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 3.
Q36 (2025)
In Dumas' method for estimation of nitrogen $0.4\mathrm{~g}$ of an organic compound gave $60\mathrm{~mL}$ of nitrogen collected at $300\mathrm{~K}$ temperature and $715\mathrm{~mm~Hg}$ pressure. The percentage composition of nitrogen in the compound is :
(Given: Aqueous tension at $300\mathrm{~K} = 15\mathrm{~mm~Hg}$)
- 15.71%
- 20.95%
- 17.46%
- 7.85%
### Related Formula
Pressure of dry nitrogen gas:
$$P_{\mathrm{N}_2} = P_{\text{total}} - \text{Aqueous tension}$$
Using Ideal Gas Law:
$$n_{\mathrm{N}_2} = \frac{P_{\mathrm{N}_2} V}{R T}$$
$$\%\mathrm{N} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Given parameters:
- Mass of compound $m = 0.4\mathrm{~g}$
- Volume of nitrogen $V = 60\mathrm{~mL} = 0.060\mathrm{~L}$
- Total pressure $P_{\text{total}} = 715\mathrm{~mm~Hg}$
- Temperature $T = 300\mathrm{~K}$
- Aqueous tension $= 15\mathrm{~mm~Hg}$
### Step 1: Calculate dry nitrogen pressure
$$P_{\mathrm{N}_2} = 715\mathrm{~mm~Hg} - 15\mathrm{~mm~Hg} = 700\mathrm{~mm~Hg}$$
$$P_{\mathrm{N}_2} = \frac{700}{760}\mathrm{~atm} \approx 0.921\mathrm{~atm}$$$
### Step 2: Calculate moles of nitrogen gas
Using $R = 0.0821\mathrm{~L\cdot atm\cdot K^{-1}\cdot mol^{-1}}$:
$$n_{\mathrm{N}_2} = \frac{\left(\frac{700}{760}\right) \times 0.060}{0.0821 \times 300} = \frac{0.05526}{24.63} \approx 2.2436 \times 10^{-3}\mathrm{~mol}$$
Mass of $\mathrm{N}_2$ gas:
$$\text{Mass} = 2.2436 \times 10^{-3} \times 28\mathrm{~g} \approx 0.06282\mathrm{~g}$$$
### Step 3: Calculate percentage of Nitrogen
$$\%\mathrm{N} = \frac{0.06282\mathrm{~g}}{0.4\mathrm{~g}} \times 100 \approx 15.71\%$$
This matches Option (1).
### Pattern Recognition
In Dumas' method calculations, always subtract the aqueous tension to obtain the pressure of dry nitrogen gas. Do not use the raw moist gas pressure, as doing so will overestimate the nitrogen content.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q40 (2025)
Given below are two statements:
Statement I: Hyperconjugation is not a permanent effect.
Statement II: In general, greater the number of alkyl groups attached to a positively charged C-atom, greater is the hyperconjugation interaction and stabilization of the cation.
In the light of the above statements, choose the correct answer from the options given below :
- Statement I is true but Statement II is false
- Both Statement I and Statement II are false
- Statement I is false but Statement II is true
- Both Statement I and Statement II are true
### Related Formula
The number of hyperconjugation structures is directly related to the count of $\alpha$-hydrogen atoms:
$$\text{Number of hyperconjugative structures} = \text{Number of } \alpha\text{-hydrogens}$$
### Core Logic
Statement I Analysis:
- Hyperconjugation (no-bond resonance) involves the delocalization of $\sigma$ electrons of $\mathrm{C-H}$ bonds of an alkyl group directly attached to an atom of unsaturated system or a positively charged carbon atom. This is a permanent ground-state electronic effect, not dependent on external reagents. Thus, Statement I is False.
### Step 1: Analyze Statement II
- Statement II states that more alkyl groups attached to a carbocation center increase hyperconjugative stabilization. Each alkyl group brings additional $\sigma_{\mathrm{C-H}}$ bonds adjacent to the empty p-orbital, increasing the total count of $\alpha$-hydrogens and enhancing charge delocalization. Thus, Statement II is True.
### Step 2: Conclusion
Therefore, Statement I is False but Statement II is True, matching Option (3).
### Pattern Recognition
Permanent organic effects include: Inductive, Mesomeric (Resonance), and Hyperconjugation effects. Temporary electronic effects include: Electromeric and Inductomeric effects (which require an attacking reagent to manifest).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q33 (2025)
Which of the following is the correct IUPAC name of given organic compound (X)?
