Hydrocarbons MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Ordinary chloromethane formation:}\) Photochemical chlorination of methane proceeds through a free-radical chain mechanism. The net substitution reaction is: \[ \mathrm{CH_4+Cl_2\xrightarrow{h\nu}CH_3Cl+HCl} \] One mole of chloromethane requires one mole of methane and one mole of chlorine. The given chloromethane production is: \[ n(\mathrm{CH_3Cl})=0.18\,\mathrm{mol} \] Therefore, methane consumed through this pathway is: \[ n(\mathrm{CH_4})_P=0.18\,\mathrm{mol} \] Chlorine consumed through this pathway is: \[ n(\mathrm{Cl_2})_P=0.18\,\mathrm{mol} \] \(\textbf{Ethane formation through radical termination:}\) A chlorine radical abstracts hydrogen from methane: \[ \mathrm{CH_4+Cl^\bullet\longrightarrow CH_3^\bullet+HCl} \] Two methyl radicals can combine in a termination reaction: \[ \mathrm{2CH_3^\bullet\longrightarrow C_2H_6} \] Producing one ethane molecule therefore requires two methyl radicals. These two radicals originate from two methane molecules under the stated assumptions. The two hydrogen-abstraction steps require two chlorine radicals. Those chlorine radicals can be generated by photochemical cleavage of one chlorine molecule: \[ \mathrm{Cl_2\xrightarrow{h\nu}2Cl^\bullet} \] Combining initiation, hydrogen abstraction and radical termination gives the net side-product reaction: \[ \mathrm{2CH_4+Cl_2\longrightarrow C_2H_6+2HCl} \] \(\textbf{Determine methane consumed in ethane formation:}\) The amount of ethane produced is: \[ n(\mathrm{C_2H_6})=0.06\,\mathrm{mol} \] Thus: \[ n(\mathrm{CH_4})_Q=2(0.06) \] \[ n(\mathrm{CH_4})_Q=0.12\,\mathrm{mol} \] \(\textbf{Determine chlorine consumed in ethane formation:}\) One mole of chlorine supplies the two radicals required for formation of one mole of ethane through this idealised pathway. Therefore: \[ n(\mathrm{Cl_2})_Q=0.06\,\mathrm{mol} \] \(\textbf{Combine the two pathways:}\) The total methane consumption is: \[ n(\mathrm{CH_4})_{\mathrm{total}}=0.18+0.12 \] \[ n(\mathrm{CH_4})_{\mathrm{total}}=0.30\,\mathrm{mol} \] The total chlorine consumption is: \[ n(\mathrm{Cl_2})_{\mathrm{total}}=0.18+0.06 \] \[ n(\mathrm{Cl_2})_{\mathrm{total}}=0.24\,\mathrm{mol} \] \(\textbf{Verify hydrogen chloride formation:}\) The chloromethane pathway produces \(0.18\,\mathrm{mol}\) HCl. The ethane pathway produces: \[ n(\mathrm{HCl})_Q=2(0.06)=0.12\,\mathrm{mol} \] Thus: \[ n(\mathrm{HCl})_{\mathrm{total}}=0.18+0.12=0.30\,\mathrm{mol} \] The total hydrogen atoms supplied by methane are: \[ 4(0.30)=1.20\,\mathrm{mol\ H\ atoms} \] The products contain: \[ 3(0.18)+6(0.06)+0.30=1.20\,\mathrm{mol\ H\ atoms} \] The hydrogen balance is therefore satisfied. The calculation demonstrates why ethane formation during methane chlorination must not be treated as ordinary one-for-one halogen substitution. Ethane is formed through radical–radical termination and incorporates carbon atoms from two methane molecules. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{CH_4})=0.30\,\mathrm{mol};\quad n(\mathrm{Cl_2})=0.24\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the equilibrium limitation:}\) Iodination of methane produces iodomethane and hydrogen iodide. The formation of HI makes the ordinary reaction reversible and generally unfavourable without suitable additional conditions. Iodic acid can oxidise HI, thereby removing it from the reacting system. This removal helps shift the iodination equilibrium towards iodomethane formation. \(\textbf{Stage 1 — Iodomethane production:}\) The initial substitution reaction is: \[ \mathrm{CH_4+I_2\rightleftharpoons CH_3I+HI} \] Formation of \(0.50\,\mathrm{mol}\) iodomethane corresponds to: \[ n(\mathrm{HI})_{\mathrm{formed}}=0.50\,\mathrm{mol} \] The gross amount of iodine entering the substitution reaction is: \[ n(\mathrm{I_2})_{\mathrm{gross}}=0.50\,\mathrm{mol} \] \(\textbf{Stage 2 — Oxidation of HI:}\) The supplied oxidation equation is: \[ \mathrm{HIO_3+5HI\longrightarrow3I_2+3H_2O} \] Five moles of HI consume one mole of iodic acid. Therefore: \[ n(\mathrm{HIO_3})=\frac{0.50}{5} \] \[ n(\mathrm{HIO_3})=0.10\,\mathrm{mol} \] \(\textbf{Determine recycled iodine:}\) Each mole of iodic acid produces three moles of iodine. Thus: \[ n(\mathrm{I_2})_{\mathrm{regenerated}}=3(0.10) \] \[ n(\mathrm{I_2})_{\mathrm{regenerated}}=0.30\,\mathrm{mol} \] \(\textbf{Calculate net iodine consumption:}\) The iodination reaction initially uses \(0.50\,\mathrm{mol}\) iodine. The HI-oxidation reaction regenerates \(0.30\,\mathrm{mol}\) iodine. Therefore: \[ n(\mathrm{I_2})_{\mathrm{net}}=0.50-0.30 \] \[ n(\mathrm{I_2})_{\mathrm{net}}=0.20\,\mathrm{mol} \] \(\textbf{Verify through the overall reaction:}\) Multiplying the methane iodination equation by five and adding the HI-oxidation equation gives: \[ \mathrm{5CH_4+5I_2\longrightarrow5CH_3I+5HI} \] \[ \mathrm{HIO_3+5HI\longrightarrow3I_2+3H_2O} \] Cancelling HI and the regenerated iodine gives: \[ \mathrm{5CH_4+2I_2+HIO_3\longrightarrow5CH_3I+3H_2O} \] The net stoichiometric ratio is: \[ \mathrm{CH_3I:I_2:HIO_3}=5:2:1 \] This independently confirms the consumption values. The key distinction is between gross iodine participation in substitution and net iodine consumption after recycling. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{I_2})_{\mathrm{net}}=0.20\,\mathrm{mol};\quad n(\mathrm{HIO_3})=0.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify competing hydrogen-abstraction pathways:}\) Ethane has the structure \(\mathrm{CH_3CH_3}\). All six hydrogen atoms of ethane are primary and constitutionally equivalent. Propane has the structure \(\mathrm{CH_3CH_2CH_3}\). It contains six primary hydrogen atoms and two secondary hydrogen atoms. Because the mixture is equimolar, the relative initial product contributions may be obtained directly from hydrogen multiplicities and the supplied per-hydrogen substitution rates. \(\textbf{Ethane contribution:}\) The relative primary-hydrogen substitution rate is one. Ethane has six primary hydrogen atoms. Therefore: \[ W_{\mathrm{ethane}}=6(1) \] \[ W_{\mathrm{ethane}}=6 \] All monochlorination of ethane produces chloroethane. \(\textbf{Propane primary-hydrogen contribution:}\) Propane contains six primary hydrogen atoms. Their combined contribution is: \[ W_{\mathrm{propane},1^\circ}=6(1) \] \[ W_{\mathrm{propane},1^\circ}=6 \] Substitution at these hydrogens produces 1-chloropropane. \(\textbf{Propane secondary-hydrogen contribution:}\) The two secondary hydrogen atoms are five times as reactive per hydrogen as the primary hydrogen atoms under the supplied conditions. Therefore: \[ W_{\mathrm{propane},2^\circ}=2(5) \] \[ W_{\mathrm{propane},2^\circ}=10 \] This contribution corresponds to formation of 2-chloropropane. \(\textbf{Determine total product weight:}\) \[ W_{\mathrm{total}}=6+6+10 \] \[ W_{\mathrm{total}}=22 \] The fraction of 2-chloropropane among all monochlorinated products is: \[ f(2\text{-chloropropane})=\frac{10}{22} \] \[ f(2\text{-chloropropane})=\frac{5}{11} \] \(\textbf{Determine total propane-derived fraction:}\) Both primary and secondary substitutions of propane must be included. \[ W_{\mathrm{propane}}=6+10=16 \] Therefore: \[ f(\mathrm{propane\text{-}derived})=\frac{16}{22} \] \[ f(\mathrm{propane\text{-}derived})=\frac{8}{11} \] \(\textbf{Interpret the competition:}\) Although ethane and propane are present in equal molecular amounts, they need not contribute equally to the monochlorinated products. Propane offers secondary hydrogen atoms that are more reactive under the specified conditions. These secondary sites increase its total contribution. The fraction of 2-chloropropane must be calculated relative to products from both hydrocarbons, not merely relative to the two chloropropane isomers. The result therefore requires combining intermolecular competition with intramolecular substitution-site selectivity. \(\textbf{Final answer:}\) \[ \boxed{f(2\text{-chloropropane})=\frac{5}{11};\quad f(\mathrm{propane\text{-}derived})=\frac{8}{11}} \]

Explanation: \(\textbf{Apply the bond-enthalpy relation:}\) The approximate enthalpy change of a reaction can be calculated from: \[ \Delta H=\sum D(\mathrm{bonds\ broken})-\sum D(\mathrm{bonds\ formed}) \] Breaking a covalent bond requires energy, whereas forming a covalent bond releases energy. Each propagation step must be analysed using the particular bonds broken and formed. \(\textbf{Step P — Hydrogen abstraction from methane:}\) The reaction is: \[ \mathrm{Cl^\bullet+CH_4\longrightarrow HCl+CH_3^\bullet} \] One carbon–hydrogen bond is broken. One hydrogen–chlorine bond is formed. Therefore: \[ \Delta H_P=D(\mathrm{CH_3-H})-D(\mathrm{H-Cl}) \] Substituting: \[ \Delta H_P=439-431 \] \[ \Delta H_P=+8\,\mathrm{kJ\,mol^{-1}} \] The positive sign indicates that the hydrogen-abstraction step is slightly endothermic according to the supplied bond energies. \(\textbf{Step Q — Chloromethane formation:}\) The reaction is: \[ \mathrm{CH_3^\bullet+Cl_2\longrightarrow CH_3Cl+Cl^\bullet} \] One chlorine–chlorine bond is broken. One carbon–chlorine bond is formed. Thus: \[ \Delta H_Q=D(\mathrm{Cl-Cl})-D(\mathrm{CH_3-Cl}) \] \[ \Delta H_Q=243-351 \] \[ \Delta H_Q=-108\,\mathrm{kJ\,mol^{-1}} \] The negative sign indicates that Step Q is exothermic. \(\textbf{Combine the propagation steps:}\) Adding the two reactions cancels the methyl radical and chlorine radical appearing on opposite sides. The net reaction is: \[ \mathrm{CH_4+Cl_2\longrightarrow CH_3Cl+HCl} \] The total enthalpy change is: \[ \Delta H_{\mathrm{net}}=\Delta H_P+\Delta H_Q \] \[ \Delta H_{\mathrm{net}}=+8-108 \] \[ \Delta H_{\mathrm{net}}=-100\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the mechanistic consequence:}\) One propagation step is slightly endothermic, but the other is substantially exothermic. Their sum gives an overall exothermic substitution reaction. The fact that Step Q releases more energy than Step P consumes does not eliminate the need for radical initiation. Photochemical cleavage of chlorine molecules is still needed to generate initial chain-carrying radicals. Once formed, those radicals can participate in repeated propagation cycles until termination occurs. The calculation also demonstrates why the sign of the overall reaction enthalpy must not be assigned automatically to every elementary mechanistic step. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_P=+8\,\mathrm{kJ\,mol^{-1}};\quad\Delta H_{\mathrm{net}}=-100\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: 2,2-Dimethylpropane has the structure \(\mathrm{C(CH_3)_4}\). Its central carbon is quaternary and carries no hydrogen atoms. The four surrounding methyl groups are constitutionally equivalent in the original molecule. Consequently, replacing any one hydrogen atom initially gives the same monochlorination product. The first substitution produces 1-chloro-2,2-dimethylpropane. \(\textbf{Identify the changed hydrogen environments:}\) After the first substitution, one methyl group becomes a chloromethyl group, \(\mathrm{-CH_2Cl}\). This group retains two hydrogen atoms. The other three methyl groups remain equivalent to one another and contain nine hydrogen atoms in total. The first substitution therefore changes the symmetry relationships relevant to the second substitution. \(\textbf{Second substitution on the chloromethyl group:}\) Replacing one of the remaining hydrogens on \(\mathrm{-CH_2Cl}\) produces a \(\mathrm{-CHCl_2}\) group. Both chlorine atoms are now attached to the same carbon atom. This gives a geminal dichlorinated product. \(\textbf{Second substitution on another methyl group:}\) Alternatively, chlorine may replace a hydrogen atom on one of the three remaining methyl groups. Because these three methyl groups are equivalent, the choice among them does not generate different constitutional products. The resulting molecule contains two separate \(\mathrm{-CH_2Cl}\) groups attached to the quaternary central carbon. Its two chlorine atoms are attached to different carbon atoms. \(\textbf{Count the possible constitutional products:}\) The second substitution can therefore produce either the geminal arrangement or the arrangement with chlorine atoms on two different methyl-derived carbons. No third constitutional arrangement is possible through substitution without carbon-skeleton rearrangement. The quaternary central carbon cannot undergo direct hydrogen substitution because it contains no hydrogen. The important distinction is that the first chlorine atom breaks the original equivalence between the chlorinated methyl group and the remaining methyl groups. Thus, a hydrocarbon giving only one constitutional monochlorination product can nevertheless give two different constitutional dichlorination products.

Explanation: \(\textbf{Infer the unknown tertiary selectivity:}\) 2-Methylpropane has the structure: \[ \mathrm{(CH_3)_3CH} \] It contains nine primary hydrogen atoms and one tertiary hydrogen atom. Let the relative substitution rates per hydrogen be: \[ r_{1^\circ}=1,\qquad r_{3^\circ}=r \] The contribution from primary substitution is: \[ W_{1^\circ}=9(1)=9 \] The contribution from tertiary substitution is: \[ W_{3^\circ}=1(r)=r \] The observed product ratio is: \[ \frac{W_{3^\circ}}{W_{1^\circ}}=\frac{4}{3} \] Therefore: \[ \frac{r}{9}=\frac43 \] \[ 3r=36 \] \[ r=12 \] Thus, the relative tertiary-hydrogen substitution rate is twelve times that of a primary hydrogen. \(\textbf{Construct the complete selectivity relationship:}\) The question separately states that secondary hydrogen atoms react five times as rapidly as primary hydrogen atoms. Therefore: \[ r_{1^\circ}:r_{2^\circ}:r_{3^\circ}=1:5:12 \] \(\textbf{Determine hydrogen environments in 3-methylpentane:}\) Its structure is: \[ \mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3} \] The three methyl groups collectively provide nine primary hydrogen atoms. The two methylene groups provide four secondary hydrogen atoms. The central branched carbon carries one tertiary hydrogen atom. Thus: \[ N_{1^\circ}=9,\qquad N_{2^\circ}=4,\qquad N_{3^\circ}=1 \] \(\textbf{Calculate the product contributions:}\) Primary substitution: \[ W_{1^\circ}=9(1)=9 \] Secondary substitution: \[ W_{2^\circ}=4(5)=20 \] Tertiary substitution: \[ W_{3^\circ}=1(12)=12 \] \(\textbf{Determine the total weight:}\) \[ W_{\mathrm{total}}=9+20+12 \] \[ W_{\mathrm{total}}=41 \] The required secondary-substitution fraction is: \[ f_{2^\circ}=\frac{20}{41} \] \(\textbf{Interpret the transferred selectivity:}\) The observed product ratio for 2-methylpropane cannot be used directly as a product ratio for 3-methylpentane. The two compounds have different numbers of primary, secondary and tertiary hydrogen atoms. The correct method is first to infer the intrinsic per-hydrogen relative reactivity from the initial experiment. That inferred rate can then be applied to the hydrogen multiplicities of the second hydrocarbon. The calculation therefore combines reverse inference with a changed-substrate prediction. \(\textbf{Final answer:}\) \[ \boxed{f_{2^\circ}=\frac{20}{41}} \]

Explanation: \(\textbf{Identify the product of the first substitution:}\) 2-Methylbutane has the structure: \[ \mathrm{CH_3-CH(CH_3)-CH_2-CH_3} \] Its tertiary carbon initially carries one hydrogen atom. Replacing that hydrogen with chlorine gives: \[ \mathrm{CH_3-C(Cl)(CH_3)-CH_2-CH_3} \] The product is 2-chloro-2-methylbutane. The carbon bearing chlorine remains directly attached to three carbon atoms. However, it no longer possesses a hydrogen atom. \(\textbf{Identify the remaining primary sites:}\) Two methyl groups are directly attached to the carbon bearing chlorine. These methyl groups are constitutionally equivalent. Substitution at either produces the same constitutional dichlorinated product. The methyl group at the opposite end of the ethyl fragment is not equivalent to those two methyl groups. Substitution there gives a second constitutional product. Therefore, the remaining primary hydrogen atoms provide two distinct constitutional substitution pathways. \(\textbf{Identify the remaining secondary site:}\) The methylene group adjacent to the chlorine-bearing carbon contains two hydrogen atoms. This carbon is directly attached to two other carbon atoms and is secondary. Replacing one of its hydrogens gives a third constitutional dichlorinated product. \(\textbf{Check whether tertiary substitution remains possible:}\) The original tertiary hydrogen was removed during the first chlorination. The chlorine-bearing carbon now has four covalent bonds: three carbon–carbon bonds and one carbon–chlorine bond. It therefore has no remaining hydrogen available for another substitution. No other tertiary carbon exists in the carbon skeleton. Consequently, the second chlorination cannot proceed through tertiary-hydrogen substitution. \(\textbf{Count the products:}\) The two nonequivalent primary substitution pathways give two products. The secondary substitution pathway gives one further product. Therefore: \[ N_{\mathrm{constitutional}}=2+1=3 \] A stereochemical difference that may arise from substitution at the methylene carbon does not increase the constitutional-product count under the stated convention. The key reasoning is that substitution changes the available hydrogen environments even though the carbon skeleton remains unchanged. Thus, primary and secondary sites remain available, producing three constitutionally distinct dichlorinated products.

