An optically active alkyl halide mathrmC_4H_9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic mathrmNaNH_2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333mathrmK to form compound [E]. The IUPAC name of compound [E] is :

Solution & Explanation

### Related Formula Dehydrohalogenation via alcoholic KOH follows E2 elimination mechanism: mathrmR-CH_2-CH(Br)-R' xrightarrowtextalc. KOH R-CH=CH-R' Hydration of alkynes using mathrmHgSO_4/H_2SO_4 yields ketones via keto-enol tautomerism. ### Core Logic Let's trace the full sequence line-by-row: 1. **[A]** is an optically active halide with formula mathrmC_4H_9Br rightarrow mathrmCH_3-CH(Br)-CH_2-CH_3 (2-Bromobutane). 2. Reaction of [A] with hot ethanolic KOH produces **[B]** as the major product: mathrmCH_3-CH=CH-CH_3 (But-2-ene). 3. Treatment of [B] with mathrmBr_2 yields a vicinal dibromide **[C]**: mathrmCH_3-CH(Br)-CH(Br)-CH_3 (2,3-Dibromobutane). 4. Reaction of [C] with alcoholic mathrmNaNH_2 converts it via double dehydrohalogenation into gas **[D]**: mathrmCH_3-Cequiv C-CH_3 (But-2-yne). 5. Hydration of 1 mole of [D] with mathrmH_2O in the presence of mathrmHg^2+/H^+ forms an enol intermediate that rapidly tautomerizes to compound **[E]**: mathrmCH_3-CO-CH_2-CH_3 (Butan-2-one). ### Step 1: Visualization
Reaction roadmap step verification for Q27
Reaction roadmap step verification for Q27
### Pattern Recognition Whenever you see a 4-carbon chain undergoing terminal/internal dehydrohalogenation followed by hydration of the resulting alkyne, look closely at the configuration: symmetric or unsymmetric alkyne hydration both systematically lead to Butan-2-one because a stable ketone cannot form on position 1 via standard Kucherov hydration of an internal chain. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Haloalkanes and Haloarenes Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

More Haloalkanes and Haloarenes Previous-Year Questions — Page 2

Q30 jee_main_2025_29_jan_evening Nucleophilic Substitution Mechanisms
Which among the following halides will generate the most stable carbocation in Nucleophilic substitution reaction?
  • A. Allylic halide option (1)
  • B. Secondary halide option (2)
  • C. Secondary benzylic halide option (3)
  • D. Triphenylmethyl halide option (4)

Solution

### Core Logic The mechanism of S_N1 substitution proceeds via carbocation intermediate formation. Option (4) gives a triphenylmethyl carbocation (Ph_3C^+), which is exceptionally stable due to extensive delocalization of positive charge across three phenyl rings (resonance stabilization via 9 canonical structures).
Nucleophilic Substitution Mechanisms diagram for Q30 - JEE Main 2025 Evening
Nucleophilic Substitution Mechanisms diagram for Q30 - JEE Main 2025 Evening
### Step 1: Stability Comparison Stability sequence: Ph_3C^+ > textbenzylic > textallylic > textalkyl carbocations ### Pattern Recognition Look for maximum phenyl groups attached directly to the carbon bearing the leaving group to maximize resonance contribution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q29 jee_main_2025_28_jan_morning Alkaline Hydrolysis and NGP
Given below are two statements : Statement I: mathrmEt_2mathrmN-mathrmCH_2-mathrmCH_2-mathrmCl will undergo alkaline hydrolysis at a faster rate than mathrmEt_2mathrmCH-mathrmCH_2-mathrmCl. Statement II: In mathrmEt_2mathrmN-mathrmCH_2-mathrmCH_2-mathrmCl, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth Statement I and Statement II are incorrect
  • B. textStatement I is incorrect but statement II is correct
  • C. textBoth Statement I and Statement II are correct
  • D. textStatement I is correct but Statement II is incorrect

Solution

### Core Logic Statement I is correct because the nitrogen atom contains a lone pair situated at the beta-position relative to the chlorine atom, promoting Neighboring Group Participation (NGP). Statement II is correct because the lone pair on nitrogen attacks internally to kick out the chloride ion, forming a cyclic aziridinium ion intermediate. This quick intramolecular cyclization leads to an exceptionally rapid hydrolysis rate compared to standard aliphatic substitution. ### Pattern Recognition Sees: Nitrogen with lone pair beta to a leaving group. Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q39 jee_main_2025_28_jan_morning Ambident Nucleophiles Reactions
The products A and B in the following reactions, respectively are mathrm A xleftarrow mathrm A g - mathrm N O _ 2 mathrm C H _ 3 - mathrm C H _ 2 - mathrm C H _ 2 - mathrm B r xrightarrow mathrm A g C N mathrm B
  • A. mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmONO, mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmNC
  • B. mathrmCH_3mathrm-CH_2mathrm-CH_2mathrm-ONO, mathrmCH_3mathrm-CH_2mathrm-CH_2mathrm-CN
  • C. mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmNO_2, mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmCN
  • D. mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmNO_2, mathrmCH_3 - mathrmCH_2 - mathrmCH_2 - mathrmNC

