Organic Chemistry - Some Basic Principles and Techniques Previous Year Questions — JEE Main Chemistry
101 past-year Organic Chemistry - Some Basic Principles and Techniques questions from JEE Main (Chemistry) — page 20.
Q71 (2026)
0.25 g of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $CaCl_2$ tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is ____%. (Nearest integer)
### Related Formula
$$\text{Mass of C} = \frac{12}{44} \times \text{Mass of } CO_2$$
$$\text{Mass of H} = \frac{2}{18} \times \text{Mass of } H_2O$$
### Core Logic
Combustion equation:
$\mathrm{C_xH_yO_z + O_2 \rightarrow CO_2 + H_2O}$
Potash (KOH) tube absorbs $CO_2$. So, mass of $CO_2$ produced = 0.1837 g (approximated as 0.18 g in solution data for simplicity, but strictly 0.1837 based on prompt. The solution explicitly uses 0.18 for C calculation, let's trace: Mass of 'C' $= \frac{0.18}{44} \times 12$).
$CaCl_2$ tube absorbs $H_2O$. So, mass of $H_2O$ produced = 0.15 g.
Mass of Carbon (C) $= \frac{12}{44} \times 0.18 \simeq 0.049 \simeq 0.05 \text{ gm}$
Mass of Hydrogen (H) $= \frac{2}{18} \times 0.15 = 0.0166 \simeq 0.017 \text{ gm}$
### Step 1: Calculate Mass of Oxygen
Since the total mass of compound A is 0.25 gm:
Mass of Oxygen (O) $= 0.25 - (\text{Mass of C} + \text{Mass of H})$
Mass of 'O' $= 0.25 - 0.05 - 0.017 = 0.183 \text{ gm}$
### Step 2: Calculate Percentage
Mass % of 'O' $= \frac{0.1833}{0.25} \times 100 = 73.32\%$
Rounding to the nearest integer gives 73.
### Pattern Recognition
In Liebig's combustion method, the $CaCl_2$ U-tube maps strictly to $H_2O$ mass, and the Potash bulb maps strictly to $CO_2$ mass. Find carbon and hydrogen masses, subtract from total sample mass to find the third element (Oxygen).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q52 (2026)
CORRECT order of stability for the following is $CH_{2}=CH^{-}$, $CH_{3}-CH_{2}^{-}$, $CH\equiv C^{-}$
- $CH_{3}-CH_{2}^{-}>CH_{2}=CH^{-}>CH\equiv C^{-}$
- $CH_{2}=CH^{-}>CH\equiv C^{-}>CH_{3}-CH_{2}^{-}$
- $CH\equiv C^{-}>CH_{2}=CH^{-}>CH_{3}-CH_{2}^{-}$
- $CH\equiv C^{-}>CH_{3}-CH_{2}^{-}>CH_{2}=CH^{-}$
### Core Logic
The stability of a carbanion is directly proportional to the electronegativity of the carbon atom bearing the negative charge. The electronegativity of carbon increases with the increase in s-character of its hybridization state.
### Step 1: Hybridization Analysis
$\bullet \; CH\equiv C^{-}$ (sp hybridized, $50\%$ s-character) $\rightarrow$ Highest electronegativity, most stable.
$\bullet \; CH_{2}=CH^{-}$ (sp$^{2}$ hybridized, $33.3\%$ s-character) $\rightarrow$ Intermediate electronegativity.
$\bullet \; CH_{3}-CH_{2}^{-}$ (sp$^{3}$ hybridized, $25\%$ s-character) $\rightarrow$ Lowest electronegativity, least stable.
### Final Conclusion
Order of stability: $CH\equiv C^{-} > CH_{2}=CH^{-} > CH_{3}-CH_{2}^{-}$
### Pattern Recognition
Stability of carbanion $\propto \% \; s\text{-character}$. More s-character pulls the electron pair closer to the nucleus, stabilizing the negative charge.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q55 (2026)
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
{{IMG1}}
- Simple distillation
- sublimation
- steam distillation
- fractional distillation
### Core Logic
The reaction of Benzene with $Br_2 / FeBr_3$ yields Bromobenzene (A).