Explanation: \(\textbf{Identify the elementary transformation:}\) A halogen radical abstracts a hydrogen atom from an alkane. The original carbon–hydrogen bond is broken, and a new hydrogen–halogen bond is formed. The remaining carbon fragment becomes a carbon-centred radical. The approximate reaction enthalpy is: \[ \Delta H=D(\mathrm{C-H})-D(\mathrm{H-X}) \] The calculation must use the bond dissociation enthalpy appropriate to the hydrogen environment being abstracted. \(\textbf{Chlorine abstraction at a tertiary hydrogen:}\) The tertiary carbon–hydrogen bond dissociation enthalpy is \(400\,\mathrm{kJ\,mol^{-1}}\). The hydrogen–chlorine bond dissociation enthalpy is \(431\,\mathrm{kJ\,mol^{-1}}\). Therefore: \[ \Delta H_{\mathrm{Cl},3^\circ}=400-431 \] \[ \Delta H_{\mathrm{Cl},3^\circ}=-31\,\mathrm{kJ\,mol^{-1}} \] The tertiary-hydrogen abstraction by chlorine is exothermic according to these supplied values. \(\textbf{Bromine abstraction at a tertiary hydrogen:}\) The hydrogen–bromine bond dissociation enthalpy is \(366\,\mathrm{kJ\,mol^{-1}}\). Thus: \[ \Delta H_{\mathrm{Br},3^\circ}=400-366 \] \[ \Delta H_{\mathrm{Br},3^\circ}=+34\,\mathrm{kJ\,mol^{-1}} \] This tertiary-hydrogen abstraction by bromine is endothermic. \(\textbf{Compare bromine abstraction at a primary hydrogen:}\) The primary C–H bond is stronger than the supplied tertiary C–H bond. Its bond dissociation enthalpy is \(423\,\mathrm{kJ\,mol^{-1}}\). Therefore: \[ \Delta H_{\mathrm{Br},1^\circ}=423-366 \] \[ \Delta H_{\mathrm{Br},1^\circ}=+57\,\mathrm{kJ\,mol^{-1}} \] Comparing bromine-abstraction pathways: \[ \Delta H_{\mathrm{Br},1^\circ}=+57\,\mathrm{kJ\,mol^{-1}} \] \[ \Delta H_{\mathrm{Br},3^\circ}=+34\,\mathrm{kJ\,mol^{-1}} \] Primary-hydrogen abstraction by bromine is therefore more endothermic by: \[ 57-34=23\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the mechanistic significance:}\) The weaker tertiary C–H bond is consistent with the relative stability of the tertiary carbon radical produced by abstraction. Bromine abstraction is endothermic for both hydrogen environments using the given data. The primary pathway is energetically less favourable than the tertiary pathway. This helps explain why radical bromination is generally more selective towards substitution pathways producing more stable carbon-centred radicals. However, reaction enthalpy alone does not provide an exact activation energy or a numerical product distribution. Such quantitative selectivity would require additional kinetic information. The supplied bond enthalpies support the stated energetic comparison without requiring any unstated rate constants. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_{\mathrm{Cl},3^\circ}=-31\,\mathrm{kJ\,mol^{-1}};\quad\Delta H_{\mathrm{Br},3^\circ}=+34\,\mathrm{kJ\,mol^{-1}};\quad\Delta H_{\mathrm{Br},1^\circ}=+57\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: Statement I is correct because ultraviolet light can initiate radical chlorination through homolytic cleavage of the chlorine–chlorine bond. The initiation reaction is: \[ \mathrm{Cl_2\xrightarrow{h\nu}2Cl^\bullet} \] Each chlorine atom retains one electron from the original shared electron pair. The chlorine radicals produced can begin the hydrogen-abstraction steps required for alkane substitution. Statement II is incorrect because radical chlorination operates through a chain mechanism. After initiation, propagation reactions can regenerate chlorine radicals without requiring a new photolysis event for every chloromethane molecule produced. The first propagation reaction is: \[ \mathrm{Cl^\bullet+CH_4\longrightarrow HCl+CH_3^\bullet} \] The second propagation reaction is: \[ \mathrm{CH_3^\bullet+Cl_2\longrightarrow CH_3Cl+Cl^\bullet} \] The chlorine radical regenerated in the second step can react with another methane molecule. Thus, repeated propagation cycles may produce many chloromethane molecules following the formation of an initial radical. Statement III is correct because radical-chain propagation depends on the continued availability of chain-carrying radicals. Termination removes radicals through reactions such as: \[ \mathrm{Cl^\bullet+Cl^\bullet\longrightarrow Cl_2} \] or \[ \mathrm{CH_3^\bullet+Cl^\bullet\longrightarrow CH_3Cl} \] Such reactions consume radicals without producing new radical carriers. If termination becomes sufficiently rapid relative to propagation and initiation, the continuing substitution chain is strongly suppressed. Statement IV is incorrect because hydrogen abstraction alone does not complete the methane-to-chloromethane conversion. The first propagation step produces a methyl radical and hydrogen chloride, not chloromethane. Reaction of the methyl radical with molecular chlorine is required to form chloromethane and regenerate the chlorine radical. Without this complementary step, the ordinary propagation cycle cannot continue as described. The distinction between initiation and propagation is therefore essential: initiation supplies initial radicals, while propagation uses and regenerates radicals through connected elementary reactions. Termination interrupts the radical chain by removing its carriers. Consequently, only Statements I and III are correct.

