Organic Compounds Containing Halogens MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Haloarenes contain a halogen atom directly attached to an aromatic-ring carbon, which is \(sp^2\)-hybridised. In chlorobenzene, \(\mathrm{C_6H_5Cl}\), chlorine is bonded directly to a carbon atom of the benzene ring. This satisfies the structural requirement for an aryl halide or haloarene. Benzyl chloride, \(\mathrm{C_6H_5CH_2Cl}\), also contains a benzene ring, but its chlorine atom is attached to the side-chain carbon rather than to the ring. That carbon is \(sp^3\)-hybridised, so benzyl chloride belongs to the benzylic-halide category. In \(\mathrm{CH_2{=}CHCH_2Cl}\), chlorine is attached to an \(sp^3\)-hybridised carbon adjacent to a carbon–carbon double bond, making it an allylic halide. In \(\mathrm{CH_2{=}CHCl}\), chlorine is directly attached to a carbon of the double bond, making it a vinylic halide. Thus, merely having an aromatic ring or an unsaturated bond somewhere in a molecule does not determine its halide classification. The decisive feature is the bonding environment of the carbon directly attached to the halogen.

Explanation: The primary, secondary or tertiary character of an alkyl halide is determined by the number of carbon atoms directly attached to the carbon bearing the halogen. It is not determined by the total number of carbon atoms or by the overall branching of the molecule. In compound P, bromine is attached to the terminal \(\mathrm{CH_2}\) group. That carbon is bonded to only one other carbon atom, so P is a primary alkyl bromide. The neighbouring carbon happens to be highly branched, but that does not make the bromine-bearing carbon tertiary. Compound P is commonly known as neopentyl bromide. In compound Q, bromine is attached to the central carbon of the tert-butyl group. This carbon is directly bonded to three methyl groups and is therefore tertiary. Compound Q is tert-butyl bromide. This distinction is especially important because a highly branched primary halide can exhibit reaction behaviour very different from an ordinary unbranched primary halide. Classification must always begin at the carbon directly bonded to the halogen.

Explanation: The classification depends on both the hybridisation of the chlorine-bearing carbon and its position relative to a neighbouring \(\pi\)-system. In P, chlorine is attached to the \(\mathrm{CH_2}\) carbon directly connected to the benzene ring. This carbon is \(sp^3\)-hybridised and occupies the benzylic position. Therefore, P satisfies both required conditions. In R, chlorine is bonded to the saturated \(\mathrm{CH_2}\) carbon immediately adjacent to the carbon–carbon double bond. This carbon is also \(sp^3\)-hybridised and occupies the allylic position. Compound Q is chlorobenzene, where chlorine is attached directly to an \(sp^2\)-hybridised aromatic carbon. Compound S is a vinylic halide because its chlorine-bearing carbon is part of the carbon–carbon double bond and is consequently \(sp^2\)-hybridised. The presence of a benzene ring or double bond elsewhere in the molecule does not by itself establish the required classification. Both the immediate bonding environment and the position of the halogen must be checked. Hence, the qualifying compounds are the benzylic halide P and the allylic halide R.

Explanation: The conversion of an alcohol into an alkyl chloride involves replacement of its hydroxyl group by chlorine. However, alcohols do not all undergo this conversion with equal ease. Compound X is 2-methylpropan-2-ol, a tertiary alcohol, whereas Y is butan-1-ol, a primary alcohol. Under the specified conditions, tertiary alcohols react readily with concentrated hydrochloric acid at room temperature. Protonation of the hydroxyl group makes departure as water possible, and the resulting tertiary carbocation is relatively stable. The three alkyl groups attached to the positively charged carbon help stabilise it through their electron-releasing effects and hyperconjugation. Consequently, chloride can replace the hydroxyl group relatively rapidly in X. A primary alcohol such as Y does not readily form a primary carbocation under the same conditions. NCERT therefore specifies a catalyst such as anhydrous \(\mathrm{ZnCl_2}\) for reactions of primary and secondary alcohols with HCl. The general reactivity order of alcohols towards a given hydrogen halide is \(3^\circ \gt 2^\circ \gt 1^\circ\). The conclusion concerns the specified reaction conditions and must not be interpreted as a universal reactivity order for every reaction of alcohols.

