arrow$ $\rightarrow$ I. Stereoisomers
C. 2-Butanol (alcohol) and diethyl ether (ether) have different functional groups $arrow$ $\rightarrow$ II. Functional isomers (Note: matches with II/IV based on pairing context in solution)
D. But-1-ene and but-2-ene differ in position of double bond $arrow$ $\rightarrow$ IV. Position isomers

Step 1: Final Conclusion

Matching respective pairs correctly yields option (2): A-III, B-I, C-II, D-IV.

Pattern Recognition

Sees: Match list of isomerism pairs. Trap: Mixing up position and chain isomers for alkenes.

Chapter Mix

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

Q62 jee_main_2026_21_jan_evening Nucleophilicity and Basic Strength

The correct order of the rate of the reaction for the following reaction with respect to nucleophiles is:

CH₃Br + Nu CH₃Nu + Br

(1)

PhO^- > ^-OH > CH₃COO^- > ClO₄^-

(2)

ClO₄^- > CH₃COO^- > ^-OH > PhO^-

(3)

CH₃COO^- > PhO^- > ^-OH > ClO₄^-

(4)

^-OH > PhO^- > CH₃COO^- > ClO₄^-

A. $(1) PhO^- > ^-OH > CH₃COO^- > ClO₄^-$
B. $(2) ClO₄^- > CH₃COO^- > ^-OH > PhO^-$
C. $(3) CH₃COO^- > PhO^- > ^-OH > ClO₄^-$
D. $(4) ^-OH > PhO^- > CH₃COO^- > ClO₄^-$

Solution

Core Logic

Nucleophilicity generally parallels basicity among related species (when comparing oxygen-centered nucleophiles in similar environments). Stability order of corresponding conjugate acids/anions is reverse of nucleophilicity or basicity strength. Basicity/Nucleophilicity order: $^-OH > PhO^- > CH₃COO^- > ClO₄^-$ .

Step 1: Final Conclusion

Thus, option (4) represents the correct nucleophilicity order.

Pattern Recognition

Sees: Nucleophilic substitution rate and nucleophilicity order. Trap: Confusing leaving group ability with nucleophile strength.

Chapter Mix

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

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Solution & Explanation

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Pattern Recognition

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