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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.

Year	2026	2025	2024	Total
Questions	22	49	30	101

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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 value Answer: 111 to 111 +4 marks

Solution & Explanation

Related Formula

Using the Principle of Atom Conservation (POAC) for Nitrogen:

ncompound \times (atoms of N per molecule) = 2 \times nN₂

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:

Molar Mass = (4 \times 12) + (10 \times 1) + (2 \times 14) = 86 g/mol

Stoichiometric parsing matrix step for Q46 — 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

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:

nN₂ = ncompound = (0.42)/(86) mol

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) \times 22700 mL \approx 110.86 mL \approx 111 mL

(Note: If calculated using $22400 mL/mol$ $22400\text{ mL/mol}$, $Volume = (0.42)/(86) \times 22400 \approx 109.4 mL \approx 109 mL$ .)

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 ( $2 N atoms 1 mol N₂ per mol of compound$ ). 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

Stoichiometric parsing matrix step for Q46 — 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.

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 5

Q60 jee_main_2026_28_january_evening Hyperconjugation

The cyclic cations having the same number of hyperconjugation are : A.

Hyperconjugation

B.

Hyperconjugation

C.

Hyperconjugation

D.

Hyperconjugation

Choose the correct answer from the options given below :

A. $(1) A and C Only$ $(1)\text{ A and C Only}$
B. $(2) B and C Only$ $(2)\text{ B and C Only}$
C. $(3) A and B Only$ $(3)\text{ A and B Only}$
D. $(4) A, C and D only$ $(4)\text{ A, C and D only}$

Solution

Core Logic

Count the number of alpha hydrogens ( $α-H$ $\alpha-H$) adjacent to the carbocation in each structure. (A)

Hyperconjugation

α-H = 6

(B)

Hyperconjugation

α-H = 7

(C)

Hyperconjugation

α-H = 6

(D)

Hyperconjugation

α-H = 5

Step 1: Final Conclusion

Both cations (A) and (C) have 6 $α$ $\alpha$-hydrogens, meaning they share the same number of hyperconjugative structures.

Pattern Recognition

Hyperconjugation count corresponds strictly to the number of $C-H$ $C-H$ bonds on the carbon atoms immediately adjacent ( $sp³$ $sp^3$) to the positive center.

Chapter Mix

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

Q70 jee_main_2026_28_january_evening Quantitative Analysis Of Elements

A student has been given

0.314 g

of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained

0.4813 g

of barium sulphate. The percentage of sulphur present in the compound is (Given Molar mass in

g mol⁻¹

S:32

,

BaSO₄:233

)

A. $(1) 42.10%$ $(1)\text{ 42.10\%}$
B. $(2) 63.15%$ $(2)\text{ 63.15\%}$
C. $(3) 21.05%$ $(3)\text{ 21.05\%}$
D. $(4) 48.24%$ $(4)\text{ 48.24\%}$

Solution

Related Formula

% of S = (32)/(233) \times Mass of BaSO₄Mass of organic compound \times 100

Core Logic

Mass of organic compound = $0.314 g$ $0.314\text{ g}$ Mass of $BaSO₄$ $BaSO_4$ formed = $0.4813 g$ $0.4813\text{ g}$ Substituting the values:

% S = (32)/(233) \times (0.4813)/(0.314) \times 100

% S = 0.1373 \times 1.5328 \times 100 \approx 21.052%

Step 1: Final Conclusion

The percentage of sulphur is approximately $21.05%$ $21.05\%$.

Pattern Recognition

Straight application of Carius method for Sulphur estimation formula.

