A hydrocarbon ‘P’ (C_4H_8$C_{4}H_{8}$) on reaction with HCl gives an optically active compound ‘Q’ (C_4H_9Cl$C_{4}H_{9}Cl$) which on reaction with one mole of ammonia gives compound ‘R’ (C_4H_11N$C_{4}H_{11}N$). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.
### Core Logic
Since P (C_4H_8$C_4H_8$) reacts with HCl to give an optically active compound Q (C_4H_9Cl$C_4H_9Cl$), 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)$$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{HCl}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C}^*\mathrm{H}(\mathrm{Cl})-\mathrm{CH}_3 \quad \text{(P is But-2-ene, Q is 2-chlorobutane)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Compound Q reacts with mathrmNH_3$\mathrm{NH}_3$ 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)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Cl})-\mathrm{CH}_3 \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \quad \text{(R is Butan-2-amine)}$$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)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HCl} / \mathrm{H}_2\mathrm{O}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \quad \text{(S is Butan-2-ol)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
### Pattern Recognition
An optically active alkyl halide formed from a C_4H_8$C_4H_8$ 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$R-NH_2$) with NaNO_2/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Amines
Keywords:#hydrocarbon P C4H8#JEE Main 2026 Morning Q54#Haloalkanes and Haloarenes JEE Main 2026#Reactions of Haloalkanes JEE Main 2026
More Haloalkanes and Haloarenes Previous-Year Questions — Page 6
Q67jee_main_2024_31_jan_eveningIUPAC Nomenclature of Haloalkanes
Identify structure of 2,3-dibromo-1-phenylpentane.
A.text(1) Structure A$\text{(1) Structure A}$
B.text(2) Structure B$\text{(2) Structure B}$
C.text(3) Structure C$\text{(3) Structure C}$
D.text(4) Structure D$\text{(4) Structure D}$
Solution
### Core Logic
Decode the IUPAC name: 2,3-dibromo-1-phenylpentane.
1) Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5).
2) Substituents:
- Phenyl group at position 1.
- Bromo groups at positions 2 and 3.
Option (3) correctly displays a 5-carbon straight chain. The first carbon attaches to the benzene ring (phenyl group), and the second and third carbons each hold a bromine atom.
IUPAC Nomenclature of Haloalkanes diagram for Q67 - JEE Main 2024 Evening
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Q68jee_main_2024_31_jan_morningElimination and Addition Reactions
The product (C) in the below mentioned reaction is:
CH_3-CH_2-CH_2-Br xrightarrow[Delta]KOH_(alc) A xrightarrow[Delta]HBr B xrightarrow[Delta]KOH_(aq) C$$CH_3-CH_2-CH_2-Br \xrightarrow[\Delta]{KOH_{(alc)}} A \xrightarrow[\Delta]{HBr} B \xrightarrow[\Delta]{KOH_{(aq)}} C$$
Q86jee_main_2024_31_jan_morningElimination and Substitution
CH_3CH_2Br + NaOH rightarrow textProduct A$$CH_3CH_2Br + NaOH \rightarrow \text{Product A}$$CH_3CH_2Br + NaOH / H_2O rightarrow textProduct B$$CH_3CH_2Br + NaOH / H_2O \rightarrow \text{Product B}$$
The total number of hydrogen atoms in product A and product B is
Numerical Answer.Answer: 10 to 10
Solution
### Core Logic
Reaction 1:
If the reagent is alcoholic NaOH (implied due to formation of a distinct product A to contrast B), elimination occurs:
CH_3CH_2Br + NaOH (textalc) rightarrow CH_2=CH_2 text (Ethene)$$CH_3CH_2Br + NaOH (\text{alc}) \rightarrow CH_2=CH_2 \text{ (Ethene)}$$
Hydrogen atoms in ethene (C_2H_4$C_2H_4$) = 4.
Reaction 2:
If the reagent is aqueous NaOH (NaOH / H_2O$NaOH / H_2O$), nucleophilic substitution (S_N2$S_N2$) occurs:
CH_3CH_2Br + NaOH (textaq) rightarrow CH_3CH_2OH text (Ethanol)$$CH_3CH_2Br + NaOH (\text{aq}) \rightarrow CH_3CH_2OH \text{ (Ethanol)}$$
Hydrogen atoms in ethanol (C_2H_6O$C_2H_6O$) = 6.
Total hydrogen atoms = 4 + 6 = 10.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
More Haloalkanes and Haloarenes Questions — jee_main_2026_21_jan_morning
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