Explanation: Thionyl chloride, \(\mathrm{SOCl_2}\), is an especially useful reagent for converting alcohols into alkyl chlorides. In this reaction, the hydroxyl group of the alcohol is replaced by chlorine while the carbon skeleton is retained. For propan-1-ol, the overall reaction is: \[ \mathrm{CH_3CH_2CH_2OH+SOCl_2\rightarrow CH_3CH_2CH_2Cl+SO_2\uparrow+HCl\uparrow} \] The desired organic product is 1-chloropropane. The other two products are sulphur dioxide and hydrogen chloride, both of which are gases under ordinary reaction conditions. Their escape from the reaction mixture reduces contamination by inorganic reaction products and helps isolate the alkyl chloride. Phosphorus chlorides can also convert alcohols into alkyl chlorides, but their reactions may leave phosphorus-containing inorganic products that require separation. The advantage of thionyl chloride is therefore related to product isolation rather than a claim that no other chlorinating reagent can perform the conversion. It is also important that thionyl chloride does not introduce another carbon atom or change the location of the carbon bearing the original hydroxyl group. The reaction provides a direct and convenient alcohol-to-alkyl-chloride preparation route.

Explanation: \(\textbf{Evaluate the Assertion:}\) Cyclohexanol is an aliphatic alcohol containing a hydroxyl group bonded to an \(sp^3\)-hybridised carbon. Its hydroxyl group can be replaced by chlorine using a suitable chlorinating reagent such as thionyl chloride. Phenol has a different bonding environment because its hydroxyl group is directly attached to the aromatic ring. Ordinary reagents used for converting aliphatic alcohols into alkyl chlorides are not suitable for preparing chlorobenzene directly from phenol. Therefore, the Assertion is true. \(\textbf{Evaluate the Reason:}\) In phenol, a lone pair of electrons on oxygen can participate in resonance with the aromatic ring. This delocalisation gives the bond between the aromatic carbon and oxygen partial double-bond character. The bond is consequently stronger and more difficult to break than an ordinary carbon–oxygen single bond. Cyclohexanol does not possess the corresponding aromatic resonance interaction because its hydroxyl-bearing carbon is \(sp^3\)-hybridised. Therefore, the Reason is also true. \(\textbf{Establish the causal relationship:}\) The resonance-strengthened carbon–oxygen bond accounts for the failure of ordinary alcohol-to-halide methods in phenol. The Reason directly explains the contrast stated in the Assertion. This illustrates why reagents suitable for preparing haloalkanes cannot automatically be applied to the preparation of haloarenes.

Explanation: Both transformations require replacement of an alcoholic hydroxyl group by a particular halogen without changing the number or arrangement of carbon atoms. The first target is an alkyl bromide, whereas the second is an alkyl iodide. Phosphorus tribromide is a suitable reagent for preparing alkyl bromides from alcohols. It can be generated in situ by reacting red phosphorus with bromine: \[ \mathrm{2P+3Br_2\rightarrow 2PBr_3} \] The resulting phosphorus tribromide converts propan-1-ol into 1-bromopropane: \[ \mathrm{3CH_3CH_2CH_2OH+PBr_3\rightarrow 3CH_3CH_2CH_2Br+H_3PO_3} \] For the second conversion, phosphorus triiodide is generated using red phosphorus and iodine: \[ \mathrm{2P+3I_2\rightarrow 2PI_3} \] It then replaces the hydroxyl group of butan-2-ol by iodine: \[ \mathrm{3CH_3CH(OH)CH_2CH_3+PI_3\rightarrow 3CH_3CH(I)CH_2CH_3+H_3PO_3} \] \(\textbf{Chemical consistency check:}\) The first reaction introduces bromine at carbon 1 of the original propanol chain, while the second introduces iodine at carbon 2 of the original butanol chain. Neither transformation requires a carbon–carbon bond to be formed or broken. Thionyl chloride would introduce chlorine rather than the requested bromine or iodine. Thus, the appropriate in situ reagent systems are red phosphorus with bromine and red phosphorus with iodine, respectively.

