Organic Compounds Containing Halogens MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Reaction P — Substitution at a stereogenic reaction centre:}\) The stipulated \(S_\mathrm{N}2\) pathway proceeds through backside nucleophilic attack. As the incoming group forms a bond and the leaving group departs, the tetrahedral arrangement at the reacting centre inverts. When that carbon is stereogenic, the product can therefore display a corresponding stereochemical inversion. The actual \(R/S\) descriptors must still be assigned from the product's substituent priorities. \(\textbf{Reaction Q — Carbocation-containing substitution:}\) The \(S_\mathrm{N}1\) mechanism includes formation of an approximately planar carbocation. A nucleophile can approach the positively charged carbon from either face. If the resulting centre remains stereogenic, products having both configurations may form. Perfect racemisation is possible under sufficiently symmetric attack conditions but is not compulsory in every real system. \(\textbf{Reaction R — Reaction at a different carbon:}\) The molecule contains an existing stereogenic centre, but the substitution occurs at a separate nonstereogenic carbon. Backside attack changes the bonding at the actual reaction centre. It does not automatically invert the spatial arrangement of a stereogenic carbon elsewhere in the molecule. Therefore, the pre-existing stereogenic centre can retain its configuration. \(\textbf{Compare local and molecular stereochemistry:}\) Inversion is a local mechanistic consequence at the carbon undergoing \(S_\mathrm{N}2\) substitution. Racemisation describes the composition of a mixture of enantiomeric products. These concepts are not interchangeable. \(\textbf{Integrated conclusion:}\) The stereochemical result depends on both the mechanism and the location of the reacting carbon. An \(S_\mathrm{N}2\) substitution at a stereogenic centre can invert it, an \(S_\mathrm{N}1\) pathway can permit both configurations, and an \(S_\mathrm{N}2\) reaction elsewhere in the molecule need not alter an untouched stereogenic centre.

Explanation: \(\textbf{Identify the bonding framework:}\) Both representations describe 1,2-dichloroethane, \(\mathrm{ClCH_2CH_2Cl}\). The carbon–chlorine bonds are polar because chlorine is more electronegative than carbon. However, molecular polarity depends on the vector sum of bond-dipole contributions rather than merely on their number. \(\textbf{Examine the anti conformation:}\) In P, the two carbon–chlorine bonds occupy opposite positions around the central carbon–carbon bond. The molecule possesses a symmetry arrangement in which the corresponding bond-dipole contributions cancel. Its ideal anti conformation therefore has zero resultant dipole moment. \(\textbf{Examine the gauche conformation:}\) In Q, the Cl–C–C–Cl dihedral angle is \(60^\circ\). The carbon–chlorine bond-dipole contributions are no longer arranged for complete cancellation. Their resultant gives the gauche conformation a non-zero molecular dipole moment. \(\textbf{Distinguish conformations from configurations:}\) The carbon–carbon bond joining the two methylene groups is a single bond. Rotation about this bond changes the relative orientations of the chlorine atoms without breaking a covalent bond. Therefore, P and Q are conformations of the same compound rather than constitutional isomers. \(\textbf{Connect molecular polarity with molecular motion:}\) At an ordinary temperature, the substance generally contains an interconverting population of conformations. Its measured bulk dipole behaviour depends on this conformational distribution. One should not automatically treat the anti and gauche conformations as separately isolable substances. \(\textbf{Integrated conclusion:}\) Changing molecular geometry can change the resultant dipole even when the chemical formula and individual bond types remain unchanged. The anti conformation has the smaller resultant dipole, while rotation about the carbon–carbon single bond permits interconversion with the gauche conformation.

Explanation: \(\textbf{Identify the two competing electrophilic sites:}\) Both halogen-bearing carbons are primary and belong to the same four-carbon chain. The difference between the sites is the identity of the leaving halogen. Cyanide ion can attack either end through its carbon atom under the specified substitution conditions. \(\textbf{Pathway through the bromine-bearing carbon:}\) Displacement of bromide introduces a nitrile group at that end while leaving the terminal chlorine intact: \[ \mathrm{Br(CH_2)_4Cl+CN^-\rightarrow NC(CH_2)_4Cl+Br^-} \] The resulting product is 5-chloropentanenitrile. The original four-carbon skeleton gains the carbon of the cyanide group, so the nitrile contains five carbon atoms. \(\textbf{Pathway through the chlorine-bearing carbon:}\) Displacement of chloride instead produces: \[ \mathrm{Br(CH_2)_4Cl+CN^-\rightarrow Br(CH_2)_4CN+Cl^-} \] This product is 5-bromopentanenitrile. Its five-carbon skeleton contains bromine at the end opposite the nitrile group. \(\textbf{Apply the supplied kinetic preference:}\) For the same molecular substrate, the two initial rates are proportional to their respective rate constants: \[ r_{\mathrm{Br}}=k_{\mathrm{Br}}[\mathrm{S}][\mathrm{CN^-}] \] \[ r_{\mathrm{Cl}}=k_{\mathrm{Cl}}[\mathrm{S}][\mathrm{CN^-}] \] The common substrate and nucleophile concentrations cancel in the ratio: \[ \frac{r_{\mathrm{Br}}}{r_{\mathrm{Cl}}} =\frac{k_{\mathrm{Br}}}{k_{\mathrm{Cl}}} \] Given: \[ k_{\mathrm{Br}}=5k_{\mathrm{Cl}} \] Therefore: \[ \frac{r_{\mathrm{Br}}}{r_{\mathrm{Cl}}}=5 \] \(\textbf{Relate pathway to product identity:}\) The faster pathway removes bromine, so the principal product retains chlorine. It would be incorrect to associate the faster bromide displacement pathway with the product that still contains bromine. \(\textbf{Check carbon conservation:}\) Both products contain five carbon atoms, not four. The cyanide carbon becomes carbon 1 of the nitrile chain regardless of which terminal halogen is replaced. \(\textbf{Final answer:}\) \[ \boxed{\frac{n(\text{5-chloropentanenitrile})}{n(\text{5-bromopentanenitrile})}=5:1} \]

Explanation: \(\textbf{Statement I — Photochemical homolysis:}\) This statement is correct under the stipulated idealised cleavage. Homolytic cleavage divides the bonding electron pair so that each fragment receives one electron. The elementary event is: \[ \mathrm{CH_3Br\xrightarrow{h\nu}CH_3^\bullet+Br^\bullet} \] The products are neutral radicals, not a methyl carbocation and bromide ion. \(\textbf{Statement II — Concerted substitution:}\) This statement is incorrect. The question explicitly specifies an \(S_\mathrm{N}2\) mechanism for Q. Hydroxide attacks the methyl carbon while the carbon–bromine bond breaks in the same elementary step. No freely separated methyl carbocation is required. Formation of such a carbocation would be especially unfavourable because it lacks stabilising alkyl substituents. \(\textbf{Statement III — Polarity versus cleavage mechanism:}\) This statement is correct. Bromine is more electronegative than carbon, so the carbon–bromine bond is polarised towards bromine. However, the existence of a polar bond does not mean that every process breaking it produces the same fragments. Homolysis produces radicals, whereas heterolytic bond breaking accompanies the ionic substitution pathway. \(\textbf{Statement IV — Identity of the leaving fragment:}\) This statement is incorrect. Bromide ion departs in the \(S_\mathrm{N}2\) substitution: \[ \mathrm{CH_3Br+OH^-\rightarrow CH_3OH+Br^-} \] In the stipulated homolytic photochemical cleavage, the bromine-containing fragment is \(\mathrm{Br^\bullet}\), not \(\mathrm{Br^-}\). \(\textbf{Integrated conclusion:}\) Bond polarity, bond-dissociation energy and the electron distribution during cleavage are related but distinct chemical ideas. Identical starting bonds can participate in different processes when the reacting conditions and mechanisms differ. Only Statements I and III are correct.

Explanation: \(\textbf{Identify the original bonding environment:}\) In chloroethene, chlorine is attached directly to a carbon of the carbon–carbon double bond. This carbon is \(sp^2\)-hybridised, and the compound is a vinylic chloride. Ordinary backside substitution at this vinylic carbon is strongly unfavourable. \(\textbf{Apply the stipulated selective hydrogenation:}\) The question explicitly states that the carbon–carbon double bond is hydrogenated without breaking the carbon–chlorine bond: \[ \mathrm{CH_2{=}CHCl+H_2\rightarrow CH_3CH_2Cl} \] The intermediate is chloroethane. The two-carbon skeleton and chlorine atom are retained, but the carbon–carbon double bond becomes a single bond. \(\textbf{Identify the changed carbon hybridisation:}\) The carbon bearing chlorine in chloroethane is \(sp^3\)-hybridised. It is an accessible primary carbon rather than a vinylic carbon. Therefore, it is suitable for ordinary nucleophilic displacement by hydroxide under appropriate conditions. \(\textbf{Predict the second reaction:}\) \[ \mathrm{CH_3CH_2Cl+OH^-\rightarrow CH_3CH_2OH+Cl^-} \] The final product is ethanol. The hydroxyl group occupies the carbon that retained chlorine during the first stage. \(\textbf{Distinguish the roles of the two steps:}\) Hydrogenation does not itself introduce oxygen or remove chlorine in the stipulated sequence. It modifies the carbon framework's bond order and converts the carbon–chlorine bond into an ordinary alkyl-halide bonding environment. Hydroxide subsequently performs the substitution. \(\textbf{Integrated conclusion:}\) A change in carbon hybridisation can fundamentally alter the suitability of a carbon–halogen bond for ordinary substitution. The sequence changes an initially resistant vinylic chloride into a reactive saturated primary chloride before forming the alcohol.

