Haloalkanes and Haloarenes appeared 38 times across 3 years — 4.4% of Chemistry.
This question is from Preparation and Reactions of Styrene derivatives.
Choose the correct set of reagents for the following conversion:
Ethyl benzene 4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$
{{Q_IMG1}}
The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.
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. The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.
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
Which among the following halides will generate the most stable carbocation in Nucleophilic substitution reaction?
A. Allylic halide option (1)
B. Secondary halide option (2)
C. Secondary benzylic halide option (3)
D. Triphenylmethyl halide option (4)
Solution
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).
Nucleophilic Substitution Mechanisms diagram for Q30 - JEE Main 2025 Evening
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
Q29jee_main_2025_28_jan_morningAlkaline Hydrolysis and NGP
Given below are two statements :
Statement I: Et₂N-CH₂-CH₂-Cl$\mathrm{Et}_2\mathrm{N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}$ will undergo alkaline hydrolysis at a faster rate than Et₂CH-CH₂-Cl$\mathrm{Et}_2\mathrm{CH}-\mathrm{CH}_2-\mathrm{Cl}$.
Statement II: In Et₂N-CH₂-CH₂-Cl$\mathrm{Et}_2\mathrm{N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}$, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.Both Statement I and Statement II are incorrect$\text{Both Statement I and Statement II are incorrect}$
B.Statement I is incorrect but statement II is correct$\text{Statement I is incorrect but statement II is correct}$
C.Both Statement I and Statement II are correct$\text{Both Statement I and Statement II are correct}$
D.Statement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
Solution
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.
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$$\mathrm {A} \xleftarrow {\mathrm {A g} - \mathrm {N O} _ {2}} \mathrm {C H} _ {3} - \mathrm {C H} _ {2} - \mathrm {C H} _ {2} - \mathrm {B r} \xrightarrow {\mathrm {A g C N}} \mathrm {B}$$
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:
A = CH₃-CH₂-CH₂-NO₂$$\mathrm{A} = \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$$
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:
B = CH₃-CH₂-CH₂-NC$$\mathrm{B} = \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NC}$$
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
Qjee_main_2025_04_april_eveningSubstitution versus Elimination
Given below are two statements :
Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.
Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the β$\beta$-carbon.
In the light of the above statements, choose the most appropriate answer from the options given below:
A. Both Statement I and Statement II are incorrect
B. Statement I is incorrect but Statement II is correct
C. Statement I is correct but Statement II is incorrect
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).
Rate of SNAr ∝ Number of electron-withdrawing groups (-I, -M) at ortho/para positions$$\text{Rate of } S_N\text{Ar} \propto \text{Number of electron-withdrawing groups (-I, -M) at ortho/para positions} $$
Core Logic
Aryl halides are generally unreactive towards nucleophilic substitution due to resonance stabilization of the C-Cl$\text{C-Cl}$ bond. However, the presence of strong electron-withdrawing groups (-NO₂$-\text{NO}_2$) at ortho and para positions dramatically increases reactivity by stabilizing the intermediate carbanion:
- (A) Chlorobenzene: Needs extreme conditions: NaOH at 623 K, 300 atm$\text{NaOH at } 623\text{ K, } 300\text{ atm}$ (Dow's Process) arrow$\rightarrow$ (IV)
- (B) p-Nitrochlorobenzene: One para -NO₂$-\text{NO}_2$ group softens required temperature to 443 K$443\text{ K}$arrow$\rightarrow$ (III)
- (C) 2,4-Dinitrochlorobenzene: Two electron-withdrawing groups lower needed temperature further to 368 K$368\text{ K}$arrow$\rightarrow$ (II)
- (D) 2,4,6-Trinitrochlorobenzene: Highly activated picryl chloride hydrolyzes smoothly with just warm water arrow$\rightarrow$ (I)
The more -NO₂$-\text{NO}_2$ groups present on the ring, the less aggressive the reagent/temperature setup required. Count -NO₂$-\text{NO}_2$ groups: 0 arrow 623K$0 \rightarrow 623\text{K}$, 1 arrow 443K$1 \rightarrow 443\text{K}$, 2 arrow 368K$2 \rightarrow 368\text{K}$, 3 arrow warm water$3 \rightarrow \text{warm water}$.
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.
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