Organic Compounds Containing Halogens MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Identify the starting alkyl group:}\) 2-Bromobutane has the structure: \[ \mathrm{CH_3CH(Br)CH_2CH_3} \] Bromine is attached to carbon 2. The corresponding sec-butyl group contains four carbon atoms, and coupling occurs through the carbon that originally bore bromine. \(\textbf{Reaction P — Carbon–carbon coupling:}\) The Wurtz reaction joins two suitable alkyl groups in the presence of sodium metal and dry ether. For the specified idealised coupling, two sec-butyl groups become connected through their original bromine-bearing carbons: \[ \mathrm{2CH_3CH(Br)CH_2CH_3+2Na \xrightarrow{dry\ ether} CH_3CH_2CH(CH_3)CH(CH_3)CH_2CH_3+2NaBr} \] The product contains eight carbon atoms. \(\textbf{Determine the constitutional name:}\) The longest continuous chain contains six carbon atoms. Methyl substituents occupy positions 3 and 4 of this chain. The coupled hydrocarbon is therefore 3,4-dimethylhexane, not n-octane. \(\textbf{Reaction Q — Hydrogen replacement:}\) Lithium aluminium hydride can reduce a suitable alkyl halide by replacing its halogen atom with hydrogen. The carbon skeleton remains unchanged: \[ \mathrm{CH_3CH(Br)CH_2CH_3\xrightarrow{LiAlH_4}CH_3CH_2CH_2CH_3} \] The product is butane. \(\textbf{Compare the carbon-count changes:}\) Ideal Wurtz coupling joins two four-carbon groups to form an eight-carbon product. Reduction converts one four-carbon bromide into a four-carbon hydrocarbon. The location of bromine is especially important in Wurtz coupling because it determines where the two alkyl fragments are joined. \(\textbf{Select the correct structures:}\) Simply doubling the number of carbon atoms does not justify naming the coupled product n-octane. The branching pattern inherited from sec-butyl groups must also be retained. Thus, the intended Wurtz product is 3,4-dimethylhexane, whereas hydride reduction produces butane. Actual Wurtz reactions of secondary halides can have competing pathways, but these are excluded by the specified idealised coupling conditions.

Explanation: \(\textbf{Identify the coupling reaction:}\) Allyl bromide contains an allyl group attached to bromine. Under the specified idealised Wurtz conditions, two allyl groups are coupled through their bromine-bearing carbon atoms: \[ \mathrm{2CH_2{=}CHCH_2Br+2Na \rightarrow CH_2{=}CHCH_2CH_2CH{=}CH_2+2NaBr} \] The coupled hydrocarbon P is hexa-1,5-diene. \(\textbf{Determine the amount of P formed:}\) The balanced coupling equation shows that two moles of allyl bromide form one mole of the diene. Given: \[ n(\mathrm{allyl\ bromide})=0.40\ \mathrm{mol} \] Therefore: \[ n(P)=\frac{0.40}{2} \] \[ n(P)=0.20\ \mathrm{mol} \] \(\textbf{Count the remaining double bonds:}\) Each molecule of hexa-1,5-diene contains two carbon–carbon double bonds. Wurtz coupling forms a new carbon–carbon single bond between the allyl fragments without consuming their terminal double bonds. \(\textbf{Apply bromine-addition stoichiometry:}\) One mole of \(\mathrm{Br_2}\) is consumed per mole of carbon–carbon double bonds undergoing complete addition. Thus: \[ \mathrm{CH_2{=}CHCH_2CH_2CH{=}CH_2+2Br_2 \rightarrow BrCH_2CHBrCH_2CH_2CHBrCH_2Br} \] The bromine requirement is: \[ n(\mathrm{Br_2})=2n(P) \] \[ n(\mathrm{Br_2})=2(0.20) \] \[ n(\mathrm{Br_2})=0.40\ \mathrm{mol} \] \(\textbf{Check the chemical interpretation:}\) The initial coupling halves the number of organic molecules but retains two double bonds in each newly formed molecule. Therefore, using the number of coupled molecules alone would underestimate the bromine requirement. Conversely, treating every allyl bromide molecule as producing one diene molecule would overestimate the amount of coupled product. Both stoichiometric stages must be considered. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Br_2})=0.40\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the starting geometry:}\) Trans-but-2-ene contains a carbon–carbon double bond with its two methyl groups on opposite sides. Its molecular formula is \(\mathrm{C_4H_8}\), and the double bond lies between carbons 2 and 3. \(\textbf{Recognise the bromine-addition pathway:}\) Under ordinary electrophilic bromination conditions, the alkene reacts with molecular bromine through a bromonium-ion intermediate. Bromide attacks from the side opposite the bridging bromine atom. Consequently, the reaction normally shows anti stereoselectivity. \(\textbf{Track the constitutional change:}\) The carbon–carbon double bond becomes a single bond. One bromine atom attaches to carbon 2, and the other attaches to carbon 3: \[ \mathrm{CH_3CH{=}CHCH_3+Br_2 \rightarrow CH_3CHBrCHBrCH_3} \] The constitutional product is 2,3-dibromobutane. \(\textbf{Determine the stereochemical relationship:}\) For trans-but-2-ene, anti addition creates the opposite absolute configurations at the two newly formed stereogenic centres. The resulting molecule has the \((2R,3S)\) configuration, equivalent by symmetry to the \((2S,3R)\) description. \(\textbf{Identify internal symmetry:}\) The resulting molecule possesses an internal plane of symmetry in an appropriate conformation. It is therefore a meso compound rather than one member of an enantiomeric pair. The fact that the molecule contains two stereogenic centres does not automatically make it optically active. \(\textbf{Distinguish the alternative outcome:}\) Anti addition to cis-but-2-ene instead leads to an enantiomeric pair of 2,3-dibromobutane under achiral conditions. The geometrical configuration of the starting alkene therefore determines whether the anti-addition product is meso or racemic. This illustrates the need to integrate alkene geometry, addition stereochemistry and molecular symmetry rather than predict stereochemistry from the product formula alone.

Explanation: \(\textbf{Identify the two chemically different chlorine atoms:}\) One chlorine atom is directly attached to an aromatic \(sp^2\)-hybridised carbon. The other is attached to the saturated carbon of a chloromethyl side chain adjacent to the benzene ring. The two carbon–chlorine bonds therefore differ in geometry, resonance interactions and accessibility to nucleophilic substitution. \(\textbf{First stage — Mild hydroxide substitution:}\) The benzylic carbon is suitable for ordinary nucleophilic substitution. Hydroxide can replace the benzylic chlorine under the specified mild conditions, forming a hydroxymethyl group: \[ \mathrm{ArCH_2Cl+OH^-\rightarrow ArCH_2OH+Cl^-} \] The aromatic chlorine remains attached to the ring during the intended selective first stage. \(\textbf{Examine the aromatic chlorine:}\) An ordinary aryl carbon–chlorine bond is resistant to the usual substitution pathways of haloalkanes. However, the molecule contains a nitro group at the para position relative to the ring-bound chlorine. This electron-withdrawing group can stabilise the negatively charged intermediate associated with an appropriate nucleophilic aromatic substitution pathway. \(\textbf{Second stage — Activated aryl substitution:}\) Under the subsequently imposed more forcing conditions, hydroxide can replace the nitro-activated aromatic chlorine. The oxygen-containing substituent remains attached to the aromatic carbon after chloride elimination and restoration of aromaticity. Acidification gives the corresponding phenolic group. \(\textbf{Integrate the regioselectivity:}\) The first substitution exploits the accessibility of the benzylic \(sp^3\)-hybridised carbon. The second exploits electronic activation of the aromatic carbon bearing chlorine. The same nucleophile can therefore replace different chlorine atoms at different stages because the appropriate reaction conditions differ. \(\textbf{Avoid the principal misconception:}\) The presence of a benzene ring does not make every carbon–chlorine bond in the molecule an aryl bond. Equally, the presence of a nitro group does not make ordinary benzylic substitution impossible. The intended sequence is benzylic substitution first, followed by activated aromatic substitution.

Explanation: \(\textbf{Identify the activated aromatic substrates:}\) Both compounds contain a nitro group para to the aromatic carbon bearing the halogen. A suitably positioned nitro group stabilises the negatively charged addition intermediate formed during nucleophilic aromatic substitution. Therefore, both substrates can undergo the appropriate addition–elimination pathway under suitable conditions. \(\textbf{Identify the nucleophile:}\) Sodium methoxide provides methoxide ions, \(\mathrm{CH_3O^-}\). The oxygen atom attacks the aromatic carbon bearing the halogen. After addition and subsequent elimination of the halide ion, an aromatic carbon–oxygen bond is formed. For P: \[ \mathrm{4\!-\!O_2NC_6H_4F+NaOCH_3 \rightarrow 4\!-\!O_2NC_6H_4OCH_3+NaF} \] For Q: \[ \mathrm{4\!-\!O_2NC_6H_4Cl+NaOCH_3 \rightarrow 4\!-\!O_2NC_6H_4OCH_3+NaCl} \] The organic product in both cases is 4-nitroanisole. \(\textbf{Compare the carbon–halogen bonds:}\) The aryl carbon–fluorine bond is stronger than the corresponding aryl carbon–chlorine bond. If bond cleavage alone controlled the rate, one might expect the chloride to react more rapidly. \(\textbf{Recognise the distinctive aromatic pathway:}\) In the classical activated nucleophilic aromatic addition–elimination mechanism, formation of the anionic addition intermediate is often the rate-determining stage. The strong electron-withdrawing inductive effect of fluorine can favour this addition process. Consequently, activated aryl fluorides can react faster than the corresponding aryl chlorides even though fluorine is a poorer leaving group in ordinary alkyl-halide substitution. \(\textbf{Resolve the mechanistic distinction:}\) Leaving-group trends cannot be transferred indiscriminately from one mechanism to another. Ordinary alkyl \(S_\mathrm{N}2\) reactions often favour iodide and bromide leaving groups, whereas activated aromatic addition–elimination reactions can show the opposite characteristic halogen trend. The decisive comparison therefore involves both the substrate's electronic activation and the mechanism controlling the rate, not carbon–halogen bond strength alone.

Explanation: \(\textbf{Identify the two alkyl fragments:}\) 1-Bromopropane contains the n-propyl group, \(\mathrm{CH_3CH_2CH_2-}\). 2-Bromopropane contains the isopropyl group, \(\mathrm{(CH_3)_2CH-}\). Both fragments contain three carbon atoms, but the carbon through which coupling occurs occupies a different position. \(\textbf{First coupling — n-Propyl with n-propyl:}\) Two n-propyl groups join through their terminal carbon atoms: \[ \mathrm{CH_3CH_2CH_2-CH_2CH_2CH_3} \] The product is n-hexane, containing an unbranched six-carbon chain. \(\textbf{Second coupling — n-Propyl with isopropyl:}\) The terminal carbon of an n-propyl group joins the central carbon of an isopropyl group: \[ \mathrm{CH_3CH_2CH_2CH(CH_3)_2} \] The longest chain contains five carbon atoms, with one methyl substituent at carbon 2. The product is 2-methylpentane. \(\textbf{Third coupling — Isopropyl with isopropyl:}\) Two isopropyl groups join through their central carbon atoms: \[ \mathrm{(CH_3)_2CH-CH(CH_3)_2} \] The longest chain contains four carbon atoms, with methyl groups at carbons 2 and 3. The product is 2,3-dimethylbutane. \(\textbf{Check the number of coupling combinations:}\) The two distinct propyl fragments can combine as n-propyl–n-propyl, n-propyl–isopropyl and isopropyl–isopropyl. Reversing the order of the fragments in a cross-coupled product does not create a new constitutional isomer. \(\textbf{Interpret the synthetic limitation:}\) All three products have the formula \(\mathrm{C_6H_{14}}\), but they differ in their carbon–carbon connectivity. Mixed Wurtz coupling therefore creates an important selectivity problem: identical carbon counts do not guarantee identical products. The three possible constitutional hydrocarbons are n-hexane, 2-methylpentane and 2,3-dimethylbutane.