Explanation: The intended transformation is replacement of the hydroxyl group of propan-1-ol by iodine. An iodide source must therefore be available in a medium suitable for alcohol-to-iodide conversion. NCERT describes the preparation of alkyl iodides by heating alcohols with sodium or potassium iodide in 95% orthophosphoric acid. The acidic medium helps make the alcoholic hydroxyl group more suitable for replacement, while iodide supplies the required halogen. Concentrated sulphuric acid presents an additional difficulty because it can act as an oxidising agent. Iodide ions are readily oxidised to molecular iodine under suitable oxidising conditions. This consumes some of the iodide needed for the desired organic substitution and can introduce unwanted side reactions. Orthophosphoric acid is preferred because it does not present the same oxidising interference. The desired product retains the three-carbon chain of propan-1-ol and has iodine attached to the carbon that originally carried the hydroxyl group. The issue is therefore not a rearrangement from carbon 1 to carbon 2, but the chemical compatibility of the iodide reagent with the chosen acid. A suitable preparation method must account for the reactions of the reagents with one another, not merely whether the required halogen is present.

Explanation: \(\textbf{Identify the target structure:}\) 1-Chlorobutane contains a four-carbon saturated chain with chlorine attached to its terminal carbon. The preparation should therefore introduce chlorine at carbon 1 without producing substantial amounts of other constitutional haloalkanes. \(\textbf{Evaluate the direct alcohol route:}\) Butan-1-ol already possesses the required carbon skeleton and has its hydroxyl group at carbon 1. Treatment with thionyl chloride replaces that hydroxyl group by chlorine: \[ \mathrm{CH_3CH_2CH_2CH_2OH+SOCl_2\rightarrow CH_3CH_2CH_2CH_2Cl+SO_2+HCl} \] The reaction retains the original carbon connectivity and gives the required constitutional structure. \(\textbf{Compare the hydrocarbon routes:}\) Free-radical chlorination of butane can replace hydrogen at different carbon positions. Consequently, both 1-chlorobutane and 2-chlorobutane can form, and further chlorination is also possible. Ordinary addition of HCl to but-1-ene follows Markovnikov orientation, favouring 2-chlorobutane rather than the requested terminal chloride. Addition of molecular chlorine to but-1-ene does not give a monochloroalkane; chlorine adds across the double bond to produce 1,2-dichlorobutane. \(\textbf{Preparation decision:}\) The alcohol route achieves the desired carbon–chlorine attachment directly, whereas the hydrocarbon routes either favour a different structure or produce additional products. The most suitable choice is therefore butan-1-ol with thionyl chloride.

Explanation: The number of constitutional monochloro products depends on the number of chemically nonequivalent hydrogen-bearing carbon positions in the starting hydrocarbon. Replacing hydrogens at positions related by molecular symmetry gives the same constitutional product and must not be counted repeatedly. \(\textbf{Analyse the selected structure:}\) 2,3-Dimethylbutane has the condensed structure: \[ \mathrm{(CH_3)_2CH-CH(CH_3)_2} \] Its four methyl groups are constitutionally equivalent because of the symmetry of the carbon skeleton. Replacement of a hydrogen from any one of these methyl groups therefore gives the same constitutional monochloro product. The two central methine carbons are also equivalent to one another. Replacement of a hydrogen attached to either central carbon gives a second constitutional monochloro product. Hence, the molecule contains only two distinct hydrogen environments relevant to constitutional monochlorination. \(\textbf{Check the other structures:}\) n-Hexane has three distinct carbon positions for monochlorination because carbons 1 and 6, 2 and 5, and 3 and 4 form equivalent pairs. In 2,2-dimethylbutane, three distinct hydrogen-bearing carbon environments occur: the equivalent methyl groups attached to the quaternary carbon, the adjacent methylene group and the terminal methyl group. 3-Methylpentane has four distinct hydrogen-bearing carbon environments when its molecular symmetry is considered. Thus, only 2,3-dimethylbutane produces exactly two constitutional monochloro derivatives. The conclusion is about structural products, not their relative amounts or stereochemical forms.

