Hydrocarbons MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Separate kinetic and thermodynamic comparisons:}\) The principal product under kinetic control is associated with the more accessible reaction pathway. Its formation depends on the relevant activation barrier. The thermodynamic stability of a final product instead depends on its energy relative to other products. These are different comparisons. \(\textbf{Calculate the activation barrier for X:}\) Both pathways begin at zero potential energy. The highest transition state for X is at \(+76\,\mathrm{kJ\,mol^{-1}}\). Thus: \[ E_{a,X}=76-0 \] \[ E_{a,X}=76\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Calculate the activation barrier for Y:}\) The corresponding transition state is at \(+92\,\mathrm{kJ\,mol^{-1}}\). Therefore: \[ E_{a,Y}=92-0 \] \[ E_{a,Y}=92\,\mathrm{kJ\,mol^{-1}} \] The barrier difference is: \[ E_{a,Y}-E_{a,X}=92-76 \] \[ E_{a,Y}-E_{a,X}=16\,\mathrm{kJ\,mol^{-1}} \] \(\textbf{Compare the product energies:}\) The final product from X lies at \(-18\,\mathrm{kJ\,mol^{-1}}\). The final product from Y lies at \(-35\,\mathrm{kJ\,mol^{-1}}\). Their energy difference is: \[ E_{P,X}-E_{P,Y}=(-18)-(-35) \] \[ E_{P,X}-E_{P,Y}=17\,\mathrm{kJ\,mol^{-1}} \] Thus, Y is the lower-energy final product. \(\textbf{Apply the specified reaction conditions:}\) The experiment is conducted under kinetic control. The final products do not interconvert during the experiment. Under the supplied comparable-prefactor assumption, the pathway with the lower activation barrier is kinetically favoured. That is pathway X. Its product may therefore predominate even though the alternative product Y is thermodynamically lower in energy. \(\textbf{Connect the profile to aromatic substitution:}\) An electrophilic aromatic substitution pathway includes electrophile attachment, arenium-ion formation and aromaticity restoration. Different positional pathways can possess different transition-state energies because the corresponding intermediates and molecular arrangements are not equivalent. The final substituted products can also differ in energy. The student has incorrectly used final-product energy as the sole predictor of a kinetically controlled distribution. \(\textbf{Final answer:}\) \[ E_{a,X}=76\,\mathrm{kJ\,mol^{-1}};\quad E_{a,Y}=92\,\mathrm{kJ\,mol^{-1}};\quad E_{P,Y}=E_{P,X}-17\,\mathrm{kJ\,mol^{-1}} \] Pathway X is kinetically favoured under the supplied assumptions.

Explanation: \(\textbf{Apply the aromaticity conditions:}\) Aromaticity requires continuous cyclic p-orbital overlap, an appropriate geometry and a permitted delocalised \(\pi\)-electron count. The Hückel aromatic electron count is expressed as \(4n+2\), where \(n\) is a non-negative integer. The presence of several alternating double bonds is not by itself sufficient. \(\textbf{Examine P:}\) Benzene is planar and contains six \(sp^2\)-hybridised ring carbons. Their p orbitals overlap continuously around the ring. Its six \(\pi\) electrons satisfy the \(4n+2\) rule with \(n=1\). Therefore, benzene is aromatic. \(\textbf{Examine the actual cyclooctatetraene geometry:}\) Cycloocta-1,3,5,7-tetraene contains four double bonds and eight \(\pi\) electrons. Its ordinary structure adopts a nonplanar tub-like arrangement. This geometry avoids the continuously overlapping planar cyclic p-orbital arrangement required for the stipulated antiaromatic comparison. The ordinary molecule is therefore classified as nonaromatic rather than as a planar antiaromatic ring. \(\textbf{Examine the constrained hypothetical structure R:}\) R is explicitly assumed to be planar with continuous p-orbital overlap. Its eight \(\pi\) electrons correspond to a \(4n\) count with \(n=2\). They do not satisfy the \(4n+2\) aromatic condition. Such a hypothetical planar, continuously conjugated eight-\(\pi\)-electron circuit would be antiaromatic. \(\textbf{Determine the hydrogenation requirement:}\) Both Q and R contain four carbon–carbon double bonds in the stipulated bonding descriptions. Each ordinary double bond requires one hydrogen molecule for complete conversion into a single bond. Hence: \[ n(\mathrm{H_2})=4(1) \] \[ n(\mathrm{H_2})=4\,\mathrm{mol\ per\ mol\ substrate} \] The product of complete hydrogenation with ring retention is cyclooctane. \(\textbf{Interpret the structural distinction:}\) Q and R have the same carbon and hydrogen counts and the same number of ordinary double bonds. Their aromaticity classifications nevertheless differ because their stipulated geometries differ. Hydrogen uptake measures hydrogenatable unsaturation but does not independently establish aromaticity. The comparison demonstrates why planarity, cyclic conjugation and electron count must all be examined. \(\textbf{Final answer:}\) \[ n(\mathrm{H_2})_Q=n(\mathrm{H_2})_R=4\,\mathrm{mol\ per\ mol} \] P is aromatic, Q is nonaromatic and the hypothetical planar R is antiaromatic.