Chapter Mix

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

Q jee_main_2025_02_april_evening Hybridization in Organic Compounds

In 3, 3-dimethylhex-1-ene-4-yne, there are

sp³

,

sp²

and

sp

hybridised carbon atoms respectively:

A. $4, 2, 2$ $4, 2, 2$
B. $3, 3, 2$ $3, 3, 2$
C. $2, 4, 2$ $2, 4, 2$
D. $2, 2, 4$ $2, 2, 4$

Solution

Related Formula

Hybridization of Carbon = cases sp³ & 4~σ-bonds sp² & 3~σ-bonds, 1~π-bond sp & 2~σ-bonds, 2~π-bonds cases

Core Logic

Let's first draw the structural formula of 3,3-dimethylhex-1-ene-4-yne:

Hybridization in Organic Compounds

6CH₃ - 5C \equiv 4C - 3C(CH₃)₂ - 2CH = 1CH₂

Step 1: Identify Hybridization of Each Carbon

We count the sigma ( $σ$ $\sigma$) bonds or pi ( $π$ $\pi$) bonds on each carbon:

$C₁$ $\mathrm{C}_1$: Involved in a double bond ( $CH₂ =$ $\mathrm{CH_2 =}$) $sp²$ $\implies \mathrm{sp^2}$
$C₂$ $\mathrm{C}_2$: Involved in a double bond ( $=CH-$ $=\mathrm{CH-}$) $sp²$ $\implies \mathrm{sp^2}$
$C₃$ $\mathrm{C}_3$: Single bonds only (bonded to $C₂$ $\mathrm{C}_2$, $C₄$ $\mathrm{C}_4$, and two methyl carbons) $sp³$ $\implies \mathrm{sp^3}$
Two methyl carbons attached to $C₃$ $\mathrm{C}_3$: Single bonds only $2 \times sp³$ $\implies 2 \times \mathrm{sp^3}$
$C₄$ $\mathrm{C}_4$: Involved in a triple bond ( $-C \equiv$ $-\mathrm{C} \equiv$) $sp$ $\implies \mathrm{sp}$
$C₅$ $\mathrm{C}_5$: Involved in a triple bond ( $\equiv C-$ $\equiv \mathrm{C}-$) $sp$ $\implies \mathrm{sp}$
$C₆$ $\mathrm{C}_6$: Single bonds only ( $-CH₃$ $-\mathrm{CH_3}$) $sp³$ $\implies \mathrm{sp^3}$

Step 2: Calculate the Count

Summing the hybridization counts:

$sp³$ $\mathrm{sp^3}$ carbons: $C₃$ $\mathrm{C}_3$, $C₆$ $\mathrm{C}_6$, and $2 \times CH₃$ $2 \times \mathrm{CH_3}$ on $C₃$ $\mathrm{C}_3$ $= 4$ $= 4$ carbon atoms
$sp²$ $\mathrm{sp^2}$ carbons: $C₁$ $\mathrm{C}_1$ and $C₂$ $\mathrm{C}_2$ $= 2$ $= 2$ carbon atoms
$sp$ $\mathrm{sp}$ carbons: $C₄$ $\mathrm{C}_4$ and $C₅$ $\mathrm{C}_5$ $= 2$ $= 2$ carbon atoms

Thus, the number of $sp³$ $\mathrm{sp^3}$, $sp²$ $\mathrm{sp^2}$ and $sp$ $\mathrm{sp}$ hybridized carbons is $4, 2, 2$ $4, 2, 2$ respectively.

Pattern Recognition

Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as $sp³$ $\mathrm{sp^3}$.

Chapter Mix

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

Q jee_main_2025_02_april_evening Methods of Purification of Organic Compounds

Match List-I with List-II:

array|l|l| arrayc List-I (Purification technique) array & arrayc List-II (Mixture of organic compounds) array (A) Distillation (simple) & (I) Diesel + Petrol (B) Fractional distillation & (II) Aniline + Water (C) Distillation under reduced pressure & (III) Chloroform + Aniline (D) Steam distillation & (IV) Glycerol + Spent-lye array

Choose the correct answer from the options given below:

A. $(A)-(II), (B)-(III), (C)-(IV), (D)-(I)$
B. $(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$
C. $(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$
D. $(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$

