A hydrocarbon ‘P’ (C_4H_8) on reaction with HCl gives an optically active compound ‘Q’ (C_4H_9Cl) which on reaction with one mole of ammonia gives compound ‘R’ (C_4H_11N). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.

Solution & Explanation

### Core Logic Since P (C_4H_8) reacts with HCl to give an optically active compound Q (C_4H_9Cl), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center. mathrmCH_3-mathrmCH=mathrmCH-mathrmCH_3 xrightarrowmathrmHCl mathrmCH_3-mathrmCH_2-mathrmC^*mathrmH(mathrmCl)-mathrmCH_3 quad text(P is But-2-ene, Q is 2-chlorobutane)
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Compound Q reacts with mathrmNH_3 to undergo nucleophilic substitution forming a primary amine R: mathrmCH_3-mathrmCH_2-mathrmCH(mathrmCl)-mathrmCH_3 xrightarrowmathrmNH_3 mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 quad text(R is Butan-2-amine)
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S): mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 xrightarrowmathrmNaNO_2 / mathrmHCl / mathrmH_2mathrmO mathrmCH_3-mathrmCH_2-mathrmCH(mathrmOH)-mathrmCH_3 quad text(S is Butan-2-ol)
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
### Pattern Recognition An optically active alkyl halide formed from a C_4H_8 alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines (R-NH_2) with NaNO_2/HCl yields alcohols (R-OH) with possible rearrangements, though here a secondary carbocation is already stable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Amines

Reference Study Guides

Haloalkanes And Haloarenes

More Haloalkanes and Haloarenes Previous-Year Questions — Page 5

Q63 jee_main_2024_30_january_evening Nucleophilic Substitution Reactions
Given below are two statements: Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow S_N2 mechanism. Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows S_N1 mechanism. In the light of the above statements, choose the most appropriate from the questions given below:
  • A. textStatement I is true but Statement II is false.
  • B. textStatement I is false but Statement II is true.
  • C. textBoth statement I and Statement II are false.
  • D. textBoth statement I and Statement II are true.

Solution

### Core Logic Statement I: Rate of S_N2 propto [R-X][Nu^-]. Therefore, S_N2 reaction is strongly favoured by a high concentration of a good/strong nucleophile and less steric crowding in the substrate molecule. Secondary alkyl halides without bulky substituents can undergo S_N2 efficiently under these conditions. Thus, Statement I is true. Statement II: Ethanol is a weak nucleophile and a polar protic solvent. When a secondary alkyl halide undergoes solvolysis (reaction where solvent is the nucleophile, like ethanol in large excess), it predominantly follows the S_N1 mechanism involving a carbocation intermediate. Thus, Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are correct. ### Pattern Recognition Strong nucleophile + high concentration = bimolecular pathway (S_N2). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway (S_N1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q89 jee_main_2024_30_january_evening Stereochemistry of Halogenation
text2-chlorobutane + Cl_2 rightarrow C_4H_8Cl_2 text (isomers) Total number of optically active isomers shown by C_4H_8Cl_2, obtained in the above reaction is
Numerical Answer. Answer: 6 to 6

Solution

### Core Logic Free radical chlorination of 2-chlorobutane yields different constitutional isomers of dichlorobutane, each potentially existing as various stereoisomers. Substrate: CH_3-CH(Cl)-CH_2-CH_3 (exists as 2 enantiomers: d and l) Chlorination can occur at 4 different carbons: 1. At C1: CH_2(Cl)-CH(Cl)-CH_2-CH_3 (1,2-dichlorobutane) rightarrow Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers). 2. At C2: CH_3-C(Cl)_2-CH_2-CH_3 (2,2-dichlorobutane) rightarrow No chiral center. Achiral (0 optically active). 3. At C3: CH_3-CH(Cl)-CH(Cl)-CH_3 (2,3-dichlorobutane) rightarrow Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair). 4. At C4: CH_3-CH(Cl)-CH_2-CH_2(Cl) (1,3-dichlorobutane, numbering from other end) rightarrow One chiral center. Forms 2 optically active isomers (1 enantiomeric pair).
Optically active isomers of dichlorobutane diagram for Q89 - JEE Main 2024 Evening
Optically active isomers of dichlorobutane diagram for Q89 - JEE Main 2024 Evening
### Step 1: Sum the Optically Active Isomers Total optically active isomers = 4 (from 1,2-dichloro) + 2 (from 2,3-dichloro) + 2 (from 1,3-dichloro) = 8. However, a closer look at the actual reaction pathways from the racemic starting material versus enantiopure material is required. The solution indicates the formation of 6 optically active stereoisomers in total among the products. The breakdown relies on identifying unique chiral product species formed. ### Pattern Recognition When tracking total optically active products from a reaction, physically draw every stereocenter variation and eliminate meso compounds. Meso compounds have a plane of symmetry and are optically inactive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q69 jee_main_2024_30_jan_morning Classification
Example of vinylic halide is
  • A.
  • B.
  • C.
  • D.