Explanation: \(\textbf{Identify the proposed alkyl fragment:}\) Tert-butyl bromide has the formula \(\mathrm{(CH_3)_3CBr}\). The carbon bearing bromine is tertiary and attached to three methyl groups. If two tert-butyl fragments could be joined at their original bromine-bearing carbons, the resulting molecule would have the structure: \[ \mathrm{(CH_3)_3C-C(CH_3)_3} \] This structure is 2,2,3,3-tetramethylbutane. \(\textbf{Examine the intended coupling:}\) The formal Wurtz coupling equation is: \[ \mathrm{2(CH_3)_3CBr+2Na \rightarrow (CH_3)_3C-C(CH_3)_3+2NaBr} \] The equation identifies the desired carbon connectivity but does not establish that the reaction is a selective or efficient preparative method. \(\textbf{Consider substrate structure:}\) Tert-butyl bromide is a tertiary alkyl halide. Its reaction centre is highly crowded, and tertiary substrates are generally susceptible to competing elimination under appropriate strongly reactive conditions. The corresponding formation of 2-methylpropene is a significant competing possibility. \(\textbf{Compare with ordinary Wurtz suitability:}\) The Wurtz reaction is most conveniently employed with suitable primary alkyl halides for preparing symmetrical alkanes. Secondary and tertiary halides are less satisfactory when clean coupling is required because side reactions can substantially reduce the desired product yield. \(\textbf{Resolve the structural misconception:}\) The existence of beta-hydrogen atoms must be checked at carbons adjacent to the bromine-bearing carbon. Tert-butyl bromide contains nine beta-hydrogens on its three methyl groups. It is therefore incorrect to argue that elimination is impossible because the molecule is tertiary. \(\textbf{Final assessment:}\) The proposed coupling connectivity is consistent with the target molecule, but chemical feasibility and product selectivity must be evaluated separately. A formally balanced coupling equation does not guarantee a good synthetic route. Tertiary alkyl bromides are not preferred for a reliable, selective Wurtz preparation of this highly branched alkane.

Explanation: \(\textbf{Identify the reaction restriction:}\) The specified hydroxide treatment replaces only one bromine atom in each dibromide. Consequently, each product retains one carbon–bromine bond and acquires one carbon–oxygen bond. The number of constitutional products depends on whether substitution at different bromine-bearing positions produces distinct structures. \(\textbf{Analyse compound P:}\) P is 1,3-dibromopropane: \[ \mathrm{BrCH_2CH_2CH_2Br} \] The two bromine-bearing terminal carbons are related by molecular symmetry. Replacing bromine at either end gives the same constitutional molecule: \[ \mathrm{HOCH_2CH_2CH_2Br} \] This compound is 3-bromopropan-1-ol. Therefore, only one constitutional monohydroxylated product is possible from P. \(\textbf{Analyse compound Q:}\) Q is 1,3-dibromobutane: \[ \mathrm{BrCH_2CH_2CH(Br)CH_3} \] The bromine-bearing carbons occupy different structural environments. One is primary, while the other is secondary. They are not symmetry-equivalent. \(\textbf{Substitution at the primary carbon:}\) Replacing the terminal bromine gives: \[ \mathrm{HOCH_2CH_2CH(Br)CH_3} \] This is 3-bromobutan-1-ol. \(\textbf{Substitution at the secondary carbon:}\) Replacing the other bromine gives: \[ \mathrm{BrCH_2CH_2CH(OH)CH_3} \] This is 4-bromobutan-2-ol. \(\textbf{Interpret the product possibilities:}\) The two constitutional products from Q need not form in equal amounts. Steric effects and reaction conditions can favour substitution at one carbon over the other. However, both products are constitutionally possible under the specified reaction model. \(\textbf{Integrated conclusion:}\) The presence of two identical leaving groups does not automatically imply the formation of two constitutional products. Their structural equivalence must first be examined. P contains symmetry-equivalent bromine-bearing carbons, whereas Q contains nonequivalent primary and secondary bromine-bearing carbons. Hence, P gives one possible constitutional monohydroxylated product, while Q can give two.

