Hydrocarbons MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: The three compounds belong to the acyclic monoalkene homologous series. They each contain one carbon–carbon double bond and follow the general formula \(\mathrm{C}_{n}\mathrm{H}_{2n}\). However, their physical properties change substantially as molecular size increases. \(\textbf{Apply the NCERT physical-state trend:}\) NCERT describes the first three members of the alkene series as gases. The following fourteen members are generally liquids, while higher members are solids. P contains four carbon atoms and belongs to the initial gaseous members. Q contains eight carbon atoms and belongs to the ordinarily liquid members. R contains nineteen carbon atoms and lies beyond the fourteen liquid members following the first three. It therefore belongs to the higher solid members under the stated general classification. \(\textbf{Interpret the molecular-size effect:}\) As the hydrocarbon chain becomes longer, the electron cloud and opportunities for intermolecular dispersion interactions increase. This generally raises melting and boiling points. The compounds may possess the same type of functional unsaturation without occupying the same physical state. \(\textbf{Assess water solubility:}\) Alkenes are predominantly non-polar hydrocarbons. Their carbon–carbon double bonds do not make them sufficiently polar to become generally soluble in water. They are more compatible with non-polar organic solvents such as petroleum ether. Thus, all three are generally insoluble in water. \(\textbf{Assess catalytic hydrogenation:}\) Each compound is specified to be a monoalkene. Every molecule therefore contains one carbon–carbon double bond. Complete catalytic hydrogenation consumes one molecule of dihydrogen per alkene molecule: \[ \mathrm{C}_{n}\mathrm{H}_{2n}+\mathrm{H}_{2}\xrightarrow{\mathrm{Ni}}\mathrm{C}_{n}\mathrm{H}_{2n+2} \] The required molar hydrogen equivalent is the same even though the molecular masses differ. The resulting alkanes retain the original carbon skeletons. The comparison shows that physical state depends strongly on molecular size, whereas hydrogen equivalents depend on the number of hydrogenatable double bonds. Therefore, P, Q and R are ordinarily gaseous, liquid and solid, respectively, under the stated NCERT classification, and each consumes one hydrogen equivalent.

Explanation: Statement I is correct because a catalyst provides an alternative reaction pathway having a different activation-energy profile. A suitable catalyst may therefore increase the hydrogenation rate at a specified temperature. Nickel, palladium and platinum are recognised catalysts for addition of dihydrogen to suitable alkene double bonds. Statement II is incorrect because reaction enthalpy is determined by the initial and final thermodynamic states. If reactants, products and specified thermodynamic conditions are the same, changing the catalyst does not change the overall reaction enthalpy. The catalytic pathway may contain different intermediate stages, but their enthalpy changes combine to the same overall value. Statement III is correct because but-2-ene has the structure \(\mathrm{CH_3CH=CHCH_3}\). Complete hydrogenation adds one molecule of hydrogen across the double bond. \[ \mathrm{CH_3CH=CHCH_3+H_2\xrightarrow{Ni/Pd/Pt}CH_3CH_2CH_2CH_3} \] The product is butane. Changing among suitable catalysts does not alter the four-carbon connectivity in this specified complete hydrogenation. Statement IV is incorrect because activation energy and reaction enthalpy describe different energetic quantities. Lowering the activation-energy barrier changes the ease of reaching the transition state, not the energy difference between reactants and products. A catalyst cannot reverse the sign of the same reaction enthalpy simply by providing a faster pathway. The decisive distinction is therefore between kinetic influence and thermodynamic invariance. Only Statements I and III are correct.

Explanation: Dehydrohalogenation of an alkyl halide requires removal of a halogen atom from the carbon bearing the halogen and a hydrogen atom from an adjacent carbon. The halogen-bearing carbon is commonly called the \(\alpha\)-carbon, while an adjacent carbon is a \(\beta\)-carbon. A carbon–carbon double bond is created between these two positions. The availability of a \(\beta\)-hydrogen is therefore essential to the ordinary elimination under consideration. \(\textbf{Analyse P — Absence of an eligible adjacent hydrogen:}\) P has the structure: \[ \mathrm{BrCH_2-C(CH_3)_3} \] The bromine-bearing carbon is a terminal methylene carbon. Its only adjacent carbon is bonded to four other carbon atoms and is quaternary. That adjacent carbon carries no hydrogen atom. Consequently, there is no \(\beta\)-hydrogen available for ordinary dehydrohalogenation. Under the stipulated no-rearrangement condition, P cannot produce an alkene by the specified elimination. \(\textbf{Analyse Q — Two nonequivalent elimination directions:}\) Q has the structure: \[ \mathrm{CH_3-C(Br)(CH_3)-CH_2CH_3} \] The bromine-bearing carbon is tertiary. Two equivalent methyl groups are attached to it, and a methylene carbon belonging to an ethyl group is also adjacent. Hydrogen removal from either equivalent methyl group gives: \[ \mathrm{CH_2=C(CH_3)CH_2CH_3} \] This is 2-methylbut-1-ene. Hydrogen removal from the methylene group instead gives: \[ \mathrm{(CH_3)_2C=CHCH_3} \] This is 2-methylbut-2-ene. These are constitutionally distinct positional alkenes. Neither contributes additional geometrical isomers because one double-bonded carbon in each structure carries two identical substituents. \(\textbf{Compare the substrates:}\) Both P and Q possess bromine atoms and carbon atoms adjacent to the C–Br bond. However, adjacency alone does not establish the availability of a removable \(\beta\)-hydrogen. P lacks that requirement altogether, whereas Q has two nonequivalent classes of eligible adjacent hydrogen. Thus, P gives no direct elimination alkene under the stated conditions, while Q can give two constitutional alkenes.

