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 4.
Q27 (2025)
Match the LIST-I with LIST-II:
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">LIST-I</th><th style="border: 1px solid #888; padding: 8px;">LIST-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Carbocation</td><td style="border: 1px solid #888; padding: 8px;">I. Species that can supply a pair of electrons.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. C-Free radical</td><td style="border: 1px solid #888; padding: 8px;">II. Species that can receive a pair of electrons.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Nucleophile</td><td style="border: 1px solid #888; padding: 8px;">III. $sp^2$ hybridized carbon with empty p-orbital.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Electrophile</td><td style="border: 1px solid #888; padding: 8px;">IV. $sp^2/sp^3$ hybridized carbon with one unpaired electron.</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-IV, B-II, C-III, D-I}$
- $\text{A-II, B-III, C-I, D-IV}$
- $\text{A-III, B-IV, C-II, D-I}$
- $\text{A-III, B-IV, C-I, D-II}$
### Core Logic
Let us analyze each term carefully:
* **A. Carbocation**: Features a positively charged trivalent carbon atom. It represents an $sp^2$ hybridized carbon with an empty unhybridized p-orbital. {{SOL_IMG1}}
* **B. Carbon Free Radical**: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits $sp^2$ or $sp^3$ hybridization depending on structural environments. {{SOL_IMG2}}
* **C. Nucleophile**: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons.
* **D. Electrophile**: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons.
### Step 1: Alignment Matrix
Matching each item yields:
* $\text{A} \rightarrow \text{III}$
* $\text{B} \rightarrow \text{IV}$
* $\text{C} \rightarrow \text{I}$
* $\text{D} \rightarrow \text{II}$
This sequence aligns flawlessly with Option (4).
### Pattern Recognition
Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q38 (2025)
On complete combustion, $0.210 \text{ g}$ of an organic compound containing C, H, and O yielded $0.127 \text{ g}$ of $\text{H}_2\text{O}$ and $0.307 \text{ g}$ of $\text{CO}_2$. The mass percentages of hydrogen and oxygen in the given organic compound respectively are:
- 53.41, 39.6
- 6.72, 53.41
- 7.55, 43.85
- 6.72, 39.87
### Related Formula
Percentage of Hydrogen in organic analysis:
$$\%\text{H} = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$$
Percentage of Carbon:
$$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$$
Percentage of Oxygen:
$$\%\text{O} = 100 - (\%\text{C} + \%\text{H})$$
### Execution
Step 1: Compute the mass percent of Hydrogen:
$$\%\text{H} = \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = \frac{0.254}{3.78} \approx 6.72\%$$
Step 2: Compute the mass percent of Carbon:
$$\%\text{C} = \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = \frac{3.684}{9.24} \approx 39.87\%$$
Step 3: Deduce the remaining mass percent of Oxygen:
$$\%\text{O} = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$$
Thus, the values of hydrogen and oxygen percentage are $6.72\%$ and $53.41\%$, matches with Option (2).
### Pattern Recognition
Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q42 (2025)
Match the reagents in LIST-I with the corresponding chemical functional groups they detect in LIST-II:
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">LIST-I (Reagent)</th><th style="border: 1px solid #888; padding: 8px;">LIST-II (Functional Group detected)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Sodium bicarbonate solution</td><td style="border: 1px solid #888; padding: 8px;">I. double bond / unsaturation</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Neutral ferric chloride</td><td style="border: 1px solid #888; padding: 8px;">II. carboxylic acid</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Ceric ammonium nitrate</td><td style="border: 1px solid #888; padding: 8px;">III. phenolic - OH</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Alkaline $\text{KMnO}_4$</td><td style="border: 1px solid #888; padding: 8px;">IV. alcoholic - OH</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-II, B-III, C-IV, D-I}$
- $\text{A-II, B-III, C-I, D-IV}$
- $\text{A-III, B-II, C-IV, D-I}$
- $\text{A-II, B-IV, C-III, D-I}$
### Core Logic
Let us review the chemical basis for each qualitative test:
* **A. Sodium bicarbonate ($\text{NaHCO}_3$) solution**: Carboxylic acids are sufficiently acidic to decompose $\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, $\text{A} \rightarrow \text{II}$.
* **B. Neutral ferric chloride ($\text{FeCl}_3$)**: Phenols react with neutral $\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, $\text{B} \rightarrow \text{III}$.
* **C. Ceric ammonium nitrate (CAN)**: Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, $\text{C} \rightarrow \text{IV}$.
* **D. Alkaline $\text{KMnO}_4$ (Baeyer's Reagent)**: Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown $\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, $\text{D} \rightarrow \text{I}$.
### Step 1: Assembly
Combining the validated relationships gives:
$$\text{A-II, B-III, C-IV, D-I}$$
This maps perfectly to Option (1).
### Pattern Recognition
Baeyer's test (alkaline $\text{KMnO}_4$) always tests for alkenes/alkynes. $\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q28 (2025)
Given below are two statements:
Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support.
Statement (II): In paper chromatography, the material of paper acts as a stationary phase.
In the light of the above statements, choose the correct answer from the options given below:
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
- Statement I is false but Statement II is true
### Core Logic
Statement I is true: In partition chromatography, the stationary phase is indeed a thin film of liquid held on the surface of an inert solid support.
Statement II is false: In paper chromatography, the water molecules trapped inside the cellulose network of the paper act as the stationary phase, not the paper material itself.
### Pattern Recognition
Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q38 (2025)
Total number of sigma $(sigma)$ and $pi(pi)$ bonds respectively present in hex-1-en-4-yne are:
- 13 and 3
- 11 and 3
- 3 and 13
- 14 and 3
### Core Logic
The structural formula of hex-1-en-4-yne is given by:
$$CH_2 = CH - CH_2 - C equiv C - CH_3$$
Let's count the chemical bonds chronologically:
* Number of $C-H$ $sigma$ bonds = $2 + 1 + 2 + 3 = 8$
* Number of $C-C$ $sigma$ bonds = $5$
Total $sigma$ bonds = $8 + 5 = 13$.
{{SOL_IMG38}}
* Number of $pi$ bonds: $1$ from double bond + $2$ from triple bond = $3$ $pi$ bonds.
### Pattern Recognition
Every single bond is $1sigma$, every double bond contains $1sigma + 1pi$, and every triple bond contains $1sigma + 2pi$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