Explanation: \(\textbf{Identify the benzylic radical structure:}\) The benzyl radical has the structure: \[ \mathrm{C_6H_5CH_2^\bullet} \] Its unpaired electron is associated with the carbon adjacent to the aromatic ring. The radical can be stabilised through resonance interaction with the phenyl group. The experiment specifies how the radical population is distributed between two termination pathways. \(\textbf{Calculate radicals forming benzyl bromide:}\) The total radical amount is: \[ n_0=0.240\,\mathrm{mol} \] Sixty percent forms benzyl bromide. Therefore: \[ n_{\mathrm{radicals},P}=0.60(0.240) \] \[ n_{\mathrm{radicals},P}=0.144\,\mathrm{mol} \] Each benzyl radical combines with one bromine atom: \[ \mathrm{C_6H_5CH_2^\bullet+Br^\bullet\longrightarrow C_6H_5CH_2Br} \] Thus: \[ n(\mathrm{benzyl\ bromide})=0.144\,\mathrm{mol} \] \(\textbf{Calculate radicals entering coupling:}\) Forty percent enters Pathway Q: \[ n_{\mathrm{radicals},Q}=0.40(0.240) \] \[ n_{\mathrm{radicals},Q}=0.096\,\mathrm{mol} \] Two benzyl radicals are required per coupling event: \[ \mathrm{2C_6H_5CH_2^\bullet\longrightarrow C_6H_5CH_2CH_2C_6H_5} \] Therefore: \[ n(\mathrm{1,2\text{-}diphenylethane})=\frac{0.096}{2} \] \[ n(\mathrm{1,2\text{-}diphenylethane})=0.048\,\mathrm{mol} \] \(\textbf{Calculate the benzyl bromide mass:}\) Benzyl bromide has formula \(\mathrm{C_7H_7Br}\). Its molar mass is: \[ M_P=7(12)+7(1)+80 \] \[ M_P=171\,\mathrm{g\,mol^{-1}} \] Hence: \[ m_P=0.144(171) \] \[ m_P=24.624\,\mathrm{g} \] \(\textbf{Calculate the coupled-product mass:}\) 1,2-Diphenylethane has formula \(\mathrm{C_{14}H_{14}}\). Its molar mass is: \[ M_Q=14(12)+14(1) \] \[ M_Q=182\,\mathrm{g\,mol^{-1}} \] Thus: \[ m_Q=0.048(182) \] \[ m_Q=8.736\,\mathrm{g} \] \(\textbf{Combine the isolated masses:}\) \[ m_{\mathrm{total}}=24.624+8.736 \] \[ m_{\mathrm{total}}=33.360\,\mathrm{g} \] \(\textbf{Verify benzyl-fragment conservation:}\) Benzyl bromide contains one benzyl fragment per molecule. 1,2-Diphenylethane contains two. Therefore: \[ n_{\mathrm{benzyl\ fragments}}=0.144+2(0.048) \] \[ n_{\mathrm{benzyl\ fragments}}=0.240\,\mathrm{mol} \] All original radicals have been accounted for. \(\textbf{Interpret the product-count distinction:}\) A one-radical termination product forms one molecule per benzyl radical used. A two-radical coupling product forms only half as many molecules as the radicals consumed. It would therefore be incorrect to convert the \(40\%\) radical fraction directly into \(0.096\,\mathrm{mol}\) of 1,2-diphenylethane. The product calculation requires following the radical stoichiometry rather than merely applying the supplied percentages. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{benzyl\ bromide})=0.144\,\mathrm{mol};\quad n(\mathrm{1,2\text{-}diphenylethane})=0.048\,\mathrm{mol};\quad m_{\mathrm{total}}=33.360\,\mathrm{g}} \]

Explanation: Statement I is correct. Ethylbenzene contains an ethyl group attached to the aromatic ring through a saturated carbon atom. An alkyl group generally donates electron density towards the aromatic ring. This increases the relative ease of electrophilic substitution compared with unsubstituted benzene under suitable comparable conditions. It also favours electrophilic attack at ortho and para positions. Statement II is correct. Acetophenone contains the aromatic ketone group: \[ \mathrm{C_6H_5COCH_3} \] The benzene ring is directly attached to the carbonyl carbon. The carbonyl-containing substituent withdraws electron density from the aromatic ring. It therefore deactivates the ring and ordinarily favours meta substitution. Statement III is incorrect. A directing group determines relative positional preference among possible substitution pathways. It does not guarantee that every proposed electrophilic substitution is experimentally feasible. Strongly deactivating carbonyl-containing substituents can make ordinary Friedel–Crafts alkylation unsuitable. The fact that such a substituent is meta-directing therefore cannot be used to declare a successful or facile meta alkylation. Statement IV is correct. The ethyl group of P is an activating ortho/para director. The acetyl group of Q is a deactivating meta director. Under suitable nitration conditions, their predominant substitution positions consequently differ. \(\textbf{Identify the key synthetic distinction:}\) P and Q both contain two carbon atoms outside the benzene ring. However, one is an alkylbenzene and the other an aromatic ketone. The different electronic effects arise from their bonding and functional groups, not simply from substituent carbon count. A valid reaction prediction must consider both relative position preference and overall ring reactivity. Thus, Statements I, II and IV are correct.

Explanation: \(\textbf{Assign the original composition:}\) Let the amounts of benzene and cyclohexa-1,4-diene both be \(x\). Let the cyclohexene amount be \(y\). The total molecular amount gives: \[ 2x+y=0.50 \] \(\textbf{Determine hydrogen equivalents:}\) Complete hydrogenation of benzene requires three hydrogen molecules per molecule. Cyclohexa-1,4-diene contains two ordinary carbon–carbon double bonds and requires two. Cyclohexene contains one double bond and requires one. The measured hydrogen uptake therefore gives: \[ 3x+2x+y=1.00 \] \[ 5x+y=1.00 \] \(\textbf{Solve the two material balances:}\) Subtracting the total-molecule equation: \[ (5x+y)-(2x+y)=1.00-0.50 \] \[ 3x=0.50 \] \[ x=\frac16\,\mathrm{mol} \] Then: \[ y=0.50-2\left(\frac16\right) \] \[ y=\frac16\,\mathrm{mol} \] Thus, the three components are present in equal amounts. \(\textbf{Calculate the bromine requirement:}\) The stated bromination conditions exclude ordinary addition to benzene. Cyclohexa-1,4-diene consumes two bromine equivalents. Cyclohexene consumes one. Therefore: \[ n(\mathrm{Br_2})=2x+y \] \[ n(\mathrm{Br_2})=2\left(\frac16\right)+\frac16 \] \[ n(\mathrm{Br_2})=\frac12\,\mathrm{mol} \] \(\textbf{Calculate the final saturated product:}\) Every original molecule contains a six-membered carbon ring. The stipulated complete hydrogenation converts each into one cyclohexane molecule. Thus: \[ n(\mathrm{cyclohexane})=0.50\,\mathrm{mol} \] \(\textbf{Check the combined conclusion:}\) The computed bromine consumption is \(0.50\,\mathrm{mol}\), not \(0.75\,\mathrm{mol}\). The initial hydrogen uptake and equal-amount condition fix all three starting quantities uniquely. Therefore, option A matches both the calculated bromine uptake and the total cyclohexane formed. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Br_2})=0.50\,\mathrm{mol};\quad n(\mathrm{cyclohexane})=0.50\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the intermediate controlling orientation:}\) Electrophilic aromatic substitution proceeds through an arenium-ion intermediate. The intermediate is formed when the aromatic \(\pi\) system bonds to the incoming electrophile. Its positive charge is delocalised over particular ring positions through resonance. The relative energies of those contributing arrangements affect the preferred substitution pathway. \(\textbf{Examine ortho and para attack:}\) For an existing substituent at C1, the supplied ortho and para arenium-ion patterns include a resonance contributor with positive charge at C1. This is the carbon directly bonded to the original substituent. The electronic influence of that substituent therefore matters strongly for the stability of this contributor. \(\textbf{Apply the methyl-group effect:}\) A methyl group donates electron density towards the aromatic system through its ordinary alkyl-group effects. It can help stabilise positively charged sigma-complex arrangements involving the substituted ring carbon. This favours ortho and para attack relative to the corresponding meta pathway. Methylbenzene is therefore activating and ortho/para-directing in the usual electrophilic-substitution comparison. \(\textbf{Apply the nitro-group effect:}\) A nitro group withdraws electron density strongly from the aromatic system. A resonance contributor placing positive charge directly on the carbon bearing this electron-withdrawing group is especially unfavourable. Ortho and para attack both generate such an arrangement. Meta attack avoids that particular contributor. Consequently, nitrobenzene is deactivated overall but favours meta attack relative to its other available positions. \(\textbf{Distinguish electron donation from resonance counting:}\) The number of resonance contributors alone cannot establish their relative stabilising influence. Two mechanisms may admit the same count of formal drawings while the electronic energies of those drawings differ. The nature of the existing substituent determines whether a particular positive-charge arrangement is stabilised or destabilised. \(\textbf{Interpret the contrasting directions:}\) The methyl and nitro groups act on the same general arenium-ion framework but influence its charged contributors in opposite ways. The meta preference of nitrobenzene does not indicate that its ring is activated at meta positions relative to benzene. It means that meta attack is less disfavoured among the competing routes on an already deactivated ring. The differing orientations therefore arise from electronic stabilisation, not merely from the geometric distance between substituents.

