Hydrocarbons MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: The three compounds possess identical molecular formulas and the same three methyl substituents. Their differences arise entirely from the positions of those substituents on the aromatic ring. The two experiments test different types of hydrogen-bearing positions. Experiment I tests ring positions, whereas Experiment II tests the environments of the methyl side chains. \(\textbf{Analyse the symmetry of P:}\) In 1,2,3-trimethylbenzene, positions 4, 5 and 6 bear aromatic hydrogens. A reflection through the axis passing through ring positions 2 and 5 interchanges positions 4 and 6. Position 5 remains a distinct environment. Consequently, P has two constitutional ring-monobromination products. The same reflection interchanges the methyl groups at positions 1 and 3. The methyl group at position 2 is different. Thus, P also has two constitutional benzylic-monochlorination products. \(\textbf{Analyse the symmetry of Q:}\) In 1,2,4-trimethylbenzene, positions 3, 5 and 6 bear aromatic hydrogens. The unequal spacing of the three methyl substituents removes the symmetry that would identify any two of these vacant positions. Each can therefore give a different ring-monobrominated constitution. Similarly, the methyl groups attached at positions 1, 2 and 4 occupy three constitutionally different environments. Monochlorination at each methyl group gives a different side-chain-chlorinated compound. Q therefore gives three constitutional products in each experiment. \(\textbf{Analyse the symmetry of R:}\) In 1,3,5-trimethylbenzene, the three methyl substituents are evenly distributed around the ring. The vacant aromatic positions 2, 4 and 6 are symmetry-equivalent. Ring monobromination therefore gives only one constitutional product. The three methyl groups are also symmetry-equivalent. Replacing one benzylic hydrogen on any methyl group gives the same constitutional side-chain chloride. R gives one product in each experiment. \(\textbf{Distinguish the reaction pathways:}\) Ring bromination proceeds through electrophilic aromatic substitution under suitable conditions. Benzylic chlorination proceeds through a radical substitution pathway under the stipulated photochemical conditions. The product counts happen to be equal within each substrate, but the reactions do not occur at the same atoms. The equality follows from the particular molecular symmetries of these three trimethylbenzene isomers. It is not a universal rule for every substituted aromatic hydrocarbon. \(\textbf{Conclude the independent counts:}\) P has two distinct ring positions and two distinct methyl environments. Q has three of each. R has one of each. The required ordered pairs are therefore \((2,2)\), \((3,3)\) and \((1,1)\), respectively.

Explanation: \(\textbf{Identify the ring-position multiplicities:}\) The chlorine-bearing carbon is designated position 1. Positions 2 and 6 are ortho to chlorine and are equivalent. Positions 3 and 5 are meta and equivalent. Position 4 is the only para position. Thus, an ortho product can form from two available hydrogens, whereas the para product arises from only one. \(\textbf{Translate product fractions into per-site tendencies:}\) Let \(k_o\), \(k_m\) and \(k_p\) denote the intrinsic substitution tendencies at an individual ortho, meta and para position, respectively. The total constitutional-product weights are: \[ W_o=2k_o \] \[ W_m=2k_m \] \[ W_p=k_p \] Under the stated proportionality assumption: \[ 2k_o:2k_m:k_p=30:2:68 \] \(\textbf{Extract the individual ortho tendency:}\) The combined ortho percentage corresponds to two equivalent positions. Its per-position relative value is: \[ k_o\propto\frac{30}{2} \] \[ k_o\propto15 \] \(\textbf{Extract the individual para tendency:}\) There is one para position. Therefore: \[ k_p\propto\frac{68}{1} \] \[ k_p\propto68 \] \(\textbf{Calculate the requested ratio:}\) \[ \frac{k_p}{k_o}=\frac{68}{15} \] \[ k_p:k_o=68:15 \] \(\textbf{Interpret the meta result:}\) The total meta fraction is small despite the availability of two meta positions. The observation is consistent with chlorine's ordinary ortho/para-directing behaviour. The experiment also shows that the actual percentages of ortho and para products need not be equal. \(\textbf{Separate positional preference from activation:}\) Chlorine withdraws electron density inductively and deactivates the aromatic ring relative to benzene. Its lone-pair resonance contribution favours ortho and para attack relative to meta attack. The supplied distribution measures relative outcomes within chlorobenzene. It does not provide an absolute rate comparison between chlorobenzene and benzene. Consequently, the predominance of para substitution cannot be used as evidence of overall ring activation. \(\textbf{Final answer:}\) \[ \boxed{\frac{k_p}{k_o}=\frac{68}{15}\approx4.53;\quad k_p:k_o=68:15} \]

Explanation: Ordinary electrophilic aromatic nitration requires an electron-deficient species capable of accepting electron density from benzene's \(\pi\) system. The concentrated acid mixture generates that electrophile before attack on the aromatic ring. The overall substitution then proceeds through a nonaromatic sigma complex. \(\textbf{Identify the nitrating electrophile:}\) Concentrated nitric acid interacts with concentrated sulphuric acid to generate the nitronium ion. An appropriate net representation of this acid-assisted process is: \[ \mathrm{HNO_3+H_2SO_4\rightleftharpoons NO_2^++HSO_4^-+H_2O} \] The nitronium ion is positively charged and electron-deficient at nitrogen. It is the principal electrophilic species in the ordinary nitration mechanism. Nitrate ion is not interchangeable with nitronium ion. Thus, proposal P is incorrect. \(\textbf{Evaluate the effect of water:}\) Water appears on the product side of the represented nitronium-ion-generation equilibrium. Increasing the amount of water does not automatically increase the nitronium-ion concentration. Under otherwise comparable equilibrium conditions, added water can oppose the net dehydration process involved in generating the electrophile. The high acidity and dehydrating action of concentrated sulphuric acid are therefore chemically important. A claim based only on the polarity of water ignores the relevant acid–base equilibrium. Proposal Q is incorrect. \(\textbf{Examine electrophilic attack:}\) Benzene donates electron density from its aromatic \(\pi\) system towards the nitronium ion. A new carbon–nitrogen sigma bond is formed. The attacked carbon still carries its original hydrogen. That carbon becomes approximately \(sp^3\)-hybridised in the arenium-ion intermediate. Continuous cyclic \(\pi\)-electron delocalisation is consequently interrupted. \(\textbf{Evaluate final proton removal:}\) The sigma complex is not the final neutral aromatic product. A base removes the hydrogen from the ring carbon that received the nitro group. The associated electron rearrangement restores the fully conjugated aromatic system. The product is nitrobenzene. Proposal R correctly describes this aromaticity-restoring step. \(\textbf{Interpret the combined mechanism:}\) Nitronium-ion formation, electrophilic attack and deprotonation have different chemical functions. The acid mixture establishes the availability of the electrophile. The ring then forms a temporary sigma complex. Finally, proton loss restores aromaticity while retaining the newly introduced nitro group. The valid evaluation is therefore that only R is correct.

Explanation: \(\textbf{Identify the common electrophilic group:}\) Both reagents can introduce an acetyl group, \(\mathrm{CH_3CO-}\), into the benzene ring under appropriate Friedel–Crafts conditions. The new carbon–carbon bond forms between a benzene ring carbon and the acyl carbonyl carbon. The common organic product is acetophenone: \[ \mathrm{C_6H_5COCH_3} \] The carbonyl oxygen is retained in the aromatic ketone. \(\textbf{Write the acyl chloride net reaction:}\) \[ \mathrm{C_6H_6+CH_3COCl\longrightarrow C_6H_5COCH_3+HCl} \] One benzene molecule reacts with one acyl chloride molecule. \(\textbf{Identify the limiting reagent in P:}\) \[ n(\mathrm{benzene})_P=0.180\,\mathrm{mol} \] \[ n(\mathrm{CH_3COCl})_P=0.120\,\mathrm{mol} \] The acyl chloride is limiting. Therefore: \[ n(\mathrm{acetophenone})_P=0.120\,\mathrm{mol} \] The associated HCl amount is: \[ n(\mathrm{HCl})_P=0.120\,\mathrm{mol} \] \(\textbf{Write the anhydride net reaction:}\) \[ \mathrm{C_6H_6+(CH_3CO)_2O\longrightarrow C_6H_5COCH_3+CH_3COOH} \] The anhydride contains two acetyl groups, but the stipulated net reaction transfers one acetyl group to benzene and releases ethanoic acid from the remaining portion. It must therefore not be counted as producing two acetophenone molecules per anhydride molecule. \(\textbf{Identify the limiting reagent in Q:}\) \[ n(\mathrm{benzene})_Q=0.180\,\mathrm{mol} \] \[ n(\mathrm{anhydride})_Q=0.150\,\mathrm{mol} \] The anhydride is limiting. Thus: \[ n(\mathrm{acetophenone})_Q=0.150\,\mathrm{mol} \] The ethanoic acid coproduct amount is: \[ n(\mathrm{CH_3COOH})_Q=0.150\,\mathrm{mol} \] \(\textbf{Combine the coproduct amounts:}\) \[ n_{\mathrm{coproducts}}=n(\mathrm{HCl})_P+n(\mathrm{CH_3COOH})_Q \] \[ n_{\mathrm{coproducts}}=0.120+0.150 \] \[ n_{\mathrm{coproducts}}=0.270\,\mathrm{mol} \] \(\textbf{Check the chemical distinction:}\) The two routes produce the same aromatic ketone through electrophilic acylation. However, the leaving-group-derived coproducts differ. The reaction with ethanoyl chloride produces HCl, while the reaction with ethanoic anhydride produces ethanoic acid. The number of acetyl groups present in an anhydride formula must not replace the actual balanced reaction stoichiometry. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{acetophenone})_P&=0.120\,\mathrm{mol}\\ n(\mathrm{acetophenone})_Q&=0.150\,\mathrm{mol}\\ n(\mathrm{coproducts})_{\mathrm{total}}&=0.270\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the two distinct requirements:}\) A suitable synthesis must not only place the incoming substituent at the desired aromatic ring position. Its proposed reactions must also be chemically feasible under the stated conditions. A directing-group classification alone does not guarantee a successful Friedel–Crafts reaction. \(\textbf{Evaluate the first stage of P:}\) Nitration of benzene introduces the nitro group through electrophilic aromatic substitution. The product is nitrobenzene: \[ \mathrm{C_6H_5NO_2} \] The nitro group withdraws electron density strongly from the aromatic ring. It is deactivating and meta-directing. \(\textbf{Examine P's second reaction:}\) Ordinary Friedel–Crafts alkylation requires an aromatic ring sufficiently reactive towards the alkyl-derived electrophile. The strongly deactivated ring of nitrobenzene does not normally undergo the proposed Friedel–Crafts alkylation. Therefore, its meta-directing tendency cannot be used to claim that 3-nitroethylbenzene is readily obtained by P. The route is unsuitable under the specified ordinary reaction conditions. \(\textbf{Evaluate the first stage of Q:}\) Benzene can undergo Friedel–Crafts alkylation with an appropriate ethyl halide and anhydrous aluminium chloride. The organic product is ethylbenzene: \[ \mathrm{C_6H_5CH_2CH_3} \] The ethyl group is an activating alkyl substituent. \(\textbf{Examine Q's second reaction:}\) Ethylbenzene can undergo electrophilic nitration under suitable conditions. The pre-existing ethyl group directs the incoming nitro group principally towards ortho and para positions. Consequently, the favoured products are 2-nitroethylbenzene and 4-nitroethylbenzene. The desired meta isomer is not the ordinary favoured product. \(\textbf{Distinguish absence from lack of preference:}\) A minor meta product from a successful nitration is not identical to a synthetically preferred meta-directed route. Likewise, an electronically meta-directing substrate need not be suitable for every electrophilic substitution. P has a fundamental Friedel–Crafts feasibility problem. Q is chemically plausible but has an unfavourable directing preference for the stated target. \(\textbf{Interpret the synthetic decision:}\) Neither proposed route is a suitable straightforward method of principally preparing 3-nitroethylbenzene under the restrictions provided. The conclusion requires checking both ring deactivation and the actual directing group present at the nitration stage. Changing reaction order solves neither problem automatically.

Explanation: \(\textbf{Identify the electrophile's carbon framework:}\) Ethanoyl chloride has the structure: \[ \mathrm{CH_3COCl} \] Reaction with a suitable Lewis acid facilitates formation of the acetyl acylium ion. That ion contains the same methyl and carbonyl carbon framework as the starting acyl group. \(\textbf{Examine contributor P:}\) The structure \(\mathrm{CH_3-C\equiv O^+}\) contains a carbon–oxygen triple bond. The carbonyl-derived carbon has a complete octet. The positive formal charge is shown on oxygen in this contributor. It is an important resonance representation of the acylium ion. \(\textbf{Examine contributor Q:}\) The structure \(\mathrm{CH_3-C^+=O}\) contains a carbon–oxygen double bond. The carbonyl carbon carries the formal positive charge. This represents another resonance contributor of the same acylium ion. P and Q possess the same atom connectivity and differ only in electron placement. They are not separate acylating reagents. \(\textbf{Reject contributor R:}\) The proposed formula inserts a hydrogen atom on the positively charged acyl carbon and a negative oxygen. It therefore changes the hydrogen distribution and is not an ordinary resonance contributor of \(\mathrm{CH_3CO^+}\). Resonance cannot create or relocate atomic nuclei. \(\textbf{Reject contributor S:}\) The ethyl cation \(\mathrm{CH_3CH_2^+}\) contains no oxygen atom. It is an alkyl electrophile, not a resonance contributor of an acetyl acylium ion. Formation of an aromatic ketone must not be confused with alkylation to ethylbenzene. \(\textbf{Follow the carbon–carbon bond formation:}\) Benzene attacks the electrophilic acyl carbon, forming a new aromatic-ring-to-carbonyl-carbon bond. A sigma complex forms and subsequently loses a proton to restore aromaticity. After appropriate work-up, the product is: \[ \mathrm{C_6H_5COCH_3} \] This is acetophenone. Its ketone carbonyl is retained. \(\textbf{Interpret resonance and product selection:}\) Resonance stabilisation of the acylium ion helps distinguish Friedel–Crafts acylation from pathways involving ordinary rearrangeable alkyl carbocations. The electron distribution may be represented in more than one valid way, but the acyl group's atom connectivity is preserved. The correct representations are P and Q, and the resulting organic functional group is a ketone.

