Given below are two statements: Statement I: Compound (X), shown below, dissolves in NaHCO3 solution and has two chiral carbon atoms. Statement II: Compound (Y), shown below, has two carbons with sp3 hybridization, one carbon with sp2 and one carbon with sp hybridization. In the light of the above statements, choose the correct answer from the options given below:

- Statement I: Compound X contains a carboxylic acid group (which dissolves in NaHCO3) and possesses two chiral centers. Thus Statement I is true. - Statement II: Compound Y has specific carbon hybridization states matching sp3, sp2, and sp carbons. Thus Statement II is true. Both Statement I and Statement II are true, matching option (3). Sees: identification of chiral centers and carbon hybridization in organic molecules. Trap: Overlooking carboxylic acid solubility requirements in NaHCO3. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement I: Compound (X), shown below, dissolves in NaHCO3 solution and has two chiral carbon atoms. Statement II: Compound (Y), shown below, has two carbons with sp3

- Statement I: Compound X contains a carboxylic acid group (which dissolves in NaHCO3) and possesses two chiral centers. Thus Statement I is true. - Statement II: Compound Y has specific carbon hybridization states matching sp3, sp2, and sp carbons. Thus Statement II is true. Both Statement I and Statement II are true, matching option (3). Sees: identification of chiral centers and carbon hybridization in organic molecules. Trap: Overlooking carboxylic acid solubility requirements in NaHCO3. Cla

Given below are two statements: Statement I: Compound (X), shown below, dissolves in NaHCO3 solution and has two chiral carbon atoms. Statement II: Compound (Y), shown below, has two carbons with sp3 hybridization, one carbon with sp2 and one carbon with sp hybridization. In the light of the above statements, choose the correct answer from the options given below:

- Statement I: Compound X contains a carboxylic acid group (which dissolves in NaHCO3) and possesses two chiral centers. Thus Statement I is true. - Statement II: Compound Y has specific carbon hybridization states matching sp3, sp2, and sp carbons. Thus Statement II is true. Both Statement I and Statement II are true, matching option (3). Sees: identification of chiral centers and carbon hybridization in organic molecules. Trap: Overlooking carboxylic acid solubility requirements in NaHCO3. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement I: Compound (X), shown below, dissolves in NaHCO3 solution and has two chiral carbon atoms. Statement II: Compound (Y), shown below, has two carbons with sp3

- Statement I: Compound X contains a carboxylic acid group (which dissolves in NaHCO3) and possesses two chiral centers. Thus Statement I is true. - Statement II: Compound Y has specific carbon hybridization states matching sp3, sp2, and sp carbons. Thus Statement II is true. Both Statement I and Statement II are true, matching option (3). Sees: identification of chiral centers and carbon hybridization in organic molecules. Trap: Overlooking carboxylic acid solubility requirements in NaHCO3. Cla

Given below are two statements: Statement I: Compound (X), shown below, dissolves in NaHCO3 solution and has two chiral…

Year	2026	2025	2024	Total
Questions	22	49	30	101

Year

2026

2025

2024

Total

Questions

101

Solution & Explanation

Core Logic

Statement I: Compound X contains a carboxylic acid group (which dissolves in $NaHCO₃$ $\text{NaHCO}_3$) and possesses two chiral centers. Thus Statement I is true.
Statement II: Compound Y has specific carbon hybridization states matching $sp³$ $sp^3$, $sp²$ $sp^2$, and $sp$ $sp$ carbons. Thus Statement II is true.

Step 1: Final Conclusion

Both Statement I and Statement II are true, matching option (3).

Pattern Recognition

Sees: identification of chiral centers and carbon hybridization in organic molecules. Trap: Overlooking carboxylic acid solubility requirements in $NaHCO₃$ $\text{NaHCO}_3$.

Chapter Mix

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

Detailed solution structures for Q68 - JEE Main 2026 Evening — Structures of compound X and compound Y containing functional groups and chiral centers.

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 2

Q74 jee_main_2026_22_january_morning Quantitative Analysis

Sodium fusion extract of an organic compound (Y) with

CHCl₃

and chlorine water gives violet color to the

CHCl₃

layer. 0.15g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is ____. (Nearest integer). (Given: molar mass

g mol⁻¹

C: 12, H: 1, Cl: 35.5, Br: 80, I: 127)

Numerical Answer. Answer: 43 to 43

Solution

Related Formula

% of Halogen (I) = Atomic weight of IMolecular weight of AgI \times Mass of AgIMass of organic compound \times 100

Core Logic

Identify the halogen: The violet color in the chloroform layer upon addition of chlorine water is the classic test for Iodine ( $I₂$ $I_2$). Chlorine oxidizes $I^-$ $I^-$ to $I₂$ $I_2$, which dissolves in $CHCl₃$ $CHCl_3$ with a violet/purple color.
Molar mass calculations:

Atomic weight of I = 127 Molar mass of AgI = $108 (Ag) + 127 (I) = 235 g mol⁻¹$ . (Since Ag is not provided in data, assume standard 108. The PDF explicitly uses 235).

