Hydrocarbons MCQs with Answers – Part 7 (NEET Chemistry)

Explanation: \(\textbf{Identify the labelled oxygen source:}\) The supplied water contains oxygen-18. The question explicitly excludes subsequent exchange of this isotope with other oxygen-containing species. Therefore, the position of oxygen-18 in the isolated organic product identifies the fate of the water-derived oxygen atom. \(\textbf{Follow hydration of propyne:}\) Propyne contains a terminal carbon–carbon triple bond: \[ \mathrm{CH_3C\equiv CH} \] Under suitable mercuric-ion-assisted acidic conditions, ordinary Markovnikov hydration introduces the water-derived oxygen at the internal triple-bonded carbon. The initial organic species is an enol. Its oxygen is bonded to the carbon that will become the ketone carbonyl carbon. \(\textbf{Identify the tautomerisation step:}\) The enol is converted into its more stable keto form. During this rearrangement, the carbon–oxygen bond changes from the enolic C–O arrangement to a carbonyl C=O arrangement. Under the stipulated no-exchange conditions, the oxygen atom is retained. The final organic product is oxygen-18-labelled propanone: \[ \mathrm{CH_3C(=^{18}O)CH_3} \] \(\textbf{Follow hydration of propene:}\) Propene contains a terminal C=C bond: \[ \mathrm{CH_3CH=CH_2} \] Ordinary acid-catalysed Markovnikov hydration favours addition of hydroxyl at the central carbon. The resulting structure is: \[ \mathrm{CH_3CH(^{18}OH)CH_3} \] The product is oxygen-18-labelled propan-2-ol. \(\textbf{Explain why Q does not need the same tautomerisation:}\) The ordinary alkene hydration already gives a saturated alcohol. No C=C bond remains in the hydrated carbon framework. There is therefore no corresponding enol-to-ketone conversion in the specified pathway. \(\textbf{Compare the functional-group outcomes:}\) P produces a ketone containing a labelled carbonyl oxygen. Q produces a secondary alcohol containing a labelled hydroxyl oxygen. The isotope does not determine whether the final organic compound is an alcohol or ketone; the underlying hydration mechanism and the starting bond order do. \(\textbf{Interpret the atom-tracking evidence:}\) The oxygen-18 label remains attached to the carbon framework in both experiments. The difference lies in the final bond order to oxygen and the arrangement of hydrogen atoms after hydration. This comparison distinguishes simple alkene hydration from alkyne hydration followed by tautomerisation without relying only on the names of the reagents.

Explanation: \(\textbf{Identify the carboxylate carbon framework:}\) The sodium salt has the structure: \[ \mathrm{(CH_3)_2CHCOONa} \] It contains four carbon atoms. One is the carboxylate carbon, while the remaining three constitute an isopropyl group. Both decarboxylation routes remove the carboxylate carbon, but they handle the remaining carbon fragment differently. \(\textbf{Follow soda-lime decarboxylation in P:}\) Heating the sodium salt with soda lime replaces the carboxylate group by hydrogen. The net equation is: \[ \mathrm{(CH_3)_2CHCOONa+NaOH\xrightarrow{CaO,\Delta}(CH_3)_2CH_2+Na_2CO_3} \] The organic product contains three carbon atoms. Its structure is propane: \[ \mathrm{CH_3CH_2CH_3} \] No joining of two organic fragments is involved. \(\textbf{Follow Kolbe electrolysis in Q:}\) At the anode, the carboxylate ion loses an electron and carbon dioxide: \[ \mathrm{(CH_3)_2CHCOO^-\longrightarrow(CH_3)_2CH^\bullet+CO_2+e^-} \] The remaining organic species is an isopropyl radical. Two identical radicals are stipulated to combine: \[ \mathrm{2(CH_3)_2CH^\bullet\longrightarrow(CH_3)_2CHCH(CH_3)_2} \] The product is 2,3-dimethylbutane. It contains six carbon atoms because two three-carbon radical fragments have joined. \(\textbf{Follow Wurtz coupling in R:}\) 2-Bromopropane has the structure: \[ \mathrm{CH_3CHBrCH_3} \] Ideal Wurtz self-coupling forms a new bond between the two carbons originally bearing bromine: \[ \mathrm{2(CH_3)_2CHBr+2Na\longrightarrow(CH_3)_2CHCH(CH_3)_2+2NaBr} \] The product is again 2,3-dimethylbutane. \(\textbf{Compare the carbon-attachment sites:}\) Q and R generate the same overall six-carbon structure. In Q, the reactive carbon centre is created by carboxylate decarboxylation. In R, the carbon–carbon bond forms at the carbon originally bonded to bromine. Both routes preserve the branched structure of the two three-carbon fragments. \(\textbf{Interpret the contrasting preparations:}\) Soda-lime decarboxylation decreases the original carboxylate's carbon count by one per molecule. Kolbe electrolysis also removes the carboxylate carbon, but the remaining fragments subsequently combine. Wurtz coupling instead joins alkyl-halide fragments without a decarboxylation step. The identity of the organic product must therefore be inferred from both the carbon fragment generated and the way that fragment reacts. The three routes produce propane, 2,3-dimethylbutane and 2,3-dimethylbutane, respectively.

Explanation: \(\textbf{Distinguish the reaction intermediate from the final product:}\) During electrophilic aromatic substitution, an incoming electrophile forms a new sigma bond with the aromatic ring. The resulting arenium ion temporarily loses full aromatic delocalisation. The positive charge can be represented at several ring positions through resonance. Its relative stability contributes to the preference for particular substitution positions. \(\textbf{Identify the special resonance contributor for ortho attack:}\) When attack occurs at C2, the supplied charge-bearing positions include C1. That is the carbon directly bonded to substituent X. Consequently, the electronic influence of X can be particularly important for the contributor having positive charge at C1. \(\textbf{Identify the corresponding contributor for para attack:}\) Attack at C4 also gives a set containing C1. The substituted carbon therefore again participates in a positively charged resonance contributor. This resemblance helps explain why some substituents favour both ortho and para orientations. \(\textbf{Examine the meta intermediate:}\) The supplied meta-attack charge positions are C2, C4 and C6. The carbon bearing X is absent from this list. Thus, the distinctive resonance interaction involving a positive charge at C1 is not available in the same way for the meta sigma complex. \(\textbf{Apply the methoxy group's electronic effect:}\) The oxygen atom in \(\mathrm{-OCH_3}\) possesses a lone pair. It can donate electron density into the aromatic system. Where a sigma-complex contributor has positive charge at the substituent-bearing carbon, that donation can provide an additional stabilising resonance interaction. Ortho- and para-attack pathways benefit from this interaction. Methoxybenzene is therefore activating and ordinarily ortho/para-directing. \(\textbf{Apply the nitro group's electronic effect:}\) The nitro group strongly withdraws electron density. It does not stabilise the positively charged ring arrangements in the same manner as methoxy. Ortho- and para-attack sigma complexes include particularly unfavourable contributors involving positive charge at the nitro-bearing carbon. The meta pathway avoids that specific contributor. Nitrobenzene therefore ordinarily favours meta substitution. \(\textbf{Separate relative direction from absolute rate:}\) Methoxybenzene is activated towards electrophilic substitution. Nitrobenzene is strongly deactivated. The table identifies a resonance-based reason for their different preferred positions, but does not supply numerical rates or product percentages. An ortho/para preference also need not imply identical ortho and para yields because steric and statistical factors may differ. \(\textbf{Interpret the two contrasting groups:}\) The charge-position sets are determined by where the electrophile attacks the ring. The substituent's electronic properties determine whether particular contributors are stabilised or destabilised. An electron donor can favour ortho/para attack through additional resonance stabilisation. A strong electron withdrawer can favour meta attack by avoiding especially unfavourable positive-charge arrangements. This directly connects directing effects with the mechanism of electrophilic aromatic substitution.