{{IMG1}}
- $\text{2-Bromo-2-methylbut-2-ene}$
- $\text{3-Bromo-3-methylprop-2-ene}$
- $\text{1-Bromo-2-methylbut-2-ene}$
- $\text{4-Bromo-3-methylbut-2-ene}$
### Core Logic
To determine the IUPAC name of the compound shown in {{IMG1}}:
1. Identify the principal functional group, which is the double bond (alkene).
2. Find the longest carbon chain containing the double bond:
$$\mathrm{C1(H_2Br) - C2(CH_3) = C3(H) - C4(H_3)}$$
The longest chain has $4$ carbons, which means the parent alkane is butane, and with a double bond it's "but-2-ene".
3. Number the chain from the end that gives lower locants to the double bond. Starting from left or right both give the double bond at position $2$. However, starting from left gives substituent locants as $1$ (for bromo) and $2$ (for methyl), whereas starting from right gives substituent locants as $3$ and $4$.
4. Hence, correct numbering is:
- $\text{C1}$: bonded to Bromine ($-\text{Br}$)
- $\text{C2}$: bonded to Methyl ($-\text{CH}_3$)
- $\text{C3}$: alkene carbon
- $\text{C4}$: terminal methyl group
{{SOL_IMG1}}
Combining these rules, the name is: **1-Bromo-2-methylbut-2-ene**.
### Pattern Recognition
Double bond takes precedence over halogen substituent in numbering direction. If double bond is symmetrical (at position $2$ in a $4$-carbon chain), use the substituent positions to break the tie, choosing lowest possible locants ($1$ and $2$ vs $3$ and $4$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Haloalkanes and Haloarenes
Q46 (2025)
An organic compound weighing $500\mathrm{\ mg}$, produced $220\mathrm{\ mg}$ of $\mathrm{CO}_2$ on complete combustion. The percentage composition of carbon in the compound is ______ %. (nearest integer)
(Given molar mass in $\mathrm{g\ mol}^{-1}$ of $\mathrm{C}: 12$, $\mathrm{O}: 16$)
### Related Formula
$$\% \mathrm{C} = \frac{12}{44} \times \frac{\text{Mass of } \mathrm{CO}_2 \text{ produced}}{\text{Mass of organic compound taken}} \times 100$$
### Core Logic
Given:
- Mass of organic compound taken $= 500 \text{ mg} = 500 \times 10^{-3} \text{ g}$
- Mass of $\mathrm{CO}_2$ produced $= 220 \text{ mg} = 220 \times 10^{-3} \text{ g}$
Using the formula:
$$\% \mathrm{C} = \frac{12}{44} \times \frac{220 \times 10^{-3}}{500 \times 10^{-3}} \times 100$$
$$\% \mathrm{C} = \frac{12}{44} \times \frac{220}{500} \times 100$$
$$\% \mathrm{C} = \frac{12}{44} \times 44 = 12 \%$$
Thus, the percentage of carbon is $12$.
### Pattern Recognition
Carbon dioxide has exactly $12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of $\mathrm{CO}_2$ ($220/500 = 0.44$) by $12/44$ to directly get $0.12$ or $12\%$.
### Evaluation Rubric / Model Answer
A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise $12$ percent carbon composition.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q585 (2025)
What is the correct IUPAC name of the following organic compound?
- 4-Ethyl-1-hydroxycyclopent-2-ene
- 1-Ethyl-3-hydroxycyclopent-2-ene
- 1-Ethylcyclopent-2-en-3-ol
- 4-Ethylcyclopent-2-en-1-ol
### Core Logic
Let us apply official IUPAC priority indexing rules:
1. **Principal Functional Group**: The hydroxyl group ($-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the $-\text{OH}$ group is assigned position **C-1**.
2. **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**.
3. **Locating Side Chains**: Proceeding with this direction puts the ethyl group at position **C-4**. {{SOL_IMG1}}
Assembling the structural parts alphabetically:
* Substituent: `4-Ethyl`
* Parent root: `cyclopent-2-en`
* Suffix: `1-ol`
Combined IUPAC format: **4-Ethylcyclopent-2-en-1-ol**.
### Pattern Recognition
Principal suffix priority hierarchy: $-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
