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 7.
Q33 (2025)
The IUPAC name of the following compound is -
$$\begin{array}{c} \mathrm {O H} \\ \mid \\ \mathrm {H C} \equiv \mathrm {C} - \mathrm {C H} _ {2} - \mathrm {C H} - \mathrm {C H} _ {2} - \mathrm {C H} = \mathrm {C H} _ {2} \end{array}$$
- 4-Hydroxyhept-1-en-6-yne
- 4-Hydroxyhept-6-en-1-yne
- Hept-6-en-1-yn-4-ol
- Hept-1-en-6-yn-4-ol
### Related Formula
$$\text{Principal Functional Group Priority: } -OH > \text{Double bond (=)} \ge \text{Triple bond (}\equiv\text{)}$$
### Core Logic
Number the seven-carbon parent chain from the side that gives the principal functional group ($-OH$) and double bond the lowest locants:
- Numbering from right-to-left gives locants: alkene at position 1, alcohol at 4, and alkyne at 6.
- Numbering from left-to-right gives locants: alkyne at 1, alcohol at 4, alkene at 6.
According to IUPAC rules, when choices are symmetric for the principal group, the lower locant is given to the double bond over the triple bond. Hence, right-to-left numbering is correct:
$$\overset{7}{\text{H}}\overset{6}{\text{C}}\equiv \overset{5}{\text{C}}-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H(OH)}-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{H}=\text{CH}_2$$
This gives: **Hept-1-en-6-yn-4-ol**.
### Pattern Recognition
When terminal unsaturations are tied symmetrically (positions 1 and 6), the alkene ('en') takes prefix allocation priority over the alkyne ('yn'). The secondary alcohol suffix '-ol' forms the principal name ending.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q34 (2025)
Match 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 (Separation of)</th><th style="border: 1px solid #888; padding: 8px;">List-II (Separation Technique)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) Aniline from aniline-water mixture</td><td style="border: 1px solid #888; padding: 8px;">(I) Simple distillation</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) Glycerol from spent-lye in soap industry</td><td style="border: 1px solid #888; padding: 8px;">(II) Fractional distillation</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) Different fractions of crude oil in petroleum industry</td><td style="border: 1px solid #888; padding: 8px;">(III) Distillation at reduced pressure</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) Chloroform-Aniline mixture</td><td style="border: 1px solid #888; padding: 8px;">(IV) Steam distillation</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
### Core Logic
Evaluating standard NCERT laboratory purification matches:
- **(A) Aniline from aniline-water mixture:** Aniline is steam volatile and immiscible with water, so it is separated via **Steam distillation (IV)**.
- **(B) Glycerol from spent-lye in soap industry:** Glycerol decomposes at or below its boiling point, hence it is separated via **Distillation at reduced pressure (III)**.
- **(C) Different fractions of crude oil:** Separated using their small differences in boiling points via **Fractional distillation (II)**.
- **(D) Chloroform-Aniline mixture:** Separated due to a substantial boiling point difference via **Simple distillation (I)**.
### Pattern Recognition
Match key words directly: Glycerol/spent-lye always links to reduced pressure (vacuum distillation). Crude oil always couples to fractional columns. Aniline + water implies steam injection.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q43 (2025)
Given below are two statements.
Statement I: The dipole moment of $\overset{4}{C}H_3-\overset{3}{C}H=\overset{2}{C}H-\overset{1}{C}H=O$ is greater than $CH_3-CH_2-CH_2-CH=O$.
Statement II: $C_1-C_2$ bond length of $CH_3-CH=CH-CH=O$ is greater than $C_1-C_2$ bond length of $CH_3-CH_2-CH_2-CH=O$.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Statement I is false but Statement II is true}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are true}$
### Core Logic
* **Statement I is true:** In the conjugated system ($CH_3-CH=CH-CH=O$), extended resonance delocalization occurs, shifting electron density toward the carbonyl oxygen:
$$CH_3-\overset{\oplus}{C}H-CH=CH-\overset{\ominus}{O}$$
This configuration increases both the partial charge separation $q$ and the dipole distance d, resulting in a significantly larger net dipole moment $\mu = q \times d$ compared to the non-conjugated saturated aldehyde.