Explanation: \(\textbf{Identify the allylic reaction centre:}\) In the starting compound, bromine is attached to a saturated carbon adjacent to a carbon–carbon double bond. Departure of bromide forms an allylic carbocation rather than a vinylic carbocation. \(\textbf{Represent the resonance system:}\) The cation can be represented by two principal contributors: \[ \mathrm{CH_2{=}CHCH^+CH_3 \leftrightarrow {}^+CH_2CH{=}CHCH_3} \] The two structures are resonance contributors of the same delocalised cation. They are not two molecules that must be produced in separate ionisation reactions. \(\textbf{Attack at the original bromine-bearing carbon:}\) Water can attack the carbon that originally carried bromine. Subsequent deprotonation gives: \[ \mathrm{CH_2{=}CHCH(OH)CH_3} \] Numbering the four-carbon chain to give the hydroxyl group the appropriate lowest locant gives but-3-en-2-ol. \(\textbf{Attack at the other allylic terminus:}\) The alternative resonance contributor places positive-charge character at the original terminal alkene carbon. Attack at this end gives, after proton transfer: \[ \mathrm{HOCH_2CH{=}CHCH_3} \] This compound is but-2-en-1-ol. \(\textbf{Compare the carbon frameworks:}\) Both products contain four carbon atoms and retain one carbon–carbon double bond. They differ in the position of the hydroxyl group and in the location of the double bond. Therefore, they are constitutional isomers, not merely stereoisomers of one structure. \(\textbf{Consider the stereochemical information:}\) Formation of the delocalised carbocation destroys the original tetrahedral arrangement at the bromine-bearing carbon. Capture at the two different resonance-related positions gives distinct connectivity outcomes. A precise major-product ratio cannot be inferred solely from the two resonance contributors. \(\textbf{Integrated conclusion:}\) The key inference is that an allylic carbocation can undergo nucleophilic capture at more than one position because its positive charge is delocalised. The possible constitutional products are but-3-en-2-ol and but-2-en-1-ol.

Explanation: \(\textbf{Identify the two reactive groups:}\) Each haloalcohol contains a terminal hydroxyl group and a terminal carbon–bromine bond. Under the stipulated basic conditions, the hydroxyl group can be converted into an alkoxide nucleophile. The alkoxide oxygen and the bromine-bearing carbon belong to the same molecule. \(\textbf{Establish the intramolecular pathway:}\) A suitable intramolecular Williamson reaction involves nucleophilic attack by the alkoxide oxygen at the saturated carbon bearing bromine. Bromide leaves as the new carbon–oxygen bond forms. Because the nucleophile and electrophilic carbon are linked by the original carbon chain, this creates a ring. \(\textbf{Analyse P:}\) P contains four methylene groups: \[ \mathrm{HO(CH_2)_4Br} \] Formation of the alkoxide followed by ring closure gives: \[ \mathrm{^-O(CH_2)_4Br\rightarrow (CH_2)_4O+Br^-} \] The ring contains four carbon atoms and one oxygen atom. It is therefore a five-membered cyclic ether known as tetrahydrofuran. \(\textbf{Analyse Q:}\) Q contains five methylene groups: \[ \mathrm{HO(CH_2)_5Br} \] Intramolecular substitution joins the alkoxide oxygen to the terminal carbon: \[ \mathrm{^-O(CH_2)_5Br\rightarrow (CH_2)_5O+Br^-} \] The ring contains five carbon atoms and one oxygen atom. It is a six-membered cyclic ether known as tetrahydropyran. \(\textbf{Apply the ring-counting principle:}\) The number of atoms in the new cyclic ether equals the number of carbon atoms in the original chain plus the oxygen atom that joins the two ends. Thus, a four-carbon haloalcohol gives a five-membered ring, while a five-carbon haloalcohol gives a six-membered ring. \(\textbf{Integrated conclusion:}\) These reactions combine alkoxide formation, nucleophilic substitution and intramolecular ring closure. No carbon atoms are added or removed. The different ring sizes follow from the original chain lengths, giving tetrahydrofuran from P and tetrahydropyran from Q.

Explanation: \(\textbf{Identify the substrate and the product:}\) Tert-butyl bromide contains carbon, hydrogen and bromine but no oxygen. Its hydrolysis produces tert-butanol, which contains an oxygen atom in the hydroxyl group. Therefore, that oxygen must be introduced by an oxygen-containing reagent or solvent during the transformation. \(\textbf{Follow the first mechanistic step:}\) Under the stipulated \(S_\mathrm{N}1\) conditions, the carbon–bromine bond ionises: \[ \mathrm{(CH_3)_3CBr\rightarrow (CH_3)_3C^++Br^-} \] The tertiary carbocation contains no oxygen. This is consistent with the mechanism because oxygen has not yet been introduced. \(\textbf{Identify the oxygen-transfer step:}\) A molecule of labelled water attacks the carbocation through its oxygen atom: \[ \mathrm{(CH_3)_3C^++H_2^{18}O \rightarrow (CH_3)_3C-{}^{18}OH_2^+} \] Subsequent loss of a proton gives the labelled alcohol: \[ \mathrm{(CH_3)_3C-{}^{18}OH_2^+ \rightarrow (CH_3)_3C-{}^{18}OH+H^+} \] The oxygen–carbon bond in the product is therefore formed from the oxygen atom of water. \(\textbf{Interpret the bromine-label experiment:}\) When only bromine is isotopically labelled, the label is associated with the leaving bromide-containing species. Bromine does not become the oxygen atom of the product's hydroxyl group. Ordinary water supplies the oxygen in that experiment. \(\textbf{Distinguish nucleophile identity from mechanism:}\) A nucleophile may deliver the oxygen atom through either a concerted substitution or a suitable carbocation-capture pathway. The isotope's origin is determined by which atom forms the new carbon–oxygen bond, not by assigning the reaction exclusively to \(S_\mathrm{N}1\) or \(S_\mathrm{N}2\). \(\textbf{Integrated conclusion:}\) Isotope tracing follows atoms through a reaction rather than simply identifying product formulas. In this hydrolysis, the hydroxyl oxygen originates from water, so labelled water gives oxygen-labelled tert-butanol under the stated no-exchange conditions.