Explanation: \(\textbf{Identify the competing sites:}\) Toluene contains both an aromatic ring and a methyl side chain. Chlorination may occur at different positions depending on the reaction conditions, so identifying the reagent alone is insufficient. \(\textbf{Experiment I — Photochemical conditions:}\) Ultraviolet light promotes free-radical chlorination of the benzylic side chain. A hydrogen of the methyl group is replaced by chlorine, giving benzyl chloride: \[ \mathrm{C_6H_5CH_3+Cl_2\xrightarrow{h\nu}C_6H_5CH_2Cl+HCl} \] The carbon carrying chlorine in the product remains outside the aromatic ring and is \(sp^3\)-hybridised. Benzyl chloride is consequently a benzylic halide, not a haloarene. \(\textbf{Experiment II — Lewis-acid conditions:}\) Iron(III) chloride facilitates electrophilic chlorination of the aromatic ring. The methyl group already attached to the benzene ring directs incoming electrophiles mainly towards the ortho and para positions. The products are therefore principally 2-chlorotoluene and 4-chlorotoluene, generally as a mixture. \(\textbf{Distinguish the products:}\) Both pathways replace a hydrogen atom by chlorine, but they do so at chemically different sites. Photochemical chlorination targets the side chain, whereas Lewis-acid-catalysed aromatic chlorination targets the ring. The reaction conditions, rather than the molecular formula of the starting hydrocarbon, determine the principal product family.

Explanation: \(\textbf{Examine the unsymmetrical alkene:}\) But-1-ene has the structure: \[ \mathrm{CH_2{=}CHCH_2CH_3} \] The double bond connects a terminal carbon bearing two hydrogen atoms and an internal carbon bearing one hydrogen atom. The orientation of hydrogen halide addition therefore determines whether the halogen appears at carbon 1 or carbon 2. \(\textbf{Reaction I — HBr with peroxides:}\) In the presence of suitable peroxides, HBr undergoes radical addition with anti-Markovnikov orientation. Bromine is introduced at the less substituted terminal carbon, while hydrogen is added to the adjacent carbon: \[ \mathrm{CH_2{=}CHCH_2CH_3+HBr\xrightarrow{ROOR}CH_3CH_2CH_2CH_2Br} \] The product is 1-bromobutane. This orientation is associated with formation of the more stable secondary carbon radical during the addition sequence. \(\textbf{Reaction II — HCl with peroxides:}\) The peroxide effect is not generally observed for HCl because the radical-chain propagation steps are not energetically favourable in the required manner. Consequently, the ordinary electrophilic addition orientation remains applicable. Hydrogen adds preferentially to the terminal carbon and chlorine to the more substituted carbon: \[ \mathrm{CH_2{=}CHCH_2CH_3+HCl\rightarrow CH_3CH(Cl)CH_2CH_3} \] The major product is therefore 2-chlorobutane. \(\textbf{Chemical check:}\) The presence of peroxides must not be interpreted as a universal instruction to reverse Markovnikov orientation for every hydrogen halide. HBr exhibits the characteristic peroxide effect, whereas HCl does not. This difference explains why the two reactions give halogens at different positions.