Explanation: Statement I is correct. Chlorobenzene contains one chlorine substituent on a benzene ring. A reflection plane passing through the chlorine-bearing carbon and the opposite ring carbon interchanges positions C2 and C6. Therefore, bonds C1–C2 and C1–C6 are symmetry-equivalent. Statement II is correct. The same reflection operation interchanges bonds C2–C3 and C5–C6. It also interchanges bonds C3–C4 and C4–C5. Consequently, the six ring edges form three symmetry classes under the available molecular symmetry. This is different from unsubstituted benzene, where the six ring bonds are mutually equivalent. Statement III is incorrect. Aromaticity does not require every aromatic molecule to possess precisely the same complete symmetry as benzene. It requires suitable cyclic conjugation, p-orbital overlap and the appropriate \(\pi\)-electron arrangement. Chlorine substitution changes the symmetry of the molecule but does not remove the six-membered conjugated ring framework. Chlorobenzene remains aromatic. Statement IV is correct. Symmetry establishes which bonds are required to be equivalent by the molecule's structural operations. It does not establish that bonds belonging to different symmetry classes must necessarily have substantially different observed lengths. Nor does reduced symmetry imply that aromatic \(\pi\)-electron delocalisation has disappeared. The chlorine substituent may perturb the electronic distribution, but the benzene-derived aromatic system remains. \(\textbf{Interpret the distinction:}\) Equal bond lengths in benzene are important evidence for its delocalised bonding. However, aromatic compounds can contain nonequivalent ring positions or bonds when substituents or fused-ring structures reduce molecular symmetry. The full symmetry of benzene and the existence of aromaticity must not be treated as interchangeable requirements. The correct statements are I, II and IV.

Explanation: \(\textbf{Identify the electrophilic acyl group:}\) Acetyl chloride is an acyl chloride, not an alkyl chloride. In the presence of anhydrous aluminium chloride, it supplies the corresponding acetyl-derived acylium electrophile. The relevant group contains the carbonyl oxygen originally present in acetyl chloride. The isotope label makes it possible to track this oxygen atom through the reaction. \(\textbf{Determine the carbon–carbon bond formed:}\) Benzene attacks the acyl electrophile at its carbonyl carbon. Following arenium-ion formation, proton removal and suitable work-up, the organic product is acetophenone. Its labelled structure is: \[ \mathrm{C_6H_5C(=^{18}O)CH_3} \] The carbonyl oxygen is retained under the stipulated no-exchange conditions. The reaction does not reduce the acyl group to ethyl or convert the ketone into an alcohol. \(\textbf{Write the labelled net reaction:}\) \[ \mathrm{C_6H_6+CH_3C(=^{18}O)Cl\longrightarrow C_6H_5C(=^{18}O)CH_3+HCl} \] The net organic stoichiometry is one acetophenone molecule per benzene molecule. \(\textbf{Calculate the labelled product molar mass:}\) Acetophenone contains eight carbon atoms and eight hydrogen atoms. The labelled oxygen contributes eighteen mass units. Therefore: \[ M=8(12)+8(1)+18 \] \[ M=96+8+18 \] \[ M=122\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine product amount and mass:}\) Quantitative monosubstitution gives: \[ n(\mathrm{acetophenone})=0.250\,\mathrm{mol} \] Thus: \[ m=0.250(122) \] \[ m=30.5\,\mathrm{g} \] \(\textbf{Calculate hydrogen chloride formation:}\) The net reaction produces one HCl molecule for every acetophenone molecule. Therefore: \[ n(\mathrm{HCl})=0.250\,\mathrm{mol} \] \(\textbf{Check isotope and functional-group conservation:}\) The labelled oxygen starts bonded to the acetyl carbonyl carbon. It remains bonded to that same carbon in the aromatic ketone. The carbon–chlorine bond is replaced by a bond between the acyl carbon and the benzene ring. The HCl by-product reflects loss of an aromatic ring hydrogen during electrophilic substitution. The product identity therefore distinguishes acylation from alkylation while the mass calculation independently tracks the oxygen isotope. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_6H_5C(=^{18}O)CH_3};\quad m=30.5\,\mathrm{g};\quad n(\mathrm{HCl})=0.250\,\mathrm{mol}} \]