Explanation: \(\textbf{Establish the relative rotational geometry:}\) A staggered ethane molecule has a \(60^\circ\) angular offset between the projected carbon–hydrogen bonds of the front and rear carbons. An eclipsed molecule has zero relative angular offset, modulo the \(120^\circ\) rotational symmetry of ethane. The conformation depends on the relative rotation of the two carbon atoms rather than their individual rotations with respect to an external observer. Let the initial front-carbon bond direction be \(0^\circ\) and the corresponding rear-carbon direction be \(60^\circ\). If the front and rear carbons rotate through \(\alpha\) and \(\beta\), respectively, the final relative angle is: \[ \theta_{\mathrm{relative}}=60^\circ+\beta-\alpha \] \(\textbf{Sample P — Opposite rotations of }30^\circ\textbf{:}\) \[ \theta_P=60^\circ-30^\circ-30^\circ \] \[ \theta_P=0^\circ \] The carbon–hydrogen bonds become eclipsed in projection. Therefore, P has maximum torsional strain among the three final arrangements. \(\textbf{Sample Q — Equal rotations in the same direction:}\) \[ \theta_Q=60^\circ+45^\circ-45^\circ \] \[ \theta_Q=60^\circ \] Both carbons rotate equally, so their relative angular arrangement remains unchanged. Q therefore remains staggered and has minimum torsional energy. \(\textbf{Sample R — Smaller opposite rotations:}\) \[ \theta_R=60^\circ-15^\circ-15^\circ \] \[ \theta_R=30^\circ \] R has an intermediate or skew conformation. Its torsional strain is greater than that of staggered ethane but smaller than that of eclipsed ethane. \(\textbf{Compare the final energies:}\) The final relative offsets are: \[ \theta_P=0^\circ,\quad\theta_R=30^\circ,\quad\theta_Q=60^\circ \] According to the supplied energy behaviour, torsional energy decreases as the angular offset increases from \(0^\circ\) to \(60^\circ\). Consequently: \[ E_P\gt E_R\gt E_Q \] This result also demonstrates that rotating both carbon atoms does not necessarily change the molecular conformation. Only their relative angular displacement determines the torsional arrangement. \(\textbf{Final answer:}\) \[ E_P\gt E_R\gt E_Q \]

Explanation: A Newman projection represents a molecule as viewed directly along a selected bond axis. In ethane, each carbon atom is bonded to three hydrogen atoms and the other carbon atom. The three projected carbon–hydrogen bonds belonging to a single carbon are separated by \(120^\circ\) around the viewing axis. Rotation about the carbon–carbon bond changes the relative orientations of the two sets of bonds without changing the \(120^\circ\) separation within either set. \(\textbf{Examine projection P:}\) The rear bonds occupy \(40^\circ\), \(160^\circ\) and \(280^\circ\). The successive angular separations are all \(120^\circ\). Therefore, P is geometrically consistent with ethane. Its rear bonds are displaced by \(40^\circ\) from the front-bond directions. P represents an intermediate conformation rather than a fully staggered or eclipsed arrangement. \(\textbf{Examine projection Q:}\) The rear bonds are specified at \(0^\circ\), \(110^\circ\) and \(240^\circ\). Their successive angular separations are \(110^\circ\), \(130^\circ\) and \(120^\circ\). They do not maintain the required equal \(120^\circ\) projected separation. Thus, Q is inconsistent with the assumed ordinary tetrahedral ethane geometry. The apparent alignment of some bonds cannot compensate for this inconsistency. \(\textbf{Examine projection R:}\) The rear bonds occur at \(60^\circ\), \(180^\circ\) and \(300^\circ\). They are mutually separated by \(120^\circ\) and lie midway between consecutive front-bond directions. R is therefore a valid staggered conformation. It has minimum torsional strain. \(\textbf{Examine projection S:}\) The rear bonds occur at \(120^\circ\), \(240^\circ\) and \(360^\circ\). These directions are mutually separated by \(120^\circ\). Because \(360^\circ\) is equivalent to \(0^\circ\), every rear bond aligns with a front bond. S is therefore a valid eclipsed conformation. It has maximum torsional strain and the highest energy among the valid projections. The important reasoning is that a valid Newman projection must satisfy the bond geometry of each carbon separately before its overall conformation is classified. Therefore, P, R and S are consistent, and S has the highest torsional potential energy.

Explanation: The Assertion is correct because stopping ultraviolet irradiation does not instantaneously destroy every radical already present in the reaction mixture. Ultraviolet light is used primarily to initiate the radical chain by promoting homolytic cleavage of chlorine molecules. The initiation reaction is: \[ \mathrm{Cl_2\xrightarrow{h\nu}2Cl^\bullet} \] Once chlorine radicals are available, they can abstract hydrogen atoms from methane molecules. The hydrogen-abstraction propagation step is: \[ \mathrm{Cl^\bullet+CH_4\longrightarrow HCl+CH_3^\bullet} \] The methyl radical can subsequently react with another chlorine molecule: \[ \mathrm{CH_3^\bullet+Cl_2\longrightarrow CH_3Cl+Cl^\bullet} \] This second propagation step regenerates the chlorine radical, which can react with another methane molecule. Consequently, an existing radical can participate in repeated propagation cycles without requiring a fresh photon for every chloromethane molecule produced. The Reason is therefore true. The Reason also explains the Assertion because regeneration of chain carriers makes temporary continuation possible after external photochemical initiation ceases. However, the chain does not continue indefinitely. Radical–radical termination reactions progressively remove active chain carriers, and their absence eventually stops further ordinary radical propagation. Thus, continuation after irradiation stops is possible but depends on the lifetime of the remaining radicals and the prevailing reaction conditions. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Determine the number of initial radical chains:}\) Photochemical initiation involves homolytic cleavage of chlorine: \[ \mathrm{Cl_2\xrightarrow{h\nu}2Cl^\bullet} \] Every chlorine molecule generates two chlorine radicals. The amount of chlorine subjected to initial homolysis is: \[ n(\mathrm{Cl_2})=0.0020\,\mathrm{mol} \] Therefore: \[ n(\mathrm{Cl^\bullet})=2(0.0020) \] \[ n(\mathrm{Cl^\bullet})=0.0040\,\mathrm{mol} \] The problem specifies that each initial chlorine radical begins one independent chain. Hence, the number of initiated chains is represented by \(0.0040\,\mathrm{mol}\) of initial radical carriers. \(\textbf{Identify the chloromethane-forming cycle:}\) The two ordinary propagation steps are: \[ \mathrm{Cl^\bullet+CH_4\longrightarrow HCl+CH_3^\bullet} \] \[ \mathrm{CH_3^\bullet+Cl_2\longrightarrow CH_3Cl+Cl^\bullet} \] One complete propagation cycle forms one chloromethane molecule and regenerates a chlorine radical. The same chain carrier can therefore initiate another propagation cycle. \(\textbf{Calculate the original average chain length:}\) The total chloromethane produced is: \[ n(\mathrm{CH_3Cl})=0.240\,\mathrm{mol} \] The average number of product-forming cycles per initial radical chain is: \[ L=\frac{n(\mathrm{CH_3Cl})}{n(\mathrm{initial\ radical\ chains})} \] Substituting: \[ L=\frac{0.240}{0.0040} \] \[ L=60 \] Therefore, each initially generated radical sustains an average of sixty chloromethane-forming cycles under the specified idealisation. \(\textbf{Apply the inhibitor-induced change:}\) The second experiment has the same number of initially generated radicals. However, the average chain length is reduced to one-fourth. Thus: \[ L_{\mathrm{inhibited}}=\frac{60}{4} \] \[ L_{\mathrm{inhibited}}=15 \] \(\textbf{Determine the inhibited product amount:}\) The amount of chloromethane formed is: \[ n(\mathrm{CH_3Cl})_{\mathrm{inhibited}}=0.0040\times15 \] \[ n(\mathrm{CH_3Cl})_{\mathrm{inhibited}}=0.060\,\mathrm{mol} \] \(\textbf{Interpret the mechanistic consequence:}\) The initial number of radical chains is unchanged between experiments. The reduction in product amount results from fewer completed propagation cycles per chain, not from fewer initial radicals. The calculation illustrates the importance of radical regeneration in chain reactions and the effect of radical removal on the overall amount of product formed. It does not require an assumption that one photon must be absorbed for every chloromethane molecule. The specified chain-length relation is sufficient to determine both requested quantities. \(\textbf{Final answer:}\) \[ \boxed{L=60;\quad n(\mathrm{CH_3Cl})_{\mathrm{inhibited}}=0.060\,\mathrm{mol}} \]

Explanation: The photochemical chlorination mixture contains radicals generated through hydrogen-abstraction reactions. A chlorine radical can abstract hydrogen from methane to produce a methyl radical. \[ \mathrm{Cl^\bullet+CH_4\longrightarrow HCl+CH_3^\bullet} \] The methyl radical contains one carbon atom and has the structure \(\mathrm{CH_3^\bullet}\). 2-Methylpropane has the structure \(\mathrm{(CH_3)_3CH}\). Its central carbon carries one tertiary hydrogen atom. Hydrogen abstraction at this position produces the tertiary alkyl radical: \[ \mathrm{Cl^\bullet+(CH_3)_3CH\longrightarrow HCl+(CH_3)_3C^\bullet} \] The resulting radical is the 2-methylpropan-2-yl radical, commonly called the tert-butyl radical. \(\textbf{Reconstruct the coupling product:}\) The methyl radical contributes one carbon atom. The tert-butyl radical contributes four carbon atoms. The radicals combine by forming a new carbon–carbon \(\sigma\) bond between their radical centres: \[ \mathrm{CH_3^\bullet+^\bullet C(CH_3)_3\longrightarrow CH_3-C(CH_3)_3} \] The product contains five carbon atoms. Its central carbon is attached to four methyl groups and is therefore quaternary. The preferred IUPAC name is 2,2-dimethylpropane. The molecular formula is \(\mathrm{C_5H_{12}}\), consistent with a saturated acyclic hydrocarbon containing five carbon atoms. \(\textbf{Classify the radical reaction:}\) Two carbon-centred radicals are consumed during the coupling. The only product is a stable non-radical hydrocarbon. No new chain-carrying radical is generated. Therefore, this elementary reaction is a termination step. The product is not a normal monohalogenation product because it contains no chlorine atom. It is also not a propagation step merely because a new covalent bond is formed. The decisive mechanistic feature is the loss of two radicals through combination. Thus, 2,2-dimethylpropane is formed through radical-chain termination.

Explanation: \(\textbf{Identify the distinct hydrogen environments:}\) 3-Methylpentane has the structure: \[ \mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3} \] The central carbon is tertiary and carries one hydrogen atom. The two methylene carbons are constitutionally equivalent and together carry four secondary hydrogen atoms. The two terminal methyl groups are equivalent and together carry six primary hydrogen atoms. The methyl branch on the tertiary carbon contains three additional primary hydrogen atoms, but it is not equivalent to the two terminal methyl groups. Therefore, primary substitution gives two distinct constitutional products. \(\textbf{Chlorination — Calculate individual product contributions:}\) For substitution at the equivalent terminal methyl groups: \[ W_{P_1}=6(1)=6 \] For substitution at the methyl branch: \[ W_{P_2}=3(1)=3 \] For substitution at the equivalent secondary carbons: \[ W_S=4(4)=16 \] For substitution at the tertiary carbon: \[ W_T=1(6)=6 \] The largest individual contribution is sixteen, corresponding to secondary substitution. Because the two secondary carbon positions are constitutionally equivalent, they produce one constitutional monochlorination product. The preferred name of this product is 2-chloro-3-methylpentane. \(\textbf{Bromination — Apply the greater tertiary selectivity:}\) The primary contributions remain: \[ W_{P_1}=6,\qquad W_{P_2}=3 \] The secondary contribution becomes: \[ W_S=4(8)=32 \] The tertiary contribution becomes: \[ W_T=1(70)=70 \] The tertiary-substitution pathway now has the largest contribution. The corresponding organic product is 3-bromo-3-methylpentane. \(\textbf{Interpret the change in major product:}\) Under the chlorination conditions, the four secondary hydrogens collectively outweigh the one tertiary hydrogen despite the latter's greater individual reactivity. Under the bromination conditions, the much larger tertiary per-hydrogen reactivity overcomes the statistical advantage of the secondary sites. Therefore, the most abundant individual product changes from a secondary halide to a tertiary halide. The exact result depends on the supplied selectivity data; the qualitative radical-stability trend alone would not justify the numerical comparison. This demonstrates how hydrogen multiplicity and halogen-dependent selectivity jointly determine product distribution.

