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 10.
Q1301 (2025)
Total number of nucleophiles from the following is :-
$$\mathrm {N H} _ {3}, \mathrm {P h S H}, (\mathrm {H} _ {3} \mathrm {C}) _ {2} \mathrm {S}, \mathrm {H} _ {2} \mathrm {C} = \mathrm {C H} _ {2}, \stackrel {\ominus} {\mathrm {O}} \mathrm {H}, \mathrm {H} _ {3} \mathrm {O} ^ {\oplus}, (\mathrm {C H} _ {3}) _ {2} \mathrm {C O}, > = \mathrm {N C H} _ {3}$$
- 5
- 4
- 7
- 6
### Related Formula
Nucleophiles are electron-rich species containing lone pairs of electrons or $\pi$-bonds that can donate an electron pair to an electrophilic center.
### Core Logic
Let us examine each species:
* $\mathrm{NH}_3$: Contains a lone pair on nitrogen $\rightarrow$ Nucleophile
* $\mathrm{PhSH}$: Contains lone pairs on sulfur $\rightarrow$ Nucleophile
* $(\mathrm{H}_3\mathrm{C})_2\mathrm{S}$: Contains lone pairs on sulfur $\rightarrow$ Nucleophile
* $\mathrm{H}_2\mathrm{C}=\mathrm{CH}_2$: Contains a nucleophilic $\pi$-bond $\rightarrow$ Nucleophile
* $\stackrel{\ominus}{\mathrm{O}}\mathrm{H}$: Negatively charged with lone pairs $\rightarrow$ Nucleophile
* $\mathrm{H}_3\mathrm{O}^{\oplus}$: Electron deficient, positively charged oxygen cannot donate electrons $\rightarrow$ Electrophile
* $(\mathrm{CH}_3)_2\mathrm{CO}$: Carbonyl carbon is electrophilic
* $>=\mathrm{NCH}_3$: Imine carbon is electrophilic
Thus, the total number of nucleophiles is 5.
### Pattern Recognition
Neutral molecules with lone pairs (N, S) or alkenes/alkynes with available $\pi$-electrons operate as good nucleophiles, along with full anions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q1305 (2025)
Match List-I with List-II.
Choose the correct answer from the options given below:
{{MAIN_IMG}}
- (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
- (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
### 1. RELATED FORMULA
IUPAC Rules select the longest principal carbon chain and number it to give substituents the lowest possible locants.
### 2. EXECUTION
**CORE LOGIC**
Let's systematic decode each name :
* (A) Longest chain contains 7 carbons (heptane) with an ethyl group at position 3 and a methyl group at position 5 $\rightarrow$ 3-Ethyl-5-methylheptane (II) .
* (B) C expanded yields a 7 carbon main chain with two methyl groups at carbon-4 $\rightarrow$ 4,4-Dimethylheptane (III) .
* (C) 5-carbon diene numbered from the left double bond side $\rightarrow$ 2-Methyl-1,3-pentadiene (IV) .
* (D) 5-carbon alkene starting from the double bond end $\rightarrow$ 4-Methylpent-1-ene (I) .
Therefore, matching sequence: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
### 3. PATTERN RECOGNITION
Expanding compressed groupings such as $\mathrm{(C_3H_7)_2}$ prevents errors regarding parent chain carbon counts.
### 4. EVALUATION RUBRIC / MODEL ANSWER
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### 5. CHAPTER MIX
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q1319 (2025)
The steam volatile compounds among the following are:
Choose the correct answer from the options given below:
- (B) and (D) only
- (A) and (C) only
- (A) and (B) only
- (A), (B) and (C) only
### Related Formula
$$\text{Intramolecular H-bonding} \implies \text{Lower boiling point} \implies \text{Steam Volatile}$$
### Core Logic
Purification via steam distillation requires a high relative vapor pressure at the boiling point of water. Let us assess the isomers :
* (A) o-Nitrophenol contains an $-\mathrm{OH}$ group right next to an $-\mathrm{NO}_2$ group, allowing for strong intramolecular hydrogen bonding. This minimizes external interactions, lowering the boiling point and making it steam volatile .
* (B) o-Nitroaniline similarly stabilizes itself via internal intramolecular hydrogen bonding between the amine and nitro components, making it steam volatile .
* (C) & (D) The para isomers form extensive intermolecular networks with neighboring molecules, which significantly elevates their boiling points and prevents steam volatility .
{{SOL_IMG1}} {{SOL_IMG1}}
Hence, compounds (A) and (B) are steam volatile, matching option (3).
### Pattern Recognition
ortho-substituted functional networks form self-contained internal loops through hydrogen bonding, preventing external pairing and maximizing volatility.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q1323 (2025)
The sum of sigma $(\sigma)$ and $\mathrm{\pi}(\pi)$ bonds in Hex-1,3-dien-5-yne is ________.
### Related Formula
$$\text{Single bond} = 1\sigma, \quad \text{Double bond} = 1\sigma + 1\pi, \quad \text{Triple bond} = 1\sigma + 2\pi$$
### Core Logic
Let us map out the complete structural bond geometry of Hex-1,3-dien-5-yne :
$$\mathrm{H_2C=CH-CH=CH-C\equiv CH}$$
Counting the individual bonds :
* Carbon-Hydrogen ($\mathrm{C-H}$) single bonds $= 2 + 1 + 1 + 1 + 1 = 6 \, \sigma$ bonds
* Carbon-Carbon single, double, and triple linkages:
* $\mathrm{C}_1=\mathrm{C}_2
ightarrow 1\sigma + 1\pi$
* $\mathrm{C}_2-\mathrm{C}_3
ightarrow 1\sigma$
* $\mathrm{C}_3=\mathrm{C}_4
ightarrow 1\sigma + 1\pi$
* $\mathrm{C}_4-\mathrm{C}_5
ightarrow 1\sigma$
* $\mathrm{C}_5\equiv\mathrm{C}_6
ightarrow 1\sigma + 2\pi$
Total values :
$$\text{Total } \sigma \text{ bonds} = 6 + 5 = 11$$
$$\text{Total } \pi \text{ bonds} = 1 + 1 + 2 = 4$$
$$\text{Sum } (\sigma + \pi) = 11 + 4 = 15$$
### Pattern Recognition
Do not skip implicit $\mathrm{C-H}$ single bonds when scanning skeletal formulas, as each counts as a full $\sigma$ bond.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q1396 (2024)
In Kjeldahl's method for estimation of nitrogen, $CuSO_4$ acts as:
- Reducing agent
- Catalytic agent
- Hydrolysis agent
- Oxidising agent
### Core Logic
Kjeldahl's method is used for the quantitative estimation of Nitrogen in organic compounds.
The organic compound is heated with concentrated $H_2SO_4$ to convert nitrogen into ammonium sulfate.
In this digestion step, a catalyst is required to speed up the decomposition and oxidation of the organic matter.
### Step 1: Identify Role of Reagent
$CuSO_4$ (or sometimes Mercury/Selenium) is added to the digestion mixture. It acts solely as a catalytic agent to increase the rate of digestion (breakdown of organic material).
### Pattern Recognition
In Kjeldahl's digestion: $H_2SO_4$ = oxidizing agent / reactant, $K_2SO_4$ = raises boiling point, $CuSO_4$ = catalyst.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