Explanation: \(\textbf{Compare the original carbon frameworks:}\) Benzene contains a six-membered carbon ring with a delocalised aromatic \(\pi\)-electron system. Hexa-1,3,5-triene contains an open chain of six carbon atoms with three ordinary double bonds. The two compounds therefore have different molecular formulas and different carbon connectivities. However, each possesses three carbon–carbon \(\pi\)-bond equivalents. \(\textbf{Calculate hydrogen uptake by P:}\) Under suitable vigorous catalytic conditions, benzene undergoes hydrogenation: \[ \mathrm{C_6H_6+3H_2\longrightarrow C_6H_{12}} \] The product is cyclohexane because the original six-membered ring is retained. The required hydrogen amount is: \[ n(\mathrm{H_2})_P=3\,\mathrm{mol} \] \(\textbf{Calculate hydrogen uptake by Q:}\) Hexa-1,3,5-triene possesses three ordinary carbon–carbon double bonds. Each consumes one hydrogen molecule during complete saturation. Therefore: \[ \mathrm{C_6H_8+3H_2\longrightarrow C_6H_{14}} \] The saturated product is hexane. Thus: \[ n(\mathrm{H_2})_Q=3\,\mathrm{mol} \] \(\textbf{Predict benzene's ozonolysis fragments:}\) Under the stipulated triozonide and reductive-cleavage pathway, the six-carbon ring is divided into three two-carbon fragments. Each becomes ethanedial: \[ \mathrm{OHC-CHO} \] Consequently, one benzene molecule produces three ethanedial molecules. \(\textbf{Trace the three double bonds in Q:}\) The triene structure is: \[ \mathrm{C_1H_2=C_2H-C_3H=C_4H-C_5H=C_6H_2} \] Cleavage occurs between C1–C2, C3–C4 and C5–C6. The two terminal carbons C1 and C6 each become methanal: \[ \mathrm{HCHO} \] The connected carbon pairs C2–C3 and C4–C5 each become ethanedial: \[ \mathrm{OHC-CHO} \] Thus, Q gives two methanal and two ethanedial molecules. \(\textbf{Verify carbon conservation:}\) For P: \[ 3(2)=6\ \mathrm{carbon\ atoms} \] For Q: \[ 2(1)+2(2)=6\ \mathrm{carbon\ atoms} \] Both fragmentation patterns preserve all six starting carbon atoms. \(\textbf{Interpret the independent measurements:}\) Equal hydrogen uptake does not establish identical carbon skeletons. The benzene sample retains a ring after hydrogenation, whereas the triene remains open-chain. Reductive ozonolysis then produces different carbonyl-fragment distributions because their original carbon connectivities differ. The distinction demonstrates why hydrogen consumption and bond-cleavage products provide complementary structural evidence.

Explanation: \(\textbf{Identify the original alkene structure:}\) Pent-1-ene has the structure: \[ \mathrm{CH_2=CHCH_2CH_2CH_3} \] Its double bond is terminal. The two double-bonded carbon atoms are constitutionally different, so the addition orientation affects the position of the incoming hydroxyl or bromine group. The resulting product must then be checked independently for a stereogenic carbon. \(\textbf{Route P — Apply acid-catalysed hydration:}\) The principal ordinary hydration orientation places the hydroxyl group at carbon 2. The product is: \[ \mathrm{CH_3CH(OH)CH_2CH_2CH_3} \] This is pentan-2-ol. Its hydroxyl-bearing carbon is attached to hydrogen, hydroxyl, methyl and propyl groups. These are four different substituents. Therefore, carbon 2 is stereogenic. Under the specified achiral conditions and ordinary planar carbocation pathway, both enantiomeric configurations are formed without an imposed enantioselective preference. The resulting racemic mixture has no net optical rotation. \(\textbf{Route Q — Apply ordinary HBr addition:}\) In the absence of peroxide, the preferred electrophilic-addition orientation again places the incoming heteroatom at carbon 2. The product is: \[ \mathrm{CH_3CH(Br)CH_2CH_2CH_3} \] This is 2-bromopentane. The bromine-bearing carbon has four different substituents. It is stereogenic and can exist as two enantiomers. The stipulated achiral conditions give a racemic product mixture through the ordinary non-stereoselective pathway. \(\textbf{Route R — Apply the peroxide effect:}\) Peroxide-assisted HBr addition proceeds through a radical-chain pathway. The orientation favours bromine addition to the terminal carbon because that produces the more stable secondary carbon radical. Hydrogen abstraction from HBr then gives: \[ \mathrm{BrCH_2CH_2CH_2CH_2CH_3} \] This is 1-bromopentane. Its bromine-bearing carbon is a methylene carbon with two identical hydrogen atoms. The molecule contains no stereogenic carbon introduced by this reaction. \(\textbf{Distinguish molecular chirality from measured optical rotation:}\) Pentan-2-ol and 2-bromopentane each have a stereogenic carbon. However, a racemic mixture of either compound is optically inactive overall because the rotations of the two enantiomers cancel. 1-Bromopentane is itself achiral. The product comparisons therefore require both regiochemical reasoning and a separate stereochemical analysis. The presence of a halogen or hydroxyl group alone does not determine optical activity.

Explanation: \(\textbf{Identify the first-stage bond-order change:}\) A carbon–carbon triple bond can add one HBr molecule and become a carbon–carbon double bond. The product is a vinyl bromide. Whether that vinyl bromide exhibits geometrical isomerism depends on the substituents attached to its two double-bonded carbons. The original alkyne's linear geometry does not automatically determine the geometrical possibilities of the product alkene. \(\textbf{Add one HBr equivalent to ethyne:}\) Ethyne has the structure: \[ \mathrm{HC\equiv CH} \] One HBr equivalent produces: \[ \mathrm{CH_2=CHBr} \] The product is bromoethene. One double-bonded carbon carries two identical hydrogen atoms. Therefore, bromoethene cannot display ordinary geometrical isomerism. \(\textbf{Add the second HBr equivalent to bromoethene:}\) Under the stated ordinary electrophilic addition conditions, the second bromine attaches to the carbon already bearing bromine. The product is: \[ \mathrm{CH_3CHBr_2} \] This is 1,1-dibromoethane. Its two bromine atoms are attached to the same carbon atom, making it a geminal dibromide. \(\textbf{Add one HBr equivalent to but-2-yne:}\) But-2-yne has the structure: \[ \mathrm{CH_3C\equiv CCH_3} \] The two triply bonded carbons are constitutionally equivalent before addition. One HBr equivalent produces 2-bromobut-2-ene, represented by: \[ \mathrm{CH_3C(Br)=CHCH_3} \] The carbon bearing bromine also carries a methyl group. The other double-bonded carbon carries hydrogen and a methyl group. Each therefore possesses two different substituents. Consequently, this vinyl bromide can exhibit geometrical isomerism. \(\textbf{Add the second HBr equivalent:}\) The stipulated ordinary second addition places another bromine at the already brominated internal carbon. The final product is: \[ \mathrm{CH_3CBr_2CH_2CH_3} \] This is 2,2-dibromobutane. Both bromine atoms are geminal. The original carbon–carbon double bond has disappeared, so the earlier alkene geometrical distinction does not remain in the saturated product. \(\textbf{Interpret the stage-dependent stereochemistry:}\) Ethyne produces a terminal vinyl bromide that has no geometrical isomers. But-2-yne produces an internal vinyl bromide capable of existing in different geometrical configurations. Both reactions converge to saturated geminal dibromides after the second HBr addition. The distinction depends on the substituents present after the first addition rather than solely on whether the starting alkyne was symmetrical.