Explanation: \(\textbf{Identify the starting functional groups:}\) 1,3-Dibromopropane has the structure: \[ \mathrm{BrCH_2CH_2CH_2Br} \] It contains three carbon atoms and two bromine atoms attached to terminal saturated carbons. Both bromine atoms are available for the specified complete nucleophilic substitution. \(\textbf{First stage — Double cyanide substitution:}\) Potassium cyanide supplies cyanide ions, which attack suitable primary alkyl bromides predominantly through carbon. Each replacement introduces one nitrile carbon into the organic skeleton. The balanced reaction is: \[ \mathrm{BrCH_2CH_2CH_2Br+2KCN \rightarrow NCCH_2CH_2CH_2CN+2KBr} \] The product is pentanedinitrile, containing five carbon atoms. \(\textbf{Determine the KCN requirement:}\) One mole of 1,3-dibromopropane requires two moles of KCN for complete double substitution. Given: \[ n(\mathrm{dibromide})=0.100\ \mathrm{mol} \] Therefore: \[ n(\mathrm{KCN})=2n(\mathrm{dibromide}) \] \[ n(\mathrm{KCN})=2(0.100) \] \[ n(\mathrm{KCN})=0.200\ \mathrm{mol} \] \(\textbf{Second stage — Hydrolysis of both nitrile groups:}\) Each nitrile group can be hydrolysed into a carboxylic acid group under suitable acidic conditions. Complete hydrolysis of both nitrile groups converts the dinitrile into a dicarboxylic acid: \[ \mathrm{NCCH_2CH_2CH_2CN+4H_2O+2H^+ \rightarrow HOOCCH_2CH_2CH_2COOH+2NH_4^+} \] The product is pentanedioic acid, also known as glutaric acid. \(\textbf{Verify carbon conservation and extension:}\) The original three carbon atoms remain in the central chain. Each cyanide group introduces one additional carbon, producing five carbon atoms in the dinitrile. During hydrolysis, the two nitrile carbons become the two carboxyl carbons. No carbon atom is lost. \(\textbf{Check the stoichiometric distinction:}\) The amount of KCN required is twice the amount of starting dibromide because two substitution events occur per molecule. However, the amount of final dicarboxylic acid under complete conversion remains \(0.100\ \mathrm{mol}\), because one molecule of starting dibromide produces one molecule of the final dicarboxylic acid. \(\textbf{Final answer:}\) \[ \boxed{\text{Pentanedioic acid};\quad n(\mathrm{KCN})=0.200\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the common reaction feature:}\) Both reactions involve electrophilic substitution on the aromatic ring of chlorobenzene. The original chlorine atom is already bonded directly to an aromatic carbon. Under ordinary electrophilic substitution conditions, this carbon–chlorine bond is retained while a ring hydrogen is replaced. \(\textbf{Reaction P — Aromatic sulphonation:}\) Fuming sulphuric acid provides the sulphonating species required for introducing the \(\mathrm{-SO_3H}\) group into an aromatic ring. The transformation can be represented as: \[ \mathrm{C_6H_5Cl\xrightarrow{fuming\ H_2SO_4}ClC_6H_4SO_3H} \] Chlorine is an ortho/para-directing substituent. Consequently, sulphonation occurs preferentially at positions 2 and 4 relative to chlorine. The products are therefore 2-chlorobenzenesulphonic acid and 4-chlorobenzenesulphonic acid, with the less sterically crowded para product commonly favoured. \(\textbf{Reaction Q — Aromatic bromination:}\) Bromine reacts with chlorobenzene in the presence of a suitable Lewis-acid catalyst through electrophilic substitution. \[ \mathrm{C_6H_5Cl+Br_2\xrightarrow{FeBr_3}BrC_6H_4Cl+HBr} \] The chlorine substituent again directs the incoming electrophile towards the ortho and para positions. Thus, the principal products are 1-bromo-2-chlorobenzene and 1-bromo-4-chlorobenzene. \(\textbf{Resolve the selectivity distinction:}\) Chlorine withdraws electrons through its inductive effect, causing overall deactivation relative to benzene. However, resonance donation makes ortho and para attack comparatively favourable. The same directing principle therefore operates in both reactions despite the different electrophiles. Neither reaction ordinarily requires displacement of the original aromatic chlorine atom. The correct conclusion is that both reactions give principally ortho/para substitution products while retaining the original carbon–chlorine bond.

