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 5.
Q49 (2025)
In the sulphur estimation, $0.20\text{ g}$ of a pure organic compound gave $0.40\text{ g}$ of barium sulphate.
The percentage of sulphur in the compound is $x \times 10^{-1}\%$, where $x$ = ________.
(Molar mass: $O=16$, $S=32$, $Ba=137\text{ in g mol}^{-1}$)
### Related Formula
$$%S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Let's substitute the given values into the formula:
$$\text{Mass of } BaSO_4 = 0.40\text{ g}$$
$$\text{Mass of organic compound} = 0.20\text{ g}$$
$$\text{Molar mass of } BaSO_4 = 137 + 32 + (4 \times 16) = 233\text{ g/mol}$$
$$%S = \frac{32}{233} \times \frac{0.40}{0.20} \times 100 = \frac{32 \times 2 \times 100}{233} approx 27.468%$$
### Step 1: Match with the Question Layout
Rounding to the standard value given in the official key:
$$%S = 27.5% = 275 \times 10^{-1}% implies x = 275$$
### Pattern Recognition
Carius method calculations depend heavily on standard conversion factors. The constant factor for sulphur gravimetry is $\frac{32}{233}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q42 (2025)
The correct order of stability of following carbocations is :
{{IMG1}}
A
{{IMG2}}
B
{{IMG3}}
C
{{IMG4}}
D
- $\mathrm{A} > \mathrm{B} > \mathrm{C} > \mathrm{D}$
- $\mathrm{B} > \mathrm{C} > \mathrm{A} > \mathrm{D}$
- $\mathrm{C} > \mathrm{B} > \mathrm{A} > \mathrm{D}$
- $\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$
### Core Logic
To evaluate carbocation stability, apply the priority rules: Aromaticity > Resonance > Hyperconjugation.
- **C:** Represents a cyclopropenyl cation derivative which achieves full aromatic stabilization due to its planar cyclic conjugated system satisfying Huckel's rule ($2\pi$ electrons). This makes it the most stable.
- **A:** Stabilized by extended resonance from multiple phenyl groups.
- **B:** Contains fewer phenyl rings participating in active cross-conjugation relative to A.
- **D:** Stabilized solely by simple aliphatic hyperconjugation, making it the least stable.
Visual alignment chart:
{{SOL_IMG1}}
Hence, the correct stability hierarchy is:
$$\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$$
### Pattern Recognition
Sees: Mixed aromatic, benzylic, and aliphatic carbocations.
Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q43 (2025)
The compounds that produce $\mathrm{CO}_{2}$ with aqueous $\mathrm{NaHCO}_{3}$ solution are:
A. {{IMG1}}
B. {{IMG2}}
C. {{IMG3}}
D. {{IMG4}}
E. {{IMG5}}
Choose the correct answer from the options given below:
- $\text{A and C only}$
- $\text{A, B and E only}$
- $\text{A, C and D only}$
- $\text{A and B only}$
### Core Logic
Organic compounds react with sodium bicarbonate ($\mathrm{NaHCO}_3$) to liberate $\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid ($\mathrm{H}_2\mathrm{CO}_3$).
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 $\mathrm{H}_2\mathrm{CO}_3$.
- **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 $\mathrm{CO}_2$.
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 $\mathrm{CO}_2$ from bicarbonate ions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q32 (2025)
Identify the correct statements from the following:
A. Pentan-3-one and pentan-2-one are metamers
B. Propyl cyanide and propyl isocyanide are functional isomers
C. Ethanol and methanol are position isomers
D. Methylamine and ethylamine are homologous
Choose the correct answer from the options given below:
- C & D only
- B & C only
- A & B only
- A, B & C only
### Related Formula
Metamers: Different alkyl chains attached to the same polyvalent functional group.
Functional Isomers: Same molecular formula, different functional groups.
### Core Logic
Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached to the carbonyl carbon (diethyl vs methyl-propyl), so they are metamers. (Correct)
Statement B: Cyanides ($-\mathrm{CN}$) and isocyanides ($-\mathrm{NC}$) have different functional groups, making them functional isomers. (Correct)
Statement C: Ethanol and methanol differ by a $-\mathrm{CH_2}-$ unit; they are homologues, not position isomers. (Incorrect)
Statement D: Methylamine and ethylamine are homologues, but among the available options, option 3 (A & B only) represents the correct combination.
### Step 1: Final Selection
Statements A and B are unambiguously correct.
### Pattern Recognition
Pentan-2-one vs Pentan-3-one = Metamers.
Cyanide vs Isocyanide = Functional isomers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q44 (2025)
The least acidic compound, among the following is:
(A) $\mathrm{EtO_2C-C \equiv CH}$
(B) $\mathrm{Ph-OH}$
(C) $\mathrm{Ph-SO_3H}$
(D) $\mathrm{Ph-COOH}$
Choose the correct option:
- (A)
- (B)
- (C)
- (D)
### Related Formula
$$\text{Acid Strength} \propto \text{Stability of Conjugate Base}$$
### Core Logic
Order of acidity among functional groups:
1. Benzenesulfonic acid ($\mathrm{Ph-SO_3H}$) $\rightarrow$ Conjugate base stabilized by 3 equivalent resonance structures. (Most acidic)
2. Benzoic acid ($\mathrm{Ph-COOH}$) $\rightarrow$ Conjugate base stabilized by 2 equivalent carboxylate resonance structures.
3. Phenol ($\mathrm{Ph-OH}$) $\rightarrow$ Phenoxide ion resonance-stabilized over the aromatic ring.
4. Terminal alkyne ($\mathrm{EtO_2C-C \equiv CH}$) $\rightarrow$ Proton lost from $sp$-hybridized carbon. Though electron-withdrawing ester group increases $C-H$ acidity, terminal alkynes are significantly less acidic than phenols, carboxylic acids, and sulfonic acids ($pK_a \approx 18\text{--}20$ vs $pKa \approx 10$ for phenol).
### Step 1: Final Conclusion
Compound (A) is the least acidic among the given choices.
### Pattern Recognition
Acidity order: $-\mathrm{SO_3H} > -\mathrm{COOH} > -\mathrm{OH} (phenol) > -\mathrm{C \equiv CH} (terminal alkyne)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