Solution

### Core Logic Both silver reagents exhibit significantly covalent bond characters: - Reaction with mathrmAgNO_2: The bond between silver and oxygen is covalent, making the lone pair on the nitrogen atom the primary nucleophilic site. Attack via nitrogen yields a nitroalkane product: mathrmA = mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmNO_2 - Reaction with mathrmAgCN: The covalent mathrmAg-mathrmC bond directs the nucleophilic attack to proceed through the lone pair on nitrogen, yielding an isocyanide compound: mathrmB = mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmNC Hence, option (4) represents the correct combination. ### Pattern Recognition Sees: Alkyl halide reacting with covalent silver salts of ambident anions. Shortcut: Silver reagents (mathrmAgCN or mathrmAgNO_2) drive bond formatting via the nitrogen center, producing isocyanides and nitroalkanes respectively. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q31 jee_main_2025_04_april_evening Substitution versus Elimination
Given below are two statements : Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction. Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the beta-carbon. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Both Statement I and Statement II are incorrect
  • B. Statement I is incorrect but Statement II is correct
  • C. Statement I is correct but Statement II is incorrect
  • D. Both Statement I and Statement II are correct.

Solution

### Related Formula textR-Cl + KOH_text(aq) rightarrow textR-OH + KCl quad (S_Ntext Nucleophilic Substitution) textR-CH_2text-CH_2text-Cl + KOH_text(alc) rightarrow textR-CH=textCH_2 + KCl + H_2O quad (E2text Elimination) ### Core Logic - **Statement I is incorrect:** Treatment of alkyl chlorides with aqueous KOH yields alcohols via a **nucleophilic substitution (S_N) reaction**, not an elimination reaction. - **Statement II is correct:** Alcoholic KOH acts as a strong base (R-O^- ions present), which preferentially abstracts a proton from the beta-carbon atom, leading to dehydrohalogenation to form an alkene via an elimination pathway. ### Pattern Recognition Remember: Aqueous medium = substitution (nucleophilic attack dominates due to highly hydrated, less basic hydroxide ions). Alcoholic medium = elimination (alkoxide acts as a bulky strong base to capture beta-hydrogens). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q32 jee_main_2025_07_april_evening Nucleophilic Aromatic Substitution
Match List-I with List-II.
List-I (Conversion) List-II (Reagents, Conditions used) [cite: 248, 249]
(A) Chlorobenzene ightarrow Phenol(I) textWarm, textH_2textO
(B) p-Nitrochlorobenzene ightarrow p-Nitrophenol(II) (a) textNaOH, 368text K; (b) textH3textO^+ (C) 2,4-Dinitrochlorobenzene ightarrow 2,4-Dinitrophenol(III) (a) textNaOH, 443text K; (b) textH_3textO^+ (D) 2,4,6-Trinitrochlorobenzene ightarrow 2,4,6-Trinitrophenol(IV) (a) textNaOH, 623text K, 300text atm; (b) textH_3textO^+ Choose the correct answer from the options given below:
  • A. text(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  • B. text(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  • C. text(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  • D. text(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Solution

### Related Formula textRate of S_NtextAr propto textNumber of electron-withdrawing groups (-I, -M) at ortho/para positions ### Core Logic Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the textC-Cl bond. However, the presence of strong electron-withdrawing groups (-textNO_2) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion: - (A) Chlorobenzene: Needs extreme conditions: textNaOH at 623text K, 300text atm (Dow's Process) ightarrow (IV) - (B) p-Nitrochlorobenzene: One para -textNO_2 group softens required temperature to 443text K ightarrow (III) - (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368text K ightarrow (II) - (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water ightarrow (I) ### Step 1: Final Match Alignment Matching sequences cleanly yields: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). ### Pattern Recognition The more -textNO_2 groups present on the ring, the less aggressive the reagent/temperature setup required. Count -textNO_2 groups: 0 ightarrow 623textK, 1 ightarrow 443textK, 2 ightarrow 368textK, 3 ightarrow textwarm water. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)