Explanation: \(\textbf{Identify the starting structure:}\) 2-Bromo-3-methylbutane has the structure: \[ \mathrm{CH_3CH(Br)CH(CH_3)CH_3} \] Bromine is attached to a secondary carbon. The adjacent carbon bears a hydrogen that can migrate if a carbocation forms under suitable conditions. \(\textbf{Reaction P — Direct replacement:}\) The stipulated direct substitution replaces bromine with hydroxide without changing the carbon skeleton: \[ \mathrm{CH_3CH(Br)CH(CH_3)CH_3 \xrightarrow{OH^-} CH_3CH(OH)CH(CH_3)CH_3} \] The product is 3-methylbutan-2-ol. Its hydroxyl-bearing carbon is attached to two carbon-containing groups, so it is a secondary alcohol. \(\textbf{Oxidation of P:}\) A secondary alcohol possesses a hydrogen on the carbon bearing the hydroxyl group. Under suitable oxidation conditions, it can form a ketone: \[ \mathrm{CH_3CH(OH)CH(CH_3)CH_3 \xrightarrow{[O]} CH_3COCH(CH_3)CH_3} \] The oxidation product is 3-methylbutan-2-one. \(\textbf{Reaction Q — Carbocation rearrangement:}\) Departure of bromide initially gives: \[ \mathrm{CH_3CH^+CH(CH_3)CH_3} \] A 1,2-hydride shift from the adjacent carbon transfers the positive charge to a more substituted carbon: \[ \mathrm{CH_3CH_2C^+(CH_3)_2} \] Water attacks the tertiary carbocation, and proton transfer produces: \[ \mathrm{CH_3CH_2C(OH)(CH_3)_2} \] The rearranged product is 2-methylbutan-2-ol. \(\textbf{Oxidation of Q:}\) The hydroxyl-bearing carbon of this alcohol is tertiary and possesses no directly attached hydrogen. It therefore resists the ordinary oxidation that converts secondary alcohols into ketones. \(\textbf{Integrated conclusion:}\) Direct substitution preserves the original position of the reacting carbon, whereas carbocation rearrangement changes which carbon becomes hydroxyl-bearing. The resulting alcohol classes differ, and their oxidation behaviour provides an independent chemical test of the proposed pathways. P gives an oxidisable secondary alcohol, while Q gives a tertiary alcohol resistant to ordinary oxidation.

Explanation: \(\textbf{Identify the two stereochemical features:}\) The starting cyclohexane derivative contains bromine at carbon 1 and methyl at carbon 2. The cis designation means that the two substituents project towards the same face of the ring. The question specifies an enantiomerically pure starting material, so the initial configurations of the stereogenic centres are defined. \(\textbf{Examine ionisation at carbon 1:}\) Departure of bromide produces a carbocation at carbon 1. The positively charged carbon is approximately planar, so its original tetrahedral configuration is lost during ionisation. However, carbon 2 remains tetrahedral and retains its stipulated configuration. \(\textbf{Consider water attack from the first face:}\) If water attacks from the face corresponding to the original bromine orientation, the hydroxyl group can occupy that face after deprotonation. The product then has the hydroxyl and methyl groups on the same face of the ring. This is cis-2-methylcyclohexan-1-ol. \(\textbf{Consider water attack from the opposite face:}\) Attack from the other face places the hydroxyl group opposite the methyl group. The resulting product is trans-2-methylcyclohexan-1-ol. \(\textbf{Determine the stereoisomeric relationship:}\) The two products differ in configuration at carbon 1, but they share the same unchanged configuration at carbon 2. They are therefore not mirror images of one another. Instead, they are diastereomers. \(\textbf{Distinguish the result from complete racemisation:}\) Formation of a planar carbocation allows attack from both faces, but the presence of a second retained stereogenic centre changes the relationship between the products. A cis/trans product pair from the same enantiomerically pure starting material is not automatically an enantiomeric pair. \(\textbf{Integrated conclusion:}\) The product relationship depends on how many stereogenic centres exist and which configurations are retained. Loss of configuration at carbon 1 permits both cis and trans products, while retention at carbon 2 makes them diastereomers.