Step 1: Apply Carius Method Formula

Given: Mass of organic compound ( $w$ $w$) = 0.15 g Mass of AgI precipitate ( $m$ $m$) = 0.12 g

% of I = (127)/(235) \times (0.12)/(0.15) \times 100

% of I = (127)/(235) \times 0.8 \times 100

% of I = 0.5404 \times 80 = 43.234%

Step 2: Rounding

The nearest integer to 43.23 is 43.

Pattern Recognition

Always use the qualitative test (chloroform layer color) to definitively identify the halogen before executing the quantitative Carius formula.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Q56 jee_main_2026_22_january_evening Kjeldahl Method Nitrogen Estimation

When

1 g

of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen,

15 mL

1M H₂SO₄

was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is:

A. 21
B. 0.42
C. 42
D. 0.21

Solution

Related Formula

% N = 1.4 \times M \times 2 \times VMass of compound (g)

Alternatively:

Meq of H₂SO₄ = V \times M \times nfactor

Core Logic

Step 1: Calculate equivalents of acid neutralized:

Equivalents of H₂SO₄ = (15 \times 1 \times 2)/(1000) = 0.03

Step 2: Calculate moles of nitrogen atoms:

Moles of NH₃ = 0.03 Moles of N = 0.03

Mass of N = 0.03 \times 14 = 0.42 g

Step 3: Calculate percentage of nitrogen:

% N = (0.42)/(1) \times 100 = 42%

Pattern Recognition

Sees: Kjeldahl's method with $H₂SO₄$ $\text{H}_2\text{SO}_4$ volume and molarity. Shortcut: Use standard formula $% N = (1.4 \times M \times 2 \times V)/(W) = (1.4 \times 1 \times 2 \times 15)/(1) = 42%$ .

Chapter Mix

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

Q58 jee_main_2026_22_january_evening IUPAC Nomenclature of Esters

The IUPAC name of the following compound is:

Organic ester structure for Q58 - JEE Main 2026 Evening — Displays a branched halogenated ester compound requiring IUPAC systematic numbering.

A. n-propyl-2-bromo-5-methylheptanoate
B. 2-bromo-5-methylhexylpropanoate
C. 2-bromo-5-methylpropanoate
D. n-propyl-1-bromo-4-methylhexanoate

Solution

Related Formula

Ester IUPAC Nomenclature: Alkyl + [Substituents] + alkanoate

Core Logic

Step 1: Identify the alkyl group attached to oxygen: $n-propyl$ $n\text{-propyl}$.

Step 2: Number the longest carbon chain containing the ester carbonyl carbon from C1 to C7.

Step 3: Substituents present: Br at C2 and Methyl at C5.

Step 4: Combine into IUPAC name: n-propyl-2-bromo-5-methylheptanoate.

Numbered carbon chain for IUPAC naming for Q58 - JEE Main 2026 Evening — Displays a branched halogenated ester compound requiring IUPAC systematic numbering.

Pattern Recognition

Sees: Ester esterified with propyl alcohol and heptanoate chain. Shortcut: Always start naming ester with alkyl group attached to oxygen (n-propyl), followed by numbered parent carboxylate chain.

Chapter Mix

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

Q66 jee_main_2026_23_january_morning Qualitative Analysis of Organic Compounds

Match List-I with List-II.

List-I (Functional group detection)	List-II (Change observed during detection)
(A) Unsaturation (Baeyer's test)	(I) Red colour Appears
(B) Alcoholic group (Ceric ammonium nitrate test)	(II) Silver mirror appears
(C) Aldehyde group (Tollen's reagent)	(III) Violet colour appears
(D) Phenolic group ( $FeCl₃$ $FeCl_3$ test)	(IV) Discharge of pink colour

Choose the correct answer from the options given below:

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

Solution

Core Logic

Recall standard qualitative laboratory tests for functional groups: (A) Unsaturation (Baeyer's Test): Dilute alkaline $KMnO₄$ $KMnO_4$ is pink. When it reacts with an alkene/alkyne, it gets reduced to $MnO₂$ $MnO_2$ (brown ppt), discharging the pink colour. Matches with (IV). (B) Alcoholic group: Primary and secondary alcohols react with Ceric ammonium nitrate to form a red-coloured coordination complex. Matches with (I). (C) Aldehyde group (Tollens' Test): Aldehydes reduce Tollens' reagent ( $[Ag(NH₃)₂]^+$ $[Ag(NH_3)_2]^+$) to metallic silver, forming a silver mirror. Matches with (II). (D) Phenolic group: Phenols form strongly colored (often violet) complexes with neutral $FeCl₃$ $FeCl_3$. Matches with (III).

Step 1: Final Mapping

(A) - (IV) (B) - (I) (C) - (II) (D) - (III)

Pattern Recognition

Baeyer's = Pink to colorless. Tollens' = Silver mirror. $FeCl₃$ $FeCl_3$ = Violet/Purple. Ceric Ammonium Nitrate = Red.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Apply Carius Method Formula

Step 2: Rounding

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Mapping

Pattern Recognition

Chapter Mix

More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2026_21_jan_evening

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