Explanation: \(\textbf{Establish the structural comparison:}\) Hexane and 2,2-dimethylbutane have the same molecular formula, \(\mathrm{C_6H_{14}}\). They therefore contain the same total number of hydrogen atoms. However, their hydrogen atoms belong to different primary and secondary environments because their carbon skeletons differ. The question specifies equal molecular concentrations and identical initial bromine-radical conditions. Consequently, the relative initial hydrogen-abstraction rates can be compared using hydrogen multiplicities and the supplied per-hydrogen reactivities. \(\textbf{Hydrogen environments in P:}\) Hexane has the structure: \[ \mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3} \] Its two terminal methyl groups contain six primary hydrogen atoms. Its four methylene groups contain eight secondary hydrogen atoms. There are no tertiary hydrogen atoms. The total relative abstraction contribution is: \[ W_P=6(1)+8(8) \] \[ W_P=6+64 \] \[ W_P=70 \] \(\textbf{Hydrogen environments in Q:}\) 2,2-Dimethylbutane has the structure: \[ \mathrm{CH_3-C(CH_3)_2-CH_2-CH_3} \] The second carbon is quaternary and carries no hydrogen atom. The four methyl groups collectively contain twelve primary hydrogen atoms. The single methylene group contains two secondary hydrogen atoms. Thus: \[ W_Q=12(1)+2(8) \] \[ W_Q=12+16 \] \[ W_Q=28 \] \(\textbf{Compare the initial rates:}\) Under the stated assumptions: \[ \frac{v_P}{v_Q}=\frac{W_P}{W_Q} \] Substituting: \[ \frac{v_P}{v_Q}=\frac{70}{28} \] \[ \frac{v_P}{v_Q}=\frac52 \] \(\textbf{Interpret the result:}\) Both compounds contain fourteen hydrogen atoms, but hexane has a substantially larger number of secondary hydrogen atoms. The supplied secondary-hydrogen reactivity is eight times the primary-hydrogen reactivity. Consequently, the larger secondary-hydrogen contribution makes hexane more reactive overall under the specified initial conditions. The presence of a quaternary carbon in Q does not generate a quaternary hydrogen-abstraction pathway because that carbon possesses no hydrogen. Therefore, equal molecular formulas and equal concentrations do not require equal overall radical-halogenation rates. \(\textbf{Final answer:}\) \[ \boxed{v_P:v_Q=5:2} \]

Explanation: \(\textbf{Examine the connectivity of P:}\) 1-Chloropropane has the structure: \[ \mathrm{ClCH_2-CH_2-CH_3} \] The chlorine atom is attached to one terminal carbon. Consequently, the two ends of the propane skeleton are no longer constitutionally equivalent. The remaining hydrogen atoms belong to three distinct carbon positions. \(\textbf{Second substitution on carbon 1:}\) Replacing another hydrogen on the already chlorinated carbon produces: \[ \mathrm{CHCl_2-CH_2-CH_3} \] The product is 1,1-dichloropropane. \(\textbf{Second substitution on carbon 2:}\) Substitution at the central carbon produces: \[ \mathrm{ClCH_2-CHCl-CH_3} \] The product is 1,2-dichloropropane. \(\textbf{Second substitution on carbon 3:}\) Substitution at the opposite terminal carbon produces: \[ \mathrm{ClCH_2-CH_2-CH_2Cl} \] The product is 1,3-dichloropropane. Therefore: \[ N_P=3 \] \(\textbf{Examine the connectivity of Q:}\) 2-Chloropropane has the structure: \[ \mathrm{CH_3-CHCl-CH_3} \] The two terminal methyl groups are constitutionally equivalent. Replacing a hydrogen on either methyl group gives the same constitutional dichlorinated product, 1,2-dichloropropane. Alternatively, replacing the remaining hydrogen on the central carbon gives 2,2-dichloropropane. There is no third distinct carbon environment. Therefore: \[ N_Q=2 \] \(\textbf{Interpret the difference:}\) P and Q initially have the same molecular formula, \(\mathrm{C_3H_7Cl}\). They differ in chlorine position and in the symmetry of their remaining hydrogen environments. The first chlorine substituent therefore affects the number of distinct products available in the second substitution. A carbon atom bearing chlorine can still undergo further substitution if it possesses another hydrogen atom. The correct numbers are three products from P and two products from Q.