Explanation: \(\textbf{Identify the common mechanistic feature:}\) Both substrates are allylic bromides. Their bromine atoms are attached to saturated carbons adjacent to carbon–carbon double bonds. Under the stipulated ionising conditions, bromide departure produces resonance-stabilised allylic carbocations. \(\textbf{Examine the carbocation from P:}\) Allyl bromide gives the allyl carbocation: \[ \mathrm{CH_2{=}CHCH_2^+ \leftrightarrow {}^+CH_2CH{=}CH_2} \] Although positive-charge character appears at either terminal carbon in the resonance descriptions, the two termini are equivalent by molecular symmetry. \(\textbf{Predict capture of P:}\) Water attack at either allylic terminus, followed by deprotonation, produces the same constitutional alcohol: \[ \mathrm{CH_2{=}CHCH_2OH} \] The product is prop-2-en-1-ol. Reversing the displayed chain direction does not produce a second constitutional isomer. \(\textbf{Examine the carbocation from Q:}\) The second substrate gives: \[ \mathrm{CH_2{=}CHCH^+CH_3 \leftrightarrow {}^+CH_2CH{=}CHCH_3} \] The terminal positions are now nonequivalent because one end is associated with a methyl-substituted carbon environment. \(\textbf{Predict capture of Q:}\) Attack at the original bromine-bearing carbon gives: \[ \mathrm{CH_2{=}CHCH(OH)CH_3} \] Attack at the other allylic terminus gives: \[ \mathrm{HOCH_2CH{=}CHCH_3} \] These are but-3-en-2-ol and but-2-en-1-ol, respectively. \(\textbf{Apply the symmetry distinction:}\) The existence of two resonance contributors does not automatically imply two constitutional products. The attack positions must be examined for structural equivalence. In P they are equivalent; in Q they are nonequivalent. \(\textbf{Integrated conclusion:}\) Resonance identifies possible sites of nucleophilic capture, while molecular symmetry determines whether those sites lead to distinct constitutional products. P gives one constitutional alcohol, whereas Q can give two.

Explanation: \(\textbf{Locate the isotopic label:}\) The starting compound contains bromine at carbon 2 and deuterium at the adjacent carbon 3: \[ \mathrm{CH_3CH(Br)C(D)(CH_3)_2} \] Deuterium is an isotope of hydrogen. Its presence permits the migrating atom to be followed through the rearrangement without changing the fundamental carbon skeleton. \(\textbf{Generate the initial carbocation:}\) Departure of bromide under the stipulated ionising conditions gives: \[ \mathrm{CH_3CH^+C(D)(CH_3)_2} \] The positive charge is initially located at carbon 2. This carbocation is secondary. \(\textbf{Identify the specified migrating group:}\) Carbon 3 carries the deuterium label. The problem explicitly states that a 1,2-deuteride shift occurs from this carbon to the adjacent carbocation centre. The migrating deuteride carries the bonding electron pair to carbon 2. \(\textbf{Track the rearrangement:}\) The rearranged carbocation is: \[ \mathrm{CH_3CH(D)C^+(CH_3)_2} \] Deuterium now occupies the carbon that originally bore bromine. Positive charge has moved to carbon 3, which is attached to three carbon-containing groups and forms a tertiary carbocation. \(\textbf{Apply nucleophilic capture:}\) Water attacks the new tertiary carbocation. Subsequent loss of a proton gives: \[ \mathrm{CH_3CH(D)C(OH)(CH_3)_2} \] The hydroxyl group is attached to carbon 3 of the original displayed skeleton, while the deuterium label is now on carbon 2. \(\textbf{Reject alternative label placements:}\) Direct substitution without rearrangement would retain deuterium at its original carbon and introduce hydroxyl at the former bromine-bearing carbon. An O–D bond would instead require deuterium transfer to oxygen, which is not the stipulated deuteride-migration pathway. \(\textbf{Integrated conclusion:}\) A 1,2-shift changes both the location of positive charge and the position of the migrating atom. Isotopic labelling makes this movement directly traceable. The specified pathway gives \(\mathrm{CH_3CH(D)C(OH)(CH_3)_2}\).

Explanation: \(\textbf{Identify the stereogenic centre:}\) Carbon 2 of the starting compound is attached to hydrogen, methyl, bromomethyl and chloromethyl groups. These four substituents are different, so carbon 2 is stereogenic. Its starting absolute configuration is specified as \((R)\). \(\textbf{Identify the substitution site:}\) Hydroxide replaces bromide at terminal carbon 1: \[ \mathrm{BrCH_2CH(CH_3)CH_2Cl+OH^- \rightarrow HOCH_2CH(CH_3)CH_2Cl+Br^-} \] The product is 3-chloro-2-methylpropan-1-ol. No bond at the stereogenic carbon 2 is broken during the stipulated reaction. \(\textbf{Assign the starting CIP priorities:}\) At carbon 2, three substituents are attached through carbon atoms, and the fourth is hydrogen. To distinguish the carbon-containing groups, compare the atoms attached to their first carbons: \[ \mathrm{CH_2Br:\ (Br,H,H)} \] \[ \mathrm{CH_2Cl:\ (Cl,H,H)} \] \[ \mathrm{CH_3:\ (H,H,H)} \] Since bromine has a higher atomic number than chlorine, the priority order is: \[ \mathrm{CH_2Br\gt CH_2Cl\gt CH_3\gt H} \] \(\textbf{Reassign priorities after substitution:}\) The bromomethyl group becomes hydroxymethyl. The relevant comparison is now: \[ \mathrm{CH_2OH:\ (O,H,H)} \] The unchanged chloromethyl group still has the comparison set \((Cl,H,H)\). Chlorine has a higher atomic number than oxygen. Therefore, the product priority order becomes: \[ \mathrm{CH_2Cl\gt CH_2OH\gt CH_3\gt H} \] \(\textbf{Distinguish descriptor change from spatial inversion:}\) The groups occupying priorities 1 and 2 have exchanged their ranking, although their spatial positions at carbon 2 have not changed. This reverses the CIP descriptor assigned to the unchanged tetrahedral arrangement. \(\textbf{Integrated stereochemical conclusion:}\) Retention of physical configuration at a stereogenic centre does not guarantee retention of its \(R/S\) descriptor when a reaction elsewhere changes substituent priorities. The product is therefore \((S)\)-3-chloro-2-methylpropan-1-ol, despite the absence of inversion at carbon 2.