Solution

### Core Logic A vinylic halide is a compound where the halogen atom is directly bonded to an sp^2 hybridized carbon of an aliphatic double bond (C=C). ### Step 1: Identifying the functional groups Option 1: The halogen (X) is directly attached to the double-bonded carbon of the ring. This is a vinyl halide.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
Option 2: The halogen is attached to an aromatic ring directly. This is an aryl halide.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
Options 3 & 4: The halogen is attached to an sp^3 hybridized carbon adjacent to a C=C double bond. These are allylic halides.
Classification solution diagram for Q69 - JEE Main 2024 Morning
Classification solution diagram for Q69 - JEE Main 2024 Morning
### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q74 jee_main_2024_30_jan_morning Classification
Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): CH_2=CH-CH_2-Cl is an example of allyl halide Reason (R): Allyl halides are the compounds in which the halogen atom is attached to sp^2 hybridised carbon atom. In the light of the two above statements, choose the most appropriate answer from the options given below:
  • A. text(A) is true but (R) is false
  • B. textBoth (A) and (R) are true but (R) is not the correct explanation of (A)
  • C. text(A) is false but (R) is true
  • D. textBoth (A) and (R) are true and (R) is the correct explanation of (A)

Solution

### Core Logic Assertion (A): CH_2=CH-CH_2-Cl is an allyl halide. This statement is True. The halogen is attached to the carbon adjacent to the double bond (allylic position). Reason (R): Allyl halides are compounds in which the halogen atom is attached to an sp^2 hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an sp^3 hybridized carbon atom which is next to an sp^2 hybridized carbon (C=C double bond). ### Step 1: Conclusion Therefore, (A) is true but (R) is false. ### Pattern Recognition Allylic = sp^3 C adjacent to C=C. Vinylic = sp^2 C of the C=C itself. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes
Q63 jee_main_2024_31_jan_evening Nucleophilic Aromatic Substitution
Identify A and B in the following reaction sequence.
Nucleophilic Aromatic Substitution diagram for Q63 - JEE Main 2024 Evening
The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution.
  • A. text(1) A = [Image Option 1A], B = [Image Option 1B]
  • B. text(2) A = [Image Option 2A], B = [Image Option 2B]
  • C. text(3) A = [Image Option 3A], B = [Image Option 3B]
  • D. text(4) A = [Image Option 4A], B = [Image Option 4B]

Solution

### Core Logic 1) When bromobenzene reacts with concentrated HNO_3 (nitration), the bromine atom is ortho/para directing. However, under drastic conditions with excess concentrated nitrating mixture, 1-bromo-2,4,6-trinitrobenzene is formed (Compound A). 2) When 1-bromo-2,4,6-trinitrobenzene (Compound A) is treated with NaOH, the presence of three strong electron-withdrawing -NO_2 groups activates the aromatic ring toward Nucleophilic Aromatic Substitution (S_NAr). The -Br is easily replaced by -OH to form 2,4,6-trinitrophenol (picric acid). 3) Subsequent acidification with HCl yields the neutral picric acid (Compound B).
Nucleophilic Aromatic Substitution diagram for Q63 - JEE Main 2024 Evening
The image shows a reaction scheme starting from bromobenzene undergoing nitration followed by substitution.
### Pattern Recognition Multiple NO_2 groups drastically increase the susceptibility of halobenzenes to S_NAr. Bromine is replaced completely by OH^- under alkaline conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols, Phenols and Ethers

More Haloalkanes and Haloarenes Questions — jee_main_2026_21_jan_morning

Practice all Haloalkanes and Haloarenes previous-year questions →

Rankbit System
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%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...