Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Haloalkanes react with KCN$KCN$ to form alkyl cyanides as a main product while with AgCN$AgCN$ form isocyanide as the main product. Reason (R) : KCN$KCN$ and AgCN$AgCN$ both are highly ionic compounds. In the light of the above statement, choose the most appropriate answer from the options given below:

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

Core Logic

Both silver reagents exhibit significantly covalent bond characters:

• Reaction with AgNO₂$\mathrm{AgNO}_2$: 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:

• Reaction with AgCN$\mathrm{AgCN}$: The covalent Ag-C$\mathrm{Ag}-\mathrm{C}$ bond directs the nucleophilic attack to proceed through the lone pair on nitrogen, yielding an isocyanide compound:

Hence, option (4) represents the correct combination.

Pattern Recognition

Sees: Alkyl halide reacting with covalent silver salts of ambident anions. Shortcut: Silver reagents (AgCN$\mathrm{AgCN}$ or AgNO₂$\mathrm{AgNO}_2$) drive bond formatting via the nitrogen center, producing isocyanides and nitroalkanes respectively.

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