Explanation: \(\textbf{Identify the two distinct carbon–halogen environments:}\) Bromine is directly attached to an \(sp^2\)-hybridised carbon of a double bond. It is therefore vinylic bromine. Chlorine is attached to a terminal \(sp^3\)-hybridised carbon, making that part of the molecule a primary alkyl chloride. \(\textbf{Recognise the Finkelstein reaction:}\) Sodium iodide in dry acetone supplies iodide ions suitable for displacement of chlorine or bromine from accessible saturated carbon atoms. The precipitation of sodium chloride or sodium bromide under appropriate conditions helps favour halogen exchange. \(\textbf{Evaluate the terminal chloride:}\) The chlorine-bearing terminal carbon is primary and accessible to backside attack. Iodide can therefore replace its chlorine atom: \[ \mathrm{CH_2{=}C(Br)CH_2CH_2Cl+NaI \rightarrow CH_2{=}C(Br)CH_2CH_2I+NaCl\downarrow} \] The double bond and its directly attached bromine remain intact. \(\textbf{Evaluate the vinylic bromide:}\) Although bromide is generally a better leaving group than chloride in comparable ordinary alkyl substitutions, the bromine-bearing carbon here is vinylic. The usual backside \(S_\mathrm{N}2\) geometry is unavailable at this \(sp^2\)-hybridised centre. The corresponding ordinary carbocation-forming pathway is also unfavourable. Therefore, its apparent leaving-group advantage does not make the vinylic site the preferred location for ordinary Finkelstein exchange. \(\textbf{Exclude addition:}\) Sodium iodide in dry acetone is being used as a nucleophilic halogen-exchange reagent. Under the stipulated conditions, it does not supply the molecular iodine addition pathway needed to produce a vicinal diiodide from the alkene. \(\textbf{Integrated conclusion:}\) Leaving-group identity must be considered together with the bonding environment of the carbon to which the halogen is attached. The accessible primary chloride undergoes exchange, whereas the vinylic bromide resists ordinary displacement. The principal product is \(\mathrm{CH_2{=}C(Br)CH_2CH_2I}\).

Explanation: \(\textbf{Identify the shared starting skeleton:}\) 1-Bromopropane contains three carbon atoms. Both proposed routes introduce one additional carbon-containing group, but they use chemically different mechanisms. The isotopic labels indicate which reagent supplies the extra carbon. \(\textbf{Route P — Reverse the polarity of the reactive carbon:}\) Magnesium reacts with 1-bromopropane in dry ether to produce a Grignard reagent: \[ \mathrm{CH_3CH_2CH_2Br+Mg \xrightarrow{dry\ ether} CH_3CH_2CH_2MgBr} \] The carbon bonded to magnesium possesses nucleophilic character. It can therefore attack the electrophilic carbon of labelled carbon dioxide: \[ \mathrm{CH_3CH_2CH_2MgBr+{}^{13}CO_2 \rightarrow CH_3CH_2CH_2{}^{13}COOMgBr} \] Acidic hydrolysis gives: \[ \mathrm{CH_3CH_2CH_2{}^{13}COOH} \] The labelled carbon from carbon dioxide becomes the carboxyl carbon of butanoic acid. \(\textbf{Route Q — Carbon-centred cyanide substitution:}\) In the second route, labelled cyanide attacks the terminal saturated carbon and replaces bromide: \[ \mathrm{CH_3CH_2CH_2Br+K{}^{13}CN \rightarrow CH_3CH_2CH_2{}^{13}CN+KBr} \] The product is butanenitrile labelled at the nitrile carbon. \(\textbf{Track the label through nitrile hydrolysis:}\) Complete acidic hydrolysis converts the nitrile functional group into a carboxylic acid group: \[ \mathrm{CH_3CH_2CH_2{}^{13}CN \xrightarrow{H_3O^+,\ \Delta} CH_3CH_2CH_2{}^{13}COOH} \] The nitrile carbon becomes the carboxyl carbon. The isotope therefore occupies the same position as the carbon introduced from carbon dioxide in Route P. \(\textbf{Compare the chemical pathways:}\) Route P uses formation of an organomagnesium compound followed by carbon–carbon bond formation with carbon dioxide. Route Q uses nucleophilic substitution by cyanide followed by hydrolysis. Both extend the three-carbon starting skeleton by one carbon atom. \(\textbf{Integrated conclusion:}\) Different synthetic mechanisms can converge on the same constitutional product and the same isotopic labelling pattern. The labelled carbon becomes the carboxyl carbon in both cases, and both routes yield butanoic acid.

