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 8.
Q47 (2025)
In Dumas' method $292\text{ mg}$ of an organic compound released $50\text{ mL}$ of nitrogen gas $(\text{N}_2)$ at $300\text{ K}$ temperature and $715\text{ mm Hg}$ pressure. [cite: 435, 451] The percentage composition of 'N' in the organic compound is $\dots$ % (Nearest integer) [cite: 452, 455]
(Aqueous tension at $300\text{ K} = 15\text{ mm Hg}$)
### Related Formula
$$P_{\text{dry } \text{N}_2} = P_{\text{total}} - \text{Aqueous Tension} $$
$$PV = nRT \implies n = \frac{PV}{RT} $$
$$\%\text{N} = \frac{\text{Mass of Nitrogen}}{\text{Mass of organic compound}} \times 100 $$
### Core Logic
First, isolate the pressure contribution of the dry nitrogen gas:
$$P_{\text{dry } \text{N}2} = 715 - 15 = 700\text{ mm Hg} = \frac{700}{760}\text{ atm} $$
Using the ideal gas parameters:
- $V = 50\text{ mL} = 0.050\text{ L}$ [cite: 1073, 1076]
- $T = 300\text{ K}$
- $R = 0.0821\text{ L atm mol}^{-1}\text{K}^{-1}$
### Step 1: Compute Moles and Mass
Calculate total moles of $\text{N}_2$ molecules collected:
$$n{\text{N}2} = \frac{\left(\frac{700}{760}
ight) \times 0.050}{0.0821 \times 300} \approx 1.868 \times 10^{-3}\text{ mol} $$
Compute corresponding mass of atomic Nitrogen elements ($2 \times 14 = 28\text{ g/mol}$): [cite: 1016, 1017]
$$\text{Mass of N} = n{\text{N}2} \times 28 = 1.868 \times 10^{-3} \times 28 \approx 0.0523\text{ g} = 52.3\text{ mg}$$
### Step 2: Calculate Percentage Composition
Applying the fraction formulation against total sample mass: [cite: 1021, 1022]
$$\%\text{N} = \frac{52.3\text{ mg}}{292\text{ mg}} \times 100 \approx 17.91\% \approx 18\% $$
### Pattern Recognition
Dumas analysis safety check: Always strip away the vapor pressure of water (aqueous tension) from the measured barometric value before computing the chemical molar counts. For standard conditions shortcuts, remember that $1\text{ mole} = 22400\text{ mL}$ at STP can act as an alternate path if values are normalized.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q30 (2025)
In the given structure, number of sp and $sp^{2}$ hybridized carbon atoms present respectively are :
{{IMG1}}
- \text{3 and 6}
- \text{3 and 5}
- \text{4 and 6}
- \text{4 and 5}
### Core Logic
To determine carbon atom hybridization within skeletal networks, evaluate the count of steric components ($\sigma$-bonds attached to each carbon):
* $4 \text{ } \sigma\text{-bonds} \rightarrow sp^3$
* $3 \text{ } \sigma\text{-bonds} \rightarrow sp^2$ (typically carbons forming one double bond like $\mathrm{C=O}$ or $\mathrm{C=C}$)
* $2 \text{ } \sigma\text{-bonds} \rightarrow sp$ (typically carbons forming a triple bond like $-\mathrm{C\equiv C}-$ or $-\mathrm{C\equiv N}$)
### Step 1: Specific Atom Assignment
Let's perform an audit across the skeletal sequence:
{{SOL_IMG1}}
* $sp^2$ Carbons:
1. Carbonyl carbon ($\mathrm{C=O}$)
2. Carbons sharing the first alkene motif (2 atoms)
3. Carbons sharing the second alkene motif (2 atoms)
Total $sp^2$ carbons = $1 + 2 + 2 = 5$
* $sp$ Carbons:
1. Carbons bound in the central alkyne group ($-\mathrm{C\equiv C}-$) (2 atoms)
2. Terminal nitrile carbon ($-\mathrm{C\equiv N}$) (1 atom)
Total $sp$ carbons = $2 + 1 = 3$
### Pattern Recognition
Count all carbons involved in triple bonds (alkynes, nitriles) to find $sp$ items. Count all double-bonded carbons (alkenes, ketones) to quickly isolate the $sp^2$ population.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q39 (2025)
Identify the correct statement/s from the following options:
(A) $-\mathrm{OCH}_{3}$ and $-\mathrm{NHCOCH}_{3}$ are activating groups
(B) $-\mathrm{CN}$ and $-\mathrm{OH}$ are meta directing groups
(C) $-\mathrm{CN}$ and $-\mathrm{SO}_{3}\mathrm{H}$ are meta directing groups
(D) Activating groups act as ortho- and para- directing groups
(E) Halides are activating groups
Choose the correct answer from the options given below :
- \text{(A), (C) and (D) only}
- \text{(A), (B) and (E) only}
- \text{(A) only}
- \text{(A) and (C) only}
### Core Logic
Let's evaluate each statement based on electrophilic aromatic substitution guidelines:
* Statement (A): Both $-\mathrm{OCH}_{3}$ and $-\mathrm{NHCOCH}_{3}$ have lone pairs on the atom directly attached to the benzene ring. These lone pairs undergo resonance delocalization into the ring, increasing electron density and activating it toward substitution. This statement is true.
