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).
Q406 (2025)
In 3, 3-dimethylhex-1-ene-4-yne, there are $\mathbf{sp}^3$, $\mathbf{sp}^2$ and $\mathbf{sp}$ hybridised carbon atoms respectively:
- $4, 2, 2$
- $3, 3, 2$
- $2, 4, 2$
- $2, 2, 4$
### Related Formula
$$\text{Hybridization of Carbon} = \begin{cases} \mathrm{sp^3} & 4~\sigma\text{-bonds} \\ \mathrm{sp^2} & 3~\sigma\text{-bonds}, 1~\pi\text{-bond} \\ \mathrm{sp} & 2~\sigma\text{-bonds}, 2~\pi\text{-bonds} \end{cases}$$
### Core Logic
Let's first draw the structural formula of **3,3-dimethylhex-1-ene-4-yne**:
{{SOL_IMG1}}
$$\overset{6}{\mathrm{C}}\mathrm{H}_3 - \overset{5}{\mathrm{C}} \equiv \overset{4}{\mathrm{C}} - \overset{3}{\mathrm{C}}(\mathrm{CH}_3)_2 - \overset{2}{\mathrm{C}}\mathrm{H} = \overset{1}{\mathrm{C}}\mathrm{H}_2$$
### Step 1: Identify Hybridization of Each Carbon
We count the sigma ($\sigma$) bonds or pi ($\pi$) bonds on each carbon:
- **$\mathrm{C}_1$**: Involved in a double bond ($\mathrm{CH_2 =}$) $\implies \mathrm{sp^2}$
- **$\mathrm{C}_2$**: Involved in a double bond ($=\mathrm{CH-}$) $\implies \mathrm{sp^2}$
- **$\mathrm{C}_3$**: Single bonds only (bonded to $\mathrm{C}_2$, $\mathrm{C}_4$, and two methyl carbons) $\implies \mathrm{sp^3}$
- **Two methyl carbons** attached to $\mathrm{C}_3$: Single bonds only $\implies 2 \times \mathrm{sp^3}$
- **$\mathrm{C}_4$**: Involved in a triple bond ($-\mathrm{C} \equiv$) $\implies \mathrm{sp}$
- **$\mathrm{C}_5$**: Involved in a triple bond ($\equiv \mathrm{C}-$) $\implies \mathrm{sp}$
- **$\mathrm{C}_6$**: Single bonds only ($-\mathrm{CH_3}$) $\implies \mathrm{sp^3}$
### Step 2: Calculate the Count
Summing the hybridization counts:
- $\mathrm{sp^3}$ carbons: $\mathrm{C}_3$, $\mathrm{C}_6$, and $2 \times \mathrm{CH_3}$ on $\mathrm{C}_3$ $= 4$ carbon atoms
- $\mathrm{sp^2}$ carbons: $\mathrm{C}_1$ and $\mathrm{C}_2$ $= 2$ carbon atoms
- $\mathrm{sp}$ carbons: $\mathrm{C}_4$ and $\mathrm{C}_5$ $= 2$ carbon atoms
Thus, the number of $\mathrm{sp^3}$, $\mathrm{sp^2}$ and $\mathrm{sp}$ hybridized carbons is $4, 2, 2$ respectively.
### Pattern Recognition
Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as $\mathrm{sp^3}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q411 (2025)
Match List-I with List-II:
$$ \begin{array}{|l|l|} \hline \begin{array}{c} \textbf{List-I} \\ \textbf{(Purification technique)} \end{array} & \begin{array}{c} \textbf{List-II} \\ \textbf{(Mixture of organic compounds)} \end{array} \\ \hline \text{(A)} \quad \text{Distillation (simple)} & \text{(I)} \quad \text{Diesel + Petrol} \\ \hline \text{(B)} \quad \text{Fractional distillation} & \text{(II)} \quad \text{Aniline + Water} \\ \hline \text{(C)} \quad \text{Distillation under reduced pressure} & \text{(III)} \quad \text{Chloroform + Aniline} \\ \hline \text{(D)} \quad \text{Steam distillation} & \text{(IV)} \quad \text{Glycerol + Spent-lye} \\ \hline \end{array} $$
Choose the correct answer from the options given below:
- $\text{(A)-(II), (B)-(III), (C)-(IV), (D)-(I)}$
- $\text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}$
- $\text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}$
- $\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}$
### Related Formula
$$\text{Method Selection} = f\left( \Delta T_{\mathrm{b}}, \text{decomposition threshold}, \text{volatility with steam} \right)$$
### Core Logic
Let's align each purification technique to its designated mixtures based on NCERT guidelines:
- **(A) Simple distillation**: Used for liquids having a significant difference in their boiling points ($>30~\mathrm{K}$ or $30^\circ\mathrm{C}$). Chloroform (b.p. $334~\mathrm{K}$) and aniline (b.p. $457~\mathrm{K}$) are separated easily using simple distillation $\rightarrow$ **(III)**.
- **(B) Fractional distillation**: Used if boiling point differences of the components are very close (less than $25~\mathrm{K}$). Separation of petrochemical fractions such as diesel and petrol uses this technique $\rightarrow$ **(I)**.
- **(C) Distillation under reduced pressure**: Used for liquids that tend to decompose at or below their normal boiling points. Glycerol is separated from spent-lye in soap manufacturing industry using this vacuum method to prevent glycerol decomposition $\rightarrow$ **(IV)**.
- **(D) Steam distillation**: Applied to substances which are steam-volatile and completely immiscible in water. Aniline and water are separated using this technique $\rightarrow$ **(II)**.
### Step 1: Conclusion
Thus, the correct match is:
**(A)-(III), (B)-(I), (C)-(IV), (D)-(II)**
This corresponds perfectly to option (4).