Solution

Related Formula

Method Selection = f( Δ Tb, decomposition threshold, volatility with steam )

Core Logic

Let's align each purification technique to its designated mixtures based on NCERT guidelines:

(A) Simple distillation: Used for liquids having a significant difference in their boiling points ( $>30~K$ $>30~\mathrm{K}$ or $30^$ $30^\circ\mathrm{C}$). Chloroform (b.p. $334~K$ $334~\mathrm{K}$) and aniline (b.p. $457~K$ $457~\mathrm{K}$) are separated easily using simple distillation $arrow$ $\rightarrow$ (III).
(B) Fractional distillation: Used if boiling point differences of the components are very close (less than $25~K$ $25~\mathrm{K}$). Separation of petrochemical fractions such as diesel and petrol uses this technique $arrow$ $\rightarrow$ (I).
(C) Distillation under reduced pressure: Used for liquids that tend to decompose at or below their normal boiling points. Glycerol is separated from spent-lye in soap manufacturing industry using this vacuum method to prevent glycerol decomposition $arrow$ $\rightarrow$ (IV).
(D) Steam distillation: Applied to substances which are steam-volatile and completely immiscible in water. Aniline and water are separated using this technique $arrow$ $\rightarrow$ (II).

Step 1: Conclusion

Thus, the correct match is: (A)-(III), (B)-(I), (C)-(IV), (D)-(II) This corresponds perfectly to option (4).

Pattern Recognition

Glycerol from spent-lye is a highly tested practical chemistry concept. Remember that vacuum distillation lowers the boiling point, permitting evaporation without decomposition.

Chapter Mix

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

Q jee_main_2025_02_april_evening Quantitative Estimation (Dumas Method)

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave

60~mL

of nitrogen collected at

300K

temperature and

715~mmHg

pressure. The percentage composition of nitrogen in the compound (Aqueous tension at

300K = 15~mmHg

) is

A. 1.257
B. 20.87
C. 18.67
D. 12.57

Solution

Related Formula

pN₂ = ptotal - paq

nN₂ = pN₂ VR T

% N = Mass of nitrogenMass of organic compound \times 100

Core Logic

Dumas' method estimates nitrogen by collecting dry nitrogen gas ( $N₂$ $N_2$). We must subtract the aqueous tension (vapor pressure of water) to find the pressure exerted solely by the dry nitrogen gas.

Step 1: Calculate Pressure of Dry Nitrogen

pN₂ = 715~mmHg - 15~mmHg = 700~mmHg

Converting pressure to atmospheres:

pN₂ = (700)/(760)~atm

Step 2: Calculate Moles of Nitrogen Gas

Using the ideal gas law with $R = 0.0821~ L~atm~mol⁻¹~K⁻¹$ , $T = 300~K$ $T = 300~\mathrm{K}$, and $V = 60~mL = 60 \times 10⁻³~L$ :

nN₂ = ((700)/(760)) \times 60 \times 10⁻³0.0821 \times 300

nN₂ = (0.92105 \times 0.060)/(24.63) \approx 2.244 \times 10⁻³~mol

Step 3: Calculate Mass and Percentage of Nitrogen

The molar mass of $N₂$ $\mathrm{N_2}$ is $28~ g~mol⁻¹$ $28~\mathrm{g~mol^{-1}}$:

Mass of N₂ = nN₂ \times 28 = 2.244 \times 10⁻³ \times 28 \approx 0.06283~g

Now find the percentage in $0.5~g$ $0.5~\mathrm{g}$ of organic compound:

% N = 0.06283~g0.5~g \times 100 = 12.566% \approx 12.57%

Pattern Recognition

Watch out! Always subtract the aqueous tension from the wet gas pressure first to find the dry gas pressure. Forgetting this step is the most common source of error in Dumas calculations.

Chapter Mix

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

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