Explanation: \(\textbf{Identify the two Kolbe radical precursors:}\) Sodium ethanoate contains the ethanoate ion, \(\mathrm{CH_3COO^-}\). Anodic oxidation followed by decarboxylation gives a methyl radical: \[ \mathrm{CH_3COO^-\longrightarrow CH_3^\bullet+CO_2+e^-} \] Sodium propanoate similarly produces an ethyl radical: \[ \mathrm{C_2H_5COO^-\longrightarrow C_2H_5^\bullet+CO_2+e^-} \] Each carboxylate ion gives one organic radical and one carbon dioxide molecule. \(\textbf{Determine the radical amounts:}\) The total carboxylate amount undergoing decarboxylation is: \[ n_{\mathrm{carboxylate}}=0.400\,\mathrm{mol} \] The two salts contribute equally. Therefore: \[ n(\mathrm{CH_3^\bullet})=n(\mathrm{C_2H_5^\bullet})=\frac{0.400}{2} \] \[ n(\mathrm{CH_3^\bullet})=n(\mathrm{C_2H_5^\bullet})=0.200\,\mathrm{mol} \] \(\textbf{Identify the three possible coupling products:}\) Two methyl radicals combine to produce ethane: \[ \mathrm{2CH_3^\bullet\longrightarrow C_2H_6} \] One methyl radical and one ethyl radical combine to produce propane: \[ \mathrm{CH_3^\bullet+C_2H_5^\bullet\longrightarrow C_3H_8} \] Two ethyl radicals combine to produce butane: \[ \mathrm{2C_2H_5^\bullet\longrightarrow C_4H_{10}} \] \(\textbf{Apply the stipulated random-combination model:}\) Each radical type has mole fraction one-half. The probabilities of methyl–methyl, mixed methyl–ethyl and ethyl–ethyl combinations are proportional to: \[ \left(\frac12\right)^2:2\left(\frac12\right)\left(\frac12\right):\left(\frac12\right)^2 \] \[ =\frac14:\frac12:\frac14 \] \[ =1:2:1 \] The factor two for the mixed product accounts for the two possible orders of selecting unlike radicals. \(\textbf{Determine the total number of coupling products:}\) Every coupling event consumes two radicals. Thus: \[ n(\mathrm{alkanes})_{\mathrm{total}}=\frac{0.400}{2} \] \[ n(\mathrm{alkanes})_{\mathrm{total}}=0.200\,\mathrm{mol} \] \(\textbf{Divide the products in the ratio 1:2:1:}\) \[ n(\mathrm{ethane})=\frac14(0.200)=0.050\,\mathrm{mol} \] \[ n(\mathrm{propane})=\frac24(0.200)=0.100\,\mathrm{mol} \] \[ n(\mathrm{butane})=\frac14(0.200)=0.050\,\mathrm{mol} \] \(\textbf{Calculate carbon dioxide independently:}\) Each decarboxylated carboxylate releases one carbon dioxide molecule. Therefore: \[ n(\mathrm{CO_2})=0.400\,\mathrm{mol} \] \(\textbf{Check radical-fragment conservation:}\) The calculated products consume methyl-radical equivalents: \[ 2(0.050)+0.100=0.200\,\mathrm{mol} \] They consume ethyl-radical equivalents: \[ 0.100+2(0.050)=0.200\,\mathrm{mol} \] Both balances agree with the original radical amounts. \(\textbf{Interpret the product distribution:}\) Electrolysis of a mixture of two carboxylates can produce symmetrical and mixed radical-coupling products. An equimolar mixture does not necessarily yield only the mixed hydrocarbon. The numerical \(1:2:1\) distribution follows from the explicitly stipulated random-coupling model; it is not a universal experimental product ratio for every mixed Kolbe electrolysis. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{C_2H_6})&=0.050\,\mathrm{mol}\\ n(\mathrm{C_3H_8})&=0.100\,\mathrm{mol}\\ n(\mathrm{C_4H_{10}})&=0.050\,\mathrm{mol}\\ n(\mathrm{CO_2})&=0.400\,\mathrm{mol} \end{aligned}} \]

