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 15.
Q76 (2024)
The correct stability order of carbocations is
- $(\mathrm{CH}_3)_3\mathrm{C}^+ > \mathrm{CH}_3 - \mathrm{CH}_2^+ > (\mathrm{CH}_3)_2\mathrm{CH}^+ > \mathrm{CH}_3^+$
- $\mathrm{CH}_3^+ > (\mathrm{CH}_3)_2\mathrm{CH}^+ > \mathrm{CH}_3 - \mathrm{CH}_2^+ > (\mathrm{CH}_3)_3\mathrm{C}^+$
- $(\mathrm{CH}_3)_3\mathrm{C}^+ > (\mathrm{CH}_3)_2\mathrm{CH}^+ > \mathrm{CH}_3 - \mathrm{CH}_2^+ > \mathrm{CH}_3^+$
- $\mathrm{CH}_3^+ > \mathrm{CH}_3 - \mathrm{CH}_2^+ > (\mathrm{CH}_3)_2\mathrm{CH}^+ > (\mathrm{CH}_3)_3\mathrm{C}^+$
### Core Logic
The stability of alkyl carbocations is primarily determined by the +I (inductive) effect of alkyl groups and hyperconjugation.
1. $(\mathrm{CH}_3)_3\mathrm{C}^+$ (tert-butyl carbocation) has 9 $\alpha$-hydrogens, leading to 9 hyperconjugative structures. It is the most stable.
2. $(\mathrm{CH}_3)_2\mathrm{CH}^+$ (isopropyl carbocation) has 6 $\alpha$-hydrogens.
3. $\mathrm{CH}_3-\mathrm{CH}_2^+$ (ethyl carbocation) has 3 $\alpha$-hydrogens.
4. $\mathrm{CH}_3^+$ (methyl carbocation) has 0 $\alpha$-hydrogens and is the least stable.
The greater the number of hyperconjugable hydrogens ($\alpha$-hydrogens), the more stable the carbocation.
### Pattern Recognition
Stability of alkyl carbocations: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$ due to hyperconjugation (+H) and inductive (+I) effects.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q1835 (2024)
Which of the following molecule/species is most stable?
### Core Logic
Stability of cyclic carbocations can be determined using Huckel's rule for aromaticity.
A species is exceptionally stable if it is aromatic. Aromaticity requires the system to be cyclic, planar, fully conjugated, and possess $(4n + 2) \pi$ electrons.
### Step 1: Analyze Option 1
The tropylium cation (Option 1) is a 7-membered ring with 3 double bonds and a positive charge in continuous conjugation.
{{SOL_IMG1}}
Number of $\pi$ electrons = $6$. Since $6$ satisfies $(4n+2)$ for $n=1$, it is aromatic and therefore highly stable.
### Pattern Recognition
Tropylium ion ($C_7H_7^+$) is a classic example of a stable aromatic carbocation. It frequently appears in stability comparison questions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q79 (2024)
The Lassiagne's extract is boiled with dil. $HNO_3$ before testing for halogens because,
- $\text{AgCN is soluble in }HNO_3$
- $\text{Silver halides are soluble in }HNO_3$
- $Ag_2S\text{ is soluble in }HNO_3$
- $Na_2S\text{ and NaCN are decomposed by }HNO_3$
### Core Logic
In Lassaigne's test for halogens, we add $AgNO_3$ to form a precipitate of silver halide ($AgX$).
However, if the organic compound also contains Nitrogen or Sulphur, the Lassaigne's extract will contain $NaCN$ or $Na_2S$.
### Step 1: Reason for adding HNO3
These ions ($CN^-$ and $S^{2-}$) would also react with $AgNO_3$ to form precipitates ($AgCN$ - white, $Ag_2S$ - black), which would interfere with the test for halogens.
Boiling the extract with concentrated/dilute $HNO_3$ decomposes the cyanide and sulphide to $HCN$ and $H_2S$ gases, which escape, thus removing the interference.
$$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$$
$$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q86 (2024)
On a thin layer chromatographic plate, an organic compound moved by $3.5\text{ cm}$, while the solvent moved by $5\text{ cm}$. The retardation factor of the organic compound is ________ $\times 10^{-1}$
### Related Formula
$$R_f = \frac{\text{Distance travelled by compound}}{\text{Distance travelled by solvent}}$$
### Step 1: Substitution and calculation
$$R_f = \frac{3.5}{5}$$
$R_f = 0.7$
$$R_f = 7 \times 10^{-1}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q1935 (2024)
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is -
- $\text{crystallisation}$
- $\text{distillation under reduced pressure}$
- $\text{distillation}$
- $\text{steam distillation}$
### Core Logic
Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water.
Since the essential oils in flowers are steam volatile and insoluble in water, steam distillation is the perfect technique for their extraction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