Explanation: The question provides two separate qualitative influences on alkyl-halide dehydrohalogenation. One concerns the nature of the leaving halogen. The other concerns the degree of substitution at the halogen-bearing carbon. A definite comparison is available when one substrate is more favourable in one factor and not less favourable in the other. When the factors oppose each other, their qualitative orders alone need not determine the overall ranking. \(\textbf{Pair I — Isolate the halogen effect:}\) 2-Iodopropane and 2-bromopropane are both secondary alkyl halides. They possess the same carbon skeleton and the same type of halogen-bearing carbon. Their comparison therefore isolates the supplied leaving-halogen trend. Since iodine ranks ahead of bromine, the first compound is favoured under the stated qualitative conditions. \(\textbf{Pair II — Isolate the substrate-class effect:}\) 2-Bromo-2-methylpropane is a tertiary bromide. 1-Bromobutane is a primary bromide. Both contain the same leaving halogen. The comparison therefore isolates the supplied order tertiary \(\gt \) secondary \(\gt \) primary. The first compound is favoured under the stated trend. \(\textbf{Pair III — Opposing influences:}\) 1-Iodobutane is a primary iodide. 2-Chlorobutane is a secondary chloride. The first compound benefits from the more favourable halogen identity, iodine. The second benefits from being secondary rather than primary. The two trends point in opposite directions. Without quantitative information establishing the relative strength of these influences under the actual conditions, the pair cannot be ranked conclusively from the stated trends alone. \(\textbf{Interpret the inference:}\) Two separate reactivity orders must not be combined as though they supplied exact numerical rates. A comparison involving a single changed factor can be definite, but a comparison involving competing changes may be underdetermined. This does not imply that the two real reaction rates must be equal. It means that the supplied qualitative information is insufficient to establish which is greater. Therefore, the first compound is favoured in Pairs I and II, while Pair III cannot be ranked from the stated trends alone.

Explanation: \(\textbf{Establish the structural relationships:}\) P is an unbranched six-carbon monoalkene with formula \(\mathrm{C_6H_{12}}\). Q is a branched six-carbon monoalkene having the same formula. R is an unbranched seven-carbon monoalkene with formula \(\mathrm{C_7H_{14}}\). P and Q are constitutional chain isomers, whereas R belongs to the next higher homologue of the unbranched alkene series. \(\textbf{Compare P and Q:}\) P and Q have identical molecular masses because they share the formula \(\mathrm{C_6H_{12}}\). Their principal intermolecular attractions are dispersion forces. The unbranched skeleton of P generally permits greater intermolecular contact than the branched skeleton of Q. Therefore, P has the higher expected boiling point under the stated trend. \(\textbf{Compare P and R:}\) R contains one additional methylene group relative to P. The larger unbranched hydrocarbon generally possesses stronger overall intermolecular dispersion interactions. Consequently, R has the higher expected boiling point. Combining the two comparisons gives: \[ T_b(R)\gt T_b(P)\gt T_b(Q) \] \(\textbf{Apply complete catalytic hydrogenation:}\) Each monoalkene consumes one mole of hydrogen per mole of hydrocarbon. Hydrogenation of P converts hex-1-ene into hexane. Hydrogenation of Q gives 2-methylpentane after correct numbering of the saturated carbon skeleton. Hydrogenation of R converts hept-1-ene into heptane. \(\textbf{Identify the relationships between products:}\) Hexane and 2-methylpentane both have molecular formula \(\mathrm{C_6H_{14}}\). They are constitutional chain isomers because their carbon skeletons differ. Heptane has formula \(\mathrm{C_7H_{16}}\). It is not an isomer of either six-carbon product. The boiling-point prediction depends on molecular size and branching, while the hydrogenation-product comparison depends on preservation of carbon connectivity. These are separate structural consequences of the same starting molecules. Thus, the predicted boiling-point order is R greater than P greater than Q, and the products are hexane, 2-methylpentane and heptane.

Explanation: \(\textbf{Establish the common bond-enthalpy method:}\) Both reactions convert the carbon–carbon double bond of ethene into a carbon–carbon single bond. They differ in the incoming reagent and in the new bonds formed with the two original double-bonded carbons. The approximate reaction enthalpy is: \[ \Delta H=\sum D(\mathrm{bonds\ broken})-\sum D(\mathrm{bonds\ formed}) \] The carbon–hydrogen bonds already present in ethene are unchanged and cancel from the net comparison. \(\textbf{Reaction P — Hydrogen addition:}\) One carbon–carbon double bond and one hydrogen–hydrogen bond are replaced. The products contain one carbon–carbon single bond and two new carbon–hydrogen bonds. \[ \Delta H_P=[D(\mathrm{C=C})+D(\mathrm{H-H})]-[D(\mathrm{C-C})+2D(\mathrm{C-H})] \] Substituting: \[ \Delta H_P=(614+436)-[348+2(413)] \] \[ \Delta H_P=1050-1174 \] \[ \Delta H_P=-124\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Reaction Q — Bromine addition:}\) One carbon–carbon double bond and one bromine–bromine bond are replaced. The products contain one carbon–carbon single bond and two carbon–bromine bonds. \[ \Delta H_Q=[D(\mathrm{C=C})+D(\mathrm{Br-Br})]-[D(\mathrm{C-C})+2D(\mathrm{C-Br})] \] Substituting: \[ \Delta H_Q=(614+193)-[348+2(285)] \] \[ \Delta H_Q=807-918 \] \[ \Delta H_Q=-111\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Compare the heat released:}\) Both calculated reaction enthalpies are negative. The heat released per mole is greater for P. \[ |\Delta H_P|-|\Delta H_Q|=124-111 \] \[ |\Delta H_P|-|\Delta H_Q|=13\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Interpret the chemistry:}\) Both reactions consume the same type of carbon–carbon multiple bond, but they break different reagent bonds and form different product bonds. Consequently, equal numbers of bonds changed do not imply equal reaction enthalpies. The supplied bond enthalpies support an energetic comparison, but they do not by themselves establish relative reaction rates. Kinetics also depends on reaction mechanisms and activation-energy barriers. \(\textbf{Final answer:}\) \[ \boxed{\Delta H_P=-124\,\mathrm{kJ\,mol^{-1}};\quad\Delta H_Q=-111\,\mathrm{kJ\,mol^{-1}};\quad\Delta|\Delta H|=13\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Establish the structural distinction:}\) Hexa-1,5-diene contains an open six-carbon chain with two carbon–carbon double bonds. Cyclohexene contains a six-membered ring with one carbon–carbon double bond. Both have molecular formula \(\mathrm{C_6H_{10}}\), but their hydrogen deficiencies arise from different combinations of rings and multiple bonds. \(\textbf{Determine their hydrogen requirements:}\) Hexa-1,5-diene requires two molecules of hydrogen for complete conversion of its two double bonds into single bonds. Its hydrogenation product is hexane: \[ \mathrm{C_6H_{10}+2H_2\xrightarrow{Ni}C_6H_{14}} \] Cyclohexene requires one molecule of hydrogen for its single double bond. Its ring remains intact: \[ \mathrm{C_6H_{10}+H_2\xrightarrow{Ni}C_6H_{12}} \] The product is cyclohexane. \(\textbf{Construct the composition equations:}\) Let \(x\) be the amount of hexa-1,5-diene and \(y\) that of cyclohexene. The total amount of hydrocarbon is: \[ x+y=0.20 \] The hydrogen-consumption relation is: \[ 2x+y=0.30 \] \(\textbf{Solve the simultaneous conditions:}\) Subtracting the first equation from the second: \[ (2x+y)-(x+y)=0.30-0.20 \] \[ x=0.10\,\mathrm{mol} \] Then: \[ y=0.20-0.10 \] \[ y=0.10\,\mathrm{mol} \] \(\textbf{Apply carbon-skeleton preservation:}\) The \(0.10\,\mathrm{mol}\) of diene becomes \(0.10\,\mathrm{mol}\) hexane. The \(0.10\,\mathrm{mol}\) of cyclohexene becomes \(0.10\,\mathrm{mol}\) cyclohexane. These saturated products have different molecular formulas, despite the identical formulas of the starting hydrocarbons. The additional hydrogenation equivalent required by the diene corresponds to its second double bond. The persistent hydrogen deficiency of cyclohexane after hydrogenation corresponds to its intact ring. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{hexa\text{-}1,5\text{-}diene})&=0.10\,\mathrm{mol}\ \longrightarrow\mathrm{hexane}\\ n(\mathrm{cyclohexene})&=0.10\,\mathrm{mol}\ \longrightarrow\mathrm{cyclohexane} \end{aligned}} \]

Explanation: \(\textbf{Identify the first-stage reaction:}\) Cold, dilute, alkaline potassium permanganate oxidises suitable alkenes to vicinal glycols. Ethene therefore produces ethane-1,2-diol. The reaction consumes the original carbon–carbon double bond without breaking the carbon skeleton. \(\textbf{Establish the permanganate requirement:}\) The supplied balanced equation shows that three moles of ethene react with two moles of potassium permanganate. Therefore: \[ \frac{n(\mathrm{ethene})}{n(\mathrm{KMnO_4})}=\frac32 \] The initial ethene amount is: \[ n_0(\mathrm{ethene})=0.30\,\mathrm{mol} \] The amount of permanganate required to oxidise all the ethene would be: \[ n(\mathrm{KMnO_4})_{\mathrm{required}}=\frac23(0.30) \] \[ n(\mathrm{KMnO_4})_{\mathrm{required}}=0.20\,\mathrm{mol} \] Only \(0.16\,\mathrm{mol}\) potassium permanganate is available. Thus, potassium permanganate is the limiting reagent. \(\textbf{Calculate glycol formation:}\) The amount of ethene oxidised is: \[ n(\mathrm{ethene})_{\mathrm{reacted}}=\frac32(0.16) \] \[ n(\mathrm{ethene})_{\mathrm{reacted}}=0.24\,\mathrm{mol} \] One mole of ethene produces one mole of ethane-1,2-diol. Therefore: \[ n(\mathrm{diol})=0.24\,\mathrm{mol} \] \(\textbf{Calculate the unreacted ethene:}\) \[ n(\mathrm{ethene})_{\mathrm{remaining}}=0.30-0.24 \] \[ n(\mathrm{ethene})_{\mathrm{remaining}}=0.06\,\mathrm{mol} \] \(\textbf{Apply the second-stage bromine reaction:}\) The separated ethene undergoes bromine addition: \[ \mathrm{CH_2=CH_2+Br_2\longrightarrow BrCH_2CH_2Br} \] Each mole of ethene consumes one mole of bromine. Hence: \[ n(\mathrm{Br_2})=0.06\times1 \] \[ n(\mathrm{Br_2})=0.06\,\mathrm{mol} \] \(\textbf{Interpret the two stages:}\) The first reagent oxidises the alkene double bond and introduces hydroxyl groups. The second reagent adds bromine across the double bonds remaining in the unreacted ethene. The glycol formed in the first stage no longer contains the original alkene double bond. It is therefore not included in the stated second-stage ordinary bromine-addition calculation. The important distinction is between the initial amount of alkene and the amount that remains after a limiting oxidation reaction. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{ethane\text{-}1,2\text{-}diol})=0.24\,\mathrm{mol};\quad n(\mathrm{Br_2})=0.06\,\mathrm{mol}} \]

Explanation: The given alkene is symmetrical about its carbon–carbon double bond. Each doubly bonded carbon carries two methyl substituents and no hydrogen atom. Therefore, neither carbon possesses fewer attached hydrogen atoms than the other. The elementary hydrogen-count wording of Markovnikov's rule cannot distinguish the two directions of addition for this particular substrate. \(\textbf{Ordinary electrophilic addition:}\) In peroxide-free HBr addition, protonation produces a carbocation. Protonation at either of the two equivalent double-bonded carbons leads to the same constitutional type of intermediate. The positively charged carbon is tertiary because it is attached to three other carbon atoms. Bromide attack produces a saturated bromoalkane. The product is 2-bromo-2,3-dimethylbutane. \(\textbf{Peroxide-assisted addition:}\) Under suitable peroxide conditions, bromine-radical addition can initiate the radical-chain pathway. A bromine radical may attach to either original double-bonded carbon. Because the two positions are equivalent in the starting molecule, neither orientation introduces a different carbon-skeleton connectivity. The carbon radical subsequently abstracts hydrogen from HBr. The resulting constitutional product is again 2-bromo-2,3-dimethylbutane. \(\textbf{Distinguish mechanism from product identity:}\) The peroxide-free route involves an ordinary electrophilic-addition pathway. The peroxide-assisted route involves radical addition and hydrogen abstraction. These mechanisms differ in their reactive intermediates and in the order of attachment of hydrogen and bromine. Nevertheless, the constitutional identity of the final product need not differ. A change in mechanism produces a different constitutional regioisomer only when the starting double-bond substitution pattern permits distinguishable addition orientations. The student's claim incorrectly assumes that the peroxide effect must always change the constitutional product. For this symmetrical tetrasubstituted alkene, both specified routes lead to the same constitutional bromoalkane.

Explanation: \(\textbf{Evaluate the combustion evidence:}\) Hex-1-ene is an acyclic monoalkene. Cyclohexane is a saturated monocyclic hydrocarbon. Despite their structural differences, both have molecular formula \(\mathrm{C_6H_{12}}\). Their complete-combustion equations therefore have identical stoichiometric coefficients: \[ \mathrm{C_6H_{12}+9O_2\longrightarrow6CO_2+6H_2O} \] For \(0.50\,\mathrm{mol}\) of either hydrocarbon or any mixture of the two: \[ n(\mathrm{O_2})=0.50(9) \] \[ n(\mathrm{O_2})=4.50\,\mathrm{mol} \] The combustion observation is consistent with the mixture but does not determine its composition. \(\textbf{Use selective bromine addition:}\) Hex-1-ene contains one carbon–carbon double bond. It consumes one molecule of bromine per molecule of hydrocarbon through ordinary addition. Cyclohexane contains no ordinary carbon–carbon multiple bond and is specified not to undergo radical substitution. Therefore, only hex-1-ene accounts for the observed bromine consumption. \[ n(\mathrm{hex\text{-}1\text{-}ene})=n(\mathrm{Br_2})=0.18\,\mathrm{mol} \] \(\textbf{Determine the cyclohexane amount:}\) The total hydrocarbon amount is \(0.50\,\mathrm{mol}\). Hence: \[ n(\mathrm{cyclohexane})=0.50-0.18 \] \[ n(\mathrm{cyclohexane})=0.32\,\mathrm{mol} \] \(\textbf{Calculate the initial hydrocarbon mass:}\) Both compounds have molar mass: \[ M(\mathrm{C_6H_{12}})=6(12)+12(1) \] \[ M(\mathrm{C_6H_{12}})=84\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m_{\mathrm{initial}}=0.50(84) \] \[ m_{\mathrm{initial}}=42.0\,\mathrm{g} \] \(\textbf{Account for bromine incorporated into organic products:}\) Hex-1-ene forms 1,2-dibromohexane through quantitative addition. Both atoms of every consumed bromine molecule enter its organic addition product. The molar mass of bromine is: \[ M(\mathrm{Br_2})=2(80)=160\,\mathrm{g\,mol^{-1}} \] The mass added to the organic mixture is: \[ m_{\mathrm{Br_2,incorporated}}=0.18(160) \] \[ m_{\mathrm{Br_2,incorporated}}=28.8\,\mathrm{g} \] \(\textbf{Determine the final organic mass:}\) \[ m_{\mathrm{final}}=m_{\mathrm{initial}}+m_{\mathrm{Br_2,incorporated}} \] \[ m_{\mathrm{final}}=42.0+28.8 \] \[ m_{\mathrm{final}}=70.8\,\mathrm{g} \] \(\textbf{Interpret the independent observations:}\) Combustion cannot distinguish the two hydrocarbons because their elemental compositions are identical. Ordinary bromine addition does distinguish them because only hex-1-ene possesses a carbon–carbon double bond. The second observation therefore determines the mixture composition, while mass conservation determines the mass after addition. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{cyclohexane})=0.32\,\mathrm{mol};\quad m_{\mathrm{final}}=70.8\,\mathrm{g}} \]

Explanation: Statement I is correct because polyethene is obtained by addition polymerisation of ethene. NCERT identifies polythene bags and sheets among common applications of this polymer. Polypropene is obtained from propene and is used for products including milk crates, plastic buckets and other moulded articles. Statement II is correct because the two monomers possess different substituent arrangements around their double bonds. Ethene has the structure: \[ \mathrm{CH_2=CH_2} \] Its ideal polymer repeating unit is: \[ \mathrm{[-CH_2-CH_2-]} \] There is no methyl side group attached to either backbone carbon. Propene has the structure: \[ \mathrm{CH_2=CH-CH_3} \] Its ideal polymer repeating unit is: \[ \mathrm{[-CH_2-CH(CH_3)-]} \] The methyl group remains attached to one of the backbone carbons. Statement III is incorrect because ordinary addition polymerisation consumes the original monomer carbon–carbon \(\pi\) bonds. The resulting ideal polymer backbone contains carbon–carbon single bonds. Retaining a double bond from every monomer would not describe the stated saturated addition-polymer structure. Statement IV is correct because the ideal polyethene repeating-unit composition is \(\mathrm{C_2H_4}\). The polypropene repeating-unit composition is \(\mathrm{C_3H_6}\). Both have a carbon-to-hydrogen atomic ratio of \(1:2\). Their common empirical ratio does not imply identical repeating-unit connectivity. The differences in their backbone substituents help distinguish the polymers structurally, while their uses provide an application-level connection to the NCERT examples. Thus, Statements I, II and IV are correct.

Explanation: Acid-catalysed dehydration of an alcohol removes the elements of water and can create a carbon–carbon double bond between the hydroxyl-bearing carbon and an adjacent carbon. The possible alkene products depend on the number of nonequivalent adjacent carbon positions possessing suitable hydrogen atoms. The resulting double bonds must then be checked independently for geometrical-isomerism eligibility. \(\textbf{Examine the symmetry of P:}\) 3-Ethylpentan-3-ol contains a central hydroxyl-bearing carbon attached to three identical ethyl groups. All three adjacent methylene positions are constitutionally equivalent. Elimination involving any one of them therefore produces the same alkene connectivity. The resulting structure is: \[ \mathrm{CH_3CH=C(CH_2CH_3)CH_2CH_3} \] The preferred name is 3-ethylpent-2-ene. \(\textbf{Assess geometrical isomerism in P's product:}\) One of its double-bonded carbons carries a methyl group and hydrogen. The other double-bonded carbon carries two identical ethyl groups. Because the latter carbon does not possess two different substituents, geometrical isomerism is impossible. Thus, P yields one distinguishable alkene. \(\textbf{Identify the elimination directions in Q:}\) 3-Methylpentan-3-ol contains two equivalent ethyl groups and one methyl group attached to the hydroxyl-bearing carbon. Elimination involving either of the equivalent ethyl groups produces: \[ \mathrm{CH_3CH=C(CH_3)CH_2CH_3} \] This is 3-methylpent-2-ene. Elimination involving the methyl group instead produces: \[ \mathrm{CH_2=C(CH_2CH_3)_2} \] This is 2-ethylbut-1-ene. These are two constitutionally distinct alkenes. \(\textbf{Assess geometrical isomerism in Q's products:}\) In 3-methylpent-2-ene, each double-bonded carbon has two different substituents. It therefore possesses two geometrical configurations. In 2-ethylbut-1-ene, the terminal double-bonded carbon has two hydrogen atoms. That alkene cannot exhibit geometrical isomerism. \(\textbf{Count distinguishable products:}\) For P: \[ N_P=1 \] For Q, the internal alkene contributes two geometrical isomers and the terminal alkene contributes one: \[ N_Q=2+1 \] \[ N_Q=3 \] \(\textbf{Interpret the structural difference:}\) P possesses three equivalent potential elimination directions, so they converge on one constitutional product. Q possesses two nonequivalent elimination directions, one of which produces an alkene capable of geometrical isomerism. The number of eligible adjacent hydrogens alone would not yield the correct count. Both constitutional equivalence and double-bond substituent identity must be examined. \(\textbf{Final answer:}\) \[ \boxed{N_P=1;\quad N_Q=3} \]

Explanation: Statement I is correct because trans-but-2-ene possesses a symmetrical arrangement of substituents around its carbon–carbon double bond. The two methyl groups lie on opposite sides of the double bond. The relevant bond-dipole contributions cancel because of the molecular symmetry. NCERT consequently describes trans-but-2-ene as approximately non-polar, while the cis isomer possesses a small non-zero dipole moment. Statement II is incorrect because geometrical arrangement and molecular symmetry are related but not identical concepts. The term trans describes the relative positions of specified groups across a restricted double bond. It does not guarantee that all polar-bond contributions throughout an unsymmetrically substituted molecule must cancel. Complete dipole cancellation requires the appropriate arrangement and magnitudes of the contributing bond dipoles. Statement III is correct because trans-pent-2-ene has an unsymmetrical carbon skeleton around the double bond. Its substituent pattern is not identical to that of trans-but-2-ene. One end of the double bond is associated with a methyl substituent, while the other is associated with an ethyl substituent. Therefore, the exact dipole cancellation found in symmetrical trans-but-2-ene cannot automatically be generalised to trans-pent-2-ene. Statement IV is correct because both geometrical isomers of pent-2-ene have the same carbon connectivity. Their difference concerns the spatial arrangement of substituents around the double bond. Complete catalytic hydrogenation consumes the original double bond and produces pentane: \[ \mathrm{CH_3CH=CHCH_2CH_3+H_2\xrightarrow{Ni}CH_3CH_2CH_2CH_2CH_3} \] The saturated product no longer possesses the restricted carbon–carbon double-bond geometry responsible for cis–trans isomerism. Therefore, both geometrical isomers yield the same constitutional alkane. The student's generalisation fails because a conclusion depending on the symmetry of one particular molecule cannot be applied automatically to every compound bearing a trans double-bond arrangement. Only Statements I, III and IV are correct.

Explanation: The carbon–carbon double bond connects carbons 3 and 4 of the heptane parent chain. Carbon 3 is attached to an ethyl group and a methyl group. Carbon 4 is attached to a propyl group and a methyl group. Neither doubly bonded carbon carries a hydrogen atom. \(\textbf{Test the elementary hydrogen-count rule:}\) The familiar form of Markovnikov's rule compares the numbers of hydrogen atoms attached to the two original double-bonded carbons. Here, both carbon atoms carry zero hydrogens. Consequently, that comparison cannot select one addition direction over the other. \(\textbf{Examine the possible carbocations:}\) If the proton attaches to carbon 3, the positive charge develops at carbon 4. After protonation, carbon 4 is bonded to three carbon groups, so the intermediate is tertiary. If the proton attaches to carbon 4, the positive charge develops at carbon 3. Carbon 3 is likewise bonded to three carbon groups, producing another tertiary carbocation. Thus, the usual tertiary-greater-than-secondary-greater-than-primary stability order does not distinguish these two intermediates. \(\textbf{Check whether the products are equivalent:}\) The two original double-bonded carbon atoms carry different alkyl substituents. They are not constitutionally equivalent positions. Bromide attack at carbon 3 and bromide attack at carbon 4 therefore lead to different constitutional regioisomers. Both orientations cannot be declared identical merely because both intermediates are tertiary. \(\textbf{Determine the defensible inference:}\) The two pathways might differ in finer electronic, steric or kinetic effects. However, no data quantifying those differences have been supplied. The stated elementary criteria therefore do not establish a unique major regioisomer. This illustrates the distinction between a useful introductory regiochemical rule and a complete mechanistic prediction for a more highly substituted unsymmetrical alkene. The correct conclusion is that additional evidence is needed rather than assigning a major product from an uninformative hydrogen count.

Explanation: \(\textbf{Interpret the polymer repeating units:}\) Polyethene has the ideal constitutional repeating unit: \[ \mathrm{[-CH_2-CH_2-]} \] Both its carbon atoms belong to the backbone. Polypropene has the repeating unit: \[ \mathrm{[-CH_2-CH(CH_3)-]} \] It contains two backbone carbon atoms and one methyl side-group carbon. Therefore, only the polypropene component contributes to the measured side-group carbon amount. \(\textbf{Recover the polypropene repeating-unit amount:}\) Each polypropene repeating unit contains one methyl side-group carbon. Thus: \[ n(\mathrm{propylene\ repeat\ units})=2.00\,\mathrm{mol} \] The molar mass of the repeating unit is: \[ M(\mathrm{C_3H_6})=3(12)+6(1) \] \[ M(\mathrm{C_3H_6})=42\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m(\mathrm{polypropene})=2.00(42) \] \[ m(\mathrm{polypropene})=84.0\,\mathrm{g} \] \(\textbf{Determine the polyethene component:}\) The combined polymer mass is \(98.0\,\mathrm{g}\). Hence: \[ m(\mathrm{polyethene})=98.0-84.0 \] \[ m(\mathrm{polyethene})=14.0\,\mathrm{g} \] The polyethene repeating-unit molar mass is: \[ M(\mathrm{C_2H_4})=2(12)+4(1) \] \[ M(\mathrm{C_2H_4})=28\,\mathrm{g\,mol^{-1}} \] The amount of its repeating units is: \[ n(\mathrm{ethylene\ repeat\ units})=\frac{14.0}{28} \] \[ n(\mathrm{ethylene\ repeat\ units})=0.50\,\mathrm{mol} \] \(\textbf{Count backbone carbon atoms:}\) Each repeating unit of either polymer contributes two carbon atoms to its backbone. Polypropene therefore contributes: \[ n(C)_{\mathrm{PP,backbone}}=2(2.00)=4.00\,\mathrm{mol} \] Polyethene contributes: \[ n(C)_{\mathrm{PE,backbone}}=2(0.50)=1.00\,\mathrm{mol} \] The combined backbone carbon amount is: \[ n(C)_{\mathrm{backbone,total}}=4.00+1.00 \] \[ n(C)_{\mathrm{backbone,total}}=5.00\,\mathrm{mol} \] \(\textbf{Verify the structural carbon balance:}\) The mixture contains \(2.00\,\mathrm{mol}\) side-group carbon atoms in addition to the \(5.00\,\mathrm{mol}\) backbone carbon atoms. Thus: \[ n(C)_{\mathrm{total}}=5.00+2.00=7.00\,\mathrm{mol} \] Both polymers have the ideal empirical composition \(\mathrm{CH_2}\). The corresponding total mass is: \[ 7.00(14)=98.0\,\mathrm{g} \] The result agrees with the specified mixture mass. This problem requires separating polymer composition from the location of carbon atoms within the repeating units. \(\textbf{Final answer:}\) \[ \boxed{m(\mathrm{polypropene})=84.0\,\mathrm{g};\quad n(C)_{\mathrm{backbone}}=5.00\,\mathrm{mol}} \]

Explanation: The original double bond joins two carbon atoms having different substituent environments. One double-bonded carbon is attached to two ethyl groups. The other is attached to one methyl group and one ethyl group. \(\textbf{Check geometrical isomerism before predicting products:}\) Geometrical isomerism requires two different substituents on each doubly bonded carbon. The carbon attached to two ethyl groups does not satisfy this requirement. Consequently, X cannot exhibit geometrical isomerism, despite possessing an internal, highly substituted double bond. \(\textbf{Reconstruct the ozonolysis fragments:}\) Reductive ozonolysis converts each original double-bonded carbon into a carbonyl carbon. The carbon attached to two ethyl groups becomes: \[ \mathrm{CH_3CH_2COCH_2CH_3} \] This is pentan-3-one. The other double-bonded carbon remains attached to methyl and ethyl groups. It becomes: \[ \mathrm{CH_3COCH_2CH_3} \] This is butan-2-one. Neither product is an aldehyde because neither original double-bonded carbon carries hydrogen. \(\textbf{Identify the saturated carbon skeleton:}\) Complete catalytic hydrogenation adds one hydrogen atom to each double-bonded carbon. The product has the structure: \[ \mathrm{CH_3CH_2CH(CH_2CH_3)CH(CH_3)CH_2CH_3} \] The longest continuous chain contains six carbon atoms. There are ethyl and methyl branches at positions 3 and 4. The equivalent locant sets are resolved by giving the alphabetically earlier ethyl substituent the lower number. The hydrogenation product is therefore 3-ethyl-4-methylhexane. \(\textbf{Verify the carbon accounting:}\) Pentan-3-one contains five carbon atoms and butan-2-one contains four. Their combined nine carbon atoms equal the carbon count of X. Hydrogenation also preserves all nine carbon atoms. The product evidence therefore agrees across both transformations. The key misconception is that an internal double bond does not necessarily imply geometrical isomerism. Both substituents on each double-bonded carbon must be examined individually.

Explanation: The double bond joins a carbon carrying two methyl groups to a carbon carrying hydrogen and a propan-2-yl group. The oxidation products depend on these original substituents and on the oxidising conditions. \(\textbf{Identify the ketone-forming carbon:}\) The carbon bearing two methyl groups has no attached hydrogen. Cleavage of its double bond produces a carbonyl group retaining both methyl substituents. The corresponding compound is propanone: \[ \mathrm{CH_3COCH_3} \] \(\textbf{Identify the aldehyde-forming carbon:}\) The other original double-bonded carbon possesses one hydrogen atom. Its remaining carbon substituent is \(\mathrm{CH(CH_3)_2}\). Reductive ozonolysis therefore produces: \[ \mathrm{(CH_3)_2CHCHO} \] This is 2-methylpropanal. \(\textbf{Process P — Preserve the aldehyde:}\) Zinc–water work-up gives the carbonyl products without the further oxidation stipulated for Process Q. The product pair is propanone and 2-methylpropanal. \(\textbf{Process Q — Continue oxidation at the aldehyde-forming end:}\) Under the specified stronger oxidative conditions, the carbonyl fragment derived from the hydrogen-bearing alkene carbon becomes a carboxylic acid. The corresponding acid is: \[ \mathrm{(CH_3)_2CHCOOH} \] This is 2-methylpropanoic acid. The fragment derived from the original double-bonded carbon without hydrogen gives propanone under the stated selective cleavage assumptions. \(\textbf{Verify carbon conservation:}\) The ketone fragment has three carbon atoms. The aldehyde or acid fragment has four carbon atoms. Thus: \[ N_C=3+4=7 \] The total equals the seven carbon atoms of the original alkene. \(\textbf{Interpret the comparison:}\) The difference between the product sets does not arise from a different alkene skeleton. It arises from the oxidation level attained by the fragment originally carrying hydrogen on the double-bonded carbon. Reductive ozonolysis preserves the aldehyde, whereas the stipulated strong oxidation converts it into the corresponding acid. The ketone-producing fragment retains its ketone connectivity in this comparison. Thus, Process P gives propanone and 2-methylpropanal, while Process Q gives propanone and 2-methylpropanoic acid.

Explanation: \(\textbf{Reconstruct the terminal end from methanal:}\) Methanal is produced when a terminal \(\mathrm{CH_2=}\) carbon undergoes reductive ozonolysis. Therefore, one original double-bonded carbon must have carried two hydrogen atoms. \(\textbf{Reconstruct the substituted end:}\) 3-Methylbutan-2-one has the structure: \[ \mathrm{CH_3COCH(CH_3)_2} \] Its carbonyl carbon is bonded to a methyl group and a propan-2-yl group. Therefore, the other original double-bonded carbon carried those two substituents. Rejoining the two carbonyl-derived carbons through a double bond gives: \[ \mathrm{CH_2=C(CH_3)CH(CH_3)CH_3} \] The compound is 2,3-dimethylbut-1-ene. \(\textbf{Verify the hydrogenation product:}\) Complete hydrogenation converts the terminal double bond into a single bond: \[ \mathrm{CH_2=C(CH_3)CH(CH_3)CH_3+H_2} \] \[ \mathrm{\xrightarrow{Ni}CH_3CH(CH_3)CH(CH_3)CH_3} \] The product is 2,3-dimethylbutane, matching the independent observation. \(\textbf{Predict ordinary HBr addition:}\) The double bond is unsymmetrical. Protonation at the terminal methylene carbon produces a tertiary carbocation at carbon 2. The alternative protonation direction would produce a primary carbocation. Therefore, the tertiary-carbocation pathway is favoured. Bromide attacks carbon 2, giving: \[ \mathrm{CH_3C(Br)(CH_3)CH(CH_3)CH_3} \] The product is 2-bromo-2,3-dimethylbutane. \(\textbf{Exclude the competing structures:}\) 3,3-Dimethylbut-1-ene would give an aldehyde rather than the specified substituted ketone upon reductive ozonolysis. 2,3-Dimethylbut-2-ene lacks a terminal methylene group and gives two molecules of propanone. 2-Methylpent-1-ene has a different carbon skeleton and would not hydrogenate to 2,3-dimethylbutane. Only 2,3-dimethylbut-1-ene satisfies the ozonolysis, hydrogenation and HBr-addition evidence simultaneously. The independent transformations fix both the original double-bond position and the subsequent addition orientation.

Explanation: \(\textbf{Identify the two radical propagation transformations:}\) The first step adds a halogen radical across the alkene double bond and produces a carbon-centred radical. The second transfers hydrogen from HX to that carbon radical and regenerates the halogen radical. Both steps must be examined separately because their enthalpy changes need not have the same sign. \(\textbf{Halogen-radical addition:}\) The approximate enthalpy relation is: \[ \Delta H_1=E_{\pi\text{-bond replacement}}-D(\mathrm{C-X}) \] For chlorine: \[ \Delta H_{1,\mathrm{Cl}}=266-330 \] \[ \Delta H_{1,\mathrm{Cl}}=-64\,\mathrm{kJ\,mol^{-1}} \] For bromine: \[ \Delta H_{1,\mathrm{Br}}=266-285 \] \[ \Delta H_{1,\mathrm{Br}}=-19\,\mathrm{kJ\,mol^{-1}} \] Both radical-addition steps are exothermic under the supplied approximation. \(\textbf{Hydrogen abstraction:}\) The second step breaks an H–X bond and forms a carbon–hydrogen bond. Its approximate enthalpy change is: \[ \Delta H_2=D(\mathrm{H-X})-D(\mathrm{C-H}) \] For HCl: \[ \Delta H_{2,\mathrm{HCl}}=431-410 \] \[ \Delta H_{2,\mathrm{HCl}}=+21\,\mathrm{kJ\,mol^{-1}} \] For HBr: \[ \Delta H_{2,\mathrm{HBr}}=366-410 \] \[ \Delta H_{2,\mathrm{HBr}}=-44\,\mathrm{kJ\,mol^{-1}} \] Hydrogen abstraction is therefore endothermic for HCl but exothermic for HBr. \(\textbf{Compare the combined hypothetical propagation enthalpies:}\) For HCl: \[ \Delta H_{\mathrm{combined,HCl}}=-64+21 \] \[ \Delta H_{\mathrm{combined,HCl}}=-43\,\mathrm{kJ\,mol^{-1}} \] For HBr: \[ \Delta H_{\mathrm{combined,HBr}}=-19-44 \] \[ \Delta H_{\mathrm{combined,HBr}}=-63\,\mathrm{kJ\,mol^{-1}} \] Both hypothetical overall transformations are exothermic. However, only the HBr sequence has exothermic enthalpy changes for both specified propagation steps. \(\textbf{Interpret the peroxide-effect implication:}\) An exothermic overall reaction is not sufficient to establish a readily sustained radical-chain mechanism. An unfavourable elementary propagation step may prevent efficient continuation of the chain. The endothermic hydrogen-abstraction step calculated for HCl therefore provides an energetic reason why the classical peroxide effect is not ordinarily observed for HCl. This does not imply that HCl is incapable of ordinary electrophilic addition to alkenes. The conclusion concerns the feasibility of the proposed radical propagation sequence. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} \mathrm{HBr:}\ &\Delta H_1=-19,\ \Delta H_2=-44\\ \mathrm{HCl:}\ &\Delta H_1=-64,\ \Delta H_2=+21 \end{aligned}\quad\mathrm{kJ\,mol^{-1}}} \]

Explanation: P and Q have the same molecular formula, \(\mathrm{C_5H_{10}}\), but different double-bond positions within the branched carbon skeleton. The three reactions probe different structural features of the starting alkenes. \(\textbf{Compare complete hydrogenation:}\) Hydrogenation converts each carbon–carbon double bond into a single bond without changing the carbon connectivity. For P: \[ \mathrm{CH_2=C(CH_3)CH_2CH_3+H_2\xrightarrow{Ni}CH_3CH(CH_3)CH_2CH_3} \] For Q: \[ \mathrm{CH_2=CHCH(CH_3)_2+H_2\xrightarrow{Ni}CH_3CH_2CH(CH_3)_2} \] The two saturated structures represent the same constitutional hydrocarbon, 2-methylbutane, after correct numbering. \(\textbf{Compare reductive ozonolysis:}\) Both compounds contain terminal \(\mathrm{CH_2=}\) groups. Each therefore produces methanal as one cleavage product. In P, the other double-bonded carbon is attached to methyl and ethyl groups. It consequently forms butan-2-one: \[ \mathrm{P\xrightarrow[Zn/H_2O]{O_3}HCHO+CH_3COCH_2CH_3} \] In Q, the other double-bonded carbon carries hydrogen and a propan-2-yl group. It forms 2-methylpropanal: \[ \mathrm{Q\xrightarrow[Zn/H_2O]{O_3}HCHO+(CH_3)_2CHCHO} \] Thus, the carbonyl-product sets are different. \(\textbf{Compare addition-polymer repeating units:}\) Polymerisation preserves the substituent connectivity of each monomer's original double-bonded carbons. The ideal repeating unit from P is: \[ \mathrm{[-CH_2-C(CH_3)(CH_2CH_3)-]_n} \] The ideal repeating unit from Q is: \[ \mathrm{[-CH_2-CH(CH(CH_3)_2)-]_n} \] These repeating units have identical elemental compositions but different side-group arrangements. \(\textbf{Interpret the structural convergence:}\) Complete hydrogenation removes the positional distinction that separates the two alkenes as unsaturated structures. However, reductive ozonolysis records their original double-bond substituents in different carbonyl products. Addition polymerisation also preserves those different substituent connectivities within its saturated backbone structure. Therefore, identical hydrogenation products do not imply identical ozonolysis products or polymer repeating units.