Explanation: \(\textbf{Identify the principle controlling the second substitution:}\) The position of the second incoming substituent depends on the directing influence of the group already present on the aromatic ring. Therefore, changing the order of two electrophilic substitution reactions can change the principal constitutional products. \(\textbf{Route I — Chlorination followed by nitration:}\) The first reaction introduces chlorine into benzene: \[ \mathrm{C_6H_6+Cl_2\xrightarrow{FeCl_3}C_6H_5Cl+HCl} \] Chlorobenzene is then subjected to nitration. Although chlorine deactivates the aromatic ring through its inductive effect, resonance donation favours electrophilic attack at the ortho and para positions. Consequently, the second reaction gives principally 2-nitrochlorobenzene and 4-nitrochlorobenzene. \(\textbf{Route II — Nitration followed by chlorination:}\) The first reaction gives nitrobenzene: \[ \mathrm{C_6H_6\xrightarrow{HNO_3/H_2SO_4}C_6H_5NO_2} \] The nitro group is strongly electron withdrawing through both inductive and resonance effects. It deactivates the aromatic ring and directs subsequent electrophilic substitution towards the meta position. Therefore, chlorination of nitrobenzene principally produces 3-chloronitrobenzene under suitable conditions. \(\textbf{Compare the final structures:}\) Both routes ultimately introduce one chlorine atom and one nitro group into the benzene ring. Their products can have the same molecular formula, but the relative positions of the two substituents differ. This demonstrates that equality of the final functional-group composition does not guarantee equality of the final constitutional structure. \(\textbf{Synthetic inference:}\) When two substituents possess different directing effects, reaction order becomes an important synthetic decision. The distinction is not caused by a change in the identities of chlorine or the nitro group, but by which substituent controls the second electrophilic attack. Thus, Route I favours ortho/para substitution, whereas Route II favours meta substitution.

Explanation: \(\textbf{Stage 1 — Grignard-reagent formation:}\) Bromoethane reacts with magnesium in dry ether: \[ \mathrm{C_2H_5Br+Mg\xrightarrow{dry\ ether}C_2H_5MgBr} \] The molar ratio of bromoethane to magnesium is \(1:1\). Given: \[ n(\mathrm{C_2H_5Br})=0.150\ \mathrm{mol} \] \[ n(\mathrm{Mg})=0.200\ \mathrm{mol} \] Since bromoethane is present in the smaller stoichiometrically equivalent amount, it limits Grignard-reagent formation. Therefore: \[ n(\mathrm{C_2H_5MgBr})=0.150\ \mathrm{mol} \] Magnesium remaining unconsumed: \[ n(\mathrm{Mg})_{\mathrm{remaining}}=0.200-0.150 \] \[ n(\mathrm{Mg})_{\mathrm{remaining}}=0.050\ \mathrm{mol} \] \(\textbf{Stage 2 — Carbon dioxide incorporation:}\) The Grignard reagent reacts with carbon dioxide to form a magnesium carboxylate: \[ \mathrm{C_2H_5MgBr+CO_2\rightarrow C_2H_5COOMgBr} \] The stoichiometric ratio is: \[ n(\mathrm{Grignard}):n(\mathrm{CO_2})=1:1 \] Available quantities: \[ n(\mathrm{Grignard})=0.150\ \mathrm{mol} \] \[ n(\mathrm{CO_2})=0.080\ \mathrm{mol} \] Carbon dioxide is therefore the limiting reagent in the carboxylation stage. The theoretical amount of magnesium propanoate formed is: \[ n(\mathrm{C_2H_5COOMgBr})=0.080\ \mathrm{mol} \] \(\textbf{Stage 3 — Acidic hydrolysis:}\) Acidification produces propanoic acid: \[ \mathrm{C_2H_5COOMgBr+H_3O^+\rightarrow C_2H_5COOH+MgBr^++H_2O} \] One mole of the magnesium carboxylate gives one mole of propanoic acid. Hence: \[ n(\mathrm{acid})_{\mathrm{theoretical}}=0.080\ \mathrm{mol} \] \(\textbf{Stage 4 — Theoretical mass:}\) \[ m_{\mathrm{theoretical}}=nM \] \[ m_{\mathrm{theoretical}}=0.080\times74 \] \[ m_{\mathrm{theoretical}}=5.92\ \mathrm{g} \] \(\textbf{Stage 5 — Apply the isolated yield:}\) \[ \%\mathrm{Yield}=\frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}}\times100 \] Therefore: \[ m_{\mathrm{actual}}=\frac{75}{100}\times5.92 \] \[ m_{\mathrm{actual}}=4.44\ \mathrm{g} \] \(\textbf{Chemical consistency check:}\) The original ethyl group contains two carbon atoms. Carbon dioxide introduces one additional carbon atom, giving the three-carbon carboxylic acid. The limiting reagent must be identified again after Grignard formation because the reagent controlling the first stage need not control the second. The \(75\%\) yield applies to the theoretical quantity of propanoic acid, not directly to the initial amount of magnesium. \(\textbf{Final answer:}\) \[ \boxed{m(\mathrm{propanoic\ acid})=4.44\ \mathrm{g}} \]