Nitration of Bromobenzene ($conc.\; HNO_3 / conc.\; H_2SO_4$) yields a mixture of ortho-bromonitrobenzene (B) and para-bromonitrobenzene (C).
{{SOL_IMG1}}
### Step 1: Justification of Separation Technique
The ortho and para isomers of bromonitrobenzene have differing boiling points but not sufficiently far apart for simple distillation. Thus, Fractional Distillation is the appropriate method to separate these positional isomers accurately based on slight differences in boiling points.
### Pattern Recognition
Mixtures of isomeric organic liquids (like ortho/para derivatives) that differ mildly in boiling points are classically separated by fractional distillation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Class 12 Chemistry: Haloalkanes and Haloarenes
Q74 (2026)
$0.53\mathrm{~g}$ of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave $0.75\mathrm{~g}$ of silver bromide precipitate. $1.0\mathrm{~g}$ of (x) gave $1.32\mathrm{~g}$ of $\mathrm{CO}_2$ gas on combustion. The percentage of hydrogen in the compound (x) is _____ %. [Nearest Integer]
[Given : Molar mass in $\mathrm{g~mol}^{-1}$ H : 1, C : 12, Br : 80, Ag : 108, O : 16; Compound (x) : $\mathrm{C}_{x}\mathrm{H}_{y}\mathrm{Br}_{z}$]
### Step 1: Calculate Percentage of Carbon
$1.0\mathrm{~g}$ of compound gives $1.32\mathrm{~g}$ of $\mathrm{CO}_2$.\nMoles of $\mathrm{CO}_2 = \frac{1.32}{44} = 0.03\mathrm{~mol}$.\nMass of Carbon $= 0.03 \times 12 = 0.36\mathrm{~g}$.\n$\text{Percentage of } C = \left(\frac{0.36}{1.0}\right) \times 100 = 36\%$.
### Step 2: Calculate Percentage of Bromine
$0.53\mathrm{~g}$ of compound gives $0.75\mathrm{~g}$ of $\mathrm{AgBr}$ (Molar mass = $108 + 80 = 188\mathrm{~g/mol}$).\n$\text{Mass of } Br = \left(\frac{80}{188}\right) \times 0.75 \approx 0.319\mathrm{~g}$.\n$\text{Percentage of } Br = \left(\frac{0.319}{0.53}\right) \times 100 = 60.2\%$.
### Step 3: Calculate Percentage of Hydrogen
The compound strictly consists of C, H, and Br (as formula is given as $\mathrm{C}_x\mathrm{H}_y\mathrm{Br}_z$).\n$\text{Percentage of } H = 100 - (\%\text{C} + \%\text{Br})$\n$\text{Percentage of } H = 100 - (36 + 60.2) = 100 - 96.2 = 3.8\%$.\nRounding off to the nearest integer, we get $4\%$.
### Pattern Recognition
For basic gravimetric elemental analysis: always map precipitate mass back to elemental mass using strict molar mass ratios, then find mass fractions, ensuring everything sums to 100%.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q60 (2026)
The cyclic cations having the same number of hyperconjugation are :
A. {{IMG1}}
B. {{IMG2}}
C. {{IMG3}}
D. {{IMG4}}
Choose the correct answer from the options given below :
- $(1)\text{ A and C Only}$
- $(2)\text{ B and C Only}$
- $(3)\text{ A and B Only}$
- $(4)\text{ A, C and D only}$
### Core Logic
Count the number of alpha hydrogens ($\alpha-H$) adjacent to the carbocation in each structure.
(A) {{SOL_IMG1}} $\alpha-H = 6$
(B) {{SOL_IMG2}} $\alpha-H = 7$
(C) {{SOL_IMG3}} $\alpha-H = 6$
(D) {{SOL_IMG4}} $\alpha-H = 5$
### Step 1: Final Conclusion
Both cations (A) and (C) have 6 $\alpha$-hydrogens, meaning they share the same number of hyperconjugative structures.
### Pattern Recognition
Hyperconjugation count corresponds strictly to the number of $C-H$ bonds on the carbon atoms immediately adjacent ($sp^3$) to the positive center.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