Explanation: \(\textbf{Distinguish bond enthalpy from product abundance:}\) Breaking a C–H bond requires energy; forming an H–X bond releases energy. The approximate enthalpy of the hydrogen-abstraction step is therefore the C–H bond enthalpy minus the H–X bond enthalpy. Product abundance, however, is determined here by the separately supplied per-hydrogen tendencies and the numbers of reactive hydrogens. \(\textbf{Calculate chlorine-radical abstraction enthalpies:}\) For a primary hydrogen: \[ \Delta H_{\mathrm{Cl,primary}}=423-431=-8\,\mathrm{kJ\,mol^{-1}} \] For the tertiary hydrogen: \[ \Delta H_{\mathrm{Cl,tertiary}}=404-431=-27\,\mathrm{kJ\,mol^{-1}} \] Both abstractions are exothermic under the supplied approximation, with tertiary abstraction more exothermic. \(\textbf{Calculate bromine-radical abstraction enthalpies:}\) For a primary hydrogen: \[ \Delta H_{\mathrm{Br,primary}}=423-366=+57\,\mathrm{kJ\,mol^{-1}} \] For the tertiary hydrogen: \[ \Delta H_{\mathrm{Br,tertiary}}=404-366=+38\,\mathrm{kJ\,mol^{-1}} \] Both are endothermic in this model, but the tertiary route is less endothermic. These enthalpies alone do not determine the exact rates or product ratios. \(\textbf{Count hydrogen sites and calculate relative product amounts:}\) 2-Methylpropane has nine equivalent primary hydrogens and one tertiary hydrogen. For chlorination: \[ W_{\mathrm{primary}}:W_{\mathrm{tertiary}}=9(1):1(5)=9:5 \] Hence, 1-chloro-2-methylpropane is the more abundant constitutional monochloride. For bromination: \[ W_{\mathrm{primary}}:W_{\mathrm{tertiary}}=9(1):1(70)=9:70 \] Hence, 2-bromo-2-methylpropane is the more abundant constitutional monobromide. \(\textbf{Reconcile the energetic and statistical conclusions:}\) The tertiary site has a more favourable hydrogen-abstraction enthalpy for either halogen. Nevertheless, the nine available primary hydrogens outweigh the fivefold tertiary tendency during chlorination. The much larger supplied tertiary preference during bromination reverses that product dominance. This is why the more favourable individual reaction site need not produce the most abundant constitutional product when site multiplicity differs.

Explanation: \(\textbf{Identify the two possible alkyl fragments:}\) 1-Bromopropane contains the n-propyl fragment: \[ \mathrm{CH_3CH_2CH_2-} \] 2-Bromopropane contains the isopropyl fragment: \[ \mathrm{(CH_3)_2CH-} \] Ideal Wurtz coupling forms a new carbon–carbon bond between the carbons originally bearing bromine. When two different alkyl halides are present, three coupling connectivities are possible. \(\textbf{Identify the self-coupled products:}\) Two n-propyl fragments combine to give n-hexane: \[ \mathrm{2CH_3CH_2CH_2Br+2Na\longrightarrow C_6H_{14}+2NaBr} \] Here, the product is specifically the unbranched six-carbon alkane. Two isopropyl fragments instead combine to give 2,3-dimethylbutane. Both products have formula \(\mathrm{C_6H_{14}}\) but different carbon skeletons. \(\textbf{Identify the cross-coupled product:}\) One n-propyl fragment combines with one isopropyl fragment: \[ \mathrm{CH_3CH_2CH_2-CH(CH_3)_2} \] The preferred name of this six-carbon framework is 2-methylpentane. Each cross-coupled molecule consumes one molecule of each original bromide. \(\textbf{Deduct the fragments consumed in cross-coupling:}\) The cross-coupled product amount is: \[ n(\mathrm{cross})=0.120\,\mathrm{mol} \] Thus, the n-propyl-derived fragment amount remaining for self-coupling is: \[ n(n\text{-propyl,\ remaining})=0.300-0.120 \] \[ n(n\text{-propyl,\ remaining})=0.180\,\mathrm{mol} \] The corresponding isopropyl-derived remainder is also: \[ n(\mathrm{isopropyl,\ remaining})=0.180\,\mathrm{mol} \] \(\textbf{Calculate n-hexane formation:}\) Each n-hexane molecule requires two n-propyl fragments: \[ n(n\text{-hexane})=\frac{0.180}{2} \] \[ n(n\text{-hexane})=0.090\,\mathrm{mol} \] \(\textbf{Calculate 2,3-dimethylbutane formation:}\) Each molecule requires two isopropyl fragments: \[ n(2,3\text{-dimethylbutane})=\frac{0.180}{2} \] \[ n(2,3\text{-dimethylbutane})=0.090\,\mathrm{mol} \] \(\textbf{Determine the total organic-product amount:}\) \[ n_{\mathrm{organic}}=0.090+0.120+0.090 \] \[ n_{\mathrm{organic}}=0.300\,\mathrm{mol} \] This is half the original total alkyl-bromide amount, consistent with two fragments per coupling product. \(\textbf{Calculate sodium bromide formation:}\) Every original bromide molecule loses one bromine atom, which appears in sodium bromide. The total bromide amount is: \[ n(\mathrm{RBr})=0.300+0.300 \] \[ n(\mathrm{RBr})=0.600\,\mathrm{mol} \] Therefore: \[ n(\mathrm{NaBr})=0.600\,\mathrm{mol} \] \(\textbf{Verify carbon conservation:}\) The original reagents contain: \[ 3(0.600)=1.800\,\mathrm{mol} \] carbon atoms. The products contain: \[ 6(0.300)=1.800\,\mathrm{mol} \] carbon atoms. The calculation therefore preserves the supplied carbon framework. \(\textbf{Interpret the mixture outcome:}\) Using two different alkyl halides in Wurtz coupling can produce self-coupled and cross-coupled alkanes. A single product formula cannot distinguish these three six-carbon constitutional isomers. The specified cross-coupling amount determines how much of each fragment remains available for self-coupling. The sodium bromide amount instead follows from all original bromide molecules, regardless of which alkane each ultimately forms. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(n\text{-hexane})&=0.090\,\mathrm{mol}\\ n(2\text{-methylpentane})&=0.120\,\mathrm{mol}\\ n(2,3\text{-dimethylbutane})&=0.090\,\mathrm{mol}\\ n(\mathrm{NaBr})&=0.600\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the original stereogenic centre:}\) 3-Methylhexane has the structure: \[ \mathrm{CH_3CH_2CH(CH_3)CH_2CH_2CH_3} \] Carbon 3 is attached to hydrogen, methyl, ethyl and propyl groups. These four substituents are different. The carbon is therefore stereogenic, and the pure starting material may exist as one enantiomer. \(\textbf{Follow hydrogen abstraction in P:}\) Route P specifically removes the hydrogen attached to carbon 3. The resulting carbon-centred radical retains its connections to methyl, ethyl and propyl groups. The problem stipulates that this radical centre becomes effectively planar before bromine transfer. Thus, the original tetrahedral arrangement at the radical-bearing atom is no longer retained as a fixed three-dimensional configuration. \(\textbf{Apply the two equivalent radical faces:}\) The bromine-transfer step forms a new C–Br bond at carbon 3. The carbon becomes tetrahedral again. Its four substituents are now bromine, methyl, ethyl and propyl. These remain four different groups, so the product carbon is stereogenic. Equal reaction from either planar radical face produces the two enantiomeric configurations in equal amounts. Therefore, P gives racemic 3-bromo-3-methylhexane under the stipulated conditions. \(\textbf{Follow the remote substitution in Q:}\) Q replaces a hydrogen on the original carbon-1 methyl group. The new carbon–bromine bond is formed at the terminal carbon rather than at carbon 3. The original carbon 3 does not undergo bond cleavage in the specified transformation. The conditions explicitly preserve its tetrahedral configuration. \(\textbf{Assess Q's newly brominated carbon:}\) The terminal methyl group becomes a bromomethyl group: \[ \mathrm{-CH_3\longrightarrow-CH_2Br} \] The newly chlorinated or brominated terminal carbon in such a bromomethyl group still carries two hydrogen atoms. It is therefore not a new tetrahedral stereogenic centre. The original stereogenic carbon 3 remains the relevant configurational feature. \(\textbf{Interpret what is meant by configuration retention:}\) Retaining the spatial arrangement at carbon 3 does not necessarily guarantee that its CIP letter designation remains unchanged if the substituent priority order is recalculated after remote modification. The chemically relevant statement is that the original centre has not undergone inversion or racemisation. The product remains configurationally derived from the original enantiomer. \(\textbf{Contrast local and remote radical formation:}\) A radical reaction does not automatically destroy every stereogenic centre in the molecule. Loss of configuration occurs at the stipulated planar radical centre when a stereogenic atom itself becomes the radical-bearing atom. A remote substitution can preserve an existing stereogenic centre if its bonds and spatial arrangement are unaffected. The comparison therefore depends on the site of radical generation, not merely on the classification of the overall process as radical bromination.

Explanation: \(\textbf{Identify the two structural tests:}\) Sodamide probes the presence of a sufficiently acidic hydrogen attached directly to an \(sp\)-hybridised carbon. The stipulated radical chlorination instead probes ordinary benzylic C–H bonds. These are different structural features. Their distinction becomes important when the side-chain triple bond is hydrogenated. \(\textbf{Inspect phenylacetylene before hydrogenation:}\) The original structure is: \[ \mathrm{C_6H_5-C\equiv CH} \] The terminal carbon of the triple bond carries one hydrogen. That carbon is \(sp\)-hybridised. The hydrogen is therefore an acetylenic hydrogen capable of removal by a sufficiently strong base. \(\textbf{Predict Test I before hydrogenation:}\) The reaction with sodamide is: \[ \mathrm{C_6H_5C\equiv CH+NaNH_2\longrightarrow C_6H_5C\equiv CNa+NH_3} \] A sodium alkynide and ammonia are formed. Thus, the original hydrocarbon gives a positive response to the stipulated terminal-alkyne test. \(\textbf{Inspect the original benzylic position:}\) The carbon directly attached to the benzene ring is one of the two triply bonded carbons. It is already bonded to the ring carbon and to the other triple-bond carbon. It carries no hydrogen. The terminal acetylenic hydrogen belongs to the farther carbon, not to the ring-attached carbon. Consequently, phenylacetylene has no ordinary benzylic C–H bond for the restricted Test II. \(\textbf{Follow complete side-chain hydrogenation:}\) The triple bond consumes two hydrogen molecules: \[ \mathrm{C_6H_5C\equiv CH+2H_2\longrightarrow C_6H_5CH_2CH_3} \] The product is ethylbenzene. The benzene ring is retained. The original ring-attached triple-bond carbon has become a saturated methylene carbon. \(\textbf{Apply Test I after hydrogenation:}\) Ethylbenzene has no carbon–carbon triple bond. Therefore, it has no terminal acetylenic hydrogen. It does not undergo the stipulated sodamide-to-alkynide transformation. \(\textbf{Apply Test II after hydrogenation:}\) Ethylbenzene contains the benzylic group: \[ \mathrm{C_6H_5CH_2-} \] Its methylene carbon carries two benzylic hydrogens. Under the specified photochemical conditions, one can be replaced by chlorine: \[ \mathrm{C_6H_5CH_2CH_3\longrightarrow C_6H_5CH(Cl)CH_3} \] The product is 1-chloroethylbenzene. \(\textbf{Interpret the reversal of test behaviour:}\) Hydrogenation removes the acidic terminal acetylenic hydrogen environment. At the same time, it creates an ordinary saturated benzylic methylene group with hydrogens. Thus, a chemical change involving only the side-chain bond order reverses the outcomes of the two selective tests. The result depends on the precise carbon bearing hydrogen, not merely on whether the molecule contains a phenyl group.

Explanation: \(\textbf{Identify the initial bond order:}\) Propyne contains a triple bond between its second and third carbon atoms. A carbon–carbon triple bond consists of one sigma bond and two pi bonds. Two successive halogen additions can consume the two pi-bond equivalents. The original three-carbon skeleton remains intact under the stipulated conditions. \(\textbf{Follow the bromine addition:}\) The first equivalent adds one bromine atom to each original triple-bonded carbon: \[ \mathrm{CH_3C\equiv CH+Br_2\longrightarrow CH_3C(Br)=CHBr} \] The intermediate is a dibromoalkene. Its two double-bonded carbons each have different substituents. E/Z geometrical configurations are therefore possible. \(\textbf{Follow the chlorine addition:}\) The second equivalent adds across the remaining C=C bond. One chlorine atom becomes attached to each of the two alkene carbons: \[ \mathrm{CH_3C(Br)=CHBr+Cl_2\longrightarrow CH_3C(Br)(Cl)CH(Br)(Cl)} \] The final carbon–carbon bond is a single bond. Both carbon atoms are now approximately tetrahedral. \(\textbf{Examine the carbon bearing methyl:}\) The former internal triple-bond carbon is now attached to methyl, bromine, chlorine and the neighbouring carbon. These are four different substituents. It is therefore a stereogenic centre. \(\textbf{Examine the carbon bearing hydrogen:}\) The former terminal triple-bond carbon is now attached to hydrogen, bromine, chlorine and the neighbouring carbon. These are also four different substituents. This carbon is independently stereogenic. \(\textbf{Check for meso symmetry:}\) The two stereogenic carbons have different nonhalogen substituents. One carries methyl, while the other carries hydrogen. An internal symmetry operation cannot exchange the two centres while preserving those different substituents. The saturated product consequently has no meso reduction in its possible stereoisomer count. \(\textbf{Count the configurations:}\) For two independent tetrahedral stereogenic centres, the nominal count is: \[ N=2^2 \] \[ N=4 \] Because the molecule lacks the relevant internal symmetry, all four configurations are distinct. They can be arranged as two pairs of enantiomers. \(\textbf{Interpret the successive additions:}\) The first halogen equivalent converts a triple bond into a double bond capable of geometrical isomerism. The second equivalent converts that double bond into a single bond and creates two tetrahedral stereogenic centres. The original E/Z descriptor no longer applies to the final saturated product. The resulting stereochemical possibilities must therefore be counted from the four substituents on each final tetrahedral carbon, rather than from the intermediate's double-bond geometry.

Explanation: \(\textbf{Identify the competing possible parent chains:}\) The molecule contains seven carbon atoms and two carbon–carbon double bonds. One possible chain contains six carbon atoms but does not include both double bonds. Another chain contains five carbon atoms and includes both double bonds. The preferred unsaturated parent must be selected using the appropriate multiple-bond-inclusion rule rather than choosing a longer chain that omits one of the double bonds. \(\textbf{Select the unsaturated parent:}\) The five-carbon parent chain runs from the left terminal methylene group through both double bonds to the right terminal methylene group. Its double bonds are located at positions 1 and 4. Therefore, the appropriate parent is penta-1,4-diene. \(\textbf{Locate the remaining substituent:}\) The two-carbon branch is an ethyl group. Numbering from the left gives its attachment position as C2. Numbering from the opposite end would give position C4. The double-bond locant sets are identical in both directions, so the lower substituent locant is selected. Thus, X is 2-ethylpenta-1,4-diene. \(\textbf{Follow complete hydrogenation:}\) Both carbon–carbon double bonds are saturated. The original C–C sigma-bond connectivity remains unchanged: \[ \mathrm{CH_2=C(CH_2CH_3)CH_2CH=CH_2+2H_2} \] \[ \mathrm{\longrightarrow CH_3CH(CH_2CH_3)CH_2CH_2CH_3} \] The product's longest saturated carbon chain contains six atoms. The remaining methyl group is attached at carbon 3. The saturated product is therefore 3-methylhexane. \(\textbf{Identify the terminal ozonolysis fragments:}\) Each original C=C bond has a terminal \(\mathrm{CH_2}\) carbon. Reductive ozonolysis converts each terminal methylene carbon into methanal: \[ \mathrm{HCHO} \] Two separate methanal molecules are therefore formed per molecule of X. \(\textbf{Determine the central surviving fragment:}\) The left nonterminal alkene carbon carries an ethyl substituent and has no hydrogen. It becomes a ketone carbonyl. The right nonterminal alkene carbon carries one hydrogen. It becomes an aldehyde carbonyl. The original methylene group between those two carbons remains intact. The connected fragment is: \[ \mathrm{CH_3CH_2COCH_2CHO} \] It is 3-oxopentanal. \(\textbf{Verify the carbon balance:}\) The two methanal molecules contain two carbon atoms altogether. 3-Oxopentanal contains five carbon atoms. The total recovered carbon count is: \[ 2(1)+5=7 \] This equals the carbon count of X. \(\textbf{Interpret the naming and reaction evidence together:}\) The name of the unsaturated compound is based on the chain that includes both double bonds. Hydrogenation subsequently changes the bonding priorities used for selecting its saturated parent chain, producing the name 3-methylhexane. Reductive ozonolysis independently confirms two terminal double-bond ends and a central keto-aldehyde fragment. The three decisions therefore require applying nomenclature, carbon-skeleton preservation and selective multiple-bond cleavage separately.

Explanation: \(\textbf{Identify why the molecular formula is insufficient:}\) Benzene and hexa-1,5-diyne each contain six carbon atoms and six hydrogen atoms. However, their carbon frameworks are different. P contains a six-membered ring with aromatic \(\pi\)-electron delocalisation. Q contains an open chain possessing two separate carbon–carbon triple bonds. Identical molecular formulas therefore do not imply identical chemical reactivities. \(\textbf{Count hydrogenation equivalents in P:}\) Benzene has three formal C=C bond equivalents in its aromatic system. Complete catalytic hydrogenation requires three hydrogen molecules: \[ \mathrm{C_6H_6+3H_2\longrightarrow C_6H_{12}} \] The saturated product is cyclohexane. The ring framework is preserved. \(\textbf{Count hydrogenation equivalents in Q:}\) Hexa-1,5-diyne contains two carbon–carbon triple bonds. Each requires two hydrogen molecules for complete saturation. Therefore: \[ n(\mathrm{H_2})_Q=2(2) \] \[ n(\mathrm{H_2})_Q=4\,\mathrm{mol\ per\ mol} \] Its hydrogenation is: \[ \mathrm{C_6H_6+4H_2\longrightarrow C_6H_{14}} \] The saturated product is n-hexane. \(\textbf{Apply the terminal-alkyne test to Q:}\) Both ends of Q contain a hydrogen attached directly to an \(sp\)-hybridised carbon. Sufficient sodamide can remove both acetylenic protons: \[ \mathrm{HC\equiv CCH_2CH_2C\equiv CH+2NaNH_2} \] \[ \mathrm{\longrightarrow NaC\equiv CCH_2CH_2C\equiv CNa+2NH_3} \] One molecule of Q consequently produces two ammonia molecules through the stipulated reaction. \(\textbf{Compare sodamide treatment of P:}\) Benzene contains hydrogen attached to \(sp^2\)-hybridised ring carbons. It has no terminal alkyne functionality. Its aromatic hydrogens do not undergo the stated formation of sodium acetylides. Therefore, no corresponding ammonia is formed in Test II. \(\textbf{Determine bromine uptake by Q:}\) Each terminal triple bond contains two ordinary \(\pi\)-bond equivalents. Complete bromine addition consumes two bromine molecules per triple bond. Thus: \[ n(\mathrm{Br_2})_Q=2(2) \] \[ n(\mathrm{Br_2})_Q=4\,\mathrm{mol\ per\ mol} \] \(\textbf{Determine bromine uptake by P:}\) The bromine experiment excludes electrophilic aromatic substitution and aromatic addition. Benzene therefore remains unreacted under the specified conditions. Its bromine-addition consumption is zero. \(\textbf{Interpret the complementary structural tests:}\) The two hydrocarbons have the same molecular formula but differ in ring structure, ordinary multiple bonds and acidic hydrogen environments. Complete hydrogenation gives a cyclic saturated product from benzene and an acyclic saturated product from the diyne. Sodamide detects the two terminal acetylenic hydrogens in Q. The stipulated mild bromine-addition test distinguishes Q's ordinary triple bonds from P's aromatic electronic system. The full result cannot be obtained by molecular-formula comparison alone.

Explanation: \(\textbf{Distinguish the two transformations:}\) Methane reacts with steam at high temperature over nickel to produce carbon monoxide and hydrogen. Controlled oxidation with a suitable catalyst instead introduces oxygen into the organic molecule and can produce methanol. These reactions have different stoichiometric requirements despite beginning with the same hydrocarbon. \(\textbf{Write the steam-reforming equation:}\) \[ \mathrm{CH_4+H_2O\xrightarrow[1273\,K]{Ni}CO+3H_2} \] One methane molecule requires one water molecule. Three hydrogen molecules are produced per methane molecule consumed. \(\textbf{Identify the limiting reagent in P:}\) \[ n(\mathrm{CH_4})_{\mathrm{initial}}=0.450\,\mathrm{mol} \] \[ n(\mathrm{H_2O})_{\mathrm{initial}}=0.300\,\mathrm{mol} \] The reactant mole ratio required is \(1:1\). Steam is therefore limiting. \(\textbf{Calculate hydrogen formation in P:}\) \[ n(\mathrm{CH_4})_{\mathrm{reacted}}=0.300\,\mathrm{mol} \] \[ n(\mathrm{H_2})_P=3(0.300) \] \[ n(\mathrm{H_2})_P=0.900\,\mathrm{mol} \] Carbon monoxide is formed in an amount of \(0.300\,\mathrm{mol}\). \(\textbf{Find the unreacted methane:}\) \[ n(\mathrm{CH_4})_{\mathrm{remaining}}=0.450-0.300 \] \[ n(\mathrm{CH_4})_{\mathrm{remaining}}=0.150\,\mathrm{mol} \] \(\textbf{Write the selective oxidation in Q:}\) \[ \mathrm{2CH_4+O_2\xrightarrow[523\,K,\ 100\,atm]{Cu}2CH_3OH} \] Two methane molecules consume one oxygen molecule. Two methanol molecules are formed. \(\textbf{Calculate the oxygen requirement:}\) \[ n(\mathrm{O_2})_Q=\frac{0.450}{2} \] \[ n(\mathrm{O_2})_Q=0.225\,\mathrm{mol} \] \(\textbf{Calculate methanol formation:}\) \[ n(\mathrm{CH_3OH})_Q=0.450\,\mathrm{mol} \] \(\textbf{Interpret the chemical contrast:}\) Steam reforming transfers methane's carbon into carbon monoxide and produces molecular hydrogen. Controlled oxidation retains methane's carbon in methanol. Methane does not ordinarily undergo these two reactions under interchangeable conditions. The catalyst, reactant supply and reaction temperature determine the permitted transformation. In particular, a regulated oxygen supply can favour partial oxidation rather than complete combustion under suitable conditions. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{H_2})_P&=0.900\,\mathrm{mol}\\ n(\mathrm{CH_4,\ remaining})_P&=0.150\,\mathrm{mol}\\ n(\mathrm{O_2})_Q&=0.225\,\mathrm{mol}\\ n(\mathrm{CH_3OH})_Q&=0.450\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the carbon-skeleton requirement:}\) Aromatisation of suitable higher straight-chain alkanes involves dehydrogenation and cyclisation. The stipulated reactions retain every original carbon atom. A six-carbon alkane can consequently form a benzene ring. A seven-carbon alkane can form a benzene ring bearing a methyl substituent. \(\textbf{Balance aromatisation of n-hexane:}\) The starting formula is \(\mathrm{C_6H_{14}}\). Benzene has formula \(\mathrm{C_6H_6}\). The difference is eight hydrogen atoms. Therefore: \[ \mathrm{C_6H_{14}\longrightarrow C_6H_6+4H_2} \] P produces four hydrogen molecules per molecule of n-hexane. \(\textbf{Balance aromatisation of n-heptane:}\) The starting formula is \(\mathrm{C_7H_{16}}\). Toluene has formula \(\mathrm{C_7H_8}\). The hydrogen difference is again eight atoms: \[ \mathrm{C_7H_{16}\longrightarrow C_7H_8+4H_2} \] Q also produces four hydrogen molecules per molecule of starting alkane. \(\textbf{Interpret the equal hydrogen yields:}\) Both processes create one six-membered aromatic ring from an acyclic saturated carbon framework. The additional carbon of n-heptane remains as a methyl group on benzene. It does not require an additional hydrogen-loss step under the stated product restriction. Thus, the difference in starting carbon count does not change the hydrogen evolution per molecule in these two cases. \(\textbf{Apply side-chain oxidation to benzene:}\) Benzene possesses no alkyl side chain. Consequently, it has no benzylic alkyl hydrogen available for the stipulated side-chain-to-carboxyl conversion. The benzene produced from P remains unchanged in this particular reaction. \(\textbf{Apply side-chain oxidation to toluene:}\) Toluene contains the group: \[ \mathrm{C_6H_5CH_3} \] Its ring-attached methyl carbon possesses three benzylic hydrogen atoms. The stipulated vigorous oxidation converts this methyl group into a carboxyl group: \[ \mathrm{C_6H_5CH_3\xrightarrow[\mathrm{acidification}]{KMnO_4,\ heat}C_6H_5COOH} \] The aromatic organic product is benzoic acid. \(\textbf{Interpret the sequential distinction:}\) The first reaction uses the different carbon counts to determine whether the aromatic product contains a side chain. The second reaction detects the presence of that oxidisable benzylic substituent. Both aromatisations evolve the same amount of hydrogen per mole, but their aromatic products respond differently to side-chain oxidation. Therefore, hydrogen evolution alone cannot establish whether the initial alkane was n-hexane or n-heptane.

Explanation: \(\textbf{Identify the effect of alkane isomerisation:}\) The stated reaction changes the arrangement of carbon–carbon bonds without adding or removing carbon or hydrogen atoms. The original and final alkanes therefore have the same molecular formula: \[ \mathrm{C_6H_{14}} \] They are chain isomers, not members of different homologous series. \(\textbf{Count the hydrogen environments in n-hexane:}\) The unbranched structure is: \[ \mathrm{CH_3CH_2CH_2CH_2CH_2CH_3} \] The two terminal methyl groups are equivalent. The two next-to-terminal methylene groups are equivalent. The two central methylene groups are equivalent. Therefore, substitution at C1, C2 or C3 represents every distinct monochlorination position. There are three constitutional monochloro products. \(\textbf{Inspect the branched product:}\) 2-Methylpentane has the structure: \[ \mathrm{CH_3CH(CH_3)CH_2CH_2CH_3} \] Two methyl groups attached to its second carbon are equivalent. The terminal methyl group at the opposite end is different from them. The methylene groups at C3 and C4 are also nonequivalent. The methine hydrogen at C2 occupies a fifth environment. \(\textbf{Count the branched-product substitution classes:}\) The five distinct environments are:- 1- Either of the two equivalent methyl groups attached to C2. 2- The terminal methyl group at C5. 3- The methylene group at C3. 4- The methylene group at C4. 5- The tertiary methine hydrogen at C2. Each gives a constitutionally distinct monochloro product. Therefore, 2-methylpentane gives five such products. \(\textbf{Interpret the boiling-point change:}\) n-Hexane has a comparatively extended, unbranched carbon framework. 2-Methylpentane is more compact because its skeleton is branched. Within this isomeric pair, increased branching generally reduces the effective surface of intermolecular contact and the associated dispersion interactions. The branched isomer consequently has a lower boiling point than n-hexane. \(\textbf{Relate the observations to different structural properties:}\) The unchanged molecular formula confirms that isomerisation does not alter the elemental composition. The increase from three to five monochlorination products reveals the changed pattern of nonequivalent hydrogen environments. The boiling-point tendency reflects the altered molecular shape. These independent effects distinguish chain isomerisation from a reaction that changes the degree of saturation or the carbon count. The chlorination count requires structural symmetry analysis, whereas the boiling-point prediction requires intermolecular-force reasoning.

Explanation: \(\textbf{Identify the distinguishing products of cracking:}\) Cracking breaks a higher alkane into smaller hydrocarbon molecules. The stipulated pathways both generate hexane but differ in their unsaturated fragments. P generates one butene molecule for each decane molecule cracked. Q generates two ethene molecules for each decane molecule cracked. \(\textbf{Use ethene formation to determine Q:}\) Only Q produces ethene. Let \(q\) be the amount of decane following Q. The reaction produces two ethene molecules per decane molecule: \[ 2q=0.200\,\mathrm{mol} \] Therefore: \[ q=\frac{0.200}{2} \] \[ q=0.100\,\mathrm{mol} \] \(\textbf{Calculate Q's bromine consumption:}\) Each ethene molecule contains one ordinary C=C bond. Complete addition consumes one bromine molecule per ethene molecule: \[ n(\mathrm{Br_2})_Q=0.200\,\mathrm{mol} \] \(\textbf{Use the remaining bromine uptake to determine P:}\) The total bromine consumption is: \[ n(\mathrm{Br_2})_{\mathrm{total}}=0.450\,\mathrm{mol} \] Thus, the bromine consumed by the butene generated in P is: \[ n(\mathrm{Br_2})_P=0.450-0.200 \] \[ n(\mathrm{Br_2})_P=0.250\,\mathrm{mol} \] Each butene molecule consumes one bromine molecule. Therefore: \[ n(\mathrm{butene})=0.250\,\mathrm{mol} \] \(\textbf{Recover P's cracked-decane amount:}\) One molecule of decane following P gives one butene molecule. Hence: \[ p=n(\mathrm{decane\ via\ P})=0.250\,\mathrm{mol} \] \(\textbf{Calculate total decane converted:}\) \[ n(\mathrm{decane})_{\mathrm{cracked}}=p+q \] \[ n(\mathrm{decane})_{\mathrm{cracked}}=0.250+0.100 \] \[ n(\mathrm{decane})_{\mathrm{cracked}}=0.350\,\mathrm{mol} \] \(\textbf{Calculate uncracked decane:}\) \[ n(\mathrm{decane})_{\mathrm{remaining}}=0.500-0.350 \] \[ n(\mathrm{decane})_{\mathrm{remaining}}=0.150\,\mathrm{mol} \] \(\textbf{Calculate hexane formation:}\) Both pathways give one hexane molecule per decane molecule cracked: \[ n(\mathrm{hexane})=p+q \] \[ n(\mathrm{hexane})=0.350\,\mathrm{mol} \] \(\textbf{Verify the carbon balance:}\) The initial sample contains: \[ n(C)_{\mathrm{initial}}=10(0.500)=5.000\,\mathrm{mol} \] The products and residual decane contain: \[ n(C)_{\mathrm{final}}=10(0.150)+6(0.350)+4(0.250)+2(0.200) \] \[ =1.500+2.100+1.000+0.400 \] \[ n(C)_{\mathrm{final}}=5.000\,\mathrm{mol} \] \(\textbf{Interpret the independent measurements:}\) The measured ethene amount identifies the contribution of the pathway producing two ethene molecules. The total bromine uptake measures all the ordinary alkene double bonds. Subtracting the ethene contribution reveals the butene amount and therefore the second cracking pathway. The remaining starting alkane and common saturated hexane product follow from the two pathway extents. Cracking is thus distinguished from isomerisation because it changes the number and sizes of hydrocarbon molecules. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{decane\ via\ P})&=0.250\,\mathrm{mol}\\ n(\mathrm{decane\ via\ Q})&=0.100\,\mathrm{mol}\\ n(\mathrm{decane,\ uncracked})&=0.150\,\mathrm{mol}\\ n(\mathrm{hexane})&=0.350\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the two possible carbon products:}\) Complete combustion converts methane carbon into carbon dioxide. The stipulated incomplete-combustion pathway instead deposits carbon as solid carbon black. Both pathways convert methane hydrogen into water, but they require different amounts of oxygen. \(\textbf{Balance complete combustion:}\) \[ \mathrm{CH_4+2O_2\longrightarrow CO_2+2H_2O} \] Each methane molecule following P consumes two oxygen molecules and forms two water molecules. \(\textbf{Balance carbon-black formation:}\) \[ \mathrm{CH_4+O_2\longrightarrow C+2H_2O} \] Each methane molecule following Q consumes one oxygen molecule and also forms two water molecules. \(\textbf{Define the competing reaction amounts:}\) Let \(x\) be methane consumed through P and \(y\) methane consumed through Q. All methane reacts, so: \[ x+y=0.360 \] \(\textbf{Use the oxygen balance:}\) P consumes two oxygen molecules per methane molecule. Q consumes one. Therefore: \[ 2x+y=0.540 \] \(\textbf{Determine complete-combustion methane:}\) Subtract the methane balance from the oxygen balance: \[ (2x+y)-(x+y)=0.540-0.360 \] \[ x=0.180\,\mathrm{mol} \] \(\textbf{Determine incomplete-combustion methane:}\) \[ y=0.360-0.180 \] \[ y=0.180\,\mathrm{mol} \] \(\textbf{Calculate carbon-black production:}\) Every methane molecule following Q deposits one carbon atom as carbon black. Therefore: \[ n(\mathrm{C,\ solid})=y \] \[ n(\mathrm{C,\ solid})=0.180\,\mathrm{mol} \] \(\textbf{Calculate total water production:}\) Both pathways give two water molecules per methane molecule: \[ n(\mathrm{H_2O})=2(x+y) \] \[ n(\mathrm{H_2O})=2(0.360) \] \[ n(\mathrm{H_2O})=0.720\,\mathrm{mol} \] \(\textbf{Verify the oxygen atoms:}\) Complete combustion produces: \[ n(\mathrm{CO_2})=0.180\,\mathrm{mol} \] The product oxygen atoms total: \[ 2(0.180)+0.720=1.080\,\mathrm{mol} \] The supplied oxygen contains: \[ 2(0.540)=1.080\,\mathrm{mol} \] oxygen atoms, agreeing with the product balance. \(\textbf{Interpret the incomplete combustion:}\) Carbon-black deposition is not the same as forming carbon monoxide. The stipulated reaction specifically conserves carbon as a solid. The oxygen restriction changes the distribution of carbon products but not the total water amount because both pathways oxidise all four methane hydrogens into two water molecules. The result therefore requires separate carbon, oxygen and hydrogen balances. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{CH_4\ via\ P})&=0.180\,\mathrm{mol}\\ n(\mathrm{CH_4\ via\ Q})&=0.180\,\mathrm{mol}\\ n(\mathrm{carbon\ black})&=0.180\,\mathrm{mol}\\ n(\mathrm{H_2O})&=0.720\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Establish the hypothetical reference:}\) A formal Kekulé structure of benzene contains three alternating C=C bonds. However, these are not three independently localised alkene bonds in the actual molecule. The proposed reference model deliberately treats them as independent to calculate an expected hydrogenation enthalpy in the absence of additional aromatic stabilisation. \(\textbf{Calculate the reference enthalpy:}\) Each ordinary isolated C=C hydrogenation releases approximately \(120\,\mathrm{kJ\,mol^{-1}}\). For three such bonds: \[ \Delta H^\circ_{\mathrm{reference}}=3(-120) \] \[ \Delta H^\circ_{\mathrm{reference}}=-360\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Compare the observed benzene hydrogenation:}\) The experimentally supplied enthalpy is: \[ \Delta H^\circ_{\mathrm{benzene}}=-208\,\mathrm{kJ\,mol^{-1}} \] Both the actual and hypothetical transformations consume three hydrogen molecules and end at the same cyclohexane state in the stipulated comparison. \(\textbf{Calculate the stabilisation difference:}\) The difference between the expected and observed heat releases is: \[ E_{\mathrm{stabilisation}}=\left|\Delta H^\circ_{\mathrm{reference}}\right|-\left|\Delta H^\circ_{\mathrm{benzene}}\right| \] \[ E_{\mathrm{stabilisation}}=360-208 \] \[ E_{\mathrm{stabilisation}}=152\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the enthalpy direction:}\) A more exothermic hydrogenation indicates a higher-energy starting material when the products and other reactants are common. The hypothetical localised reference releases more energy on reaching cyclohexane. Actual benzene releases less. Thus, benzene begins at lower enthalpy relative to that reference. \(\textbf{Connect the result to bonding:}\) All six carbon atoms of benzene are \(sp^2\)-hybridised. Their p orbitals permit continuous cyclic electron delocalisation. The six carbon–carbon connections are equivalent in the actual benzene molecule. A single alternating-double-bond drawing is therefore not an adequate description of fixed, independent alkene bonds. \(\textbf{Interpret chemical consequences:}\) Aromatic stabilisation helps explain why benzene ordinarily favours substitution reactions that restore or retain aromaticity instead of undergoing simple addition as readily as a typical alkene. The comparison does not imply that benzene cannot be hydrogenated. Complete hydrogenation is possible under suitable catalytic conditions. It instead shows that the initial aromatic system has additional stability relative to the hypothetical localised-bond reference. \(\textbf{Final answer:}\) \[ \boxed{E_{\mathrm{aromatic\ stabilisation}}\approx152\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Identify the function of the nitrating mixture:}\) Concentrated nitric acid supplies the nitrogen-containing fragment that ultimately becomes the nitro group. Concentrated sulphuric acid provides strongly acidic conditions that facilitate formation of the active electrophile. The resulting electrophilic species is the nitronium ion. \(\textbf{Evaluate Statement I:}\) The nitronium ion has formula: \[ \mathrm{NO_2^+} \] It is electron-deficient and can accept electron density from benzene's aromatic \(\pi\) system. The electrophile entering the substitution mechanism is therefore not simply an unchanged neutral nitric acid molecule. Statement I is correct. \(\textbf{Evaluate Statement II:}\) Nitric acid is protonated under the strongly acidic reaction conditions. The protonated species can lose water to generate the nitronium ion. A simplified representation is: \[ \mathrm{HNO_3+H^+\longrightarrow NO_2^++H_2O} \] This explains how the nitrating mixture produces a powerful electrophile. Statement II is correct. \(\textbf{Evaluate Statement III:}\) Benzene initially attacks the nitronium ion by forming a new carbon–nitrogen sigma bond. The attacked aromatic carbon temporarily carries both its original hydrogen and the incoming nitro group. A positively charged arenium-ion intermediate is formed. The original complete aromatic delocalisation is temporarily interrupted. Therefore, direct formation of fully aromatic nitrobenzene in the first bond-forming step is an incorrect mechanism description. Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) The sigma complex subsequently loses a proton from the carbon bearing the newly introduced substituent. This restores the aromatic six-\(\pi\)-electron framework. The resulting product is nitrobenzene: \[ \mathrm{C_6H_5NO_2} \] Sulphuric acid promotes formation of the electrophilic species and participates in acid–base processes during the reaction. It is not the source of the nitro group's nitrogen or oxygen atoms in the stated mechanism. Statement IV is correct. \(\textbf{Distinguish nitration from addition:}\) The overall reaction substitutes one benzene hydrogen with a nitro group. It does not permanently saturate one carbon–carbon double bond. The sigma-complex formation and proton-removal stages explain how a new C–N bond can be formed while aromaticity is ultimately recovered. The correct statements are I, II and IV.

Explanation: \(\textbf{Identify the general alkyne hydration sequence:}\) Water adds to a carbon–carbon triple bond under the stipulated acidic mercuric-ion conditions. The initial organic product is an enol containing a C=C bond and an OH group. This enol generally undergoes keto–enol tautomerisation. The carbonyl product depends on the original substituents attached to the triple-bonded carbons. \(\textbf{Follow hydration of ethyne:}\) Ethyne contains two equivalent triply bonded carbons: \[ \mathrm{HC\equiv CH} \] Each initially carries hydrogen. Hydration produces an enol with structure: \[ \mathrm{CH_2=CHOH} \] This is ethenol. Tautomerisation changes it into: \[ \mathrm{CH_3CHO} \] The final product is ethanal, an aldehyde. \(\textbf{Explain ethyne's special outcome:}\) The carbon that becomes the carbonyl carbon retains a hydrogen substituent. Consequently, the final C=O group belongs to an aldehyde. The product should not be called an ordinary methyl ketone because a ketone carbonyl carbon must be attached to two carbon substituents. \(\textbf{Follow hydration of propyne:}\) Propyne has the structure: \[ \mathrm{CH_3C\equiv CH} \] Ordinary Markovnikov hydration places the oxygen-derived group at the internal triple-bonded carbon. The corresponding enol tautomerises to: \[ \mathrm{CH_3COCH_3} \] This is propanone, a methyl ketone. \(\textbf{Follow hydration of but-2-yne:}\) But-2-yne is a symmetrical internal alkyne: \[ \mathrm{CH_3C\equiv CCH_3} \] Hydration in either direction produces an enol that tautomerises to the same constitutional carbonyl compound: \[ \mathrm{CH_3COCH_2CH_3} \] The final product is butan-2-one. \(\textbf{Compare the three outcomes:}\) P gives an aldehyde because its two-carbon triple-bond framework supplies a carbonyl carbon bearing hydrogen. Q gives a methyl ketone through ordinary terminal-alkyne Markovnikov hydration. R gives a single ketone constitution because its internal triple bond is symmetrical. The hydration reaction does not break the original carbon chain. \(\textbf{Interpret the generalisation correctly:}\) The common statement that terminal alkynes give methyl ketones applies to suitably substituted terminal alkynes. Ethyne must be considered separately. Predicting the carbonyl product requires tracking the substituents on the carbon that ultimately acquires the carbonyl oxygen rather than applying a memorised rule without its structural qualification.

Explanation: \(\textbf{Identify the amount entering each reaction:}\) The original alkyl bromide contains four carbon atoms per molecule. The sample is divided into two equal portions: \[ n(\mathrm{RBr})_P=n(\mathrm{RBr})_Q=\frac{0.400}{2} \] \[ n(\mathrm{RBr})_P=n(\mathrm{RBr})_Q=0.200\,\mathrm{mol} \] \(\textbf{Identify the reactive carbon:}\) The starting structure is: \[ \mathrm{BrCH_2CH(CH_3)_2} \] The bromine atom is attached to a primary carbon. The rest of the four-carbon framework is retained in both stipulated pathways. \(\textbf{Follow reduction in P:}\) Reduction replaces bromine by hydrogen: \[ \mathrm{BrCH_2CH(CH_3)_2\xrightarrow{Zn/HCl}CH_3CH(CH_3)_2} \] The hydrocarbon product is 2-methylpropane. One alkyl bromide molecule yields one hydrocarbon molecule. Therefore: \[ n(\mathrm{2\text{-}methylpropane})_P=0.200\,\mathrm{mol} \] \(\textbf{Follow Wurtz coupling in Q:}\) Two molecules of the alkyl bromide join through their original bromine-bearing carbons: \[ \mathrm{2BrCH_2CH(CH_3)_2+2Na\longrightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2+2NaBr} \] The product retains all eight carbon atoms from the two four-carbon reactants. Its longest chain contains six carbon atoms. Two methyl branches occur at positions 2 and 5. Thus, the product is 2,5-dimethylhexane. \(\textbf{Calculate the coupled alkane amount:}\) \[ n(\mathrm{2,5\text{-}dimethylhexane})_Q=\frac{0.200}{2} \] \[ n(\mathrm{2,5\text{-}dimethylhexane})_Q=0.100\,\mathrm{mol} \] \(\textbf{Calculate sodium bromide:}\) Every alkyl bromide molecule contributes one bromine atom to sodium bromide. Therefore: \[ n(\mathrm{NaBr})_Q=0.200\,\mathrm{mol} \] The sodium requirement is also \(0.200\,\mathrm{mol}\). \(\textbf{Verify carbon conservation in Q:}\) The starting bromide contains: \[ 4(0.200)=0.800\,\mathrm{mol} \] carbon atoms. The organic product contains: \[ 8(0.100)=0.800\,\mathrm{mol} \] carbon atoms. \(\textbf{Interpret the reaction distinction:}\) Reduction of an alkyl halide replaces the halogen substituent with hydrogen without extending the carbon skeleton. Wurtz coupling instead joins two carbon frameworks through a new carbon–carbon sigma bond. The same bromide therefore gives a four-carbon alkane in P and an eight-carbon alkane in Q. The calculation must distinguish the one-to-one reduction stoichiometry from the two-to-one coupling stoichiometry. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} P:&\quad0.200\,\mathrm{mol\ 2\text{-}methylpropane}\\ Q:&\quad0.100\,\mathrm{mol\ 2,5\text{-}dimethylhexane}\\ n(\mathrm{NaBr})_Q&=0.200\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the special coupling geometry:}\) Each starting molecule contains two bromine atoms at opposite ends of the same carbon chain. The question explicitly restricts Wurtz coupling to an intramolecular process. Both terminal carbon atoms therefore participate in forming a new bond within one molecule. No second organic molecule supplies carbon atoms to the product. \(\textbf{Follow P's ring closure:}\) 1,3-Dibromopropane contains three carbon atoms connected in an open chain: \[ \mathrm{BrCH_2CH_2CH_2Br} \] Joining the two terminal carbon atoms closes a three-membered ring: \[ \mathrm{Br(CH_2)_3Br+2Na\longrightarrow cyclo\text{-}C_3H_6+2NaBr} \] The organic product is cyclopropane. \(\textbf{Follow Q's ring closure:}\) 1,4-Dibromobutane contains a four-carbon open chain: \[ \mathrm{BrCH_2CH_2CH_2CH_2Br} \] Intramolecular coupling creates a bond between its terminal carbons: \[ \mathrm{Br(CH_2)_4Br+2Na\longrightarrow cyclo\text{-}C_4H_8+2NaBr} \] The product is cyclobutane. \(\textbf{Explain the molecular formulas:}\) Cyclopropane contains three methylene groups: \[ \mathrm{C_3H_6} \] Cyclobutane contains four: \[ \mathrm{C_4H_8} \] Both conform to: \[ \mathrm{C}_{n}\mathrm{H}_{2n} \] \(\textbf{Distinguish ring unsaturation from double-bond unsaturation:}\) An acyclic alkene with one C=C bond also follows \(\mathrm{C}_{n}\mathrm{H}_{2n}\). However, the two cyclic products have only carbon–carbon single bonds. Their deficiency of two hydrogens relative to an acyclic alkane arises from ring formation, not from a C=C bond. \(\textbf{Interpret the hydrogenation test:}\) Under the stipulated conditions, saturated rings are preserved without hydrogenolytic opening. Neither cyclopropane nor cyclobutane contains an ordinary C=C bond available for the specified addition. Therefore, neither undergoes the ordinary one-hydrogen-equivalent alkene hydrogenation reaction. This conclusion depends on the stated ring-preserving conditions; sufficiently forcing conditions can give different reactions for strained small rings. \(\textbf{Compare with ordinary intermolecular Wurtz coupling:}\) Coupling two separate alkyl bromides generally joins two carbon fragments and increases the number of carbon atoms in each product molecule. The stipulated intramolecular coupling instead retains the original molecule's carbon count and closes a ring. The result illustrates why \(\mathrm{C}_{n}\mathrm{H}_{2n}\) alone cannot establish whether a hydrocarbon contains a double bond or a saturated ring. It also shows why the carbon skeleton and reaction conditions must accompany molecular-formula classification.