Explanation: \(\textbf{Identify the structural distinction:}\) Both compounds have the molecular formula \(\mathrm{C_2H_4Br_2}\). However, P is a geminal dibromide because both bromine atoms are attached to the same carbon. Q is a vicinal dibromide because the bromine atoms are attached to adjacent carbon atoms. This difference in halogen arrangement determines the structures obtained after complete hydrolysis. \(\textbf{Hydrolysis of P:}\) P is 1,1-dibromoethane: \[ \mathrm{CH_3CHBr_2} \] Replacement of both bromine atoms by hydroxyl groups formally gives a geminal diol: \[ \mathrm{CH_3CHBr_2+2OH^-\rightarrow CH_3CH(OH)_2+2Br^-} \] The resulting ethane-1,1-diol is not the stable principal isolated form under ordinary conditions. It readily loses water to produce the corresponding carbonyl compound: \[ \mathrm{CH_3CH(OH)_2\rightarrow CH_3CHO+H_2O} \] The stable organic product is ethanal. \(\textbf{Hydrolysis of Q:}\) Q is 1,2-dibromoethane: \[ \mathrm{BrCH_2CH_2Br} \] Replacement of the two bromine atoms by hydroxyl groups gives: \[ \mathrm{BrCH_2CH_2Br+2OH^-\rightarrow HOCH_2CH_2OH+2Br^-} \] The product is ethane-1,2-diol, commonly called ethylene glycol. Its hydroxyl groups are attached to different carbon atoms, so it does not have the same geminal-diol instability. \(\textbf{Check the product identities:}\) Both transformations preserve the two-carbon skeleton. The difference arises from whether the two hydroxyl groups initially appear on one carbon or on neighbouring carbons. The geminal arrangement can lead to a carbonyl compound after water loss, whereas the vicinal arrangement produces a stable diol. Thus, P gives ethanal and Q gives ethane-1,2-diol.

Explanation: \(\textbf{Determine the required carbon-skeleton change:}\) 1-Bromobutane contains four carbon atoms. The desired pentan-1-amine contains five carbon atoms. Therefore, the synthesis must introduce exactly one additional carbon atom while ultimately forming a terminal primary amino group. \(\textbf{First transformation — Carbon-chain extension:}\) Alcoholic potassium cyanide supplies cyanide ions, which predominantly attack suitable primary alkyl halides through carbon. The resulting substitution is: \[ \mathrm{CH_3CH_2CH_2CH_2Br+KCN \rightarrow CH_3CH_2CH_2CH_2CN+KBr} \] The intermediate P is pentanenitrile. The carbon atom of the cyanide group becomes the fifth carbon atom of the organic skeleton. \(\textbf{Second transformation — Reduction of the nitrile:}\) Lithium aluminium hydride can reduce a nitrile group to a primary amine under suitable conditions. The overall organic transformation is: \[ \mathrm{CH_3CH_2CH_2CH_2CN \xrightarrow[\text{hydrolysis}]{LiAlH_4} CH_3CH_2CH_2CH_2CH_2NH_2} \] The nitrile carbon becomes a methylene carbon attached to the amino group. The final compound is pentan-1-amine. \(\textbf{Check the nitrogen connectivity:}\) In the nitrile intermediate, nitrogen is connected to carbon through the nitrile functional group. After reduction, the nitrogen becomes part of a terminal \(\mathrm{-CH_2NH_2}\) group. The product is a primary amine because nitrogen is attached to only one carbon-containing group. \(\textbf{Reject the unsuitable routes:}\) Excess ammonia would directly replace bromine to produce butan-1-amine, retaining only four carbon atoms. Silver cyanide would predominantly give an isocyanide rather than the required nitrile intermediate. Acidic hydrolysis of pentanenitrile would form pentanoic acid, not pentan-1-amine. \(\textbf{Synthetic conclusion:}\) The appropriate route first extends the carbon skeleton through cyanide substitution and then reduces the nitrile to the primary amine. The sequence therefore uses alcoholic KCN followed by lithium aluminium hydride and hydrolysis.