Explanation: Both reactions begin with the same sodium carboxylate. Its organic anion contains four carbon atoms, one of which belongs to the carboxylate group. The product carbon count depends on whether the remaining alkyl fragment receives hydrogen or combines with another such fragment. \(\textbf{Follow soda-lime decarboxylation:}\) The net reaction for a sodium carboxylate is: \[ \mathrm{RCOONa+NaOH\xrightarrow{CaO,\Delta}RH+Na_2CO_3} \] For sodium 2-methylpropanoate, the organic fragment \(\mathrm{R}\) is isopropyl, \(\mathrm{(CH_3)_2CH-}\). Replacement of the carboxylate group by hydrogen gives: \[ \mathrm{(CH_3)_2CH-H} \] Written with ordinary carbon connectivity, this is: \[ \mathrm{CH_3CH_2CH_3} \] The product is propane. One carbon atom from the original carboxylate enters the inorganic carbonate. \(\textbf{Identify the radical formed during Kolbe electrolysis:}\) At the anode, the carboxylate ion loses an electron and carbon dioxide. The remaining organic fragment becomes an isopropyl radical: \[ \mathrm{(CH_3)_2CHCOO^-\longrightarrow (CH_3)_2CH^\bullet+CO_2+e^-} \] The radical centre is the carbon previously attached to the carboxylate group. Each radical contains three carbon atoms. \(\textbf{Combine the radicals:}\) The stipulated principal termination reaction is: \[ \mathrm{2(CH_3)_2CH^\bullet\longrightarrow (CH_3)_2CH-CH(CH_3)_2} \] The coupled product contains six carbon atoms. The two radical centres become directly connected by a new carbon–carbon single bond. The resulting structure is: \[ \mathrm{CH_3CH(CH_3)CH(CH_3)CH_3} \] Its systematic name is 2,3-dimethylbutane. \(\textbf{Check why the product is not 2,2-dimethylbutane:}\) Each radical centre initially carries two methyl groups and one hydrogen. Coupling forms a bond between those two radical centres. Both central carbons consequently retain one hydrogen and two methyl substituents. Neither becomes a quaternary carbon. Thus, the connectivity corresponds to 2,3-dimethylbutane rather than 2,2-dimethylbutane. \(\textbf{Compare the two carbon balances:}\) Soda-lime decarboxylation converts each four-carbon carboxylate into one three-carbon hydrocarbon. Kolbe electrolysis first produces three-carbon radicals, then joins two of them to make one six-carbon hydrocarbon. The reactions share a decarboxylation feature but differ in their carbon–carbon bond-forming outcome. This distinction is essential when selecting a preparation method for a particular alkane skeleton.

Explanation: \(\textbf{Identify the ambiguity of the molecular formula:}\) An acyclic monoalkene follows the general formula \(\mathrm{C}_{n}\mathrm{H}_{2n}\). A saturated monocyclic hydrocarbon can have the same molecular formula. Thus, formula \(\mathrm{C_6H_{12}}\) does not uniquely establish the presence of a double bond. P contains a C=C bond. Q and R are saturated rings with no carbon–carbon multiple bonds. \(\textbf{Apply the selective hydrogenation condition:}\) Hex-1-ene consumes one hydrogen molecule per molecule: \[ \mathrm{CH_2=CH(CH_2)_3CH_3+H_2\longrightarrow CH_3(CH_2)_4CH_3} \] Its product is hexane. Cyclohexane and methylcyclopentane remain unchanged because the specified reaction does not open their rings. Therefore, only P contributes to hydrogen uptake. \(\textbf{Determine P's original amount:}\) Let the amounts of P, Q and R be \(x\), \(y\) and \(z\). The hydrogen consumption gives: \[ x=0.12\,\mathrm{mol} \] \(\textbf{Use the total-molecule balance:}\) \[ x+y+z=0.30 \] Substituting: \[ y+z=0.30-0.12 \] \[ y+z=0.18\,\mathrm{mol} \] \(\textbf{Interpret the combined saturated-product measurement:}\) After hydrogenation, P becomes hexane. Q remains cyclohexane. R remains methylcyclopentane. The measured combined amount of hexane and cyclohexane is: \[ x+y=0.23 \] Therefore: \[ y=0.23-0.12 \] \[ y=0.11\,\mathrm{mol} \] \(\textbf{Recover R's original amount:}\) \[ z=0.18-0.11 \] \[ z=0.07\,\mathrm{mol} \] \(\textbf{Verify the two experimental observations:}\) The hydrogen requirement is: \[ n(\mathrm{H_2})=1(0.12)+0(0.11)+0(0.07) \] \[ n(\mathrm{H_2})=0.12\,\mathrm{mol} \] The combined measured saturated products are: \[ n(\mathrm{hexane+cyclohexane})=0.12+0.11 \] \[ n(\mathrm{hexane+cyclohexane})=0.23\,\mathrm{mol} \] Both agree with the supplied results. \(\textbf{Interpret the complementary structural evidence:}\) The hydrogen-uptake measurement distinguishes the alkene from the saturated cyclic hydrocarbons. However, it cannot separately determine the amounts of the two rings. The hydrogenation-product distribution supplies the additional constraint required to distinguish cyclohexane from methylcyclopentane. The example illustrates why equal molecular formulas and equal hydrogen deficiencies do not imply identical reactivity or connectivity. \(\textbf{Final answer:}\) \[ \boxed{n(P)=0.12\,\mathrm{mol};\quad n(Q)=0.11\,\mathrm{mol};\quad n(R)=0.07\,\mathrm{mol}} \]

Explanation: \(\textbf{Select the parent chain:}\) The main chain must include both the carbon–carbon double bond and the carbon–carbon triple bond. The displayed structure contains a five-carbon chain satisfying that requirement. The additional carbon appears as a methyl substituent on carbon 3. \(\textbf{Compare the two possible numbering directions:}\) Numbering from the alkene end gives the double bond locant 1 and the triple bond locant 4. Numbering from the alkyne end gives the triple bond locant 1 and the double bond locant 4. Both directions produce the same overall set of multiple-bond locants, \(\{1,4\}\). When this tie occurs, the double bond receives the lower locant. The appropriate name is therefore 3-methylpent-1-en-4-yne. \(\textbf{Identify the required hydrogen uptake:}\) The molecule contains one carbon–carbon double bond and one carbon–carbon triple bond. Complete saturation of the double bond requires one hydrogen molecule. Complete saturation of the triple bond requires two hydrogen molecules. Thus, each X molecule requires: \[ n(\mathrm{H_2})=1+2 \] \[ n(\mathrm{H_2})=3\,\mathrm{mol\ per\ mol\ X} \] \(\textbf{Determine the saturated carbon skeleton:}\) Hydrogenation preserves the existing carbon–carbon connections under the specified conditions. The resulting condensed structure is: \[ \mathrm{CH_3CH_2CH(CH_3)CH_2CH_3} \] Its preferred name is 3-methylpentane. The methyl substituent remains attached to the central carbon of the five-carbon parent chain. \(\textbf{Evaluate the sodamide reaction:}\) The original triple bond is terminal. The terminal carbon carries one hydrogen directly attached to an \(sp\)-hybridised carbon. This hydrogen can be removed by sodamide: \[ \mathrm{RC\equiv CH+NaNH_2\longrightarrow RC\equiv CNa+NH_3} \] The alkene functionality remains present during the specified acid–base reaction. \(\textbf{Interpret the combined structural clues:}\) The naming decision depends on a multiple-bond numbering tie. The hydrogenation product depends on retaining the carbon skeleton while saturating both types of multiple bond. The sodamide reaction depends only on the presence of the terminal acetylenic hydrogen. These three observations distinguish parent-chain nomenclature, total unsaturation and local C–H acidity without treating them as interchangeable properties.

Explanation: \(\textbf{Reconstruct X from Route P:}\) The alkene ethene contains two carbon atoms. The stated unbranched five-carbon alkane is pentane. Adding their carbon counts gives seven carbon atoms in the original hydrocarbon. The product formulas are: \[ \mathrm{C_2H_4} \] and: \[ \mathrm{C_5H_{12}} \] Their combined hydrogen count is sixteen. Therefore, X has molecular formula \(\mathrm{C_7H_{16}}\). \(\textbf{Confirm the formula from Route Q:}\) Propene has formula \(\mathrm{C_3H_6}\). The unbranched four-carbon alkane is butane, \(\mathrm{C_4H_{10}}\). Together, they again contain seven carbon atoms and sixteen hydrogen atoms. Both cracking pathways are therefore consistent with the same starting formula. \(\textbf{Write the two balanced transformations:}\) Route P: \[ \mathrm{C_7H_{16}\longrightarrow C_2H_4+C_5H_{12}} \] Route Q: \[ \mathrm{C_7H_{16}\longrightarrow C_3H_6+C_4H_{10}} \] The stipulated unbranched hydrocarbon is n-heptane. \(\textbf{Identify the bromine-addition sites:}\) Each cracking pathway produces exactly one molecule of an ordinary monoalkene. The accompanying alkane has no carbon–carbon multiple bond. Under selective addition conditions, one bromine molecule is consumed by each alkene molecule. \(\textbf{Calculate the requirement from one mole of X:}\) For complete conversion through P: \[ n(\mathrm{Br_2})_P=1(1.00) \] \[ n(\mathrm{Br_2})_P=1.00\,\mathrm{mol} \] For complete conversion through Q: \[ n(\mathrm{Br_2})_Q=1(1.00) \] \[ n(\mathrm{Br_2})_Q=1.00\,\mathrm{mol} \] \(\textbf{Interpret the unchanged measurement:}\) The cracking products differ in the locations where the original carbon skeleton has been divided. Nevertheless, the stipulated pathways produce one monoalkene molecule per starting alkane molecule. The number of hydrogenatable or bromine-addition C=C equivalents is therefore unchanged between the two routes. Equal bromine consumption does not establish identical cracking products. The result follows from simultaneous carbon conservation, hydrogen conservation and selective unsaturation counting.

Explanation: \(\textbf{Identify the carbon changes in soda-lime decarboxylation:}\) Sodium ethanoate contains two carbon atoms. Soda-lime decarboxylation removes its carboxylate carbon as part of an inorganic carbonate. The remaining one-carbon organic fragment becomes methane: \[ \mathrm{CH_3COONa+NaOH\xrightarrow{CaO,\Delta}CH_4+Na_2CO_3} \] Thus, P gives methane. \(\textbf{Identify the radical-coupling change in Kolbe electrolysis:}\) Anodic decarboxylation of ethanoate produces methyl radicals. Two methyl radicals combine through a new carbon–carbon sigma bond. Their combination gives ethane: \[ \mathrm{2CH_3^\bullet\longrightarrow CH_3CH_3} \] Thus, Q gives ethane. \(\textbf{Analyse Wurtz coupling:}\) 1-Bromopropane has a three-carbon chain. In the stipulated symmetrical Wurtz reaction, two propyl-derived fragments couple at the original terminal halogen-bearing carbons. \[ \mathrm{2CH_3CH_2CH_2Br+2Na\xrightarrow{dry\ ether}CH_3(CH_2)_4CH_3+2NaBr} \] The resulting six-carbon alkane is hexane. Thus, R gives hexane. \(\textbf{Analyse catalytic hydrogenation:}\) Propene contains a three-carbon skeleton and one carbon–carbon double bond. Complete hydrogenation adds one hydrogen molecule without changing the number or arrangement of carbon atoms: \[ \mathrm{CH_3CH=CH_2+H_2\longrightarrow CH_3CH_2CH_3} \] The product is propane. Thus, S gives propane. \(\textbf{Compare the preparation principles:}\) Soda-lime decarboxylation reduces the number of carbon atoms by one relative to a carboxylate precursor. Kolbe electrolysis first removes the carboxylate carbon and then couples two remaining radical fragments. Wurtz coupling directly joins two alkyl fragments. Catalytic hydrogenation changes carbon–carbon bond order but preserves the carbon skeleton. The correct matching requires applying the distinct carbon-accounting rule for each reaction instead of inferring the product only from the number of carbons in one starting molecule.

Explanation: \(\textbf{Separate combustion stoichiometry from substitution-site equivalence:}\) The two hydrocarbons share molecular formula \(\mathrm{C_7H_{16}}\). They differ in carbon connectivity. Complete combustion depends on elemental composition, whereas the number of constitutional monochlorination products depends on the symmetry and hydrogen-bearing carbon environments. \(\textbf{Establish their common combustion equation:}\) \[ \mathrm{C_7H_{16}+11O_2\longrightarrow7CO_2+8H_2O} \] Both P and Q therefore consume eleven oxygen molecules per hydrocarbon molecule under complete combustion. Branching does not change that stoichiometric coefficient. \(\textbf{Analyse P's carbon skeleton:}\) 2,2,3-Trimethylbutane can be represented as: \[ \mathrm{(CH_3)_3C-CH(CH_3)_2} \] One central carbon is quaternary and carries no hydrogen. The adjoining carbon carries one hydrogen and is tertiary. The three methyl groups attached to the quaternary carbon are constitutionally equivalent. The two methyl groups attached to the tertiary carbon are also equivalent to each other, but different from the first group. \(\textbf{Count P's substitution environments:}\) Chlorination at the equivalent methyl groups attached to the quaternary carbon gives one constitutional product. Chlorination at the equivalent methyl groups attached to the tertiary carbon gives another. Chlorination at the single tertiary hydrogen gives a third. Thus, P actually has three constitutional monochlorination products. \(\textbf{Analyse Q's carbon skeleton:}\) 2,3-Dimethylpentane has the structure: \[ \mathrm{CH_3CH(CH_3)CH(CH_3)CH_2CH_3} \] Its methyl groups attached to carbon 2 are equivalent. The methyl group attached to carbon 3 is a different environment. The terminal methyl group of the pentane chain is different again. The tertiary hydrogens on carbons 2 and 3 are nonequivalent. The methylene hydrogens on carbon 4 occupy a further constitutional substitution environment. \(\textbf{Count Q's substitution environments:}\) There are three distinct primary carbon environments, two distinct tertiary carbon environments and one secondary carbon environment. Therefore, Q gives six constitutional monochloro products. \(\textbf{Combine the independent predictions:}\) P and Q have identical molecular formulas and therefore identical complete-combustion oxygen requirements. However, the three constitutionally distinct hydrogen environments in P give three monochloro products. The six constitutionally distinct hydrogen environments in Q give six monochloro products. The correct option is B, which combines unchanged combustion stoichiometry with the different substitution-site counts.