### Pattern Recognition
Glycerol from spent-lye is a highly tested practical chemistry concept. Remember that vacuum distillation lowers the boiling point, permitting evaporation without decomposition.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q420 (2025)
In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave $60~\mathrm{mL}$ of nitrogen collected at $300\mathrm{K}$ temperature and $715~\mathrm{mmHg}$ pressure. The percentage composition of nitrogen in the compound (Aqueous tension at $300\mathrm{K} = 15~\mathrm{mmHg}$) is
- 1.257
- 20.87
- 18.67
- 12.57
### Related Formula
$$p_{\mathrm{N_2}} = p_{\text{total}} - p_{\text{aq}}$$
$$n_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}} V}{R T}$$
$$\% \mathrm{N} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Dumas' method estimates nitrogen by collecting dry nitrogen gas ($N_2$). We must subtract the aqueous tension (vapor pressure of water) to find the pressure exerted solely by the dry nitrogen gas.
### Step 1: Calculate Pressure of Dry Nitrogen
$$p_{\mathrm{N_2}} = 715~\mathrm{mmHg} - 15~\mathrm{mmHg} = 700~\mathrm{mmHg}$$
Converting pressure to atmospheres:
$$p_{\mathrm{N_2}} = \frac{700}{760}~\mathrm{atm}$$
### Step 2: Calculate Moles of Nitrogen Gas
Using the ideal gas law with $R = 0.0821~\mathrm{L~atm~mol^{-1}~K^{-1}}$, $T = 300~\mathrm{K}$, and $V = 60~\mathrm{mL} = 60 \times 10^{-3}~\mathrm{L}$:
$$n_{\mathrm{N_2}} = \frac{\left(\frac{700}{760}\right) \times 60 \times 10^{-3}}{0.0821 \times 300}$$
$$n_{\mathrm{N_2}} = \frac{0.92105 \times 0.060}{24.63} \approx 2.244 \times 10^{-3}~\mathrm{mol}$$
### Step 3: Calculate Mass and Percentage of Nitrogen
The molar mass of $\mathrm{N_2}$ is $28~\mathrm{g~mol^{-1}}$:
$$\text{Mass of } \mathrm{N_2} = n_{\mathrm{N_2}} \times 28 = 2.244 \times 10^{-3} \times 28 \approx 0.06283~\mathrm{g}$$
Now find the percentage in $0.5~\mathrm{g}$ of organic compound:
$$\% \mathrm{N} = \frac{0.06283~\mathrm{g}}{0.5~\mathrm{g}} \times 100 = 12.566\% \approx 12.57\%$$
### Pattern Recognition
Watch out! Always subtract the aqueous tension from the wet gas pressure first to find the dry gas pressure. Forgetting this step is the most common source of error in Dumas calculations.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q451 (2025)
Designate whether each of the following compounds is aromatic or not aromatic:
{{IMG1}}
Choose the correct answer from the options given below:
- $\text{(1) e, g aromatic and a, b, c, d, f, h not aromatic}$
- $\text{(2) b, e, f, g aromatic and a, c, d, h not aromatic}$
- $\text{(3) a, b, c, d aromatic and e, f, g, h not aromatic}$
- $\text{(4) a, c, d, e, h aromatic and b, f, g not aromatic}$
### Related Formula
According to Huckel's Rule, a planar, monocyclic, completely conjugated system is aromatic if it contains:
$$(4n + 2)\pi \quad \text{electrons (where } n = 0, 1, 2, \dots)$$
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Classification
Hence, compounds a, c, d, e, and h follow Huckel's rule and are aromatic, whereas b, f, and g are not aromatic.
### Pattern Recognition
Quick check for aromaticity: Count the pairs of localized/delocalized $\pi$ electrons moving through the continuous loop. Odd number of pairs (1, 3, 5...) means aromatic ($2\pi, 6\pi, 10\pi$). Even pairs mean anti-aromatic/non-aromatic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Class 11 Chemistry: Hydrocarbons
Q468 (2025)
Consider the following compound (X)
$$\begin{array}{c} \mathrm {I} \\ \mathrm {H} - \mathrm {C} \equiv \mathrm {C} - \mathrm {C H} _ {2} - \mathrm {C H} - \mathrm {C H} _ {3} \\ \mathrm {I} \\ \mathrm {C H} _ {3} \end{array}$$
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding $\mathrm{C - H}$ bond are :
- $(1)\ \text{II, IV}$
- $(2)\ \text{III, II}$
- $(3)\ \text{I, IV}$
- $(4)\ \text{II, I}$
### Related Formula
Free radical stability structural hierarchy sequence:
$$\text{Resonance Stabilized (Propargyl/Allyl)} > 3^\circ > 2^\circ > 1^\circ > \text{Vinylic/Alkyne Center}$$
### Core Logic
Let's analyze individual cleavage points across the carbon backbone skeleton:
* **Position II** yields a propargyl intermediate radical directly adjacent to the alkyne bond. This allows strong resonance stabilization across the $\pi$ system, making it the most stable radical position.
* **Position I** places the radical directly on an $\mathrm{sp}$-hybridized carbon center. The high electronegativity of $\mathrm{sp}$ orbitals tightly holds the unpaired electron, making homolytic cleavage extremely difficult and rendering this intermediate the least stable radical position.
{{SOL_IMG1}}
### Step 1: Verdict
Therefore, the most stable and least stable positions are II and I, respectively.
### Pattern Recognition
Radicals located on $\mathrm{sp}$ carbons (vinylic/alkynic) are highly unstable due to poor orbital overlap, while positions next to triple bonds (propargylic) are exceptionally stable due to active resonance delocalization.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
