Choose the correct set of reagents for the following conversion: Ethyl benzene 4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$ {{Q_IMG1}}

Core Logic

Decode the IUPAC name: 2,3-dibromo-1-phenylpentane.

• Parent chain is pentane (5 carbon chain: C1-C2-C3-C4-C5).
• 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.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Step 1: Elimination to form Propene

Step 2: Electrophilic Addition of HBr

Addition of HBr$HBr$ follows Markovnikov's rule:

Step 3: Nucleophilic Substitution

Reaction with aqueous KOH leads to SN2/SN1$S_N2/S_N1$ substitution of Br^-$Br^-$ with OH^-$OH^-$:

Pattern Recognition

Alc. KOH gives elimination (alkene). Aq. KOH gives substitution (alcohol). HBr$HBr$ on unsymmetrical alkene gives Markovnikov addition.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

Reaction 1: If the reagent is alcoholic NaOH (implied due to formation of a distinct product A to contrast B), elimination occurs:

Hydrogen atoms in ethene (C₂H₄$C_2H_4$) = 4.

Reaction 2: If the reagent is aqueous NaOH (NaOH / H₂O$NaOH / H_2O$), nucleophilic substitution (SN2$S_N2$) occurs:

Hydrogen atoms in ethanol (C₂H₆O$C_2H_6O$) = 6.

Total hydrogen atoms = 4 + 6 = 10.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes