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 11.
Q1406 (2024)
Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is
{{IMG1}}
### Core Logic
Deactivating groups pull electron density away from the aromatic ring, making it less reactive towards electrophilic substitution. These groups generally have a strong -M (mesomeric/resonance) effect or a very strong -I (inductive) effect without compensating +M.
Let's evaluate each group from the image:
1. $-CO-CH_3$ (Acetyl group): Has a carbonyl double bond directly attached to the ring. Shows -M and -I effect. $\rightarrow$ Deactivating.
2. $-OCH_3$ (Methoxy group): Oxygen has a lone pair. Shows strong +M effect which dominates its -I effect. $\rightarrow$ Activating.
3. $-NH-CH_3$ (N-Methylamino group): Nitrogen has a lone pair. Shows strong +M effect. $\rightarrow$ Activating.
4. $-C\equiv N$ (Cyano group): Triple bond directly attached, nitrogen is highly electronegative. Shows strong -M and -I effect. $\rightarrow$ Deactivating.
5. Wait, looking closely at the solution image for the 5 groups:
The groups presented are:
(1) $-CO-CH_3$ (-M group, deactivating)
(2) $-OCH_3$ (+M group, activating)
(3) $-NH-CH_3$ (+M group, activating)
(4) $-C\equiv N$ (-M group, deactivating)
(Wait, the fifth group in the question image is not explicitly named here but the solution labels show 4 groups in the block? Ah, the question image actually has 5 structural parts or the text just lists them. The solution states: "-C=N, $-OCH_3$" and labels (-M group), (+M group). It counts exactly 2 deactivating groups.)
Let's analyze the groups labeled as (-M) in the solution: $-CO-CH_3$ and $-C\equiv N$.
### Step 1: Count Deactivating Groups
Deactivating groups (-M effect):
1. $-CO-CH_3$
2. $-C\equiv N$
Total number = 2.
{{SOL_IMG1}}
### Pattern Recognition
If the atom directly attached to the benzene ring has a multiple bond to a more electronegative atom (like $C=O$, $C\equiv N$, $N=O$, $S=O$), the group is deactivating (-M).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q73 (2024)
Ionic reactions with organic compounds proceed through:
(A) Homolytic bond cleavage
(B) Heterolytic bond cleavage
(C) Free radical formation
(D) Primary free radical
(E) Secondary free radical
Choose the correct answer from the options given below:
- $\text{(A) only}$
- $\text{(C) only}$
- $\text{(B) only}$
- $\text{(D) and (E) only}$
### Core Logic
Bond fission in organic chemistry can happen in two ways:
1. Homolytic cleavage: The shared pair of electrons gets distributed equally, leading to free radical formation.
2. Heterolytic cleavage: The shared pair of electrons goes completely to one of the bonded atoms, leading to the formation of positive (cation) and negative (anion) ions.
### Step 1: Match Definition
Since ionic reactions involve ions, they inherently proceed through heterolytic bond cleavage where ionic intermediates (carbocations, carbanions, or leaving groups) are formed.
### Pattern Recognition
Ionic reactions $\rightarrow$ Ions formed $\rightarrow$ Heterolytic cleavage.
Radical reactions $\rightarrow$ Free radicals formed $\rightarrow$ Homolytic cleavage.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q1471 (2024)
The ascending acidity order of the following H atoms is
{{IMG1}}
- $\text{C} < \text{D} < \text{B} < \text{A}$
- $\text{A} < \text{B} < \text{C} < \text{D}$
- $\text{A} < \text{B} < \text{D} < \text{C}$
- $\text{D} < \text{C} < \text{B} < \text{A}$
### Related Formula
$$\text{Acidic strength} \propto \text{Stability of conjugate base}$$
### Core Logic
On removing the respective hydrogen atoms as protons ($\text{H}^+$), the conjugate bases formed have the following stability order due to hybridization and inductive effect:
$$\text{CH}\equiv\text{C}^\ominus > \text{CH}_2=\text{CH}^\ominus > \text{CH}_3\text{CH}_2^\ominus > (\text{CH}_3)_3\text{C}^\ominus$$
### Step 1: Evaluation
Thus, the stability of the conjugate bases follows the order: $\text{C} < \text{D} < \text{B} < \text{A}$. Consequently, the acidic strength follows the exact same ascending order.
### Pattern Recognition
Higher $s$-character increases electronegativity, stabilizing the negative charge on the conjugate base. Thus, $sp$ carbon hydrogens are significantly more acidic than $sp^2$ or $sp^3$ counterparts.
Q1475 (2024)
According to IUPAC system, the compound
{{IMG1}}
is named as :
- Cyclohex-1-en-2-ol
- 1-Hydroxyhex-2-ene
- Cyclohex-1-en-3-ol
- Cyclohex-2-en-1-ol
### Related Formula
Principal functional group priority rules: $-\text{OH} > \text{Alkene}$.
### Core Logic
According to IUPAC nomenclature priority rules, the principal functional group is the alcohol group ($-\text{OH}$). The ring numbering starts at the carbon containing the $-\text{OH}$ group as C1.
Numbering then proceeds towards the double bond to give it the lowest possible position numbers. This designates C2 and C3 for the double bond carbons, making it a 'cyclohex-2-en-1-ol'.
### Step 1: Final Assignment
Hence, combining the root and suffix gives the correct systematic designation: Cyclohex-2-en-1-ol.
### Pattern Recognition
Always number functional cyclic systems to prioritize principal suffixes ($-\text{OH}$) first, then provide minimal numbers to unsaturation points.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Nomenclature of Organic Compounds
Q1488 (2024)
Which one of the following will show geometrical isomerism?
- Option (1)
- Option (2)
- Option (3)
- Option (4)
### Related Formula
$$\text{Condition for Geometrical Isomerism: } \text{C}=\text{C} \text{ must have distinct groups attached to each olefinic carbon.}$$
### Core Logic
For a compound containing a carbon-carbon double bond to exhibit geometrical ($cis/trans$) isomerism, neither of the two alkene carbons can have two identical groups attached to it.
Evaluating Option (3), both olefinic carbons are unsymmetrically substituted with two completely different groups, satisfying all geometric requirements for stereoisomerism.
### Step 1: Conclusion
The structural composition of Option (3) enables it to form distinct spatial structures, allowing it to exhibit geometrical isomerism.
### Pattern Recognition
Check both ends of the double bond. If either carbon contains identical groups, geometrical isomerism is structurally impossible.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Isomerism of Organic Compounds