Explanation: \(\textbf{Reaction P — Radical substitution:}\) Ethane is saturated and contains only carbon–carbon and carbon–hydrogen single bonds. Under ultraviolet irradiation, bromine can participate in a free-radical substitution reaction. Exclusive monosubstitution gives bromoethane: \[ \mathrm{C_2H_6+Br_2\xrightarrow{h\nu}C_2H_5Br+HBr} \] One mole of ethane consumes one mole of bromine. For \(0.30\,\mathrm{mol}\) ethane: \[ n(\mathrm{Br_2})_P=0.30\,\mathrm{mol} \] The reaction produces \(0.30\,\mathrm{mol}\) bromoethane and \(0.30\,\mathrm{mol}\) HBr. Only one of the two bromine atoms from each consumed bromine molecule enters the organic product. \(\textbf{Reaction Q — Addition across the double bond:}\) Ethene is unsaturated and contains one carbon–carbon double bond. Bromine adds across this double bond to form 1,2-dibromoethane. \[ \mathrm{CH_2=CH_2+Br_2\longrightarrow BrCH_2CH_2Br} \] One mole of ethene consumes one mole of bromine. Therefore: \[ n(\mathrm{Br_2})_Q=0.30\,\mathrm{mol} \] Both bromine atoms from each bromine molecule are incorporated into the organic addition product. No HBr is formed in the stated addition reaction. \(\textbf{Calculate total bromine consumption:}\) \[ n(\mathrm{Br_2})_{\mathrm{total}}=0.30+0.30 \] \[ n(\mathrm{Br_2})_{\mathrm{total}}=0.60\,\mathrm{mol} \] The corresponding amount of consumed bromine atoms is: \[ n(\mathrm{Br\ atoms})_{\mathrm{consumed}}=2(0.60) \] \[ n(\mathrm{Br\ atoms})_{\mathrm{consumed}}=1.20\,\mathrm{mol} \] \(\textbf{Calculate bromine incorporated into organic products:}\) Bromoethane contains one bromine atom per molecule: \[ n(\mathrm{Br\ atoms})_P=0.30(1)=0.30\,\mathrm{mol} \] 1,2-Dibromoethane contains two bromine atoms per molecule: \[ n(\mathrm{Br\ atoms})_Q=0.30(2)=0.60\,\mathrm{mol} \] Therefore: \[ n(\mathrm{Br\ atoms})_{\mathrm{organic}}=0.30+0.60 \] \[ n(\mathrm{Br\ atoms})_{\mathrm{organic}}=0.90\,\mathrm{mol} \] The requested fraction is: \[ f=\frac{0.90}{1.20} \] \[ f=\frac34 \] \(\textbf{Verify the bromine-atom balance:}\) The remaining \(0.30\,\mathrm{mol}\) bromine atoms occur in the \(0.30\,\mathrm{mol}\) HBr produced during radical substitution. Thus: \[ 0.90+0.30=1.20\,\mathrm{mol\ Br\ atoms} \] The bromine balance is complete. The calculation highlights the difference between substitution, which replaces hydrogen and produces hydrogen halide, and addition, which incorporates both halogen atoms across a multiple bond. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Br_2})=0.60\,\mathrm{mol};\quad f_{\mathrm{organic}}=\frac34} \]

Explanation: Staggered and eclipsed ethane are conformational arrangements of the same molecule. They have identical molecular formulas, \(\mathrm{C_2H_6}\), and identical carbon–hydrogen bond connectivity. The difference between them is the relative angular orientation of the carbon–hydrogen bonds around the carbon–carbon single-bond axis. Staggered ethane has lower torsional energy, whereas eclipsed ethane has greater torsional strain. However, the energy barrier separating these conformations is sufficiently small for rapid interconversion at ordinary temperatures. Thus, the initially prepared conformational states do not represent permanently isolated constitutional structures under the stated conditions. \(\textbf{Apply the monochlorination reaction:}\) Ultraviolet irradiation can initiate free-radical chlorination through homolytic cleavage of chlorine molecules. The net substitution reaction is: \[ \mathrm{C_2H_6+Cl_2\xrightarrow{h\nu}C_2H_5Cl+HCl} \] One hydrogen atom is replaced by chlorine. The two carbon atoms of unsubstituted ethane are constitutionally equivalent. All six hydrogen atoms consequently represent one constitutional substitution class. Replacing any one hydrogen produces chloroethane. \(\textbf{Connect conformation with chemical connectivity:}\) Conformational rotation does not alter which carbon atoms are connected or the constitutional equivalence of ethane's hydrogen positions. The initial staggered or eclipsed arrangement therefore cannot create two different constitutional monochlorination products. Both samples produce the same organic connectivity, \(\mathrm{CH_3CH_2Cl}\). The reaction conditions may influence reaction rate and product distribution in more complicated systems, but the present question restricts the comparison to constitutional monochlorination of ethane after conformational equilibration. Therefore, both samples give chloroethane, and the initial torsional-energy difference does not produce a new constitutional product.

Explanation: The molecular formula \(\mathrm{C_7H_{16}}\) belongs to the saturated acyclic alkane series. The additional structural information must be used to distinguish among constitutional isomers. A quaternary carbon is directly attached to four other carbon atoms and carries no hydrogen. A tertiary carbon is directly attached to three other carbon atoms and carries one hydrogen in an ordinary saturated hydrocarbon. \(\textbf{Reconstruct a compatible carbon skeleton:}\) 2,2,3-Trimethylbutane has the condensed structure: \[ \mathrm{CH_3-C(CH_3)_2-CH(CH_3)-CH_3} \] Carbon 2 is bonded directly to four carbon atoms. It is therefore quaternary. Carbon 3 is bonded directly to three carbon atoms and one hydrogen. It is therefore tertiary. The molecule contains seven carbon atoms and sixteen hydrogen atoms, satisfying the formula. \(\textbf{Identify the monochlorination environments:}\) Three methyl groups are attached directly to the quaternary carbon. These are constitutionally equivalent and collectively contain nine hydrogen atoms. Replacing any one of these hydrogens gives the same constitutional monochlorinated product. Two methyl groups are attached directly to the tertiary carbon. They are equivalent to each other but different from the three methyl groups attached to the quaternary carbon. Their hydrogen atoms therefore produce a second constitutional monochlorinated product. The tertiary carbon itself carries one hydrogen atom. Replacing this hydrogen gives a third constitutional product. \(\textbf{Count the products:}\) The distinct hydrogen environments are: - Methyl groups attached to the quaternary carbon. - Methyl groups attached to the tertiary carbon. - The hydrogen on the tertiary carbon. Thus: \[ N_{\mathrm{monochloro}}=1+1+1=3 \] \(\textbf{Exclude the alternative skeletons:}\) 2,2-Dimethylpentane contains a quaternary carbon but no tertiary carbon. 3-Ethylpentane contains one tertiary carbon but no quaternary carbon. 2,3-Dimethylpentane contains two tertiary carbons and no quaternary carbon. Only 2,2,3-trimethylbutane satisfies the specified carbon-degree conditions. Its three distinct hydrogen environments independently agree with the observed monochlorination-product count. The combined use of carbon connectivity and substitution-site symmetry uniquely identifies X.