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 2.
Q469 (2025)
Consider the following molecules :
{{IMG1}}
The correct order of rate of hydrolysis is :
- $(1)\ r > q > p > s$
- $(2)\ q > p > r > s$
- $(3)\ p > r > q > s$
- $(4)\ p > q > r > s$
### Related Formula
The relative rate of nucleophilic acyl substitution follows the leaving group ability:
$$\text{Rate of Hydrolysis} \propto \text{Leaving Group Ability} \propto \frac{1}{\text{Basic Strength of Leaving Group}}$$
{{SOL_IMG1}}
### Core Logic
Let's analyze the leaving groups across all choices layout-by-row:
* For **(p)**, the leaving group is $\mathrm{Cl^-}$ (Very weak base, excellent leaving group).
* For **(q)**, the leaving group is $\mathrm{RCOO^-}$ (Resonance stabilized carboxylate, good leaving group).
* For **(r)**, the leaving group is $\mathrm{RO^-}$ (Alkoxide, strong base, poor leaving group).
* For **(s)**, the leaving group is $\mathrm{NH_2^-}$ (Extremely strong base, exceptionally poor leaving group due to nitrogen lone pair resonance into the carbonyl).
This structural comparison yields the final sequence: $\mathrm{p > q > r > s}$.
### Pattern Recognition
Acyl chlorides (p) are always the most reactive acid derivatives, while amides (s) are consistently the least reactive due to strong amide resonance stabilizing the carbonyl group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q42 (2025)
On complete combustion $1.0\mathrm{~g}$ of an organic compound (X) gave $1.46\mathrm{~g}$ of $\mathrm{CO}_{2}$ and $0.567\mathrm{~g}$ of $\mathrm{H}_{2}\mathrm{O}$. The empirical formula mass of compound (X) is ________ g.
Given molar mass in $\mathrm{g \cdot mol^{-1}\ C:12,\ H:1,\ O:16}$
- $(1)\ 30$
- $(2)\ 45$
- $(3)\ 60$
- $(4)\ 15$
### Related Formula
Elemental content calculation system equations:
$$\text{Moles of C} = \frac{\text{Mass of } \mathrm{CO_2}}{44}$$
$$\text{Moles of H} = 2 \times \frac{\text{Mass of } \mathrm{H_2O}}{18}$$
### Core Logic
Let's perform the stoichiometry layout step-by-step:
* Moles of $\mathrm{C}$ inside sample system:
$$\mathrm{n_C} = \frac{1.46}{44} = 0.033\mathrm{~mol}$$
$$\text{Mass of C} = 0.033 \times 12 = 0.396\mathrm{~g}$$
* Moles of $\mathrm{H}$ inside sample system:
$$\mathrm{n_H} = 2 \times \frac{0.567}{18} = 0.063\mathrm{~mol}$$
$$\text{Mass of H} = 0.063 \times 1 = 0.063\mathrm{~g}$$
* Determine Oxygen mass by subtracting values from total starting mass:
$$\text{Mass of O} = 1.0 - (0.396 + 0.063) = 0.541\mathrm{~g}$$
$$\mathrm{n_O} = \frac{0.541}{16} = 0.033\mathrm{~mol}$$
* Find atomic whole-number ratio profile: $\mathrm{C : H : O} = 0.033 : 0.063 : 0.033 \approx 1 : 2 : 1$.
* This gives an empirical configuration of $\mathrm{CH_2O}$.
### Step 1: Evaluation
Calculating formula mass:
$$\text{Empirical Mass} = 12 + (2 \times 1) + 16 = 30\mathrm{~g}$$
### Pattern Recognition
When calculated mole properties output identical numbers for two elements ($0.033$ for both C and O), their structural subscript ratio is exactly $1:1$. This pattern significantly speeds up empirical calculations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q194 (2025)
What is the correct IUPAC name of the compound given below? {{IMG1}}
- 3-Bromo-2-hydroxy-5-nitrobenzoic acid
- 3-Bromo-4-hydroxy-1-nitrobenzoic acid
- 2-Hydroxy-3-bromo-5-nitrobenzoic acid
- 5-Nitro-3-bromo-2-hydroxybenzoic acid
### Related Formula
According to IUPAC rules for nomenclature of aromatic compounds:
- Principal functional group has highest priority:
$$-\mathrm{COOH} > -\mathrm{OH}$$
- The principal functional group carbon is designated as Carbon-1, and numbering is directed to give substituents the lowest possible locants.
### Core Logic
Assign priority and number the ring:
- Carbon-1: $-\mathrm{COOH}$ (Carboxyl carbon, parent name 'benzoic acid')
- Carbon-2: $-\mathrm{OH}$ (Hydroxyl substituent)
- Carbon-3: $-\mathrm{Br}$ (Bromo substituent)
- Carbon-5: $-\mathrm{NO}_2$ (Nitro substituent)
This numbering yields substituent locants at positions 2, 3, and 5.
### Step 1: Arrange alphabetically
List the substituents alphabetically with locants:
- 3-Bromo
- 2-Hydroxy
- 5-Nitro
Combining these names:
$$\text{3-Bromo-2-hydroxy-5-nitrobenzoic acid}$$
This matches Option (1).
### Pattern Recognition
Carboxylic acid always dictates position 1 in ring numbering over alcohol. Numbering clockwise gives 2-hydroxy, 3-bromo, and 5-nitro, whereas counterclockwise numbering would yield much higher locants (2-nitro, 4-bromo, 5-hydroxy) which violates the lowest-locant rule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q196 (2025)
$X\mathrm{~g}$ of nitrobenzene on nitration gave $4.2\mathrm{~g}$ of m-dinitrobenzene. The value of $X$ is ________ $\mathrm{g}$. (nearest integer)
[Given: molar mass (in $\mathrm{g~mol}^{-1}$ ) C: 12, H: 1, O: 16, N: 14]
### Related Formula
Balanced reaction for nitration of nitrobenzene:
$$\mathrm{C_6H_5NO_2} + \mathrm{HNO_3} \rightarrow \mathrm{C_6H_4(NO_2)_2} + \mathrm{H_2O}$$
$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$
### Core Logic
From the balanced stoichiometry:
- 1 mole of nitrobenzene yields 1 mole of m-dinitrobenzene.
### Step 1: Determine molar masses
- Molar mass of Nitrobenzene ($\mathrm{C_6H_5NO_2}$):
$$M_1 = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123\mathrm{~g/mol}$$
- Molar mass of m-Dinitrobenzene ($\mathrm{C_6H_4(NO_2)_2}$):
$$M_2 = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168\mathrm{~g/mol}$$$
{{SOL_IMG1}}
### Step 2: Calculate moles and find X
Moles of m-dinitrobenzene produced:
$$n = \frac{4.2\mathrm{~g}}{168\mathrm{~g/mol}} = 0.025\mathrm{~mol}$$
Since stoichiometry is $1:1$, the moles of nitrobenzene required is also $0.025\mathrm{~mol}$:
$$\text{Mass of nitrobenzene } X = 0.025\mathrm{~mol} \times 123\mathrm{~g/mol} = 3.075\mathrm{~g}$$
Rounding to the nearest integer gives $3$.
### Pattern Recognition
Electrophilic aromatic substitution stoichiometry is straightforward: each aromatic precursor ring converts to exactly one product ring. Finding moles from the heavier substituted product and converting back using the reactant's molecular weight quickly yields the answer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 12 Chemistry: Amines
Q200 (2025)
The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula $\mathrm{C}_9\mathrm{H}_{12}$ is ________.
### Related Formula
Degrees of Unsaturation (Double Bond Equivalents, DBE):
$$\mathrm{DBE} = C + 1 - \frac{H}{2} + \frac{N}{2}$$
For formula $\mathrm{C_9H_{12}}$:
$$\mathrm{DBE} = 9 + 1 - \frac{12}{2} = 4$$
These 4 degrees of unsaturation match a benzene ring exactly (one ring + three double bonds).
### Core Logic
Since the question specifies 'substituted benzene derivatives', we must keep the benzene core ($\mathrm{C_6H_5-}$ or similar) intact. This leaves 3 carbon atoms to be distributed as alkyl substituents.
### Step 1: Categorize by substitution patterns
1. **Mono-substituted benzene** (one propyl group containing 3 carbons):
- n-Propylbenzene: $\mathrm{C_6H_5-CH_2-CH_2-CH_3}$ (Isomer 1)
- Isopropylbenzene (Cumene): $\mathrm{C_6H_5-CH(CH_3)_2}$ (Isomer 2)
2. **Di-substituted benzene** (one ethyl group and one methyl group):
- 1-Ethyl-2-methylbenzene (ortho-ethylmethylbenzene) (Isomer 3)
- 1-Ethyl-3-methylbenzene (meta-ethylmethylbenzene) (Isomer 4)
- 1-Ethyl-4-methylbenzene (para-ethylmethylbenzene) (Isomer 5)
### Step 2: Tri-substituted benzenes
3. **Tri-substituted benzene** (three methyl groups):
- 1,2,3-Trimethylbenzene (Hemimellitene) (Isomer 6)
- 1,2,4-Trimethylbenzene (Pseudocumene) (Isomer 7)
- 1,3,5-Trimethylbenzene (Mesitylene) (Isomer 8)
### Step 3: Total Count
Summing all options:
$$\text{Total structural isomers} = 2 + 3 + 3 = 8$$
### Pattern Recognition
For alkyl benzenes with $N$ extra carbons, systematically group them as single chain substituents down to multiple methyl substituents. This hierarchical sorting prevents duplicates or missing patterns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Hydrocarbons
