In which pairs, the first ion is more stable than the second?
(A) The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D). & The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
(B) The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D). & The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
(C) The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D). & The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
(D) The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D). & The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
A.(B) & (D) only
B.(A) & (B) only
C.(B) & (C) only
D.(A) & (C) only
Solution & Explanation
### Related Formula
textStability propto textDelocalization of charge via resonance, mesomeric, and back-bonding effects$\text{Stability} \propto \text{Delocalization of charge via resonance, mesomeric, and back-bonding effects}$
### Core Logic
Evaluating each specific structural pair:
- **Pair (A):** The first carbocation is stabilized by strong +M$+M$ back-bonding from the methoxy (-OMe$-OMe$) oxygen lone pair, making it significantly more stable than the second.
- **Pair (B):** The first carbanion is strongly stabilized by the -M$-M$ and -I$-I$ electronic effects of the nitro (-NO_2$-NO_2$) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable.
- **Pair (C):** The second cation has extended allylic resonance stabilization, meaning the first is less stable.
- **Pair (D):** Cation with -OMe$-OMe$ backbonding is more stable than tertiary aliphatic carbocation, making the first less stable than the second.
Thus, only in **(A) & (B)** is the first ion more stable than the second.
### Pattern Recognition
Back-bonding from an adjacent oxygen lone pair always triumphs over standard inductive or hyperconjugative alkyl stability templates. For carbanions, ensure electron-withdrawing groups like -NO_2$-NO_2$ match the sign of the charge.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Keywords:#carbocation carbanion electronic stability#JEE Main 2025 Evening Q30#Some Basic Principles of Organic Chemistry organic mechanisms#mesomeric backbonding stabilization#backbonding#resonance effect#inductive effect
More Some Basic Principles of Organic Chemistry Previous-Year Questions
Q262025Basicity of Organic Bases
The correct order of basicity for the following molecules is:
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
(P)
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
(Q)
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
(R)
A.P > Q > R$P > Q > R$
B.R > P > Q$R > P > Q$
C.Q > P > R$Q > P > R$
D.R > Q > P$R > Q > P$
Solution
### Related Formula
textBasicity propto textAvailability of lone pair of electrons on Nitrogen$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen}$
### Core Logic
Analyzing the molecules:
- In molecule **(R)**, according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic.
- In molecule **(Q)**, the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability.
- In molecule **(P)**, the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available.
Therefore, the correct basicity order is:
R > Q > P$R > Q > P$
### Step 1: Final Identification
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct.
### Pattern Recognition
Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q332025IUPAC Nomenclature
The IUPAC name of the following compound is -
beginarrayc 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 endarray$\begin{array}{c} \mathrm {O H} \\ \mid \\ \mathrm {H C} \equiv \mathrm {C} - \mathrm {C H} _ {2} - \mathrm {C H} - \mathrm {C H} _ {2} - \mathrm {C H} = \mathrm {C H} _ {2} \end{array}$
A. 4-Hydroxyhept-1-en-6-yne
B. 4-Hydroxyhept-6-en-1-yne
C. Hept-6-en-1-yn-4-ol
D. Hept-1-en-6-yn-4-ol
Solution
### Related Formula
textPrincipal Functional Group Priority: -OH > textDouble bond (=) ge textTriple bond (equivtext)$\text{Principal Functional Group Priority: } -OH > \text{Double bond (=)} \ge \text{Triple bond (}\equiv\text{)}$
### Core Logic
Number the seven-carbon parent chain from the side that gives the principal functional group (-OH$-OH$) 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:
overset7textHoverset6textCequiv overset5textC-overset4textCtextH_2-overset3textCtextH(OH)-overset2textCtextH_2-overset1textCtextH=textCH_2$\overset{7}{\text{H}}\overset{6}{\text{C}}\equiv \overset{5}{\text{C}}-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H(OH)}-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{H}=\text{CH}_2$
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
Q342025Purification of Organic Compounds
Match List-I with List-II -
List-I (Separation of)
List-II (Separation Technique)
(A) Aniline from aniline-water mixture
(I) Simple distillation
(B) Glycerol from spent-lye in soap industry
(II) Fractional distillation
(C) Different fractions of crude oil in petroleum industry
(III) Distillation at reduced pressure
(D) Chloroform-Aniline mixture
(IV) Steam distillation
Choose the correct answer from the options given below:
A. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
B. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Solution
### 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
Q432025Organic Reactions and Mechanisms
Consider the following molecule (X).
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
The structure of X is
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
A. Structure (1)
B. Structure (2)
C. Structure (3)
D. Structure (4)
Solution
### Core Logic
Analyzing the mechanism:
1. Protonation (H^+$H^+$ attack) on the double bond occurs to generate the most stable intermediate carbocation.
2. The system forms a highly stable **tertiary (3^circ$3^\circ$) carbocation** at the bridge junction ring site.
3. Nucleophilic attack by bromide ion (Br^-$Br^-$) captures this bridge junction center selectively, yielding the major brominated structure designated as Structure (2).
### Step 1: Mechanism Breakdown
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Electrophilic addition follows Markovnikov-like stability rules, directing the halide specifically to the bridgehead junction carbocation position.
### Pattern Recognition
Whenever adding hydrohalic acids across cyclic non-conjugated dienes or isolated double bonds, always convert the alkene into the most stable, un-strained tertiary carbocation before attaching the nucleophile.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Class 11 Chemistry: Hydrocarbons
More Some Basic Principles of Organic Chemistry Questions — jee_main_2025_04_april_evening
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