Organic Chemistry - Some Basic Principles and Techniques appeared 93 times across 3 years — 10.8% of Chemistry.
This question is from Acidity of Organic Compounds.
The compounds that produce CO₂$\mathrm{CO}_{2}$ with aqueous NaHCO₃$\mathrm{NaHCO}_{3}$ solution are:
A. The prompt lists five structures labeled A through E evaluating structural acidities.
B. The prompt lists five structures labeled A through E evaluating structural acidities.
C. The prompt lists five structures labeled A through E evaluating structural acidities.
D. The prompt lists five structures labeled A through E evaluating structural acidities.
E. The prompt lists five structures labeled A through E evaluating structural acidities.
Choose the correct answer from the options given below:
A.A and C only$\text{A and C only}$
B.A, B and E only$\text{A, B and E only}$
C.A, C and D only$\text{A, C and D only}$
D.A and B only$\text{A and B only}$
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
Keywords:#produce CO2 with aqueous NaHCO3 solution#JEE Main 2025 Morning Q43#Qualitative Organic Analysis JEE Main 2025#Acidity Trends JEE Main 2025#Picric acid#Phenol derivative#Bicarbonate test
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 2
Q68jee_main_2026_21_jan_eveningHybridization, Chiral Centers, and Functional Groups
Given below are two statements:
Statement I: Compound (X), shown below, dissolves in NaHCO₃$\text{NaHCO}_3$ solution and has two chiral carbon atoms.
Structures of compound X and compound Y containing functional groups and chiral centers.
Statement II: Compound (Y), shown below, has two carbons with sp³$sp^3$ hybridization, one carbon with sp²$sp^2$ and one carbon with sp$sp$ hybridization.
Structures of compound X and compound Y containing functional groups and chiral centers.
In the light of the above statements, choose the correct answer from the options given below:
A.(1) Statement I is true but Statement II is false$(1) \text{ Statement I is true but Statement II is false}$
B.(2) Statement I is false but Statement II is true$(2) \text{ Statement I is false but Statement II is true}$
C.(3) Both Statement I and Statement II are true$(3) \text{ Both Statement I and Statement II are true}$
D.(4) Both Statement I and Statement II are false$(4) \text{ Both Statement I and Statement II are false}$
Solution
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
Sodium fusion extract of an organic compound (Y) with CHCl₃$CHCl_{3}$ and chlorine water gives violet color to the CHCl₃$CHCl_{3}$ 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$g\text{ mol}^{-1}$ 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 × Mass of AgIMass of organic compound × 100$$\% \text{ of Halogen (I)} = \frac{\text{Atomic weight of I}}{\text{Molecular weight of AgI}} \times \frac{\text{Mass of AgI}}{\text{Mass of organic compound}} \times 100$$
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$108 (Ag) + 127 (I) = 235\text{ g mol}^{-1}$. (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) × (0.12)/(0.15) × 100$$\% \text{ of I} = \frac{127}{235} \times \frac{0.12}{0.15} \times 100$$% of I = (127)/(235) × 0.8 × 100$$\% \text{ of I} = \frac{127}{235} \times 0.8 \times 100$$% of I = 0.5404 × 80 = 43.234%$$\% \text{ of I} = 0.5404 \times 80 = 43.234\%$$
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
Q58jee_main_2026_22_january_eveningIUPAC Nomenclature of Esters
The IUPAC name of the following compound is:
Displays a branched halogenated ester compound requiring IUPAC systematic numbering.
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.
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
Q66jee_main_2026_23_january_morningQualitative Analysis of Organic Compounds
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).
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q67jee_main_2026_23_january_morningMethods of Purification
Given below are two statements:
Statement-I : Sublimation is used for the separation and purification of compounds with low melting point.
Statement-II : The boiling point of a liquid increases as the external pressure is reduced.
In the light of the above statements, choose the correct answer from the options given below :
A.Statement-I is false but Statement-II is true.$\text{Statement-I is false but Statement-II is true.}$
B.Statement-I is true but Statement-II is false.$\text{Statement-I is true but Statement-II is false.}$
C.Both Statement-I and Statement-II are true.$\text{Both Statement-I and Statement-II are true.}$
D.Both Statement-I and Statement-II are false.$\text{Both Statement-I and Statement-II are false.}$
Solution
Core Logic
Assess theoretical principles of purification processes.
Statement-I: Sublimation is a process used for separating sublimable compounds from non-sublimable impurities. It does not strictly depend on a 'low melting point'. Sublimable solids bypass the liquid phase altogether when heated. (False)
Statement-II: The boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure. If external pressure is reduced, the liquid needs less vapor pressure (and thus lower temperature) to boil. Hence, boiling point decreases with reduced external pressure. (False)
Step 1: Final Conclusion
Both statements are fundamentally false based on basic thermodynamics and purification principles.
Pattern Recognition
Lower pressure = Lower boiling point (used in vacuum distillation). Sublimation relies on vapor pressure of solid overcoming external pressure without melting, independent of specifically 'low' melting points.
Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2025_28_jan_morning
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