* **Statement II is false:** Due to resonance conjugation, the single bond between $C_1$ and $C_2$ acquires partial double-bond character. This double-bond character shortens the bond length, making it smaller than the standard single bond found in $CH_3-CH_2-CH_2-CH=O$.
### Pattern Recognition
Conjugation spreads partial charges across a longer carbon chain, expanding the charge separation distance to drive up the overall dipole moment $\mu$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q46 (2025)
In Dumas' method for estimation of nitrogen, $1\mathrm{~g}$ of an organic compound gave $150\mathrm{~mL}$ of nitrogen collected at $300\mathrm{~K}$ temperature and $900\mathrm{~mmHg}$ pressure. The percentage composition of nitrogen in the compound is _______ $%$ (nearest integer).
(Aqueous tension at $300\mathrm{~K} = 15\mathrm{~mmHg}$)
### Related Formula
$$P_{\text{dry } N_2} = P_{\text{total}} - P_{\text{aqueous tension}}$$
$$PV = nRT \implies n = \frac{PV}{RT}$$
$$\% N = \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \times 100$$
### Core Logic
First, calculate the actual pressure exerted by the dry nitrogen gas:
$$P_{N_2} = 900 - 15 = 885\mathrm{~mmHg} = \frac{885}{760}\mathrm{~atm}$$
Convert volume data to liters: $V = 150\mathrm{~mL} = 0.15\mathrm{~L}$.
Using the ideal gas law to determine the moles of $N_2$ collected:
$$n = \frac{\left(\frac{885}{760}\right) \times 0.15}{0.0821 \times 300} = \frac{1.1645 \times 0.15}{24.63} \approx 0.0071\mathrm{~moles}$$
Calculate the total mass of the liberated nitrogen gas:
$$\text{Mass} = n \times M_{\text{molar}} = 0.0071 \times 28 = 0.1988\mathrm{~g}$$
Determine the percentage composition relative to the initial $1\mathrm{~g}$ sample size:
$$\% N = \frac{0.1988}{1} \times 100 = 19.88\% \approx 20\%$$
### Pattern Recognition
Always remember to subtract the aqueous tension value first. Failing to correct for water vapor pressure is the most common pitfall in Dumas' method calculations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q27 (2025)
Mixture of $1\text{ g}$ each of chlorobenzene, aniline and benzoic acid is dissolved in $50\text{ mL}$ ethyl acetate and placed in a separating funnel, $5\text{ M NaOH}$ ($30\text{ mL}$) was added in the same funnel. The funnel was shaken vigorously and then kept aside. The ethyl acetate layer in the funnel contains:
- $\text{benzoic acid}$
- $\text{benzoic acid and aniline}$
- $\text{benzoic acid and chlorobenzene}$
- $\text{chlorobenzene and aniline}$
### Related Formula
$$\text{Ph-COOH} + \text{NaOH}
ightarrow \text{Ph-COO}^-\text{Na}^+ \text{ (Water soluble salt)} [cite: 884, 885]$$
### Core Logic
When aqueous $\text{NaOH}$ is added to the mixture:
1. Benzoic acid ($\text{Ph-COOH}$) reacts to form sodium benzoate ($\text{Ph-COONa}$), which is highly water-soluble and moves completely into the aqueous layer.
2. Chlorobenzene ($\text{Ph-Cl}$) and aniline ($\text{Ph-NH}_2$) are organic compounds that do not react with aqueous $\text{NaOH}$ under normal conditions.
Therefore, they remain unreacted in the organic layer (ethyl acetate).
### Step 1: Visual Reaction Progress
The separation steps can be visualized tracking the structural components:
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
Thus, the ethyl acetate layer strictly contains chlorobenzene and aniline.
### Pattern Recognition
Acid-base separation shortcut: Strong bases ($\text{NaOH}$) pull organic acids into the aqueous phase as salts. Neutral compounds and basic amines stay behind in the organic layer unless a strong acid is added.
### 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