Explanation: \(\textbf{Identify the nucleophilic atom:}\) Thiophenoxide contains a negatively charged sulphur-containing group, \(\mathrm{C_6H_5S^-}\). Sulphur acts as the nucleophilic centre in the intended substitutions. The new bond is therefore a carbon–sulphur bond rather than a carbon–oxygen bond. \(\textbf{Examine substrate P:}\) 4-Chlorobenzyl bromide contains two different carbon–halogen bonds. Chlorine is directly attached to an aromatic carbon, whereas bromine is attached to the saturated benzylic methylene carbon. The latter carbon is suitable for ordinary nucleophilic substitution under the specified mild conditions. \(\textbf{Predict the benzylic substitution product:}\) Thiophenoxide attacks the benzylic carbon and replaces bromide: \[ \mathrm{4\!-\!ClC_6H_4CH_2Br+C_6H_5S^- \rightarrow 4\!-\!ClC_6H_4CH_2SC_6H_5+Br^-} \] The original aromatic carbon–chlorine bond remains intact. The sulphur atom becomes bonded to the benzylic methylene group. \(\textbf{Examine substrate Q:}\) In 4-nitrochlorobenzene, chlorine is bonded directly to an aromatic \(sp^2\)-hybridised carbon. Ordinary alkyl-halide backside substitution is not applicable. However, the nitro group is para to chlorine and can stabilise the negatively charged addition intermediate associated with the activated aromatic substitution pathway. \(\textbf{Predict the aromatic substitution product:}\) Under the stipulated conditions, thiophenoxide attacks the aromatic carbon bearing chlorine. Subsequent chloride elimination restores aromaticity: \[ \mathrm{4\!-\!O_2NC_6H_4Cl+C_6H_5S^- \rightarrow 4\!-\!O_2NC_6H_4SC_6H_5+Cl^-} \] The resulting compound contains sulphur directly bonded to the nitro-substituted aromatic ring. \(\textbf{Compare the two carbon–sulphur bonds:}\) In P, sulphur becomes bonded to an \(sp^3\)-hybridised benzylic carbon. In Q, sulphur becomes bonded directly to an aromatic carbon through the nitro-activated addition–elimination pathway. Both reactions use the same nucleophile but differ in the bonding environment and mechanism of the electrophilic substrate. \(\textbf{Integrated conclusion:}\) The identity of the nucleophile establishes the incoming atom, while the substrate and reaction conditions determine the carbon at which substitution occurs. The correct products are the benzylic sulphide from P and the aryl sulphide from Q.

Explanation: \(\textbf{Statement I — Dichloromethane:}\) Dichloromethane, \(\mathrm{CH_2Cl_2}\), is also known as methylene chloride. It has been used as a solvent in paint-removing formulations, metal cleaning and industrial processes. Its useful solvent properties do not make it harmless. Exposure can adversely affect the central nervous system, particularly when vapours accumulate. Therefore, Statement I is correct. \(\textbf{Statement II — Chloroform:}\) Chloroform is trichloromethane, \(\mathrm{CHCl_3}\). It was historically used as a general anaesthetic because inhalation depresses the central nervous system. However, chloroform is toxic, and safer alternatives replaced its former general-anaesthetic applications. The statement correctly distinguishes historical use from present-day safety. Thus, Statement II is correct. \(\textbf{Statement III — Freon-12:}\) The formula \(\mathrm{CCl_2F_2}\) represents dichlorodifluoromethane, commonly called Freon-12. It is a chlorofluorocarbon. Its favourable physical properties and relative chemical stability contributed to its historical use in refrigeration and aerosol applications. Its atmospheric persistence and ozone-depleting potential are important environmental disadvantages. Statement III is correct. \(\textbf{Statement IV — DDT:}\) DDT is a chlorinated organic insecticide. Its chemical stability helped provide prolonged insecticidal action, but this same persistence contributes to environmental problems. DDT can accumulate in fatty tissues and adversely affect organisms exposed through food chains. Therefore, chemical persistence does not establish environmental safety. Statement IV is incorrect. \(\textbf{Integrated conclusion:}\) Technical usefulness and environmental acceptability are separate considerations. The correct combination is Statements I, II and III only.