Explanation: \(\textbf{Identify the three five-carbon alkane skeletons:}\) Pentane has an unbranched carbon chain. 2-Methylbutane has one carbon branch. 2,2-Dimethylpropane has a compact central quaternary carbon bonded to four methyl groups. All three contain five carbon atoms and twelve hydrogen atoms. They are chain isomers rather than homologues. \(\textbf{Use the boiling-point trend:}\) For alkane isomers of similar molecular mass, increased branching generally reduces intermolecular contact and lowers the boiling point. The most compact structure, 2,2-dimethylpropane, corresponds to the lowest supplied boiling point. The moderately branched 2-methylbutane corresponds to the intermediate value. The unbranched pentane corresponds to the highest value. Therefore: \[ X=\mathrm{2,2\text{-}dimethylpropane} \] \[ Y=\mathrm{2\text{-}methylbutane} \] \[ Z=\mathrm{pentane} \] \(\textbf{Count substitution environments in X:}\) 2,2-Dimethylpropane has four equivalent methyl groups. Its central carbon carries no hydrogen. All twelve hydrogens are constitutionally equivalent. Replacing any one with chlorine therefore produces the same constitutional monochlorination product. X gives one constitutional product. \(\textbf{Count substitution environments in Y:}\) 2-Methylbutane contains a tertiary carbon with one hydrogen. It also contains a secondary methylene carbon. The two methyl groups bonded to the tertiary carbon are equivalent. The methyl group at the opposite end of the chain is constitutionally different from that pair. These environments give four constitutional monochlorination products. \(\textbf{Count substitution environments in Z:}\) Pentane has two equivalent terminal methyl groups. Its two carbons adjacent to the ends are equivalent. The central methylene carbon forms a third environment. Thus, Z gives three constitutional monochlorination products. \(\textbf{Interpret the independent trends:}\) The boiling-point ranking responds to whole-molecule shape and intermolecular interactions. The monochlorination count responds to symmetry and substitution-site connectivity. Greater branching often reduces the boiling point, but it does not establish a universal numerical relation with the number of monochlorination products. Both observations must therefore be interpreted using the actual carbon skeletons. The correct assignment is neopentane, isopentane and n-pentane in increasing boiling-point order, with product counts one, four and three.

Explanation: \(\textbf{Write the anodic oxidation:}\) The ethanoate ion loses one electron and one carbon dioxide molecule to form a methyl radical. \[ \mathrm{CH_3COO^-\longrightarrow CH_3^\bullet+CO_2+e^-} \] Two methyl radicals subsequently couple: \[ \mathrm{2CH_3^\bullet\longrightarrow C_2H_6} \] Combining these steps gives the anodic half-reaction: \[ \mathrm{2CH_3COO^-\longrightarrow C_2H_6+2CO_2+2e^-} \] \(\textbf{Relate electrons to ethanoate consumption:}\) The observed electron amount is: \[ n(e^-)=0.600\,\mathrm{mol} \] One ethanoate ion produces one electron under the specified pathway. Therefore: \[ n(\mathrm{CH_3COO^-})=0.600\,\mathrm{mol} \] \(\textbf{Determine ethane and carbon dioxide:}\) Two ethanoate ions produce one ethane molecule. Hence: \[ n(\mathrm{C_2H_6})=\frac{0.600}{2} \] \[ n(\mathrm{C_2H_6})=0.300\,\mathrm{mol} \] Each ethanoate ion produces one carbon dioxide molecule: \[ n(\mathrm{CO_2})=0.600\,\mathrm{mol} \] \(\textbf{Write the cathodic reduction:}\) Water accepts electrons at the cathode: \[ \mathrm{2H_2O+2e^-\longrightarrow H_2+2OH^-} \] Two electrons produce one hydrogen molecule and two hydroxide ions. Thus: \[ n(\mathrm{H_2})=\frac{0.600}{2} \] \[ n(\mathrm{H_2})=0.300\,\mathrm{mol} \] The hydroxide amount is: \[ n(\mathrm{OH^-})=0.600\,\mathrm{mol} \] \(\textbf{Account for sodium ions:}\) The starting sodium ethanoate provides sodium counterions. The hydroxide ions formed at the cathode correspond to \(0.600\,\mathrm{mol}\) sodium hydroxide in the ideal overall reaction. \[ n(\mathrm{NaOH})=0.600\,\mathrm{mol} \] \(\textbf{Check the overall equation:}\) Combining both electrode processes and their spectator sodium ions gives: \[ \mathrm{2CH_3COONa+2H_2O\longrightarrow C_2H_6+2CO_2+H_2+2NaOH} \] The calculated amounts obey this stoichiometry. \(\textbf{Interpret the distinct gas sources:}\) Ethane results from anodically generated organic radical coupling. Carbon dioxide results from anodic decarboxylation. Hydrogen results from cathodic water reduction. Their amounts cannot be determined correctly by treating every gas as a product of the same elementary reaction. The electron balance links the two electrodes and establishes the complete material balance. \(\textbf{Final answer:}\) \[ \boxed{\begin{aligned} n(\mathrm{C_2H_6})&=0.300\,\mathrm{mol}\\ n(\mathrm{CO_2})&=0.600\,\mathrm{mol}\\ n(\mathrm{H_2})&=0.300\,\mathrm{mol}\\ n(\mathrm{NaOH})&=0.600\,\mathrm{mol} \end{aligned}} \]