Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.8% of Chemistry.
This question is from Quantitative Analysis - Dumas Method.
During estimation of nitrogen by Dumas' method of compound X (0.42 g):
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
mL of N2$N{2}$ gas will be liberated at STP. (nearest integer)
(Given molar mass in g mol: C: 12, H: 1, N: 14)
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Numerical Answer Type:
Enter a numerical valueAnswer: 111 to 111+4 marks
Solution & Explanation
Related Formula
Using the Principle of Atom Conservation (POAC) for Nitrogen:
ncompound × (atoms of N per molecule) = 2 × nN₂$$n_{\text{compound}} \times (\text{atoms of N per molecule}) = 2 \times n_{\text{N}_2}$$
Core Logic
The molecular weight of the given heterocyclic amine organic structure X$X$ (piperazine, C₄H₁₀N₂$\text{C}_4\text{H}_{10}\text{N}_2$) is calculated as:
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Given mass of compound = 0.42 g$= 0.42\text{ g}$:
Moles of compound X = (0.42)/(86) mol$$\text{Moles of compound } X = \frac{0.42}{86}\text{ mol}$$
Step 1: Calculating STP Volume
Since each molecule contains 2$2$ nitrogen atoms, 1 mol$1\text{ mol}$ of compound produces 1 mol$1\text{ mol}$ of N₂$\text{N}_2$ gas:
Using standard molar volume at STP (22700 mL/mol$22700\text{ mL/mol}$ per IUPAC convention, or 22400 mL/mol$22400\text{ mL/mol}$ in traditional calculations):
Volume of N₂ at STP = (0.42)/(86) × 22700 mL ≈ 110.86 mL ≈ 111 mL$$\text{Volume of } \text{N}_2\text{ at STP} = \frac{0.42}{86} \times 22700\text{ mL} \approx 110.86\text{ mL} \approx 111\text{ mL}$$
(Note: If calculated using 22400 mL/mol$22400\text{ mL/mol}$, Volume = (0.42)/(86) × 22400 ≈ 109.4 mL ≈ 109 mL$\text{Volume} = \frac{0.42}{86} \times 22400 \approx 109.4\text{ mL} \approx 109\text{ mL}$.)
Rounding to the nearest integer gives 111 (official accepted range: 109 to 111).
Pattern Recognition
Shortcut: Determine the molar mass (M = 86 g/mol$M = 86\text{ g/mol}$) and nitrogen atom count (2 N atoms 1 mol N₂ per mol of compound$2\text{ N atoms} \implies 1\text{ mol } \text{N}_2\text{ per mol of compound}$). Multiply moles directly by molar volume at STP to find the liberated gas volume.
Evaluation Rubric / Model Answer
111
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.
Keywords:#Dumas method quantitative#Molar weight tracking#Nitrogen gas evolution
More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8
An organic compound weighing 500 mg$500\mathrm{\ mg}$, produced 220 mg$220\mathrm{\ mg}$ of CO₂$\mathrm{CO}_2$ on complete combustion. The percentage composition of carbon in the compound is ______ %. (nearest integer)
(Given molar mass in g mol⁻¹$\mathrm{g\ mol}^{-1}$ of C: 12$\mathrm{C}: 12$, O: 16$\mathrm{O}: 16$)
Numerical Answer.Answer: 12 to 12
Solution
Related Formula
% C = (12)/(44) × Mass of CO₂ producedMass of organic compound taken × 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{\text{Mass of } \mathrm{CO}_2 \text{ produced}}{\text{Mass of organic compound taken}} \times 100$$
Core Logic
Given:
Mass of organic compound taken = 500 mg = 500 × 10⁻³ g$= 500 \text{ mg} = 500 \times 10^{-3} \text{ g}$
Mass of CO₂$\mathrm{CO}_2$ produced = 220 mg = 220 × 10⁻³ g$= 220 \text{ mg} = 220 \times 10^{-3} \text{ g}$
Carbon dioxide has exactly 12/44 ≈ 27.27%$12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of CO₂$\mathrm{CO}_2$ (220/500 = 0.44$220/500 = 0.44$) by 12/44$12/44$ to directly get 0.12$0.12$ or 12%$12\%$.
Evaluation Rubric / Model Answer
A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise 12$12$ percent carbon composition.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Qjee_main_2025_08_april_eveningIUPAC Nomenclature
What is the correct IUPAC name of the following organic compound?
The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.