Explanation: \(\textbf{Identify the starting compound:}\) Chloroform has the molecular formula \(\mathrm{CHCl_3}\). It is a polyhalogenated methane derivative containing three chlorine atoms and one hydrogen atom. Its historical use as a solvent and anaesthetic does not imply that it remains chemically unchanged under every storage condition. \(\textbf{Analyse the first storage environment:}\) Sample P is exposed to light and atmospheric oxygen. Chloroform can undergo slow oxidation by air in the presence of light. The hazardous product is carbonyl chloride, commonly called phosgene. Its molecular formula is \(\mathrm{COCl_2}\). \(\textbf{Represent the oxidation:}\) The overall reaction is: \[ \mathrm{2CHCl_3+O_2\rightarrow 2COCl_2+2HCl} \] The reaction produces phosgene and hydrogen chloride. Phosgene is extremely poisonous, so its formation introduces an additional hazard beyond chloroform's own toxicity. \(\textbf{Analyse the second storage environment:}\) A dark-coloured bottle limits exposure to light. Keeping the container tightly closed and nearly full reduces the amount of air available to contact the liquid. These precautions reduce the conditions favouring the stated oxidation. \(\textbf{Distinguish the alternative products:}\) The oxidation equation does not convert chloroform into carbon tetrachloride, dichloromethane or chloromethane. Those changes would require different reaction pathways involving different carbon–halogen or carbon–hydrogen bond transformations. \(\textbf{Integrated conclusion:}\) Both the identity of the organic compound and its storage environment matter. The relevant concern is oxidation of chloroform to poisonous phosgene in the presence of light and oxygen. Therefore, Sample Q represents the more appropriate protective storage arrangement.

Explanation: \(\textbf{Identify the stoichiometric relationship:}\) The balanced oxidation equation is: \[ \mathrm{2CHCl_3+O_2\rightarrow 2COCl_2+2HCl} \] It shows that two moles of chloroform react with one mole of oxygen. Two moles of phosgene are produced for every mole of oxygen consumed. \(\textbf{Compare the available reactants:}\) Given: \[ n_0(\mathrm{CHCl_3})=0.300\ \mathrm{mol} \] \[ n_0(\mathrm{O_2})=0.100\ \mathrm{mol} \] The oxygen needed to oxidise all the chloroform would be: \[ n_{\mathrm{required}}(\mathrm{O_2}) =\frac{0.300}{2} \] \[ n_{\mathrm{required}}(\mathrm{O_2})=0.150\ \mathrm{mol} \] Only \(0.100\ \mathrm{mol}\) oxygen is available. Oxygen is therefore the limiting reactant. \(\textbf{Determine the chloroform consumed:}\) From the reaction stoichiometry: \[ n_{\mathrm{consumed}}(\mathrm{CHCl_3}) =2n_{\mathrm{consumed}}(\mathrm{O_2}) \] \[ n_{\mathrm{consumed}}(\mathrm{CHCl_3}) =2(0.100) \] \[ n_{\mathrm{consumed}}(\mathrm{CHCl_3})=0.200\ \mathrm{mol} \] \(\textbf{Determine the chloroform remaining:}\) \[ n_{\mathrm{remaining}}(\mathrm{CHCl_3}) =0.300-0.200 \] \[ n_{\mathrm{remaining}}(\mathrm{CHCl_3})=0.100\ \mathrm{mol} \] \(\textbf{Calculate the phosgene produced:}\) The mole ratio of oxygen to phosgene is \(1:2\): \[ n(\mathrm{COCl_2})=2(0.100) \] \[ n(\mathrm{COCl_2})=0.200\ \mathrm{mol} \] The corresponding mass is: \[ m=nM \] \[ m(\mathrm{COCl_2})=(0.200)(99.0) \] \[ m(\mathrm{COCl_2})=19.8\ \mathrm{g} \] \(\textbf{Chemical interpretation:}\) The calculation gives the theoretical quantity under the stipulated complete-reaction model, not an expected amount from an ordinary stored bottle. Actual chloroform oxidation during storage depends on exposure conditions and reaction kinetics. The important stoichiometric distinction is that oxygen limits the reaction even though more chloroform is initially present. \(\textbf{Final answer:}\) \[ \boxed{m(\mathrm{COCl_2})=19.8\ \mathrm{g},\quad n_{\mathrm{remaining}}(\mathrm{CHCl_3})=0.100\ \mathrm{mol}} \]

Explanation: \(\textbf{Statement I — Compare molecular compositions:}\) Chloroform is \(\mathrm{CHCl_3}\), while iodoform is \(\mathrm{CHI_3}\). Both are trihalomethanes containing one hydrogen and three halogen atoms. However, their halogen identities differ. Their molecular masses, physical properties and chemical behaviour are therefore not identical. Statement I is correct. \(\textbf{Statement II — Identify iodoform:}\) Iodoform is triiodomethane, \(\mathrm{CHI_3}\). It is a yellow crystalline solid with a characteristic odour. These observations distinguish it from the familiar liquid chloroform. Statement II is correct. \(\textbf{Statement III — Interpret historical antiseptic use:}\) Iodoform was historically associated with antiseptic applications. NCERT attributes the antiseptic effect in that context to the liberation of free iodine rather than to iodoform itself acting as elemental iodine. Its historical use should not be treated as evidence of unrestricted suitability for modern medical applications. Statement III is correct. \(\textbf{Statement IV — Identify carbon tetrachloride:}\) R contains four chlorine atoms and no hydrogen atom. Its systematic name is tetrachloromethane, not trichloromethane. Chloroform, rather than carbon tetrachloride, is the compound historically associated with general anaesthesia in this comparison. Statement IV is incorrect. \(\textbf{Integrated conclusion:}\) The compounds are all halogen-containing methane derivatives, but their precise compositions control their identities and applications. The first three statements correctly distinguish structure, appearance and historical use. The correct combination is I, II and III.