Explanation: Both reactions involve addition across the carbon–carbon double bond of cyclohexene, but the reacting species supply different atoms. This distinction determines both the number of bromine atoms introduced and the product's constitutional structure. \(\textbf{Reaction P — Addition of molecular bromine:}\) A molecule of \(\mathrm{Br_2}\) supplies two bromine atoms. During addition, the carbon–carbon double bond is converted into a single bond, and one bromine atom becomes attached to each of the two formerly double-bonded carbon atoms. The resulting product is a vicinal dibromide: \[ \mathrm{Cyclohexene+Br_2\xrightarrow{CCl_4}1,2\!-\!dibromocyclohexane} \] The two bromine atoms occupy adjacent carbon atoms, so a 1,1-dibromide is not produced by this ordinary addition. \(\textbf{Reaction Q — Addition of hydrogen bromide:}\) Hydrogen bromide supplies one hydrogen atom and one bromine atom across the double bond. Only one carbon–bromine bond is therefore introduced: \[ \mathrm{Cyclohexene+HBr\rightarrow Bromocyclohexane} \] The two alkene carbons in unsubstituted cyclohexene are constitutionally equivalent, so the product does not present a competing positional-isomer problem. \(\textbf{Product comparison:}\) Molecular bromine produces a vicinal dibromide, whereas hydrogen bromide produces a monobromide. The difference cannot be resolved merely by observing that both reagents contain bromine. The complete composition of the reagent and the bonding changes during addition must be considered. Hence, the products are 1,2-dibromocyclohexane and bromocyclohexane, respectively.

Explanation: \(\textbf{Evaluate the Assertion:}\) The Finkelstein reaction is a halogen-exchange method used to prepare suitable alkyl iodides from alkyl chlorides or bromides. Sodium iodide is commonly employed in dry acetone, a solvent that permits the required exchange while favouring separation of the inorganic chloride or bromide salt. In the present case, iodide replaces chlorine at the terminal carbon of 1-chloropropane: \[ \mathrm{CH_3CH_2CH_2Cl+NaI\xrightarrow{dry\ acetone}CH_3CH_2CH_2I+NaCl\downarrow} \] The carbon skeleton is preserved, and the required 1-iodopropane is formed. Thus, the Assertion is true. \(\textbf{Evaluate the Reason:}\) Sodium iodide is sufficiently soluble in dry acetone for the reaction to proceed, whereas the sodium chloride produced has very low solubility in this medium. As sodium chloride separates as a precipitate, it is effectively removed from the reacting solution. This removal favours the forward process according to Le Chatelier's principle. The Reason is therefore true. \(\textbf{Establish the explanation:}\) The solvent is not merely an inactive liquid used to dissolve the organic reactant. Its selective effect on salt solubility contributes directly to the efficiency of the exchange. This precipitation-driven behaviour explains why dry acetone is specified in the standard Finkelstein preparation. Both the Assertion and Reason are true, and the Reason correctly accounts for the Assertion.

Explanation: \(\textbf{Identify the bonding environments:}\) Compound X is benzyl bromide, in which bromine is bonded to the side-chain \(\mathrm{CH_2}\) carbon adjacent to a benzene ring. Compound Y is bromobenzene, in which bromine is bonded directly to an aromatic-ring carbon. Although both molecules contain bromine and a benzene ring, their carbon–bromine bonds have substantially different bonding environments. \(\textbf{Predict the reaction of X:}\) The bromine-bearing carbon in benzyl bromide is \(sp^3\)-hybridised and is accessible to nucleophilic attack. Iodide can displace bromide in a suitable substitution reaction. The transformation is: \[ \mathrm{C_6H_5CH_2Br+NaI\xrightarrow{dry\ acetone}C_6H_5CH_2I+NaBr\downarrow} \] Formation and precipitation of sodium bromide help favour the forward exchange. The organic product is benzyl iodide, with iodine still attached to the side-chain carbon. \(\textbf{Predict the reaction of Y:}\) Bromobenzene contains bromine directly attached to an \(sp^2\)-hybridised aromatic carbon. The aromatic carbon–bromine bond possesses partial double-bond character due to resonance and does not undergo an ordinary \(S_\mathrm{N}2\)-type displacement. The usual backside-attack pathway is not available at this aryl carbon under the stated conditions. \(\textbf{Integrated conclusion:}\) The relevant distinction is not whether a benzene ring is present, but whether bromine is attached to a benzylic \(sp^3\) carbon or an aryl \(sp^2\) carbon. Dry acetone and iodide facilitate exchange in the suitable benzylic substrate but do not make ordinary bromobenzene undergo the corresponding reaction. Therefore, X forms benzyl iodide, whereas Y does not readily undergo the exchange.

