The compounds that produce CO2 with aqueous NaHCO3 solution are: A. B. C. D. E. Choose the correct answer from the…

Q: The compounds that produce CO2 with aqueous NaHCO3 solution are: A. B. C. D. E. Choose the correct answer from the options given below:

Organic compounds react with sodium bicarbonate (NaHCO3) to liberate CO2 gas if they are stronger acids than carbonic acid (H2CO3). Evaluating the structures: - **A:** Benzoic acid, which is significantly more acidic than carbonic acid. - **C:** Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and H2CO3. - **D:** Benzenesulfonic acid, a highly strong mineral-like organic acid. - **B & E:** Standard phenols or

Year	2026	2025	2024	Total
Questions	19	49	25	93

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Organic compounds react with sodium bicarbonate ( $NaHCO₃$ $\mathrm{NaHCO}_3$) to liberate $CO₂$ $\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid ( $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$). Evaluating the structures:

A: Benzoic acid, which is significantly more acidic than carbonic acid.
C: Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$.
D: Benzenesulfonic acid, a highly strong mineral-like organic acid.
B & E: Standard phenols or weakly substituted phenols, which are less acidic than carbonic acid and do not liberate $CO₂$ $\mathrm{CO}_2$.

Therefore, structures A, C, and D give a positive test result.

Pattern Recognition

Sees: Sodium bicarbonate test for organic systems. Shortcut: Only carboxylic acids, sulfonic acids, and highly nitrated phenols like picric acid possess sufficient proton acidity to displace $CO₂$ $\mathrm{CO}_2$ from bicarbonate ions.

Chapter Mix

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

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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

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