The products A and B in the following reactions, respectively are A A g - N O _ 2 C H _ 3 - C H _ 2 - C H _ 2 - B r A g C N B$$\mathrm {A} \xleftarrow {\mathrm {A g} - \mathrm {N O} _ {2}} \mathrm {C H} _ {3} - \mathrm {C H} _ {2} - \mathrm {C H} _ {2} - \mathrm {B r} \xrightarrow {\mathrm {A g C N}} \mathrm {B}$$

Core Logic

To synthesize 4-bromostyrene starting from ethyl benzene, we must carry out ring functionalization prior to developing the side-chain double bond:

• Ring Bromination: Treatment of ethylbenzene with Br₂$\text{Br}_2$ in the presence of Fe$\text{Fe}$ (or FeBr₃$\text{FeBr}_3$) acts via electrophilic aromatic substitution. The ethyl group is an ortho/para director, yielding 1-bromo-4-ethylbenzene as the major product owing to steric mitigation.
• Side-Chain Halogenation: Free radical substitution with Cl₂$\text{Cl}_2$ under thermal conditions (Δ$\Delta$) or UV light specifically chlorinates the benzylic position because the benzylic radical is exceptionally stable via resonance.
• Elimination: Heating with alcoholic KOH$\text{KOH}$ drives an E2$E2$ elimination of the benzylic chloride, cleanly synthesizing the terminal alkene linkage of the styrene system.

  

Pattern Recognition

If you perform side-chain halogenation/alkene generation first, the ring substitution later would lack para-selectivity control and risk reacting across the alkene path. Hence, ring substitution MUST precede double bond creation.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Hydrocarbons

Core Logic

The mechanism of SN1$S_N1$ substitution proceeds via carbocation intermediate formation. Option (4) gives a triphenylmethyl carbocation (Ph₃C⁺$Ph_{3}C^{+}$), which is exceptionally stable due to extensive delocalization of positive charge across three phenyl rings (resonance stabilization via 9 canonical structures).

Step 1: Stability Comparison

Stability sequence:

Pattern Recognition

Look for maximum phenyl groups attached directly to the carbon bearing the leaving group to maximize resonance contribution.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

Statement I is correct because the nitrogen atom contains a lone pair situated at the β$\beta$-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 β$\beta$ to a leaving group. Shortcut: NGP (Neighboring Group Participation) accelerates substitution dramatically via intramolecular assistance.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

• Statement I is incorrect: Treatment of alkyl chlorides with aqueous KOH$KOH$ yields alcohols via a nucleophilic substitution (SN$S_N$) reaction, not an elimination reaction.
• Statement II is correct: Alcoholic KOH$KOH$ acts as a strong base (R-O^-$R-O^-$ ions present), which preferentially abstracts a proton from the β$\beta$-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 β$\beta$-hydrogens).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes