Hydrocarbons MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Distinguish the two ethyne polymerisation pathways:}\) Ethyne can undergo cyclic trimerisation to benzene under suitable high-temperature conditions. It can also undergo linear polymerisation to polyacetylene under other conditions. The products differ in molecular architecture and residual unsaturation. \(\textbf{Determine ethyne allocated to cyclic trimerisation:}\) The initial ethyne amount is: \[ n_0(\mathrm{C_2H_2})=0.90\,\mathrm{mol} \] One-third enters Route P. Therefore: \[ n(\mathrm{C_2H_2})_P=\frac13(0.90) \] \[ n(\mathrm{C_2H_2})_P=0.30\,\mathrm{mol} \] \(\textbf{Apply benzene-formation stoichiometry:}\) The cyclic reaction is: \[ \mathrm{3C_2H_2\xrightarrow{red\ hot\ iron,\ 873\,K}C_6H_6} \] Three ethyne molecules combine to form one benzene molecule. Hence: \[ n(\mathrm{benzene})=\frac{0.30}{3} \] \[ n(\mathrm{benzene})=0.10\,\mathrm{mol} \] \(\textbf{Calculate ethyne allocated to linear polymerisation:}\) The remaining amount is: \[ n(\mathrm{C_2H_2})_Q=0.90-0.30 \] \[ n(\mathrm{C_2H_2})_Q=0.60\,\mathrm{mol} \] This amount enters the polyacetylene backbone as ethyne-derived repeating-unit equivalents. \(\textbf{Determine the amount of polymer chains:}\) Each ideal chain contains an average of 100 repeating-unit equivalents. Thus: \[ n(\mathrm{polymer\ chains})=\frac{0.60}{100} \] \[ n(\mathrm{polymer\ chains})=0.0060\,\mathrm{mol} \] The number of complete polymer molecules is much smaller than the number of monomer-derived units. \(\textbf{Calculate residual hydrogenatable unsaturation:}\) The ideal polyacetylene backbone retains one carbon–carbon double-bond equivalent per ethyne-derived \(\mathrm{C_2H_2}\) unit, neglecting end effects. Complete hydrogenation of those double bonds therefore requires one hydrogen molecule per repeating-unit equivalent. \[ n(\mathrm{H_2})=1(0.60) \] \[ n(\mathrm{H_2})=0.60\,\mathrm{mol} \] \(\textbf{Verify conservation of carbon:}\) The initial carbon amount is: \[ n(C)_{\mathrm{initial}}=2(0.90)=1.80\,\mathrm{mol} \] Benzene contains: \[ n(C)_P=6(0.10)=0.60\,\mathrm{mol} \] The polyacetylene contains: \[ n(C)_Q=2(0.60)=1.20\,\mathrm{mol} \] The total is: \[ 0.60+1.20=1.80\,\mathrm{mol} \] The carbon balance is satisfied. \(\textbf{Interpret the chemical distinction:}\) Cyclic trimerisation uses three ethyne molecules to produce one aromatic benzene molecule. Linear polymerisation instead retains unsaturation within an extended polymer backbone. Only the polyacetylene portion is included in the stated residual-double-bond hydrogenation. The calculation therefore requires differentiating reaction stoichiometry, repeating-unit count and residual multiple-bond content. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{benzene})=0.10\,\mathrm{mol};\quad n(\mathrm{chains})=0.0060\,\mathrm{mol};\quad n(\mathrm{H_2})=0.60\,\mathrm{mol}} \]

Explanation: \(\textbf{Apply the alkyne physical-state trend:}\) The ordinary alkyne homologous series begins with ethyne, followed by propyne and butyne. NCERT describes the first three members as gases under ordinary conditions. The following eight members are generally liquids, and higher homologues are solids. Ethyne therefore belongs to the gaseous group. But-2-yne is an isomeric butyne and belongs to the four-carbon range included among the first three members. It is accordingly classified as gaseous for the stated generalisation. Hex-1-yne contains six carbon atoms and belongs to the liquid range. Tridec-1-yne contains thirteen carbon atoms and belongs to the higher solid range under the supplied trend. \(\textbf{Distinguish chain size from acidic-hydrogen availability:}\) Acetylenic acidity depends on the presence of hydrogen directly bonded to an \(sp\)-hybridised triple-bonded carbon. It does not depend simply on whether the compound is a gas, liquid or solid. \(\textbf{Examine P:}\) Ethyne has two acetylenic hydrogen atoms. It can form monosodium and, under appropriate conditions, disodium ethynide. Therefore, P gives the characteristic acidic-hydrogen response. \(\textbf{Examine Q:}\) But-2-yne is an internal alkyne. Both triply bonded carbons are attached to methyl groups. Neither carries hydrogen. It therefore does not form sodium alkynide through replacement of an acetylenic hydrogen. \(\textbf{Examine R and S:}\) Hex-1-yne and tridec-1-yne are terminal alkynes. Each possesses one hydrogen attached directly to a triply bonded carbon. Both can undergo the corresponding deprotonation under suitable strongly basic conditions. \(\textbf{Interpret the combined classification:}\) Physical state mainly reflects molecular size and the resulting intermolecular interactions. The acidic-hydrogen test reflects local bonding and hybridisation at the triple bond. Therefore, two gaseous alkynes can differ in this chemical response, while a liquid and a solid terminal alkyne can share it. The correct classification is that P and Q are gases, R is liquid and S is solid, while P, R and S possess acidic acetylenic hydrogen.

Explanation: \(\textbf{Use the absence of acidic acetylenic hydrogen:}\) An alkyne forms the characteristic sodium alkynide through deprotonation only when it possesses hydrogen attached directly to a triply bonded carbon. The unknown contains no such hydrogen. Therefore, its triple bond must be internal. \(\textbf{Apply the molecular formula and hydrogenation evidence:}\) The molecular formula \(\mathrm{C_5H_8}\) is consistent with an acyclic five-carbon monoalkyne. Complete catalytic hydrogenation to pentane establishes an unbranched five-carbon skeleton under the no-rearrangement condition. The internal monoalkyne with that skeleton is: \[ \mathrm{CH_3C\equiv CCH_2CH_3} \] This is pent-2-yne. \(\textbf{Identify the two nonequivalent triple-bonded carbons:}\) One triple-bonded carbon is attached to a methyl group. The other is attached to an ethyl group. These substituent environments are different. Therefore, the two possible orientations of water addition need not produce identical carbonyl compounds. \(\textbf{Hydration with carbonyl formation at carbon 2:}\) If the applicable enol intermediate tautomerises to a ketone having its carbonyl at carbon 2, the product is: \[ \mathrm{CH_3COCH_2CH_2CH_3} \] This is pentan-2-one. \(\textbf{Hydration with carbonyl formation at carbon 3:}\) The alternative addition orientation leads, after tautomerisation, to: \[ \mathrm{CH_3CH_2COCH_2CH_3} \] This is pentan-3-one. \(\textbf{Distinguish possible products from guaranteed equal yields:}\) Both ketones are constitutionally possible from the unsymmetrical internal alkyne. However, this observation does not establish that they must form in equal amounts. Their relative abundances would depend on the detailed reaction conditions and pathway preferences. The question requests the structurally possible carbonyl products rather than their experimental ratio. \(\textbf{Interpret the combined identification:}\) Sodamide distinguishes terminal from internal alkynes. Complete hydrogenation fixes the saturated carbon skeleton. The unsymmetrical internal triple bond then explains why two different ketone connectivities are possible upon hydration. Thus, X is pent-2-yne and the two possible products are pentan-2-one and pentan-3-one.

Explanation: Statement I is correct. Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] Propyne has the structure: \[ \mathrm{CH_3C\equiv CH} \] Both compounds contain at least one hydrogen attached directly to an \(sp\)-hybridised carbon. However, the presence or absence of an alkyl substituent can still influence their relative acidity. NCERT gives ethyne as more acidic than propyne in this qualitative comparison. Statement II is correct. But-2-yne has the structure: \[ \mathrm{CH_3C\equiv CCH_3} \] Neither triply bonded carbon carries hydrogen. Therefore, the characteristic terminal-alkyne deprotonation cannot occur at its triple bond. The presence of hydrogen atoms on its methyl groups does not make them equivalent to acidic acetylenic hydrogens. Statement III is correct. An alkyl group generally exhibits an electron-releasing inductive influence. In the comparison of ethyne and propyne, this substitution changes the electronic environment of the terminal alkyne. The corresponding electron-donating influence does not provide the same stabilisation of the conjugate base as would be expected from an electron-withdrawing substituent. It is therefore consistent with propyne being less acidic than ethyne. Statement IV is incorrect. The methyl-group hydrogen atoms of propyne are attached to an \(sp^3\)-hybridised carbon. The terminal acetylenic hydrogen is attached to an \(sp\)-hybridised carbon having substantially greater \(s\)-character. The characteristic acidic response to a sufficiently strong base occurs at the terminal acetylenic hydrogen. It is not ordinarily explained by preferential removal of hydrogen from the methyl group. \(\textbf{Combine the structural trend:}\) The presence of a triple bond alone is not enough to establish the number of acidic hydrogens. One must determine which carbon atoms carry hydrogen and how substitution affects their electronic environment. The comparison separates three considerations: hybridisation, the presence of acetylenic C–H bonds and the influence of alkyl substitution on relative acidity. Only Statements I, II and III are correct.

Explanation: \(\textbf{Identify the functional features of X:}\) The structure is: \[ \mathrm{CH_3CH=CHCH_2C\equiv CH} \] It contains one ordinary carbon–carbon double bond and one terminal carbon–carbon triple bond. The terminal triple-bonded carbon possesses one acidic acetylenic hydrogen. Its carbon skeleton contains six carbon atoms. \(\textbf{Experiment P — Calculate hydrogen evolution:}\) Sodium can replace the hydrogen attached directly to the terminal \(sp\)-hybridised carbon. Each molecule of X contains one such hydrogen. The idealised reaction is: \[ \mathrm{2RC\equiv CH+2Na\longrightarrow2RC\equiv CNa+H_2} \] For one mole of X: \[ n(\mathrm{H_2})_P=\frac12(1) \] \[ n(\mathrm{H_2})_P=0.50\,\mathrm{mol} \] The ordinary alkene hydrogen atoms are not counted as acidic acetylenic hydrogens. \(\textbf{Experiment Q — Calculate complete hydrogen uptake:}\) The double bond requires one hydrogen molecule. The triple bond requires two hydrogen molecules. Therefore: \[ n(\mathrm{H_2})_Q=1+2 \] \[ n(\mathrm{H_2})_Q=3.00\,\mathrm{mol} \] Complete hydrogenation produces hexane without changing the carbon skeleton. \(\textbf{Experiment R — First possible hydrogenation pathway:}\) Suppose both hydrogen equivalents are added to the original terminal triple bond. The triple bond becomes a single bond, while the original internal double bond remains. The resulting structure is: \[ \mathrm{CH_3CH=CHCH_2CH_2CH_3} \] This is hex-2-ene. \(\textbf{Experiment R — Second possible hydrogenation pathway:}\) Alternatively, one hydrogen equivalent can saturate the original carbon–carbon double bond, while one equivalent partially reduces the triple bond. The resulting structure is: \[ \mathrm{CH_3CH_2CH_2CH_2CH=CH_2} \] This is hex-1-ene. \(\textbf{Determine why the product is not unique:}\) Both routes consume exactly two hydrogen molecules per starting molecule. Both preserve the six-carbon skeleton. However, they leave the residual double bond at different positions. Therefore, knowing only the total hydrogen uptake does not establish a unique partially hydrogenated constitutional product. A selective catalyst or a specified order of reduction would be needed to make that prediction definite. \(\textbf{Compare the three experiments:}\) Sodium measures acidic acetylenic hydrogen. Complete hydrogenation measures the full multiple-bond hydrogen requirement. Partial hydrogenation additionally depends on which unsaturated bond is reduced and to what extent. These independent distinctions lead to the correct combined conclusion. \(\textbf{Final answer:}\) \[ n(\mathrm{H_2})_P=0.50\,\mathrm{mol};\quad n(\mathrm{H_2})_Q=3.00\,\mathrm{mol};\quad R:\ \mathrm{hex\text{-}1\text{-}ene\ or\ hex\text{-}2\text{-}ene} \]

Explanation: \(\textbf{Identify the structural change in successive elimination:}\) A vicinal dibromide contains bromine atoms attached to adjacent carbon atoms. The first elimination removes HBr and produces a vinyl bromide. The second elimination removes another HBr and creates a carbon–carbon triple bond. The position of the new triple bond is determined by the original pair of bromine-bearing carbons. \(\textbf{Follow Route P:}\) In 1,2-dibromohexane, bromine occupies carbons 1 and 2. The stipulated elimination sequence creates the triple bond between those carbons: \[ \mathrm{BrCH_2CHBr(CH_2)_3CH_3\xrightarrow[\text{then }NaNH_2]{alc.\ KOH}HC\equiv C(CH_2)_3CH_3} \] The resulting compound is hex-1-yne. Its terminal triply bonded carbon carries one hydrogen atom. Consequently, it can undergo deprotonation with sodamide: \[ \mathrm{RC\equiv CH+NaNH_2\longrightarrow RC\equiv CNa+NH_3} \] \(\textbf{Apply selective hydrogenation to P:}\) Lindlar's catalyst partially reduces the terminal triple bond to a terminal double bond. The product is hex-1-ene. Because one double-bonded carbon carries two hydrogen atoms, this alkene has no distinguishable cis–trans pair. \(\textbf{Follow Route Q:}\) In 2,3-dibromohexane, the bromine atoms occupy carbons 2 and 3. Successive elimination produces: \[ \mathrm{CH_3C\equiv CCH_2CH_2CH_3} \] This compound is hex-2-yne. Neither triply bonded carbon carries hydrogen, so the characteristic terminal-alkyne deprotonation does not occur. \(\textbf{Apply selective hydrogenation to Q:}\) Hex-2-yne is an internal alkyne. Its two triple-bonded carbons carry different substituents after reduction. Lindlar's catalyst therefore gives cis-hex-2-ene as the principal alkene. \(\textbf{Interpret the combined evidence:}\) Both starting dibromides contain six carbon atoms and undergo the same formal sequence of eliminations. Nevertheless, their different bromine positions generate different triple-bond positions. The acetylenic-hydrogen test then distinguishes terminal from internal alkynes, while selective reduction distinguishes the stereochemical possibilities of their alkene products. Thus, only option D satisfies the elimination, acidity and selective-reduction observations.

Explanation: \(\textbf{Identify the relevant hydration principle:}\) Ordinary mercury(II)-catalysed hydration of a suitable alkyne produces an enol intermediate. The enol undergoes tautomerisation to a carbonyl compound. For an internal alkyne, the possible carbonyl positions must be determined from the original triple-bonded carbons. The number of constitutionally different products depends on whether those positions are equivalent. \(\textbf{Analyse hex-2-yne:}\) P has the structure: \[ \mathrm{CH_3C\equiv CCH_2CH_2CH_3} \] One triple-bonded carbon carries a methyl group. The other carries a propyl group. These environments are not constitutionally equivalent. \(\textbf{Consider the first hydration orientation:}\) Formation of an enol with its hydroxyl-bearing carbon at position 2 leads, after tautomerisation, to: \[ \mathrm{CH_3COCH_2CH_2CH_2CH_3} \] This compound is hexan-2-one. \(\textbf{Consider the alternative orientation:}\) An enol with the hydroxyl-bearing carbon at position 3 can tautomerise to: \[ \mathrm{CH_3CH_2COCH_2CH_2CH_3} \] This is hexan-3-one. Therefore, P permits two constitutional ketone products under the supplied assumptions. \(\textbf{Analyse hex-3-yne:}\) Q has the structure: \[ \mathrm{CH_3CH_2C\equiv CCH_2CH_3} \] The two triple-bonded carbon atoms each carry an ethyl substituent. The molecule is symmetrical about the triple bond. Adding water in either formal orientation therefore produces constitutionally equivalent enol structures. Both lead to hexan-3-one after tautomerisation. \(\textbf{Distinguish symmetry from reaction multiplicity:}\) Two possible directions of approach do not necessarily mean two different chemical products. For P, the triple-bonded positions are nonequivalent and can become two distinct carbonyl positions. For Q, reversing the addition orientation does not change the final constitutional structure. The question determines possible products, not their relative yields. It is therefore unnecessary to assume equal reaction rates or equal amounts of P's two ketones. The correct comparison is two possible ketones from P and one from Q.

Explanation: \(\textbf{Identify the starting unsaturation:}\) Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] Its triple bond contains two \(\pi\) bonds. Each specified sequence adds two reagent molecules, so the final carbon–carbon bond can become a single bond. However, the intermediates depend on which reagent is added first. \(\textbf{Route P — Bromine addition first:}\) One equivalent of bromine adds across the triple bond. Each original triply bonded carbon receives one bromine atom: \[ \mathrm{HC\equiv CH+Br_2\longrightarrow BrCH=CHBr} \] The intermediate is 1,2-dibromoethene. One carbon–carbon double bond remains. \(\textbf{Complete Route P using HBr:}\) The two doubly bonded carbons in 1,2-dibromoethene are constitutionally equivalent. Addition of HBr therefore produces the same constitutional product regardless of which end receives hydrogen. One carbon becomes \(\mathrm{CH_2Br}\). The other becomes \(\mathrm{CHBr_2}\). The final structure is: \[ \mathrm{BrCH_2CHBr_2} \] Its preferred name is 1,1,2-tribromoethane. \(\textbf{Route Q — HBr addition first:}\) One equivalent of HBr adds to ethyne to produce bromoethene: \[ \mathrm{HC\equiv CH+HBr\longrightarrow CH_2=CHBr} \] This intermediate contains one carbon–carbon double bond and one bromine substituent. \(\textbf{Complete Route Q using bromine:}\) Bromine adds across the remaining double bond. One bromine atom becomes attached to each double-bonded carbon: \[ \mathrm{CH_2=CHBr+Br_2\longrightarrow BrCH_2CHBr_2} \] The final product is again 1,1,2-tribromoethane. \(\textbf{Compare the two reaction orders:}\) The intermediates are constitutionally different because bromine addition introduces two bromine atoms, whereas one HBr addition introduces only one. Nevertheless, both sequences ultimately place three bromine atoms and three hydrogen atoms across the original two-carbon skeleton. The final substituted-carbon arrangement is also identical in this particular case. This convergence should not be generalised to every unsymmetrical substituted alkyne without examining its addition orientations. The correct conclusion is that different first-stage products can converge upon the same saturated constitutional product.

Explanation: \(\textbf{Identify the bond changes in each stage:}\) An alkyne initially possesses a carbon–carbon triple bond. Addition of one hydrogen molecule reduces its bond order from triple to double. A second hydrogen molecule reduces the remaining double bond to a single bond. Each stage also forms two new carbon–hydrogen bonds. The approximate reaction enthalpy is calculated from the sum of bonds broken minus the sum of bonds formed. \(\textbf{Stage P — Partial hydrogenation:}\) The governing bond-energy relation is: \[ \Delta H_P=[D(\mathrm{C\equiv C})+D(\mathrm{H-H})]-[D(\mathrm{C=C})+2D(\mathrm{C-H})] \] Substituting: \[ \Delta H_P=(839+436)-[614+2(413)] \] \[ \Delta H_P=1275-(614+826) \] \[ \Delta H_P=1275-1440 \] \[ \Delta H_P=-165\,\mathrm{kJ\,mol^{-1}} \] Thus, the alkyne-to-alkene stage is exothermic under the given approximation. \(\textbf{Stage Q — Hydrogenation of the remaining double bond:}\) The corresponding relation is: \[ \Delta H_Q=[D(\mathrm{C=C})+D(\mathrm{H-H})]-[D(\mathrm{C-C})+2D(\mathrm{C-H})] \] Substituting: \[ \Delta H_Q=(614+436)-[348+2(413)] \] \[ \Delta H_Q=1050-1174 \] \[ \Delta H_Q=-124\,\mathrm{kJ\,mol^{-1}} \] This stage is also exothermic. \(\textbf{Combine the two stages:}\) By Hess's law, the total enthalpy change equals the sum of the stagewise changes: \[ \Delta H_{\mathrm{overall}}=\Delta H_P+\Delta H_Q \] \[ \Delta H_{\mathrm{overall}}=-165+(-124) \] \[ \Delta H_{\mathrm{overall}}=-289\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Verify using direct bond changes:}\) Complete conversion of the triple bond to a single bond consumes two hydrogen molecules and forms four carbon–hydrogen bonds. Therefore: \[ \Delta H_{\mathrm{direct}}=[839+2(436)]-[348+4(413)] \] \[ \Delta H_{\mathrm{direct}}=1711-2000 \] \[ \Delta H_{\mathrm{direct}}=-289\,\mathrm{kJ\,mol^{-1}} \] The direct and stagewise results agree. \(\textbf{Interpret the comparison:}\) Both stages release heat, but their estimated molar enthalpy changes are different. The first stage is more exothermic by \(41\,\mathrm{kJ\,mol^{-1}}\) according to the supplied values. These thermochemical calculations do not establish which catalyst is required to stop selectively at the alkene. Selective reduction also depends on the reaction pathway and catalyst. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_P=-165;\quad\Delta H_Q=-124;\quad\Delta H_{\mathrm{overall}}=-289\ \mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Distinguish the polymer backbones:}\) Polyacetylene contains conjugated carbon–carbon double bonds in its ideal backbone. Polyethene contains a saturated carbon–carbon backbone. Therefore, under the specified addition-only conditions, only polyacetylene consumes bromine through ordinary double-bond addition. \(\textbf{Recover the polyacetylene repeating-unit amount:}\) Every ideal \(\mathrm{C_2H_2}\) repeating-unit equivalent contributes one reactive double-bond equivalent. One bromine molecule is required to saturate each such equivalent. The observed bromine uptake is: \[ n(\mathrm{Br_2})=1.00\,\mathrm{mol} \] Therefore: \[ n(\mathrm{C_2H_2\ repeat\ equivalents})=1.00\,\mathrm{mol} \] \(\textbf{Calculate polyacetylene mass:}\) The repeating-unit molar mass is: \[ M(\mathrm{C_2H_2})=2(12)+2(1) \] \[ M(\mathrm{C_2H_2})=26\,\mathrm{g\,mol^{-1}} \] Thus: \[ m(\mathrm{polyacetylene})=1.00(26) \] \[ m(\mathrm{polyacetylene})=26.0\,\mathrm{g} \] \(\textbf{Calculate the remaining polymer mass:}\) The total mixture mass is \(54.0\,\mathrm{g}\). Therefore: \[ m(\mathrm{polyethene})=54.0-26.0 \] \[ m(\mathrm{polyethene})=28.0\,\mathrm{g} \] The polyethene component contains one mole of ideal \(\mathrm{C_2H_4}\) repeating units, but those units do not contain residual alkene double bonds. \(\textbf{Predict complete hydrogenation of the original mixture:}\) One hydrogen molecule adds across one carbon–carbon double bond. The polyacetylene contribution therefore requires: \[ n(\mathrm{H_2})_{\mathrm{polyacetylene}}=1.00\,\mathrm{mol} \] The saturated polyethene component requires none for ordinary double-bond hydrogenation: \[ n(\mathrm{H_2})_{\mathrm{polyethene}}=0 \] Hence: \[ n(\mathrm{H_2})_{\mathrm{total}}=1.00+0 \] \[ n(\mathrm{H_2})_{\mathrm{total}}=1.00\,\mathrm{mol} \] \(\textbf{Interpret the independent measurements:}\) Bromine addition identifies the residual unsaturation of the polyacetylene component rather than merely the total number of carbon atoms. Hydrogenation requires the same number of double-bond equivalents under the stated idealisation. The two polymers differ both in repeating-unit composition and in residual bond order. This is why total polymer mass alone would not determine hydrogen uptake, but the bromine measurement does. \(\textbf{Final answer:}\) \[ \boxed{m(\mathrm{polyacetylene})=26.0\,\mathrm{g};\quad m(\mathrm{polyethene})=28.0\,\mathrm{g};\quad n(\mathrm{H_2})=1.00\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the common starting structure:}\) Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] Its triple bond contains two \(\pi\) bonds and can undergo successive addition reactions. However, HBr and molecular bromine introduce their atoms differently. \(\textbf{Route P — First HBr addition:}\) One HBr molecule adds across the triple bond. Because ethyne is symmetrical, the two addition orientations are constitutionally equivalent. The resulting vinyl halide is: \[ \mathrm{CH_2=CHBr} \] This compound is bromoethene. \(\textbf{Route P — Second HBr addition:}\) The second HBr molecule adds across the remaining double bond. Under the stated ordinary addition conditions, the bromine atom becomes attached to the carbon already bearing bromine. The final product is: \[ \mathrm{CH_3CHBr_2} \] This is 1,1-dibromoethane. Both bromine atoms occupy the same carbon atom, so the compound is a geminal dibromide. \(\textbf{Route Q — Molecular bromine addition:}\) One bromine molecule adds across the ethyne triple bond. Each triply bonded carbon receives a bromine atom: \[ \mathrm{HC\equiv CH+Br_2\longrightarrow BrCH=CHBr} \] The intermediate is 1,2-dibromoethene. \(\textbf{Route Q — Hydrogenate the residual double bond:}\) One hydrogen molecule adds across the remaining C=C bond. The existing C–Br bonds remain intact by stipulation: \[ \mathrm{BrCH=CHBr+H_2\longrightarrow BrCH_2CH_2Br} \] The final product is 1,2-dibromoethane. Its bromine atoms occupy adjacent carbons, so it is a vicinal dibromide. \(\textbf{Compare formulas and connectivity:}\) Both final compounds contain two carbon atoms, four hydrogen atoms and two bromine atoms. Their molecular formula is: \[ \mathrm{C_2H_4Br_2} \] However, the arrangement of bromine atoms differs. The compounds are positional constitutional isomers, not chain isomers. \(\textbf{Interpret the synthetic distinction:}\) Identical starting hydrocarbons and identical final molecular formulas do not guarantee identical organic products. The identity of the reagent controls which atoms are introduced at each stage and where they become attached. The two routes therefore provide a direct distinction between geminal and vicinal dihalide formation.

Explanation: \(\textbf{Distinguish hydration from the final carbonyl-forming step:}\) Under mercury(II)-catalysed acidic conditions, an alkyne can add water to form an enol. The enol then undergoes keto–enol tautomerisation. The oxygen atom introduced into the enol becomes the carbonyl oxygen in the corresponding keto or aldehyde tautomer under the supplied no-exchange conditions. \(\textbf{Hydrate ethyne:}\) Ethyne contains two equivalent triple-bonded carbon atoms. Its hydration initially produces the enol vinyl alcohol: \[ \mathrm{CH_2=CH^{18}OH} \] The labelled oxygen is directly attached to the carbon of the enol hydroxyl group. \(\textbf{Tautomerise the ethyne-derived enol:}\) During tautomerisation, the carbon–carbon double bond shifts to a carbon–oxygen double bond, accompanied by proton transfer. The labelled oxygen remains attached to the same carbon atom. The stable final product is: \[ \mathrm{CH_3CH=^{18}O} \] This is ethanal labelled at its carbonyl oxygen. \(\textbf{Hydrate but-1-yne:}\) But-1-yne is an unsymmetrical terminal alkyne. Under the stated conditions, ordinary Markovnikov hydration places the enol hydroxyl group on carbon 2. The enol structure is: \[ \mathrm{CH_3CH_2C(^{18}OH)=CH_2} \] \(\textbf{Tautomerise the but-1-yne-derived enol:}\) The enol converts into the corresponding methyl ketone: \[ \mathrm{CH_3CH_2C(=^{18}O)CH_3} \] This is oxygen-labelled butan-2-one. \(\textbf{Check oxygen-atom conservation:}\) The label is attached to water's oxygen atom before reaction. Hydration forms a new carbon–oxygen bond. Tautomerisation changes bonding and proton locations but does not require elimination of that oxygen atom. Thus, the oxygen label remains in the carbonyl product under the explicitly specified conditions. \(\textbf{Interpret the terminal-alkyne distinction:}\) Ethyne gives an aldehyde because its two-carbon skeleton leads to ethanal. A substituted terminal alkyne such as but-1-yne instead gives a methyl ketone under ordinary mercury-catalysed hydration conditions. The isotope experiment traces the oxygen source while the substrate structure determines the carbonyl product. The correct products are labelled ethanal and labelled butan-2-one.

Explanation: \(\textbf{Identify the hydrogen deficiencies:}\) All four compounds have molecular formula \(\mathrm{C_4H_6}\). Their degrees of hydrogen deficiency are identical, but the structural origins differ. P and Q each contain one triple bond. R contains two double bonds. S contains one ring and one double bond. \(\textbf{Apply the sodamide test:}\) P is a terminal alkyne: \[ \mathrm{HC\equiv CCH_2CH_3} \] Its terminal triply bonded carbon carries an acidic acetylenic hydrogen. It can therefore form sodium butynide and ammonia through reaction with sodamide. Q is an internal alkyne whose triply bonded carbons carry methyl substituents. It has no acetylenic hydrogen. R and S contain double bonds but no terminal carbon–carbon triple bond. They do not undergo the stated characteristic terminal-alkyne deprotonation. Thus, Experiment I identifies P among these four candidates. \(\textbf{Compare complete hydrogenation:}\) P and Q each consume two hydrogen molecules to convert their triple bonds into single bonds. Both hydrogenate to butane. R contains two carbon–carbon double bonds and also consumes two hydrogen molecules. It too hydrogenates to butane. S consumes one hydrogen molecule because only its double bond is hydrogenated. Its four-membered ring remains intact, so the product is cyclobutane. \(\textbf{Compare complete bromine-addition equivalents:}\) P and Q each contain a triple bond capable of adding two bromine molecules. R contains two double bonds, each capable of adding one bromine molecule. Thus, P, Q and R each consume two bromine equivalents under the specified complete-addition conditions. S consumes one bromine equivalent across its single double bond. The ordinary ring bonds of S are stipulated not to participate in additional bromine reactions. \(\textbf{Identify the unresolved comparison:}\) Q and R give no sodium-alkynide response in Experiment I. Both consume two hydrogen equivalents in Experiment II. Both consume two bromine equivalents in Experiment III. Their complete hydrogenation products are also the same constitutional alkane, butane. Consequently, the stated measurements cannot distinguish Q from R despite their different bonding arrangements. \(\textbf{Interpret the limitation:}\) A positive acetylenic-hydrogen test establishes the presence of a terminal alkyne under suitable conditions. A negative result does not establish the absence of all triple bonds. Likewise, identical addition equivalents do not establish identical multiple-bond arrangements. The correct option recognises both the information provided by the tests and the structural ambiguity they leave unresolved.

Explanation: \(\textbf{Distinguish cyclic from linear polymerisation:}\) Ethyne can undergo different polymerisation pathways depending on the conditions. Cyclic trimerisation produces the aromatic compound benzene. Linear polymerisation produces polyacetylene, whose backbone contains conjugated carbon–carbon double bonds. These products have related elemental compositions but very different molecular structures and reactivities. \(\textbf{Calculate benzene formed in P:}\) The trimerisation reaction is: \[ \mathrm{3C_2H_2\xrightarrow{red\ hot\ iron,\ 873\,K}C_6H_6} \] Three ethyne molecules form one benzene molecule. For the specified sample: \[ n(\mathrm{benzene})=\frac{n(\mathrm{ethyne})}{3} \] \[ n(\mathrm{benzene})=\frac{0.60}{3} \] \[ n(\mathrm{benzene})=0.20\,\mathrm{mol} \] \(\textbf{Assess bromine addition to P:}\) Benzene has a delocalised aromatic \(\pi\)-electron system. Unlike an ordinary alkene, it does not readily undergo the specified bromine-addition reaction under normal carbon-tetrachloride test conditions. Consequently: \[ n(\mathrm{Br_2})_P=0 \] This statement concerns the specified ordinary addition test, not every possible bromination condition for benzene. \(\textbf{Examine the polyacetylene formed in Q:}\) Linear polymerisation retains one carbon–carbon double-bond equivalent per ethyne-derived repeating-unit equivalent. The initial ethyne amount gives: \[ n(\mathrm{C_2H_2\ repeat\ equivalents})=0.60\,\mathrm{mol} \] Under the stipulated idealisation, the same amount of reactive double-bond equivalents remains in the polymer backbone. \(\textbf{Calculate bromine consumption for Q:}\) Each double bond consumes one bromine molecule during complete addition. Therefore: \[ n(\mathrm{Br_2})_Q=0.60(1) \] \[ n(\mathrm{Br_2})_Q=0.60\,\mathrm{mol} \] \(\textbf{Check the elemental-composition argument:}\) Benzene has molecular formula \(\mathrm{C_6H_6}\). Polyacetylene has an ideal empirical composition \(\mathrm{CH}\). Both therefore have a carbon-to-hydrogen ratio of \(1:1\). However, equal empirical compositions do not imply equal addition reactivity. The aromatic structure of benzene and the conjugated alkene backbone of polyacetylene must be considered separately. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{benzene})_P=0.20\,\mathrm{mol};\quad n(\mathrm{Br_2})_P=0;\quad n(\mathrm{Br_2})_Q=0.60\,\mathrm{mol}} \]

Explanation: \(\textbf{Apply the multiple-bond counting rule:}\) A carbon–carbon single bond contains one \(\sigma\) bond. A carbon–carbon double bond contains one \(\sigma\) bond and one \(\pi\) bond. A carbon–carbon triple bond contains one \(\sigma\) bond and two \(\pi\) bonds. An ordinary carbon–hydrogen or carbon–bromine single bond contains one \(\sigma\) bond. \(\textbf{Count bonds in ethyne:}\) Ethyne has the structure: \[ \mathrm{H-C\equiv C-H} \] Its carbon–carbon triple bond contributes one \(\sigma\) and two \(\pi\) bonds. The two C–H bonds contribute two additional \(\sigma\) bonds. Therefore: \[ N_{\sigma,0}=1+2=3 \] \[ N_{\pi,0}=2 \] \(\textbf{Count bonds after Stage I:}\) One bromine equivalent converts the triple bond into a double bond: \[ \mathrm{BrCH=CHBr} \] The C=C bond contributes one \(\sigma\) and one \(\pi\) bond. There are two C–H bonds and two C–Br bonds. Thus: \[ N_{\sigma,I}=1+2+2 \] \[ N_{\sigma,I}=5 \] \[ N_{\pi,I}=1 \] \(\textbf{Count bonds after Stage II:}\) The second bromine molecule adds across the remaining double bond. The final organic structure is: \[ \mathrm{CHBr_2-CHBr_2} \] Its central C–C bond is now a single bond. The molecule contains two C–H bonds and four C–Br bonds. Therefore: \[ N_{\sigma,II}=1+2+4 \] \[ N_{\sigma,II}=7 \] \[ N_{\pi,II}=0 \] \(\textbf{Verify the progression:}\) Each bromine-addition stage consumes one carbon–carbon \(\pi\) bond. It also forms two additional C–Br \(\sigma\) bonds. Therefore, the number of \(\pi\) bonds decreases successively from two to one to zero. The number of \(\sigma\) bonds increases from three to five to seven. \(\textbf{Interpret the bond-order change:}\) The carbon–carbon \(\sigma\) bond persists throughout both additions. The two stages do not require destruction of the entire original carbon–carbon connection. Instead, successive \(\pi\)-bond consumption permits new single bonds to bromine. The resulting bond counts are consistent with the final carbon valencies. \(\textbf{Final answer:}\) \[ \boxed{(N_\sigma,N_\pi)_0=(3,2);\quad(N_\sigma,N_\pi)_I=(5,1);\quad(N_\sigma,N_\pi)_{II}=(7,0)} \]

Explanation: \(\textbf{Identify the competing uses of HBr:}\) Hydrogen bromide can act as a strong acid in proton-transfer reactions. It can also participate in electrophilic addition to an unsaturated hydrocarbon. The actual outcome depends on the chemical species present when HBr is introduced. \(\textbf{First step of Route P — Form an alkynide:}\) Propyne contains one hydrogen atom attached directly to a triply bonded carbon. This acetylenic hydrogen is sufficiently acidic to be removed by sodamide under suitable conditions. \[ \mathrm{CH_3C\equiv CH+NaNH_2\longrightarrow CH_3C\equiv CNa+NH_3} \] The sodium alkynide retains the carbon–carbon triple bond. It is the conjugate base of the terminal alkyne. \(\textbf{Second step of Route P — Protonate the alkynide:}\) The alkynide ion is a strong base. When an equimolar amount of HBr is introduced under the stipulated acid–base conditions, proton transfer occurs: \[ \mathrm{CH_3C\equiv CNa+HBr\longrightarrow CH_3C\equiv CH+NaBr} \] The organic product is propyne again. The carbon–carbon triple bond remains intact. No excess HBr is available for a subsequent addition reaction. \(\textbf{Route Q — Direct addition to propyne:}\) In this route, HBr encounters neutral propyne rather than an alkynide ion. The unsymmetrical triple bond undergoes ordinary electrophilic addition. The applicable Markovnikov orientation places bromine on the more substituted triply bonded carbon: \[ \mathrm{CH_3C\equiv CH+HBr\longrightarrow CH_3C(Br)=CH_2} \] The product is 2-bromoprop-1-ene. One carbon–carbon double bond remains. \(\textbf{Explain the changed reaction outcome:}\) The same reagent, HBr, is used in both routes. However, the first route presents it with a strongly basic carbon anion, making proton transfer the specified reaction. The second route presents it with a neutral unsaturated hydrocarbon, allowing electrophilic addition. The amount of HBr is also important: the first route consumes its entire equivalent during neutralisation. It would be incorrect to treat HBr as automatically adding across a triple bond without examining the species and conditions. Therefore, P regenerates propyne, whereas Q forms 2-bromoprop-1-ene.

Explanation: \(\textbf{Distinguish the two gas-generating reactions:}\) Calcium carbide reacts with water to produce ethyne. Calcium carbonate reacts with dilute acid to produce carbon dioxide. The sequential treatment therefore provides independent measurements of the two original solid components. \(\textbf{Recover the calcium carbide amount:}\) The hydrolysis equation is: \[ \mathrm{CaC_2+2H_2O\longrightarrow Ca(OH)_2+C_2H_2} \] One mole of calcium carbide produces one mole of ethyne. Therefore: \[ n(\mathrm{CaC_2})=n(\mathrm{C_2H_2}) \] \[ n(\mathrm{CaC_2})=0.180\,\mathrm{mol} \] \(\textbf{Calculate the carbide mass:}\) \[ M(\mathrm{CaC_2})=40+2(12) \] \[ M(\mathrm{CaC_2})=64\,\mathrm{g\,mol^{-1}} \] Hence: \[ m(\mathrm{CaC_2})=0.180(64) \] \[ m(\mathrm{CaC_2})=11.52\,\mathrm{g} \] \(\textbf{Recover the carbonate amount:}\) The acid reaction is: \[ \mathrm{CaCO_3+2HCl\longrightarrow CaCl_2+CO_2+H_2O} \] One mole of calcium carbonate produces one mole of carbon dioxide. Thus: \[ n(\mathrm{CaCO_3})=0.120\,\mathrm{mol} \] Its molar mass is: \[ M(\mathrm{CaCO_3})=40+12+3(16) \] \[ M(\mathrm{CaCO_3})=100\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m(\mathrm{CaCO_3})=0.120(100) \] \[ m(\mathrm{CaCO_3})=12.00\,\mathrm{g} \] \(\textbf{Calculate the original solid composition:}\) \[ m_{\mathrm{total}}=11.52+12.00 \] \[ m_{\mathrm{total}}=23.52\,\mathrm{g} \] The mass percentage of calcium carbide is: \[ w(\mathrm{CaC_2})=\frac{11.52}{23.52}\times100 \] \[ w(\mathrm{CaC_2})\approx48.98\% \] \[ w(\mathrm{CaC_2})\approx49.0\% \] \(\textbf{Apply cyclic trimerisation to the collected ethyne:}\) Ethyne undergoes cyclic polymerisation on passage through a red-hot iron tube at approximately \(873\,\mathrm{K}\): \[ \mathrm{3C_2H_2\longrightarrow C_6H_6} \] Three ethyne molecules form one benzene molecule. Therefore: \[ n(\mathrm{benzene})=\frac{0.180}{3} \] \[ n(\mathrm{benzene})=0.060\,\mathrm{mol} \] \(\textbf{Interpret the independent measurements:}\) The ethyne yield identifies the carbide content. The carbon dioxide yield identifies the carbonate content. Their calculated masses determine the original mixture's composition. Only the ethyne collected in the first stage contributes to the subsequent benzene synthesis. The carbonate component does not become an additional source of ethyne through the stipulated reactions. \(\textbf{Final answer:}\) \[ \boxed{w(\mathrm{CaC_2})=49.0\%;\quad n(\mathrm{benzene})=0.060\,\mathrm{mol}} \]

Explanation: Statement I is correct. Propyne has the structure: \[ \mathrm{CH_3C\equiv CH} \] It is an unsymmetrical terminal alkyne. In ordinary peroxide-free electrophilic addition, the preferred orientation places bromine at the more substituted carbon of the original triple bond. One HBr equivalent produces: \[ \mathrm{CH_3C(Br)=CH_2} \] The product is 2-bromoprop-1-ene. Statement II is correct. NCERT describes ordinary addition reactions of alkynes as predominantly electrophilic and relates the products of unsymmetrical hydrogen-halide additions to the stability of the relevant vinylic cation. This provides a mechanistic basis for the ordinary Markovnikov orientation. The intermediate must be considered in the appropriate alkyne reaction rather than simply copied from an alkene mechanism without examination. Statement III is incorrect. The classical peroxide or Kharash effect is established for suitable HBr additions to unsymmetrical alkenes such as propene. It does not follow that adding peroxide must reverse every addition orientation for every hydrogen halide and every unsaturated substrate. Different hydrogen halides have different radical propagation energetics. Alkynes also produce different intermediate bonding arrangements from ordinary alkenes. Therefore, a universal peroxide prediction for unqualified alkyne conditions is unjustified. Statement IV is correct. The first HBr equivalent gives the vinyl bromide 2-bromoprop-1-ene. The second equivalent can add across its remaining carbon–carbon double bond under the specified conditions. The resulting product is: \[ \mathrm{CH_3CBr_2CH_3} \] This is 2,2-dibromopropane. Both bromine atoms are attached to carbon 2, so the compound is a geminal dibromide. The distinction between one and two equivalents is important because the first addition does not completely saturate the original triple bond. \(\textbf{Combine the conclusions:}\) Ordinary alkyne hydrogen-halide addition is governed by the applicable electrophilic mechanism and reagent quantity. The peroxide effect of alkenes must not be transferred automatically to every alkyne or hydrogen halide without appropriate conditions and evidence. Statements I, II and IV are correct, whereas Statement III is an unsupported generalisation.

Explanation: \(\textbf{Separate acidic hydrogen counting from triple-bond counting:}\) A terminal alkyne group contains a hydrogen atom attached directly to an \(sp\)-hybridised carbon. An internal alkyne group has carbon substituents attached to both triply bonded carbon atoms and possesses no such hydrogen. The sodium-metal reaction measures the number of acidic acetylenic hydrogens. Complete catalytic hydrogenation instead measures the total number of carbon–carbon triple bonds to be saturated. \(\textbf{Analyse the terminal positions in P:}\) Hepta-1,6-diyne contains two terminal triple bonds. One occurs between carbons 1 and 2. The other occurs between carbons 6 and 7. Each terminal end carries one acidic acetylenic hydrogen. Therefore, P contains two such hydrogens per molecule. \(\textbf{Calculate P's hydrogen evolution:}\) The general sodium-replacement relationship is: \[ \mathrm{2(C\equiv C-H)+2Na\longrightarrow2(C\equiv C-Na)+H_2} \] Two acidic hydrogen atoms produce one hydrogen molecule. For one mole of P: \[ n(\mathrm{acidic\ H})_P=2.0\,\mathrm{mol} \] \[ n(\mathrm{H_2,evolved})_P=\frac{2.0}{2} \] \[ n(\mathrm{H_2,evolved})_P=1.0\,\mathrm{mol} \] \(\textbf{Analyse the terminal positions in Q:}\) Hepta-1,3-diyne contains a terminal triple bond between carbons 1 and 2. Its second triple bond, between carbons 3 and 4, is internal. Only the carbon at position 1 carries an acetylenic hydrogen. Therefore, Q contains one such hydrogen per molecule. \(\textbf{Calculate Q's hydrogen evolution:}\) For one mole of Q: \[ n(\mathrm{acidic\ H})_Q=1.0\,\mathrm{mol} \] \[ n(\mathrm{H_2,evolved})_Q=\frac{1.0}{2} \] \[ n(\mathrm{H_2,evolved})_Q=0.5\,\mathrm{mol} \] \(\textbf{Evaluate complete hydrogenation:}\) Each alkyne triple bond consumes two hydrogen molecules when converted completely into a carbon–carbon single bond. Both P and Q contain two triple bonds. Therefore, each requires: \[ n(\mathrm{H_2,consumed})=2(2) \] \[ n(\mathrm{H_2,consumed})=4.0\,\mathrm{mol} \] Both compounds have unbranched seven-carbon skeletons. Complete hydrogenation therefore produces heptane from either compound. \(\textbf{Verify molecular formulas:}\) An acyclic saturated seven-carbon hydrocarbon has formula \(\mathrm{C_7H_{16}}\). Two triple bonds contribute four degrees of hydrogen deficiency. The unsaturated formula is consequently: \[ \mathrm{C_7H_{16-8}=C_7H_8} \] Complete addition of four hydrogen molecules restores the saturated formula: \[ \mathrm{C_7H_8+4H_2\longrightarrow C_7H_{16}} \] \(\textbf{Interpret the distinction:}\) The two diynes consume identical amounts of hydrogen during complete hydrogenation because they have the same number of triple bonds. However, they evolve different amounts of hydrogen with sodium because they contain different numbers of terminal acetylenic hydrogens. A triple-bond count alone cannot determine sodium reactivity. Likewise, the identical hydrogenation requirements do not establish equivalent triple-bond positions. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} P:&\ n(\mathrm{H_2,evolved})=1.0\,\mathrm{mol},\quad n(\mathrm{H_2,consumed})=4.0\,\mathrm{mol}\\ Q:&\ n(\mathrm{H_2,evolved})=0.5\,\mathrm{mol},\quad n(\mathrm{H_2,consumed})=4.0\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Examine the starting alkyne:}\) Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] Its triply bonded carbons are \(sp\)-hybridised. The carbon–carbon triple-bond geometry is linear. There is no pair of independent substituent arrangements across the triple bond corresponding to ordinary alkene cis–trans geometrical isomerism. Therefore, ethyne does not exhibit the specified geometrical isomerism. \(\textbf{Examine the product after one bromine equivalent:}\) Addition of one bromine molecule reduces the carbon–carbon bond order from triple to double. Each original triply bonded carbon acquires one bromine substituent. The resulting constitutional structure is: \[ \mathrm{BrCH=CHBr} \] This is 1,2-dibromoethene. \(\textbf{Apply the alkene geometrical-isomerism criterion:}\) The first doubly bonded carbon carries hydrogen and bromine. The second doubly bonded carbon also carries hydrogen and bromine. Both carbons therefore possess two different substituents. Restricted rotation about the carbon–carbon double bond permits two distinguishable geometrical configurations. These correspond to cis and trans arrangements of the bromine substituents. \(\textbf{Distinguish eligibility from product selectivity:}\) The structural possibility of two geometrical isomers does not establish that the actual bromination reaction produces both in equal amounts. A reaction can preferentially form one stereoisomer depending on its mechanism and conditions. The question supplies no stereochemical product-ratio evidence. It therefore supports only the conclusion that two geometrical configurations are possible. \(\textbf{Examine the product after the second bromine equivalent:}\) A second bromine molecule adds across the remaining double bond: \[ \mathrm{BrCH=CHBr+Br_2\longrightarrow CHBr_2-CHBr_2} \] The product is 1,1,2,2-tetrabromoethane. Its carbon–carbon bond is now a single bond. The geometrical-isomerism restriction associated with the former double bond no longer exists. \(\textbf{Interpret the sequence:}\) Initially, the alkyne has linear geometry and does not possess ordinary cis–trans configurations. The first bromine addition creates an alkene whose substituent arrangements make geometrical isomerism possible. The second addition removes that carbon–carbon double bond and hence removes the corresponding geometrical distinction. This illustrates why chemical addition can first create and then eliminate the structural requirement for geometrical isomerism. The correct statement is option C.