Haloalkanes And Haloarenes – JEE Main ChemistryPractice Questions & PYQs

The Balz-Schiemann reaction is the preferred method for synthesizing fluorobenzene because direct fluorination of benzene is highly violent and uncontrollable. Aniline is first converted to a diazonium salt and then reacted with fluoroboric acid ($\mathrm{HBF_4}$) to form stable benzene diazonium fluoroborate ($\mathrm{C_6H_5N_2^+BF_4^-}$), which thermally decomposes to yield the desired product according to the equation $\mathrm{C_6H_5N_2^+BF_4^-} \xrightarrow{\Delta} \mathrm{C_6H_5F} + \mathrm{N_2} + \mathrm{BF_3}$.

Nucleophilic aromatic substitution is inherently difficult because the electron-rich aromatic $\pi$-system repels nucleophiles and the resonance effect confers partial double-bond character to the carbon-substituent bond, making it exceptionally stable. This electronic environment prevents any incoming group from effectively displacing the substituent directly attached to the aromatic nucleus. Because nuclear substitution will not take place under these conditions, the reaction pathway remains inactive, leaving the original substituent intact and resulting in the specific molecular structure shown.

Bimolecular elimination (E2) reactions require the leaving group and the beta-hydrogen to be arranged in an anti-periplanar geometry within a single plane. In the trans-2-phenyl-1-bromocyclopentane molecule, the bulky phenyl group constrains the ring conformation such that the hydrogen on the C2 carbon is not properly aligned to be anti-periplanar to the bromine atom when reacting with $\mathrm{KOH}$. Consequently, the elimination must involve a hydrogen atom on the C5 carbon, which is correctly positioned to satisfy the anti-periplanar requirement relative to the bromine. This regioselectivity leads to the formation of the double bond between the C1 and C5 carbons, resulting in the production of 3-phenylcyclopentene.

The $S_N2$ mechanism involves a concerted approach where a nucleophile attacks the electrophilic carbon atom from the backside, directly opposite the leaving group. In tertiary alkyl halides, the central carbon is shielded by three bulky alkyl groups, creating significant spatial congestion that physically prevents the nucleophile from reaching the reactive site. This strong steric hindrance renders the $S_N2$ pathway practically impossible, explaining why reactivity follows the trend $\mathrm{CH}_2-\mathrm{X} > \mathrm{CH}_3-\mathrm{CH}_2 \mathrm{X} > (\mathrm{CH}_3)_2\mathrm{CH} \cdot \mathrm{X} > (\mathrm{CH}_3)_3\mathrm{C}-\mathrm{X}$ and forces these compounds to undergo substitution via the $S_N1$ mechanism.

The Corey-House synthesis provides a precise method for coupling two alkyl groups to form a larger, saturated carbon chain. When an alkyl halide reacts with a lithium dialkyl cuprate, the alkyl group from the copper reagent replaces the halogen atom on the substrate, creating a new carbon-carbon bond and producing an alkane. This specific transformation is defined by the chemical equation $R'X + \mathrm{Li}R_2 + \mathrm{Cu} \rightarrow R'-R + R\mathrm{Cu} + \mathrm{LiX}$, illustrating how the reagents combine to construct the desired hydrocarbon framework while eliminating the halide.

The dehydrohalogenation of 2-bromobutane is governed by Saytzeff's rule, which dictates that the reaction favors the formation of the most highly substituted alkene. Removing a hydrogen atom from the carbon at the third position leads to the formation of $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$, whereas removal from the first position would yield a less substituted terminal alkene. Since the internal double bond provides greater thermodynamic stability than a terminal one, the more substituted isomer, 2-butene, is generated as the predominant product.

The synthesis of DDT relies on an electrophilic aromatic substitution reaction where chlorobenzene undergoes condensation with chloral ($CCl_3CHO$) under acidic conditions. When concentrated sulphuric acid is added as a dehydrating agent, it promotes the elimination of water between the aldehyde group of the chloral and the para-positions of the two chlorobenzene rings, resulting in the formation of the compound known as DDT.

The reactivity difference between aryl halides and benzyl halides is governed by the strength of the carbon-halogen bond. In phenyl iodide ($\mathrm{C}_6\mathrm{H}_5\mathrm{I}$), the iodine atom is bonded directly to the benzene ring, where resonance imparts partial double bond character to the bond, making it resistant to substitution by $\mathrm{NaOH}$. Conversely, benzyl iodide ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{I}$) features the iodine atom on a saturated side chain, which undergoes easy nucleophilic substitution to release iodide ions when boiled with $\mathrm{NaOH}$. Consequently, only the solution containing $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{I}$ will generate a yellow precipitate of $\mathrm{AgI}$ upon acidification and treatment with $\mathrm{AgNO}_3$, identifying it as substance B and confirming that substance A is $\mathrm{C}_6\mathrm{H}_5\mathrm{I}$.

At its heart, this process is a classic example of nucleophilic substitution, where one functional group is effectively swapped for another within the molecule. When $(\mathrm{CH_3})_3\mathrm{C-Br}$ undergoes hydrolysis, the bromine atom acts as a leaving group and departs the central carbon during a slow, rate-determining step, which generates a carbocation intermediate. This intermediate is then rapidly attacked by a water molecule in a subsequent fast step to produce $(\mathrm{CH_3})_3\mathrm{C-OH}$. Because this mechanism involves the specific replacement of the bromine atom with a hydroxyl group rather than an elimination or radical process, it is classified as a substitution reaction.

A geminal dihalide undergoes hydrolysis with aqueous $KOH$ to yield a carbonyl compound, and the subsequent reaction with a Grignard reagent followed by acidic hydrolysis produces an alcohol, which can be confirmed by the Lucas test. In this reaction sequence, compound $I$ must be 2,2-dichloropropane ($\mathrm{CH_3-CCl_2-CH_3}$) because its alkaline hydrolysis yields acetone ($\mathrm{CH_3COCH_3}$) as compound $II$. Acetone subsequently reacts with $\mathrm{CH_3MgBr}$ followed by $\mathrm{H_2O/H^+}$ to produce tert-butyl alcohol ($\mathrm{(CH_3)_3COH}$) as compound $III$. As tert-butyl alcohol is a tertiary alcohol, it reacts instantaneously with the Lucas reagent ($\mathrm{Anhy. ZnCl_2 + Conc. HCl}$) to form an alkyl chloride, which manifests as immediate turbidity in the solution.