* Statement (B): While $-\mathrm{CN}$ is a deactivating meta-directing group, $-\mathrm{OH}$ is a strongly activating ortho/para-directing group due to resonance. This statement is false.
* Statement (C): Both $-\mathrm{CN}$ and $-\mathrm{SO}_{3}\mathrm{H}$ withdraw electron density via inductive and resonance effects ($-I, -M$). This deactivates the ring and directs substitution to the meta position. This statement is true.
* Statement (D): Activating groups increase electron density primarily at the ortho and para positions via resonance, directing incoming electrophiles to those sites. This statement is true.
* Statement (E): Halogens are an exception: they are deactivating due to their strong inductive effect ($-I$), but ortho/para-directing due to resonance ($+M$). This statement is false.
Therefore, statements (A), (C), and (D) are correct, which matches Option (1).
### Pattern Recognition
Groups that activate the aromatic ring by donating electrons via resonance always direct incoming electrophiles to the ortho and para positions. Halogens are a unique exception: they deactivate the ring but still direct to the ortho and para positions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Q47 (2025)
The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is .
### Related Formula
For an unsymmetrical molecule with $n$ distinct stereogenic units (chiral centers or stereogenic double bonds):
$$\text{Total Stereoisomers} = 2^n$$
### Core Logic
Let's examine the structure of 5-phenylpent-4-en-2-ol:
$$\mathrm{Ph-CH=CH-CH_2-CH(OH)-CH_3} $$
Identify the stereogenic units:
1. **Double Bond ($\mathrm{-CH=CH-}$):** Positioned between carbons 4 and 5, this alkene group can exist in 2 distinct geometric configurations: *cis* ($Z$) or *trans* ($E$).
2. **Chiral Carbon Center ($*\mathrm{C}$):** Carbon-2 is attached to four distinct groups: $-\mathrm{H}$, $-\mathrm{OH}$, $-\mathrm{CH}_3$, and $-\mathrm{CH_2-CH=CH-Ph}$. This asymmetric carbon center can exist in 2 distinct optical configurations: ($R$) or ($S$).
Since the molecule is unsymmetrical, the two stereogenic units behave independently ($n = 2$):
$$\text{Total Stereoisomers} = 2^2 = 4$$
### Visual Mapping
The structural tracking confirms the presence of these stereogenic sites:
{{SOL_IMG1}}
### Pattern Recognition
Always break the molecule down to count chiral centers and stereogenic double bonds independently. Since the molecule has asymmetric ends, you can safely use the simplified $2^n$ formula without worrying about meso configurations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q49 (2025)
The hydrocarbon (X) with molar mass 80 g mol $^{-1}$ and 90% carbon has ____ degree of unsaturation.
### Related Formula
For a hydrocarbon with the molecular formula $\mathrm{C}_x \mathrm{H}_y$, the Double Bond Equivalent (DBE) or Degree of Unsaturation (DU) is given by:
$$\text{DU} = x + 1 - \frac{y}{2}$$
### Core Logic
1. Calculate the mass of carbon in 1 mole of the hydrocarbon:
$$\text{Mass of Carbon} = 80\text{ g} \cdot 90\% = 72\text{ g}$$
2. Find the number of carbon atoms ($x$):
$$x = \frac{72}{12} = 6$$
3. Find the mass and number of hydrogen atoms ($y$):
$$\text{Mass of Hydrogen} = 80\text{ g} - 72\text{ g} = 8\text{ g}$$
$$y = \frac{8}{1} = 8$$
Thus, the molecular formula of hydrocarbon (X) is $\mathrm{C}_{6}\mathrm{H}_{8}$.
4. Calculate the degree of unsaturation:
$$\text{DU} = 6 + 1 - \frac{8}{2} = 7 - 4 = 3$$
### Pattern Recognition
First, use the percentage composition and total molar mass to determine the exact number of carbon and hydrogen atoms. Once you have the molecular formula, plug it into the standard $\text{DU} = x + 1 - \frac{y}{2}$ equation to find the total number of rings and/or pi bonds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
