Identify the correct statement/s from the following options: (A) -mathrmOCH_3 and -mathrmNHCOCH_3 are activating groups (B) -mathrmCN and -mathrmOH are meta directing groups (C) -mathrmCN and -mathrmSO_3mathrmH are meta directing groups (D) Activating groups act as ortho- and para- directing groups (E) Halides are activating groups Choose the correct answer from the options given below :

Solution & Explanation

### Core Logic Let's evaluate each statement based on electrophilic aromatic substitution guidelines: * Statement (A): Both -mathrmOCH_3 and -mathrmNHCOCH_3 have lone pairs on the atom directly attached to the benzene ring. These lone pairs undergo resonance delocalization into the ring, increasing electron density and activating it toward substitution. This statement is true. * Statement (B): While -mathrmCN is a deactivating meta-directing group, -mathrmOH is a strongly activating ortho/para-directing group due to resonance. This statement is false. * Statement (C): Both -mathrmCN and -mathrmSO_3mathrmH withdraw electron density via inductive and resonance effects (-I, -M). This deactivates the ring and directs substitution to the meta position. This statement is true. * Statement (D): Activating groups increase electron density primarily at the ortho and para positions via resonance, directing incoming electrophiles to those sites. This statement is true. * Statement (E): Halogens are an exception: they are deactivating due to their strong inductive effect (-I), but ortho/para-directing due to resonance (+M). This statement is false. Therefore, statements (A), (C), and (D) are correct, which matches Option (1). ### Pattern Recognition Groups that activate the aromatic ring by donating electrons via resonance always direct incoming electrophiles to the ortho and para positions. Halogens are a unique exception: they deactivate the ring but still direct to the ortho and para positions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 7

Q jee_main_2025_04_april_evening Basicity of Organic Bases
The correct order of basicity for the following molecules is:
Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
  • A. P > Q > R
  • B. R > P > Q
  • C. Q > P > R
  • D. R > Q > P

Solution

### Related Formula textBasicity propto textAvailability 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 ### Step 1: Final Identification
Basicity breakdown explanation diagram for Q26
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Basicity breakdown explanation diagram for Q26
The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.
Basicity breakdown explanation diagram for Q26
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
Q jee_main_2025_04_april_evening Carbocation and Carbanion Stability
In which pairs, the first ion is more stable than the second?
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
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

### Related Formula textStability propto textDelocalization 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 back-bonding from the methoxy (-OMe) oxygen lone pair, making it significantly more stable than the second. - **Pair (B):** The first carbanion is strongly stabilized by the -M and -I electronic effects of the nitro (-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 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 match the sign of the charge. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Q jee_main_2025_04_april_evening Organic Reactions and Mechanisms
Consider the following molecule (X).
Organic reactant hydrocarbon layout for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
The structure of X is
Organic reactant hydrocarbon layout for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Organic reactant hydrocarbon layout for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Organic reactant hydrocarbon layout for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Organic reactant hydrocarbon layout for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Organic reactant hydrocarbon layout for Q43
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^+ attack) on the double bond occurs to generate the most stable intermediate carbocation. 2. The system forms a highly stable **tertiary (3^circ) carbocation** at the bridge junction ring site. 3. Nucleophilic attack by bromide ion (Br^-) captures this bridge junction center selectively, yielding the major brominated structure designated as Structure (2). ### Step 1: Mechanism Breakdown
Carbocation intermediate step illustration for Q43
The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.
Carbocation intermediate step illustration for Q43
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
Q33 jee_main_2025_04_april_evening IUPAC 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
  • 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) ### 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: overset7textHoverset6textCequiv overset5textC-overset4textCtextH_2-overset3textCtextH(OH)-overset2textCtextH_2-overset1textCtextH=textCH_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 jee_main_2025_04_april_evening Purification 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

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