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 6.
Q46 (2025)
During estimation of nitrogen by Dumas' method of compound X ($0.42\mathrm{~g}$): {{IMG1}}
____ $\mathrm{mL}$ of $\mathrm{N}_2$ gas will be liberated at STP. (nearest integer)
(Given molar mass in $\mathrm{g~mol^{-1}}$ : $\mathrm{C: 12, H: 1, N: 14}$)
### Related Formula
$$\text{Moles of Compound X} = \text{Moles of } \mathrm{N_2}$$
$$V_{\mathrm{N_2}} (\text{at STP}) = n_{\mathrm{N_2}} \times 22400\mathrm{~mL}$$
### Core Logic
From the molecular structure of compound X, its molecular formula is $\mathrm{C_5H_{10}N_2}$.
Molar Mass of X $= (5 \times 12) + (10 \times 1) + (2 \times 14) = 60 + 10 + 28 = 86\mathrm{~g~mol^{-1}}$.
### Step 1: Calculation
Moles of compound X in $0.42\mathrm{~g}$:
$$n_{\text{X}} = \frac{0.42}{86}\mathrm{~mol}$$
By Conservation of Atomic Mass (POAC) on Nitrogen atoms:
$$2 \times n_{\text{X}} = 2 \times n_{\mathrm{N_2}} \implies n_{\mathrm{N_2}} = n_{\text{X}} = \frac{0.42}{86}\mathrm{~mol}$$
Volume of $\mathrm{N_2}$ liberated at STP:
$$V_{\mathrm{N_2}} = \frac{0.42}{86} \times 22400\mathrm{~mL} \approx 109.395\mathrm{~mL}$$
Using $22.7\mathrm{~L}$ standard STP molar volume: $\frac{0.42}{86} \times 22700 \approx 110.88\mathrm{~mL} \approx 111\mathrm{~mL}$.
### Pattern Recognition
Dumas method volume calculation:
$V = \frac{m}{M_{\text{compound}}} \times 22.4\text{ L/mol} \times 1000\text{ mL} = 111\text{ mL}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q47 (2025)
$0.5\mathrm{~g}$ of an organic compound on combustion gave $1.46\mathrm{~g}$ of $\mathrm{CO}_2$ and $0.9\mathrm{~g}$ of $\mathrm{H}_2\mathrm{O}$. The percentage of carbon in the compound is _____. (Nearest integer)
[Given: Molar mass (in $\mathrm{g~mol^{-1}}$) $\mathrm{C: 12, H: 1, O: 16}$]
### Related Formula
$$\%\mathrm{C} = \frac{12}{44} \times \frac{\text{Mass of } \mathrm{CO_2}}{\text{Mass of Organic Compound}} \times 100$$
### Core Logic
Mass of $\mathrm{CO_2} = 1.46\mathrm{~g}$.
Mass of organic compound $= 0.5\mathrm{~g}$.
### Step 1: Calculation
$$\%\mathrm{C} = \frac{12}{44} \times \frac{1.46}{0.5} \times 100$$
$$\%\mathrm{C} = \frac{12 \times 1.46 \times 2}{44} \times 100 = \frac{35.04}{44} \times 100 = 79.636\%$$
Rounding off to the nearest integer gives $80\%$.
### Pattern Recognition
Combustion formula directly yields $\%C = 79.63\% \approx 80\%$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q851 (2025)
The correct order of basicity for the following molecules is:
- $P > Q > R$
- $R > P > Q$
- $Q > P > R$
- $R > Q > P$
### Related Formula
$$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen}$$
### Core Logic
Analyzing the molecules:
- In molecule **(R)**, according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic.
- In molecule **(Q)**, the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability.
- In molecule **(P)**, the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available.
Therefore, the correct basicity order is:
$R > Q > P$
### Step 1: Final Identification
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct.
### Pattern Recognition
Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q855 (2025)
In which pairs, the first ion is more stable than the second?
- (B) & (D) only
- (A) & (B) only
- (B) & (C) only
- (A) & (C) only
### Related Formula
$$\text{Stability} \propto \text{Delocalization of charge via resonance, mesomeric, and back-bonding effects}$$
### Core Logic
Evaluating each specific structural pair:
- **Pair (A):** The first carbocation is stabilized by strong $+M$ back-bonding from the methoxy ($-OMe$) oxygen lone pair, making it significantly more stable than the second.
- **Pair (B):** The first carbanion is strongly stabilized by the $-M$ and $-I$ electronic effects of the nitro ($-NO_2$) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable.
- **Pair (C):** The second cation has extended allylic resonance stabilization, meaning the first is less stable.
- **Pair (D):** Cation with $-OMe$ backbonding is more stable than tertiary aliphatic carbocation, making the first less stable than the second.
Thus, only in **(A) & (B)** is the first ion more stable than the second.
### Pattern Recognition
Back-bonding from an adjacent oxygen lone pair always triumphs over standard inductive or hyperconjugative alkyl stability templates. For carbanions, ensure electron-withdrawing groups like $-NO_2$ match the sign of the charge.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q868 (2025)
Consider the following molecule (X).
{{IMG1}}
The structure of X is
- Structure (1)
- Structure (2)
- Structure (3)
- Structure (4)
### Core Logic
Analyzing the mechanism:
1. Protonation ($H^+$ attack) on the double bond occurs to generate the most stable intermediate carbocation.
2. The system forms a highly stable **tertiary ($3^\circ$) carbocation** at the bridge junction ring site.
3. Nucleophilic attack by bromide ion ($Br^-$) captures this bridge junction center selectively, yielding the major brominated structure designated as Structure (2).
### Step 1: Mechanism Breakdown
{{SOL_IMG1}}
{{SOL_IMG2}}
Electrophilic addition follows Markovnikov-like stability rules, directing the halide specifically to the bridgehead junction carbocation position.
### Pattern Recognition
Whenever adding hydrohalic acids across cyclic non-conjugated dienes or isolated double bonds, always convert the alkene into the most stable, un-strained tertiary carbocation before attaching the nucleophile.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Class 11 Chemistry: Hydrocarbons
