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 9.
Q50 (2025)
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ____ $\times 10^{-1}$ % (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is 80 g mol $^{-1}$)
### Related Formula
$$\% \text{ Bromine} = \frac{\text{Molar Mass of Br}}{\text{Molar Mass of AgBr}} \cdot \frac{\text{Mass of AgBr formed}}{\text{Mass of Organic Sample}} \cdot 100$$
### Core Logic
1. Calculate the molar mass of silver bromide ($\mathrm{AgBr}$):
$$\text{Molar Mass of AgBr} = 108 + 80 = 188\text{ g/mol}$$
2. Substitute the given values into the Carius quantitative formula:
* $\text{Mass of AgBr} = 0.15\text{ g}$
* $\text{Mass of sample} = 0.25\text{ g}$
$$\% \mathrm{Br} = \frac{80}{188} \cdot \frac{0.15}{0.25} \cdot 100$$
$$\% \mathrm{Br} = \frac{80}{188} \cdot 0.6 \cdot 100 = \frac{48}{188} \cdot 100 = 25.5319\%$$
3. Convert the percentage to match the requested output units ($\times 10^{-1}\%$):
$$25.5319\% = 255.319 \cdot 10^{-1}\% \approx 255 \cdot 10^{-1}\%$$
Rounding to the nearest integer gives 255.
### Pattern Recognition
Pay close attention to the final multiplier units requested in the blank ($\times 10^{-1}\%$). Always calculate the raw percentage first, then adjust the decimal place to match the required format.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q1154 (2025)
Which one of the carbocations from the following is most stable?
### Core Logic
Carbocations are stabilized by structural factors such as the inductive effect ($+I$), mesomeric effect ($+M$), and hyperconjugation.
In Structure (2), the carbocation center is situated directly adjacent to a strong electronic donor methoxy group ($-OCH_3$). This configuration allows highly effective lone pair donation into the vacant p-orbital of the carbocation via the structural $+M$ mesomeric path, rendering it exceptionally stable.
### Pattern Recognition
An adjacent heteroatom with a lone pair ($O, N$) triggers dynamic back-bonding stability ($+M$), which fundamentally outweighs basic hyperconjugation trends.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q34 (2025)
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 $\mathrm{H}_2\mathrm{SO}_4$
In the light of the above statements, choose the correct answer from the options given below
- Both Statement I and Statement II are true.
- Statement I is false but Statement II is true
- Both Statement I and Statement II are false.
- Statement I is true but Statement II is false
### 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 $\mathrm{H}_2\mathrm{SO}_4$ 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 ($N_2$).
### Pattern Recognition
Dumas method collects element gas $N_2$ via volumetric analysis; Kjeldahl maps digestions via standard $(\mathrm{NH}_4)_2\mathrm{SO}_4$ pathways.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q31 (2025)
The purification method based on the following physical transformation is :
$$\text{Solid} \xrightarrow[(\text{X})]{\text{Heat}} \text{Vapour} \xrightarrow[(\text{X})]{\text{Cool}} \text{Solid}$$
- Sublimation
- Distillation
- Crystallization
- Extraction
### Related Formula
Direct phase transition without passing through an intermediate liquid phase defines sublimation:
$$\text{Solid} \rightleftharpoons \text{Vapour}$$
### Core Logic
The schematic diagram represents a solid turning directly into vapor on heating, which then reverts back to a solid phase upon cooling. This distinct behavior isolates sublimable solids from non-sublimable impurities.
### Step 1: Identification
This transformation perfectly defines the laboratory purification process known as **Sublimation**.
### Pattern Recognition
Look for the skipping of the liquid state entirely: Solid $\rightarrow$ Vapour $\rightarrow$ Solid. Common examples include camphor, naphthalene, benzoic acid, and iodine.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q44 (2025)
Given below are two statements:
Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds.
Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
### Related Formula
Isomers share an identical molecular formula but differ in structural arrangement or functional groups:
$$\text{Same } M_f \neq \text{Same structural layout}$$
### Core Logic
Evaluating the structural parameters:
- **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula $C_3H_6O$. They are functional/ring-chain isomers, so Statement I is true.
- **Statement II**: Propan-1-amine ($1^\circ$ amine) and N-methylethanamine ($2^\circ$ amine) both share the molecular formula $C_3H_9N$. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true.
### Step 1: Conclusion Match
Since both structural statements are valid, both Statement I and Statement II are true.
{{SOL_IMG1}}
### Pattern Recognition
Always remember that $1^\circ$, $2^\circ$, and $3^\circ$ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