Explanation: \(\textbf{Identify the required transformation:}\) The starting compound is ethanol, whereas the desired product is fluoroethane, \(\mathrm{CH_3CH_2F}\). The carbon skeleton contains two carbon atoms throughout the proposed synthesis. The required chemical change is replacement of the alcoholic hydroxyl group, ultimately giving a carbon–fluorine bond. \(\textbf{Stage 1 — Formation of a haloalkane:}\) Thionyl chloride converts ethanol into chloroethane by replacing the hydroxyl group with chlorine: \[ \mathrm{CH_3CH_2OH+SOCl_2\rightarrow CH_3CH_2Cl+SO_2+HCl} \] This produces a suitable alkyl chloride intermediate. The gaseous inorganic by-products can escape from the reaction mixture. \(\textbf{Stage 2 — Introduction of fluorine:}\) The Swarts reaction is used to prepare alkyl fluorides by heating suitable alkyl chlorides or bromides with metallic fluorides such as \(\mathrm{AgF}\), \(\mathrm{Hg_2F_2}\), \(\mathrm{CoF_2}\) or \(\mathrm{SbF_3}\). Using silver fluoride, the required halogen exchange is: \[ \mathrm{CH_3CH_2Cl+AgF\xrightarrow{\Delta}CH_3CH_2F+AgCl} \] The chlorine atom is replaced by fluorine, and the desired fluoroethane is obtained. \(\textbf{Distinguish the alternative exchanges:}\) Sodium iodide in dry acetone is associated with the Finkelstein reaction, which introduces iodine rather than fluorine. Therefore, the sequences involving sodium iodide lead towards alkyl iodides rather than the requested fluoride. Photochemical treatment with chlorine does not provide the required fluorine source and is not a Swarts exchange. \(\textbf{Final synthetic conclusion:}\) The correct route combines alcohol-to-chloride conversion with subsequent chloride-to-fluoride exchange. Both steps preserve the two-carbon skeleton, and the second step specifically establishes the required carbon–fluorine bond. The suitable sequence is therefore ethanol to chloroethane using thionyl chloride, followed by heating with silver fluoride.

Explanation: The suitability of a halogen for direct aromatic substitution depends on its reactivity and the conditions required for electrophilic substitution. Benzene readily undergoes chlorination in the presence of a suitable Lewis-acid catalyst such as iron(III) chloride. The catalyst facilitates the generation of an electrophilic chlorinating species, allowing chlorine to replace a ring hydrogen while preserving aromaticity. Thus, Reaction I provides chlorobenzene under the stated conditions. Iodination is more difficult because the reaction is reversible and hydrogen iodide is formed as a by-product. An oxidising agent such as nitric acid or periodic acid helps remove the hydrogen iodide by oxidising it, thereby favouring the formation of iodobenzene. Consequently, Reaction II is also a suitable preparation under the specified conditions. Fluorine, however, is extremely reactive, making ordinary direct aromatic fluorination difficult to control. The standard electrophilic halogenation procedure used for chlorine or bromine is therefore unsuitable for preparing fluorobenzene with elemental fluorine. Aryl fluorides are more conveniently obtained through an appropriate diazonium-salt route. The important distinction is that failure of an ordinary direct preparation method does not imply that the corresponding aromatic carbon–halogen bond cannot exist.