Explanation: \(\textbf{Identify the two elementary transformations:}\) The first stage forms the sigma complex through electrophilic attack on the aromatic \(\pi\) system. This temporarily interrupts the benzene-like aromatic delocalisation. The second stage removes a proton and restores the aromatic framework. Each stage has its own reactant energy and transition-state energy. \(\textbf{Calculate the first forward barrier:}\) The separated reactants start at: \[ E_R=0\,\mathrm{kJ\,mol^{-1}} \] The first transition state is at: \[ E_{\mathrm{TS1}}=95\,\mathrm{kJ\,mol^{-1}} \] Thus: \[ E_{a,1}=E_{\mathrm{TS1}}-E_R \] \[ E_{a,1}=95-0 \] \[ E_{a,1}=95\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Identify the intermediate's energetic position:}\) The arenium ion is at: \[ E_I=48\,\mathrm{kJ\,mol^{-1}} \] It lies above the separated reactants because aromaticity has been temporarily lost. This energy is the starting reference for the second elementary stage. \(\textbf{Calculate the second forward barrier:}\) The second transition state lies at: \[ E_{\mathrm{TS2}}=70\,\mathrm{kJ\,mol^{-1}} \] Therefore: \[ E_{a,2}=E_{\mathrm{TS2}}-E_I \] \[ E_{a,2}=70-48 \] \[ E_{a,2}=22\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Compare the two barriers:}\) \[ E_{a,1}-E_{a,2}=95-22 \] \[ E_{a,1}-E_{a,2}=73\,\mathrm{kJ\,mol^{-1}} \] The stipulated first stage has the higher forward activation barrier. This is consistent with the importance of disrupting aromaticity during sigma-complex formation. \(\textbf{Determine the overall energy change:}\) The final substituted aromatic product lies at: \[ E_P=-12\,\mathrm{kJ\,mol^{-1}} \] Therefore: \[ \Delta H=E_P-E_R \] \[ \Delta H=-12-0 \] \[ \Delta H=-12\,\mathrm{kJ\,mol^{-1}} \] The stipulated overall reaction is exothermic. \(\textbf{Distinguish absolute energies from activation barriers:}\) The second transition state is at \(+70\,\mathrm{kJ\,mol^{-1}}\), but the second stage does not begin at zero. It begins at the arenium-ion intermediate at \(+48\,\mathrm{kJ\,mol^{-1}}\). Using zero instead would overestimate the second forward activation barrier. Similarly, the overall enthalpy change must compare the original reactants with the final product, not compare the two transition states. \(\textbf{Final answer:}\) \[ \boxed{E_{a,1}=95\,\mathrm{kJ\,mol^{-1}};\quad E_{a,2}=22\,\mathrm{kJ\,mol^{-1}};\quad\Delta H=-12\,\mathrm{kJ\,mol^{-1}}} \]

Explanation: \(\textbf{Identify the starting side chain:}\) Toluene has the structure: \[ \mathrm{C_6H_5CH_3} \] Its methyl carbon carries three benzylic hydrogen atoms. Its aromatic ring carries five hydrogen atoms. The same reagent can replace hydrogens in either region, depending on the specified reaction conditions. \(\textbf{Follow the two benzylic substitutions in P:}\) The first radical substitution changes the methyl group into a bromomethyl group: \[ \mathrm{-CH_3\longrightarrow-CH_2Br} \] The second replacement occurs at that same benzylic carbon under the stipulated conditions: \[ \mathrm{-CH_2Br\longrightarrow-CHBr_2} \] The resulting compound is: \[ \mathrm{C_6H_5CHBr_2} \] It is benzal bromide, also named (dibromomethyl)benzene. \(\textbf{Apply hydrolysis to P:}\) Both bromine atoms are attached to the same saturated benzylic carbon. The stipulated hydrolysis replaces the geminal halogen arrangement by a carbonyl group. The carbonyl carbon remains directly attached to the phenyl ring: \[ \mathrm{C_6H_5CHBr_2+H_2O\longrightarrow C_6H_5CHO+2HBr} \] The organic product is benzaldehyde. \(\textbf{Follow aromatic substitution in Q:}\) Under suitable Lewis-acid-assisted bromination, bromine substitutes at aromatic ring positions. The methyl group remains unchanged. Because two aromatic hydrogens are replaced, the products are dibromotoluene constitutional isomers. Their precise distribution requires positional and reaction-condition information beyond the two-equivalent stoichiometry. \(\textbf{Compare the product formulas:}\) P replaces two benzylic hydrogens with bromine. Q replaces two aromatic ring hydrogens with bromine. Both therefore have the same overall formula: \[ \mathrm{C_7H_6Br_2} \] However, their carbon–bromine bonds occupy different structural environments. \(\textbf{Apply the hydrolysis selectivity to Q:}\) The bromine atoms of Q are directly attached to aromatic \(sp^2\)-hybridised carbons. They do not form a geminal benzylic dibromide. Under the stated selective hydrolysis conditions, the aromatic carbon–bromine bonds remain intact. \(\textbf{Interpret the structural evidence:}\) The two products have identical molecular formulas but different constitutional connectivities. Reagent consumption alone therefore cannot identify their reaction sites. Selective hydrolysis distinguishes the carbon–halogen bonding arrangements. Only P contains the geminal benzylic dihalide needed for the stipulated conversion to benzaldehyde.

Explanation: \(\textbf{Identify benzene's stipulated addition pathway:}\) Benzene possesses an aromatic six-electron \(\pi\) system. Under the specified ultraviolet conditions, it can undergo chlorine addition rather than ordinary electrophilic substitution. The net reaction is: \[ \mathrm{C_6H_6+3Cl_2\xrightarrow{h\nu}C_6H_6Cl_6} \] All six original hydrogen atoms remain in the organic product. Six chlorine atoms have been added. \(\textbf{Calculate the requirement for P:}\) Three chlorine molecules supply six chlorine atoms. Therefore: \[ n(\mathrm{Cl_2})_P=3\,\mathrm{mol} \] Because no hydrogen atom is displaced: \[ n(\mathrm{HCl})_P=0 \] \(\textbf{Identify cyclohexane's stipulated substitution pathway:}\) Cyclohexane is a saturated hydrocarbon. The reaction does not consume a carbon–carbon double bond. Instead, six of its original twelve hydrogen atoms are replaced by chlorine. The net stoichiometric equation is: \[ \mathrm{C_6H_{12}+6Cl_2\xrightarrow{h\nu}C_6H_6Cl_6+6HCl} \] \(\textbf{Calculate the requirement for Q:}\) Every C–H replacement consumes one chlorine molecule. Thus: \[ n(\mathrm{Cl_2})_Q=6\,\mathrm{mol} \] Six displaced hydrogens appear in hydrogen chloride: \[ n(\mathrm{HCl})_Q=6\,\mathrm{mol} \] \(\textbf{Compare the final molecular formulas:}\) P starts with six hydrogen atoms and retains all six. Q starts with twelve hydrogen atoms but loses six through substitution. Both final products consequently have formula: \[ \mathrm{C_6H_6Cl_6} \] The common molecular formula does not establish that their chlorine atoms occupy the same carbon positions or stereochemical arrangements. \(\textbf{Interpret the reaction-history distinction:}\) P undergoes addition, destroying its aromaticity under the specified conditions. Q undergoes radical substitution on an already saturated ring. The different chlorine requirements and HCl yields provide a chemical means of distinguishing the two processes. The comparison illustrates why the total number of chlorine atoms in an organic product cannot by itself reveal whether addition or substitution occurred. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} P:&\ \mathrm{C_6H_6Cl_6},\quad n(\mathrm{Cl_2})=3\,\mathrm{mol},\quad n(\mathrm{HCl})=0\\ Q:&\ \mathrm{C_6H_6Cl_6},\quad n(\mathrm{Cl_2})=6\,\mathrm{mol},\quad n(\mathrm{HCl})=6\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Identify the oxidation requirement:}\) The stipulated vigorous oxidation converts an alkylbenzene side chain into a carboxyl group if the carbon directly bonded to the benzene ring carries at least one hydrogen. This is the benzylic-hydrogen requirement. The total number of hydrogen atoms elsewhere in the side chain is not sufficient to establish eligibility. \(\textbf{Examine P's ethyl group:}\) The ethyl substituent is: \[ \mathrm{Ar-CH_2CH_3} \] Its benzylic carbon is a methylene carbon. It carries two benzylic hydrogens and is therefore eligible for oxidation. The side chain is converted into a ring-attached carboxyl group. \(\textbf{Examine P's methyl group:}\) The methyl substituent is: \[ \mathrm{Ar-CH_3} \] It carries three benzylic hydrogens. It is independently eligible for the same transformation. Because the original side chains occupy positions 1 and 4, both positions become carboxyl-substituted. The product is benzene-1,4-dicarboxylic acid, also called terephthalic acid. \(\textbf{Examine Q's tert-butyl group:}\) The tert-butyl substituent has the structure: \[ \mathrm{Ar-C(CH_3)_3} \] The carbon directly attached to the aromatic ring is bonded to four carbon atoms. It has no benzylic hydrogen. Under the explicitly restricted oxidation conditions, that substituent does not undergo the specified side-chain-to-carboxyl conversion. \(\textbf{Examine Q's methyl group:}\) The other substituent in Q is an ordinary ring-attached methyl group. It contains benzylic hydrogens and is oxidised to a carboxyl group. The tert-butyl group remains at its original para position. The resulting product is 4-tert-butylbenzoic acid. \(\textbf{Interpret the structural contrast:}\) P and Q each contain two alkyl substituents on a benzene ring. Yet the number of oxidisable benzylic positions differs. P has two eligible side chains and gives a dicarboxylic acid. Q has only one and gives a monocarboxylic acid retaining the tert-butyl group. The decisive factor is the local carbon–hydrogen bonding at each benzylic carbon, not simply the total number of alkyl substituents. The final acid products therefore provide information about side-chain structure that the designation "dialkylbenzene" alone does not supply.

Explanation: \(\textbf{Identify the complete aromaticity requirements:}\) Aromaticity is not established by counting \(\pi\) electrons alone. An appropriate system must possess a cyclic conjugated arrangement. Its orbitals must permit effective continuous overlap around the ring. For a planar continuously conjugated system, the relevant aromatic electron count follows the \(4n+2\) rule. \(\textbf{Examine the neutral seven-membered ring P:}\) Cyclohepta-1,3,5-triene contains three ordinary carbon–carbon double bonds. Its total \(\pi\)-electron count is six. However, the remaining ring carbon is tetrahedral and contains no unhybridised p orbital participating in continuous conjugation. The p-orbital network is consequently interrupted at that carbon. P does not possess a continuous aromatic cyclic \(\pi\) circuit. It is therefore nonaromatic. \(\textbf{Examine the stipulated cation Q:}\) Q possesses seven conjugated \(sp^2\)-hybridised carbon centres in a continuous ring arrangement. The stated cation has six delocalised \(\pi\) electrons. Because every ring position participates in cyclic conjugation and planarity is stipulated, the structural requirements are satisfied. The electron count follows: \[ 6=4(1)+2 \] Q therefore satisfies the aromaticity criterion. \(\textbf{Examine benzene R:}\) Benzene has six \(sp^2\)-hybridised ring carbons. Each contributes an unhybridised p orbital. The six delocalised \(\pi\) electrons are distributed around a planar continuous six-membered circuit. Benzene also satisfies \(4n+2\) with \(n=1\). Therefore, R is aromatic. \(\textbf{Interpret the shared electron count:}\) P, Q and R each possess six electrons associated with formal carbon–carbon \(\pi\) bonding or an aromatic conjugated circuit. However, P lacks continuous cyclic overlap because one carbon interrupts the p-orbital system. Q and R do not have that interruption. The fact that Q is positively charged and has seven ring atoms does not exclude aromaticity under the stipulated conditions. The comparison therefore demonstrates why electron count must be interpreted together with the actual orbital connectivity and molecular geometry.

Explanation: \(\textbf{Identify the desired carbon framework:}\) n-Propylbenzene has the structure: \[ \mathrm{C_6H_5CH_2CH_2CH_3} \] Isopropylbenzene has the structure: \[ \mathrm{C_6H_5CH(CH_3)_2} \] Both contain nine carbon atoms and twelve hydrogen atoms, but the aromatic ring is bonded to different carbons of the three-carbon substituent. \(\textbf{Follow the electrophile in P:}\) 1-Chloropropane contains chlorine on a primary carbon: \[ \mathrm{CH_3CH_2CH_2Cl} \] Under the stipulated conditions, the initially formed primary propyl-derived electrophilic arrangement rearranges. The more stable secondary propyl electrophile is then captured by benzene. The principal organic product is: \[ \mathrm{C_6H_5CH(CH_3)_2} \] This is isopropylbenzene, also called cumene. \(\textbf{Follow acylation in Q:}\) Propanoyl chloride supplies a propanoyl-derived acylium electrophile. Benzene undergoes acylation at the carbonyl carbon: \[ \mathrm{C_6H_6+CH_3CH_2COCl\longrightarrow C_6H_5COCH_2CH_3+HCl} \] The aromatic ketone is 1-phenylpropan-1-one. Its acylium-derived carbon framework has not undergone the ordinary carbocation rearrangement stipulated in P. \(\textbf{Reduce the ketone in Q:}\) Clemmensen reduction converts the carbonyl group into a methylene group: \[ \mathrm{C_6H_5COCH_2CH_3\xrightarrow{Zn(Hg)/HCl}C_6H_5CH_2CH_2CH_3} \] The product is n-propylbenzene. Thus, the acylation–reduction route reaches the desired unbranched aromatic side chain. \(\textbf{Evaluate the oxidation of P's product:}\) Isopropylbenzene contains one benzylic hydrogen on the carbon directly attached to the aromatic ring. That carbon is tertiary but is not quaternary. It therefore satisfies the stated requirement for vigorous side-chain oxidation. The aromatic organic product is benzoic acid. \(\textbf{Evaluate the oxidation of Q's product:}\) n-Propylbenzene contains two benzylic hydrogens on its ring-attached methylene carbon. It is also eligible for the stated strong side-chain oxidation. Its aromatic organic product is likewise benzoic acid. \(\textbf{Interpret the convergent oxidation:}\) The two routes initially produce different constitutional alkylbenzene isomers. However, vigorous oxidation removes the distinctions between their eligible three-carbon side chains when both give a ring-attached carboxyl group. The common benzoic acid product therefore cannot independently establish which propylbenzene isomer was originally present. The decisive synthesis choice is the acylation–reduction sequence, which avoids the specified rearrangement during direct alkylation.

Explanation: \(\textbf{Distinguish fused and linked aromatic frameworks:}\) Benzene contains one six-membered aromatic ring. Naphthalene contains two fused rings sharing a carbon–carbon edge. Biphenyl contains two benzene rings joined by an ordinary carbon–carbon single bond. The different connections affect the total number of carbon atoms, hydrogen atoms and formal \(\pi\)-bond equivalents. \(\textbf{Determine the requirement for benzene:}\) Benzene has three formal carbon–carbon double-bond equivalents. Complete hydrogenation consumes three hydrogen molecules: \[ \mathrm{C_6H_6+3H_2\longrightarrow C_6H_{12}} \] The product is cyclohexane. Therefore: \[ n(\mathrm{H_2})_P=3\,\mathrm{mol\ per\ mol} \] \(\textbf{Determine naphthalene's hydrogen deficiency:}\) Naphthalene contains ten carbon atoms and eight hydrogen atoms. Its degree of hydrogen deficiency is: \[ \mathrm{DBE}_Q=\frac{2(10)+2-8}{2} \] \[ \mathrm{DBE}_Q=7 \] Two degrees arise from the two fused rings. The remaining five correspond to formal carbon–carbon double-bond equivalents. \(\textbf{Calculate naphthalene hydrogenation:}\) Complete saturation with ring retention therefore requires five hydrogen molecules: \[ \mathrm{C_{10}H_8+5H_2\longrightarrow C_{10}H_{18}} \] The product is the corresponding saturated fused bicyclic hydrocarbon framework, commonly represented by decalin connectivity. Thus: \[ n(\mathrm{H_2})_Q=5\,\mathrm{mol\ per\ mol} \] \(\textbf{Determine biphenyl's hydrogen deficiency:}\) Biphenyl contains two benzene-derived aromatic rings. Each ring has three formal \(\pi\)-bond equivalents. The carbon–carbon single bond joining the two phenyl rings does not add another degree of unsaturation. Consequently, six hydrogen molecules are required to saturate both rings. \(\textbf{Calculate biphenyl hydrogenation:}\) \[ \mathrm{C_{12}H_{10}+6H_2\longrightarrow C_{12}H_{22}} \] The saturated product contains two cyclohexane rings connected by a single C–C bond. Thus: \[ n(\mathrm{H_2})_R=6\,\mathrm{mol\ per\ mol} \] \(\textbf{Check the saturated-product formulas:}\) A saturated bicyclic hydrocarbon with ten carbons has formula \(\mathrm{C_{10}H_{18}}\). Two saturated cyclohexane rings joined by a single carbon–carbon bond likewise constitute a bicyclic framework, but with twelve carbons and formula \(\mathrm{C_{12}H_{22}}\). The calculated products are consistent with ring retention and complete loss of the original \(\pi\)-bond equivalents. \(\textbf{Interpret the structural distinction:}\) Naphthalene possesses a ten-electron fused aromatic \(\pi\) system, whereas biphenyl contains two benzene-like aromatic rings with twelve \(\pi\) electrons in total. Their hydrogen requirements are not obtained by simply dividing an aromatic electron count by two without considering the actual sigma-bond framework. The number of formal double-bond equivalents and the original ring count must be distinguished. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} P:&\quad 3\,\mathrm{mol\ H_2}\longrightarrow\mathrm{C_6H_{12}}\\ Q:&\quad 5\,\mathrm{mol\ H_2}\longrightarrow\mathrm{C_{10}H_{18}}\\ R:&\quad 6\,\mathrm{mol\ H_2}\longrightarrow\mathrm{C_{12}H_{22}} \end{aligned}} \]

Explanation: \(\textbf{Separate two different kinds of evidence:}\) The relative reaction rate measures the overall ease of electrophilic substitution under the specified experimental conditions. The directing effect describes the preferred position of substitution within a particular aromatic ring. These properties are related to substituent electronic effects but are not identical. \(\textbf{Interpret P's methoxy group:}\) In methoxybenzene, oxygen is directly bonded to an aromatic ring carbon. Its lone pair can participate in resonance donation into the ring. This stabilises appropriate ortho- and para-attack pathways. The supplied rate is considerably greater than that of benzene. P is therefore activating and ortho/para-directing. \(\textbf{Interpret Q's methyl group:}\) The methyl group donates electron density through ordinary alkyl-group effects. This increases the rate relative to benzene in the supplied series. It also favours electrophilic substitution at ortho and para positions. Q is therefore activating and ortho/para-directing. \(\textbf{Use benzene as the reference:}\) R has no pre-existing substituent. Its rate is defined as unity. Its available ring positions are all constitutionally equivalent before the first substitution. Consequently, no ortho/meta/para preference arises for monosubstitution of unsubstituted benzene. \(\textbf{Interpret S's chlorine substituent:}\) Chlorine withdraws electron density through its negative inductive effect. This decreases overall substitution reactivity compared with benzene. However, chlorine's lone pairs can participate in resonance donation. That contribution favours ortho and para substitution relative to meta substitution. S is therefore deactivating yet ortho/para-directing. \(\textbf{Interpret T's nitro group:}\) The nitro group strongly withdraws electron density through inductive and resonance effects. It greatly decreases the ring's reactivity towards electrophilic substitution. Ortho and para sigma complexes contain particularly unfavourable positive-charge arrangements associated with the nitro-bearing ring carbon. Meta attack is comparatively preferred. Thus, T is strongly deactivating and meta-directing. \(\textbf{Order the supplied rates:}\) Reading from the smallest measured value to the largest gives: \[ T\lt S\lt R\lt Q\lt P \] The ordering is consistent with the electronic classifications of the specified substituents. \(\textbf{Interpret the halogen exception:}\) Chlorobenzene illustrates why positional direction cannot be inferred solely from whether a substituent activates or deactivates the ring. The overall rate and the preferred sites depend on different aspects of the substituent's electron-withdrawing and electron-donating effects. The stated order and directing classifications therefore agree with option A.

Explanation: \(\textbf{Identify the carbon removed during decarboxylation:}\) Sodium benzoate has the structure: \[ \mathrm{C_6H_5COONa} \] It contains six carbon atoms in the benzene ring and one additional carbon atom in the carboxylate group. Soda-lime decarboxylation removes the carboxylate carbon from the organic framework. The original six-carbon aromatic ring is retained. \(\textbf{Write the ordinary transformation:}\) \[ \mathrm{C_6H_5COONa+NaOH\xrightarrow{CaO,\Delta}C_6H_6+Na_2CO_3} \] The carboxylate carbon becomes the carbon atom of the carbonate product. It does not become a new carbon atom of the benzene ring. \(\textbf{Track the isotope in P:}\) In P, carbon-13 is located specifically at the carboxylate carbon. That atom is removed from the organic structure during decarboxylation. The corresponding labelled reaction can be represented as: \[ \mathrm{C_6H_5\,^{13}COONa+NaOH\longrightarrow C_6H_6+Na_2\,^{13}CO_3} \] The benzene product contains only the original ordinary-carbon ring atoms. Therefore, it is unlabelled under the stated isotopic assumptions. \(\textbf{Track the isotope in Q:}\) In Q, one aromatic ring carbon is labelled. The carboxylate carbon is not labelled. Decarboxylation removes the ordinary carboxylate carbon while preserving every original aromatic ring carbon. The resulting benzene therefore retains one carbon-13 atom within its ring. \(\textbf{Apply electrophilic bromination to P's benzene:}\) The unlabelled benzene undergoes substitution: \[ \mathrm{C_6H_6+Br_2\xrightarrow{FeBr_3}C_6H_5Br+HBr} \] No carbon-13 atom is introduced by the stipulated reagents. Therefore, P gives ordinary unlabelled bromobenzene. \(\textbf{Apply electrophilic bromination to Q's benzene:}\) The labelled benzene undergoes the same electrophilic aromatic substitution pathway. A bromine electrophile forms a bond with a ring carbon, and a proton is subsequently removed to restore aromaticity. No carbon atom is expelled from the ring. The ring carbon-13 label therefore remains present in bromobenzene. \(\textbf{Distinguish proton substitution from skeletal replacement:}\) Electrophilic aromatic bromination replaces an aromatic hydrogen by bromine. It does not replace a carbon atom of the aromatic ring. The isotope remains within the ring even if bromination occurs at the labelled carbon's attached hydrogen position. \(\textbf{Interpret the complementary experiments:}\) Decarboxylation selectively removes a particular carbon atom from the sodium benzoate skeleton. Electrophilic substitution instead preserves the carbon skeleton while replacing a hydrogen atom. Carbon-13 labelling makes this difference observable. The result follows directly from tracking the fate of individual atoms through both reaction mechanisms rather than merely comparing the molecular formulas of the final products.

Explanation: \(\textbf{Separate the two carbon-assembly pathways:}\) Linear polymerisation of ethyne can generate an extended conjugated carbon backbone. Cyclic trimerisation instead combines three ethyne molecules into one benzene ring. These pathways conserve carbon atoms but create different carbon frameworks and different distributions of multiple bonds. \(\textbf{Count the carbon atoms in P:}\) Each ethyne molecule contributes two carbon atoms. Twelve ethyne molecules therefore provide: \[ N_C=12(2)=24 \] The stipulated single open chain contains 24 carbon atoms. \(\textbf{Determine its hydrogen content:}\) The twelve ethyne molecules initially contain: \[ N_H=12(2)=24 \] The problem explicitly supplies two additional terminal hydrogen atoms through one H2 molecule. Therefore: \[ N_{H,\mathrm{final}}=24+2=26 \] The polyene formula is: \[ \mathrm{C_{24}H_{26}} \] \(\textbf{Determine the retained multiple bonds:}\) A saturated acyclic 24-carbon hydrocarbon would have formula: \[ \mathrm{C_{24}H_{50}} \] The stipulated polyene contains 24 fewer hydrogen atoms. For an acyclic molecule, each ordinary double bond corresponds to a deficiency of two hydrogen atoms relative to the saturated formula. Therefore: \[ N_{\mathrm{C=C}}=\frac{50-26}{2} \] \[ N_{\mathrm{C=C}}=12 \] This agrees with the specified alternating conjugated polyene structure. \(\textbf{Calculate bromine addition:}\) Each ordinary C=C bond consumes one Br2 molecule during complete addition. Thus: \[ N_{\mathrm{Br_2}}=12(1) \] \[ N_{\mathrm{Br_2}}=12 \] The polyene becomes saturated with respect to carbon–carbon multiple bonds. \(\textbf{Calculate benzene formation in Q:}\) The number of ethyne molecules remaining is: \[ N_{\mathrm{ethyne},Q}=24-12 \] \[ N_{\mathrm{ethyne},Q}=12 \] Cyclic trimerisation follows: \[ \mathrm{3C_2H_2\longrightarrow C_6H_6} \] Therefore: \[ N_{\mathrm{benzene}}=\frac{12}{3} \] \[ N_{\mathrm{benzene}}=4 \] \(\textbf{Verify carbon conservation:}\) The open-chain product contains 24 carbon atoms. Four benzene molecules contain: \[ 4(6)=24 \] The total is 48 carbon atoms, equal to the carbon contained in 24 ethyne molecules. \(\textbf{Interpret the polymerisation distinction:}\) The open-chain product retains ordinary C=C bonds and undergoes the specified bromine addition. The benzene products possess aromatic delocalisation and are excluded from ordinary bromine addition under the stated conditions. The extra capping hydrogen is essential when assigning the molecular formula of the finite open-chain molecule. An extended ideal polyacetylene backbone should not be confused with a saturated polyethene backbone, nor should ethyne's cyclic polymerisation be assumed to give the same product as its linear polymerisation. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_{24}H_{26}};\quad N_{\mathrm{benzene}}=4;\quad N_{\mathrm{Br_2}}=12} \]

Explanation: \(\textbf{Identify ethyne's two acidic hydrogens:}\) Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] Both hydrogen atoms are attached directly to \(sp\)-hybridised carbon atoms. Consequently, both can be replaced through suitable reactions with strong bases or sufficiently reactive metals. This distinguishes ethyne from terminal alkynes possessing only one acetylenic hydrogen. \(\textbf{Follow Stage I with sodamide:}\) The stipulated first reaction removes one acetylenic proton per ethyne molecule: \[ \mathrm{HC\equiv CH+NaNH_2\longrightarrow HC\equiv CNa+NH_3} \] One mole of sodamide produces one mole of ammonia in this transformation. Therefore: \[ n(\mathrm{NH_3})=n(\mathrm{NaNH_2}) \] \[ n(\mathrm{NH_3})=0.300\,\mathrm{mol} \] \(\textbf{Identify the remaining acidic site:}\) The intermediate \(\mathrm{HC\equiv CNa}\) is monosodium ethynide. One end contains the sodium acetylide linkage. The other end still contains a hydrogen bonded to an \(sp\)-hybridised carbon. Therefore, a second deprotonation or replacement remains possible. \(\textbf{Follow Stage II with sodium metal:}\) Sodium metal reacts with the remaining acetylenic hydrogen: \[ \mathrm{2HC\equiv CNa+2Na\longrightarrow2NaC\equiv CNa+H_2} \] The coefficient of sodium metal is two for every hydrogen molecule released. The specified \(0.300\,\mathrm{mol}\) sodium therefore releases: \[ n(\mathrm{H_2})=\frac{0.300}{2} \] \[ n(\mathrm{H_2})=0.150\,\mathrm{mol} \] The organic-derived salt becomes disodium ethynide. \(\textbf{Apply Stage III hydrolysis:}\) Disodium ethynide contains an acetylide centre at both ends. Excess water can supply hydrogen to regenerate the original ethyne: \[ \mathrm{NaC\equiv CNa+2H_2O\longrightarrow HC\equiv CH+2NaOH} \] Each disodium ethynide unit gives one ethyne molecule. Hence: \[ n(\mathrm{C_2H_2})_{\mathrm{recovered}}=0.300\,\mathrm{mol} \] \(\textbf{Check sodium and carbon balances:}\) The two stages introduce a total of two sodium atoms per original ethyne molecule. Hydrolysis subsequently removes both sodium equivalents as sodium hydroxide. The carbon–carbon triple bond remains intact throughout the acid–base sequence. Therefore, the final ethyne amount equals the original ethyne amount. \(\textbf{Interpret the different gaseous products:}\) Sodamide consumes a proton to form ammonia. Sodium metal converts acidic hydrogen into molecular hydrogen. The same acetylenic C–H type can therefore lead to different inorganic products depending on the reagent. The sequence must not be treated as producing hydrogen gas in both stages. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{NH_3})=0.300\,\mathrm{mol};\quad n(\mathrm{H_2})=0.150\,\mathrm{mol};\quad n(\mathrm{C_2H_2})_{\mathrm{recovered}}=0.300\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the unsymmetrical triple bond:}\) Propyne has the structure: \[ \mathrm{CH_3C\equiv CH} \] Its terminal triply bonded carbon carries hydrogen. Its internal triply bonded carbon carries a methyl group. These different substituents make the first addition regioselective under the stipulated conditions. \(\textbf{Follow the HCl addition in P:}\) Ordinary Markovnikov addition places hydrogen at the terminal carbon. Chlorine becomes attached to the internal carbon. The triple bond becomes a double bond: \[ \mathrm{CH_3C\equiv CH+HCl\longrightarrow CH_3C(Cl)=CH_2} \] The first intermediate is 2-chloroprop-1-ene. It retains one ordinary C=C bond. \(\textbf{Complete Route P with HBr:}\) The stipulated second-stage orientation attaches hydrogen to the terminal methylene carbon. Bromine becomes attached to the carbon already bearing chlorine: \[ \mathrm{CH_3C(Cl)=CH_2+HBr\longrightarrow CH_3C(Cl)(Br)CH_3} \] The final compound is 2-bromo-2-chloropropane. Both halogen atoms occupy the same carbon. \(\textbf{Follow the HBr addition in Q:}\) One equivalent of HBr adds conventionally to propyne: \[ \mathrm{CH_3C\equiv CH+HBr\longrightarrow CH_3C(Br)=CH_2} \] The first intermediate is 2-bromoprop-1-ene. It differs from P's intermediate because bromine, rather than chlorine, is attached to the internal alkene carbon. \(\textbf{Complete Route Q with HCl:}\) Under the given second-stage orientation, chlorine is introduced at the brominated alkene carbon: \[ \mathrm{CH_3C(Br)=CH_2+HCl\longrightarrow CH_3C(Br)(Cl)CH_3} \] This has the same constitutional connectivity as P's final product. \(\textbf{Verify the net addition:}\) Both sequences add one HCl and one HBr molecule to one propyne molecule. The formula changes from \(\mathrm{C_3H_4}\) to \(\mathrm{C_3H_6BrCl}\). The original carbon skeleton remains unchanged. The final two halogen atoms are geminal rather than vicinal. \(\textbf{Interpret the condition dependence:}\) The first-stage intermediates remember which hydrogen halide reacted first. Under the explicitly stipulated second-stage orientation, however, both routes converge on the same mixed geminal dihalide. This result must not be generalised automatically to every unsymmetrical alkene or to peroxide-assisted addition conditions. The intermediate and the final product must be predicted separately.

Explanation: \(\textbf{Identify the successive carbon-containing transformations:}\) Calcium carbonate is first converted into calcium oxide. The oxide then reacts with carbon to form calcium carbide. Hydrolysis of calcium carbide releases ethyne. Finally, ethyne undergoes cyclic trimerisation to benzene. Each stage has its own balanced equation and limiting-reagent condition. \(\textbf{Calculate calcium oxide from calcination:}\) \[ \mathrm{CaCO_3\xrightarrow{\Delta}CaO+CO_2} \] The mole ratio of calcium carbonate to calcium oxide is \(1:1\). Thus: \[ n(\mathrm{CaO})=0.500\,\mathrm{mol} \] The carbon dioxide evolved is also \(0.500\,\mathrm{mol}\), although it is not one of the requested quantities. \(\textbf{Identify the limiting reagent in carbide formation:}\) \[ \mathrm{CaO+3C\longrightarrow CaC_2+CO} \] One mole of calcium oxide requires three moles of carbon. Complete conversion of the available calcium oxide would require: \[ n(C)_{\mathrm{required}}=3(0.500) \] \[ n(C)_{\mathrm{required}}=1.500\,\mathrm{mol} \] Only \(1.200\,\mathrm{mol}\) carbon is available. Therefore, carbon is limiting. \(\textbf{Determine carbide formation:}\) \[ n(\mathrm{CaC_2})=\frac{1.200}{3} \] \[ n(\mathrm{CaC_2})=0.400\,\mathrm{mol} \] One carbon monoxide molecule forms per calcium carbide molecule: \[ n(\mathrm{CO})=0.400\,\mathrm{mol} \] \(\textbf{Identify the limiting reagent in hydrolysis:}\) The reaction is: \[ \mathrm{CaC_2+2H_2O\longrightarrow Ca(OH)_2+C_2H_2} \] Hydrolysis of all \(0.400\,\mathrm{mol}\) carbide would require: \[ n(\mathrm{H_2O})_{\mathrm{required}}=2(0.400) \] \[ n(\mathrm{H_2O})_{\mathrm{required}}=0.800\,\mathrm{mol} \] The available water is only \(0.600\,\mathrm{mol}\). Water is therefore limiting. \(\textbf{Calculate the ethyne amount:}\) \[ n(\mathrm{C_2H_2})=\frac{0.600}{2} \] \[ n(\mathrm{C_2H_2})=0.300\,\mathrm{mol} \] The calcium carbide remaining after this hydrolysis is: \[ 0.400-0.300=0.100\,\mathrm{mol} \] \(\textbf{Apply cyclic trimerisation:}\) \[ \mathrm{3C_2H_2\xrightarrow{red\!-\!hot\ Fe,\ 873\,K}C_6H_6} \] Therefore: \[ n(\mathrm{benzene})=\frac{0.300}{3} \] \[ n(\mathrm{benzene})=0.100\,\mathrm{mol} \] \(\textbf{Check the stage-dependent limitation:}\) Carbon limits the amount of carbide prepared. Water then limits how much of that carbide is actually converted into ethyne. The final benzene yield must be based on evolved ethyne, not on the total carbide initially formed. Carbon monoxide is produced in Stage II and is unaffected by the later hydrolysis limitation. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{CaC_2})&=0.400\,\mathrm{mol}\\ n(\mathrm{C_2H_2})&=0.300\,\mathrm{mol}\\ n(\mathrm{C_6H_6})&=0.100\,\mathrm{mol}\\ n(\mathrm{CO})&=0.400\,\mathrm{mol} \end{aligned}} \]

Explanation: \(\textbf{Establish the constitutional possibilities:}\) An acyclic six-carbon hydrocarbon containing one triple bond has formula \(\mathrm{C_6H_{10}}\). Its different alkynes arise from changing the triple-bond position or the carbon skeleton. The full set must be counted without treating opposite numbering directions as different structures. \(\textbf{Enumerate the unbranched alkynes:}\) An unbranched six-carbon chain permits three distinct triple-bond positions: - Hex-1-yne. - Hex-2-yne. - Hex-3-yne. Positions 4, 5 and 6 do not produce additional structures because the chain can be numbered from its opposite end. \(\textbf{Enumerate branched alkynes:}\) The remaining valid constitutional structures are: - 3-Methylpent-1-yne. - 4-Methylpent-1-yne. - 4-Methylpent-2-yne. - 3,3-Dimethylbut-1-yne. These four structures have different carbon connectivities from the three unbranched alkynes and from one another. Therefore: \[ N_{\mathrm{alkyne\ isomers}}=3+4 \] \[ N_{\mathrm{alkyne\ isomers}}=7 \] \(\textbf{Identify the terminal alkynes:}\) Hex-1-yne contains one hydrogen directly attached to a triply bonded carbon. The same is true of 3-methylpent-1-yne, 4-methylpent-1-yne and 3,3-dimethylbut-1-yne. These four compounds can form sodium alkynides by appropriate treatment with sodamide. Therefore: \[ N_{\mathrm{terminal}}=4 \] The other three structures have internal triple bonds. \(\textbf{Apply hydration to the terminal members:}\) The usual mercuric-sulphate-assisted hydration of a substituted terminal alkyne gives a methyl ketone after tautomerisation. Each of the four different terminal structures therefore has one principal constitutional ketone connectivity under the stipulated hydration rule. Although the identities of their ketones differ, each individual terminal alkyne contributes to the one-product category. \(\textbf{Examine the symmetrical internal member:}\) Hex-3-yne has equivalent carbon substituents on its two triple-bonded carbons. Addition in either orientation followed by tautomerisation gives hexan-3-one. Thus, this internal alkyne also gives only one constitutional ketone product. \(\textbf{Examine the unsymmetrical internal members:}\) Hex-2-yne has different groups attached to its triple-bonded carbons. Hydration in opposite orientations can give hexan-2-one and hexan-3-one. Likewise, 4-methylpent-2-yne has nonequivalent triple-bonded carbons and can give two distinct carbonyl positions under the stipulated two-orientation model. Neither belongs in the one-product category. \(\textbf{Calculate the final count:}\) Four terminal members give one principal ketone connectivity each. One symmetrical internal member also gives one. Therefore: \[ N_{\mathrm{one\ ketone}}=4+1 \] \[ N_{\mathrm{one\ ketone}}=5 \] \(\textbf{Interpret the independent tests:}\) The constitutional-isomer count depends on carbon skeleton and triple-bond position. Sodamide distinguishes terminal from internal alkynes using local C–H acidity. Hydration-product multiplicity also depends on the symmetry of an internal triple bond. These properties are related but not interchangeable, so each count requires an independent structural decision. \(\textbf{Final answer:}\) \[ \boxed{N_{\mathrm{isomers}}=7;\quad N_{\mathrm{terminal}}=4;\quad N_{\mathrm{one\ ketone}}=5} \]