The major product of the following reaction is:
Dihaloalkane starting reactant structure for Q38
The image details a multi-halogenated hydrocarbon chain reacting with excess ethanolic potassium hydroxide under heating conditions.

Solution & Explanation

### Related Formula Base-induced dehydrohalogenation follows the Zaitsev rule to maximize thermodynamic stability via conjugated double bond networks: R-CHX-CH_2-CHX-R' xrightarrowtextexcess KOH/EtOH, Delta textConjugated Diene ### Core Logic The reactant is a dihalide containing a phenyl substitution. Treating with excess alcoholic KOH and heat induces double dehydrohalogenation via successive E2 elimination pathways. The eliminations occur to yield the most stable, highly conjugated product where the double bonds are conjugated with each other and, if possible, with the aromatic phenyl ring system. ### Step 1: Eliminating and Tracking Conjugation Eliminating the first and second equivalents of HBr sets up a conjugated diene system along the heptadiene chain. Tracing carbon numbers correctly from the end closest to the phenyl ring reveals that the conjugated diene centers sit across carbons 2 and 4, producing **2-Phenylhepta-2,4-diene**.
Elimination mechanism steps for conjugated diene synthesis
The image details a multi-halogenated hydrocarbon chain reacting with excess ethanolic potassium hydroxide under heating conditions.
### Pattern Recognition When dealing with excess elimination agents on dihalides, look for options that form a *continuous conjugated diene* structure (alternating double-single-double bonds). This conjugation offers significant thermodynamic stability, especially when directly extended from a phenyl group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Elimination mechanism steps for conjugated diene synthesis
The image details a multi-halogenated hydrocarbon chain reacting with excess ethanolic potassium hydroxide under heating conditions.

Reference Study Guides

Haloalkanes And Haloarenes

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
Q 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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