Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.8% of Chemistry.
This question is from Quantitative Analysis of Organic Compounds.
Given below are two statements I and II.
Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc H₂SO₄$\mathrm{H}_2\mathrm{SO}_4$
In the light of the above statements, choose the correct answer from the options given below
A.Both Statement I and Statement II are true.
B.Statement I is false but Statement II is true
C.Both Statement I and Statement II are false.
D.Statement I is true but Statement II is false
Solution & Explanation
Core Logic
Statement I is fully accurate: Dumas method is standardly applied for estimating elemental nitrogen across structural compounds.
Statement II is incorrect: The reaction leading to ammonium sulphate generation by intense thermal heating alongside concentrated H₂SO₄$\mathrm{H}_2\mathrm{SO}_4$ describes the Kjeldahl method, not the Dumas strategy. The Dumas process instead relies on burning carbonaceous compounds explicitly with copper oxide to convert nitrogen cleanly into free gas (N₂$N_2$).
Pattern Recognition
Dumas method collects element gas N₂$N_2$ via volumetric analysis; Kjeldahl maps digestions via standard (NH₄)₂SO₄$(\mathrm{NH}_4)_2\mathrm{SO}_4$ pathways.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#Dumas method is used for estimation of Nitrogen#JEE Main 2025 Morning Q34#Organic Chemistry JEE Main 2025#Quantitative Analysis of Organic Compounds
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
When 1 g$1\text{ g}$ of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL$15\text{ mL}$, 1M H₂SO₄$1\text{M } \text{H}_2\text{SO}_4$ 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 × M × 2 × VMass of compound (g)$$\%\text{ N} = \frac{1.4 \times M \times 2 \times V}{\text{Mass of compound (g)}}$$
Alternatively:
Meq of H₂SO₄ = V × M × nfactor$$\text{Meq of } \text{H}_2\text{SO}_4 = V \times M \times n_{\text{factor}}$$
Core Logic
Step 1: Calculate equivalents of acid neutralized:
Moles of NH₃ = 0.03 Moles of N = 0.03$$\text{Moles of } \text{NH}_3 = 0.03 \implies \text{Moles of N} = 0.03$$Mass of N = 0.03 × 14 = 0.42 g$$\text{Mass of N} = 0.03 \times 14 = 0.42\text{ g}$$
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).
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