Explanation: \(\textbf{Determine the required structural change:}\) The starting compound, 3-methylaniline, contains an amino group and a methyl group at meta positions on the benzene ring. The desired product requires replacing the amino group by bromine without changing the carbon skeleton or the position of the methyl group. Direct electrophilic bromination does not selectively replace an existing amino group by bromine. Instead, the amino group strongly influences the orientation of electrophilic aromatic substitution and can promote additional bromination of the ring. \(\textbf{Convert the amino group into a replaceable group:}\) The aromatic primary amino group is first converted into a diazonium group using sodium nitrite and hydrochloric acid at approximately \(273\text{–}278\,\mathrm{K}\). This gives the corresponding 3-methylbenzenediazonium chloride. The diazonium group occupies precisely the position originally occupied by the amino group. \(\textbf{Introduce bromine through Sandmeyer substitution:}\) Treatment of the diazonium salt with cuprous bromide in the appropriate acidic medium replaces the diazonium group by bromine: \[ \mathrm{3\!-\!CH_3C_6H_4N_2^+Cl^- \xrightarrow{CuBr/HBr} 3\!-\!CH_3C_6H_4Br+N_2\uparrow} \] The bromine atom is introduced at the original amino-group position, so the required meta relationship is retained. Ultraviolet irradiation with a brominating reagent would instead favour free-radical chemistry at suitable side-chain positions rather than the required direct amino-group replacement. Bromine water is also unsuitable because the strongly activating amino group can lead to ring bromination. The diazonium route is therefore selected for its positional control and its ability to replace an aromatic amino group by bromine.

Explanation: Aromatic diazonium salts are useful intermediates for introducing halogens directly into an aromatic ring. Their diazonium group can be replaced by iodide with the liberation of molecular nitrogen. Unlike the preparation of aryl chlorides and bromides by the Sandmeyer reaction, the corresponding iodide replacement does not require a cuprous halide reagent. Potassium iodide is sufficient for this transformation under suitable conditions. Benzenediazonium chloride therefore produces iodobenzene: \[ \mathrm{C_6H_5N_2^+Cl^-+KI\rightarrow C_6H_5I+N_2\uparrow+KCl} \] In Q, the methyl substituent occupies the para position relative to the diazonium group. Replacement of the diazonium group by iodine occurs at that same aromatic carbon, leaving the methyl substituent unchanged. Consequently, Q forms 4-iodotoluene. Neither product is a benzyl iodide because iodine is not introduced onto the methyl side chain. The generation of nitrogen gas is associated with the departure of the diazonium group rather than with ordinary alkyl-halide exchange. This reaction also illustrates an important advantage of diazonium chemistry: the position of the introduced halogen is determined by the position of the original aromatic amino group. Hence, the products retain the intended aromatic substitution patterns.

Explanation: \(\textbf{First transformation — Diazotisation:}\) Aniline is a primary aromatic amine and undergoes diazotisation with nitrous acid generated in situ from sodium nitrite and hydrochloric acid. The reaction is conducted at a low temperature, approximately \(273\text{–}278\,\mathrm{K}\), to preserve the diazonium salt. The resulting intermediate X is benzenediazonium chloride: \[ \mathrm{C_6H_5NH_2+NaNO_2+2HCl\rightarrow C_6H_5N_2^+Cl^-+NaCl+2H_2O} \] \(\textbf{Second transformation — Fluoroborate formation:}\) Treatment with fluoroboric acid replaces the chloride counter-ion with the tetrafluoroborate ion, giving benzenediazonium fluoroborate: \[ \mathrm{C_6H_5N_2^+Cl^-+HBF_4\rightarrow C_6H_5N_2^+BF_4^-+HCl} \] This fluoroborate salt is intermediate Y. The aromatic carbon–nitrogen connectivity remains intact at this stage. \(\textbf{Third transformation — Thermal decomposition:}\) On heating, the diazonium fluoroborate decomposes with the liberation of nitrogen and formation of fluorobenzene: \[ \mathrm{C_6H_5N_2^+BF_4^-\xrightarrow{\Delta}C_6H_5F+BF_3+N_2\uparrow} \] This sequence is the Balz–Schiemann reaction. It differs from the Swarts reaction, which prepares alkyl fluorides by exchanging chlorine or bromine in suitable alkyl halides using metallic fluorides. The aromatic carbon–fluorine bond is formed here by replacement of an aromatic diazonium group. Thus, X is benzenediazonium chloride and Y is benzenediazonium fluoroborate, followed by thermal decomposition.