A. 4-Ethyl-1-hydroxycyclopent-2-ene
B. 1-Ethyl-3-hydroxycyclopent-2-ene
C. 1-Ethylcyclopent-2-en-3-ol
D. 4-Ethylcyclopent-2-en-1-ol
Solution
Core Logic
Let us apply official IUPAC priority indexing rules:
Principal Functional Group: The hydroxyl group (-OH$-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the -OH$-\text{OH}$ group is assigned position C-1.
Numbering Direction: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates C-2 and C-3.
Locating Side Chains: Proceeding with this direction puts the ethyl group at position C-4. The molecule contains a five-membered carbon ring with one double bond, a hydroxyl substituent, and an ethyl group.
Principal suffix priority hierarchy: -OH > Double bond > Alkyl side-chain$-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q27jee_main_2025_08_april_eveningReactive Intermediates and Reagents
Match the LIST-I with LIST-II:
LIST-I
LIST-II
A. Carbocation
I. Species that can supply a pair of electrons.
B. C-Free radical
II. Species that can receive a pair of electrons.
C. Nucleophile
III. sp²$sp^2$ hybridized carbon with empty p-orbital.
D. Electrophile
IV. sp²/sp³$sp^2/sp^3$ hybridized carbon with one unpaired electron.
Choose the correct answer from the options given below:
A. Carbocation: Features a positively charged trivalent carbon atom. It represents an sp²$sp^2$ hybridized carbon with an empty unhybridized p-orbital. Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
B. Carbon Free Radical: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits sp²$sp^2$ or sp³$sp^3$ hybridization depending on structural environments. Carbocation orbital hybridization diagram for Q27 - JEE Main 2025
C. Nucleophile: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons.
D. Electrophile: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons.
Step 1: Alignment Matrix
Matching each item yields:
A arrow III$\text{A} \rightarrow \text{III}$
B arrow IV$\text{B} \rightarrow \text{IV}$
C arrow I$\text{C} \rightarrow \text{I}$
D arrow II$\text{D} \rightarrow \text{II}$
This sequence aligns flawlessly with Option (4).
Pattern Recognition
Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
On complete combustion, 0.210 g$0.210 \text{ g}$ of an organic compound containing C, H, and O yielded 0.127 g$0.127 \text{ g}$ of H₂O$\text{H}_2\text{O}$ and 0.307 g$0.307 \text{ g}$ of CO₂$\text{CO}_2$. The mass percentages of hydrogen and oxygen in the given organic compound respectively are:
A. 53.41, 39.6
B. 6.72, 53.41
C. 7.55, 43.85
D. 6.72, 39.87
Solution
Related Formula
Percentage of Hydrogen in organic analysis:
%H = (2)/(18) × Mass of H₂OMass of Compound × 100$$\%\text{H} = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$$
Percentage of Carbon:
%C = (12)/(44) × Mass of CO₂Mass of Compound × 100$$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$$
Thus, the values of hydrogen and oxygen percentage are 6.72%$6.72\%$ and 53.41%$53.41\%$, matches with Option (2).
Pattern Recognition
Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q42jee_main_2025_08_april_eveningQualitative Analysis of Functional Groups
Match the reagents in LIST-I with the corresponding chemical functional groups they detect in LIST-II:
LIST-I (Reagent)
LIST-II (Functional Group detected)
A. Sodium bicarbonate solution
I. double bond / unsaturation
B. Neutral ferric chloride
II. carboxylic acid
C. Ceric ammonium nitrate
III. phenolic - OH
D. Alkaline KMnO₄$\text{KMnO}_4$
IV. alcoholic - OH
Choose the correct answer from the options given below:
Let us review the chemical basis for each qualitative test:
A. Sodium bicarbonate (NaHCO₃$\text{NaHCO}_3$) solution: Carboxylic acids are sufficiently acidic to decompose NaHCO₃$\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, A arrow II$\text{A} \rightarrow \text{II}$.
B. Neutral ferric chloride (FeCl₃$\text{FeCl}_3$): Phenols react with neutral FeCl₃$\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, B arrow III$\text{B} \rightarrow \text{III}$.
C. Ceric ammonium nitrate (CAN): Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, C arrow IV$\text{C} \rightarrow \text{IV}$.
D. Alkaline KMnO₄$\text{KMnO}_4$ (Baeyer's Reagent): Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown MnO₂$\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, D arrow I$\text{D} \rightarrow \text{I}$.
Baeyer's test (alkaline KMnO₄$\text{KMnO}_4$) always tests for alkenes/alkynes. NaHCO₃$\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table.
Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Alcohols, Phenols and Ethers
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.
 "Dumas method quantitative, Molar weight tracking, Nitrogen gas evolution")
 "Dumas method quantitative, Molar weight tracking, Nitrogen gas evolution")