Explanation: \(\textbf{Identify the substitution reactions:}\) Both methyl halides undergo nucleophilic substitution with hydroxide under the specified conditions. For chloromethane: \[ \mathrm{CH_3Cl+NaOH\rightarrow CH_3OH+NaCl} \] For bromomethane: \[ \mathrm{CH_3Br+NaOH\rightarrow CH_3OH+NaBr} \] Each reaction requires one mole of hydroxide per mole of methyl halide. Each mole of methyl halide consumed produces one mole of methanol. \(\textbf{Sample P — Calculate the initial amount:}\) \[ n(\mathrm{CH_3Cl})=\frac{m}{M} \] \[ n(\mathrm{CH_3Cl})=\frac{32.0}{50.5} \] \[ n(\mathrm{CH_3Cl})=0.6337\ \mathrm{mol} \] The available NaOH is: \[ n(\mathrm{NaOH})=0.500\ \mathrm{mol} \] Since the substitution requires a \(1:1\) ratio, NaOH is the limiting reagent for P. Thus: \[ n_P(\mathrm{CH_3OH})=0.500\ \mathrm{mol} \] \(\textbf{Sample Q — Calculate the initial amount:}\) \[ n(\mathrm{CH_3Br})=\frac{32.0}{95.0} \] \[ n(\mathrm{CH_3Br})=0.3368\ \mathrm{mol} \] The available NaOH is greater than the amount of bromomethane. Therefore, bromomethane is the limiting reagent for Q. Thus: \[ n_Q(\mathrm{CH_3OH})=0.3368\ \mathrm{mol} \] \(\textbf{Compare the methanol amounts:}\) \[ \Delta n=n_P-n_Q \] \[ \Delta n=0.500-0.3368 \] \[ \Delta n=0.1632\ \mathrm{mol} \] To three significant figures: \[ \Delta n=0.163\ \mathrm{mol} \] \(\textbf{Interpret the result:}\) Equal masses of two different methyl halides do not contain equal numbers of molecules because their molar masses differ. Moreover, the limiting reagent is different in the two experiments. Chloromethane is in excess relative to NaOH, whereas bromomethane is present in a smaller amount than NaOH. The calculation therefore requires separate mole conversions and separate limiting-reagent decisions. No conclusion about relative intrinsic reaction rates is needed because complete reaction of the limiting reagent has been explicitly assumed. \(\textbf{Final answer:}\) \[ \boxed{\Delta n(\mathrm{CH_3OH})=0.163\ \mathrm{mol}} \]

Explanation: \(\textbf{Identify the starting substitution pattern:}\) 4-Bromotoluene contains bromine and a methyl group at para positions on a benzene ring. The bromine atom is attached directly to an aromatic carbon, while the methyl group is attached through a separate carbon–carbon bond. These two substituents must be tracked independently through the reaction sequence. \(\textbf{First stage — Organomagnesium formation:}\) Under suitable dry-ether conditions, magnesium is inserted into the aromatic carbon–bromine bond: \[ \mathrm{4\!-\!BrC_6H_4CH_3+Mg \xrightarrow{dry\ ether} 4\!-\!CH_3C_6H_4MgBr} \] The intermediate is 4-methylphenylmagnesium bromide. The methyl group remains attached to the aromatic ring. \(\textbf{Second stage — Carbon dioxide incorporation:}\) The carbon-containing group of the Grignard reagent behaves nucleophilically towards carbon dioxide. Formation of a new carbon–carbon bond produces the corresponding magnesium carboxylate: \[ \mathrm{4\!-\!CH_3C_6H_4MgBr+CO_2 \rightarrow 4\!-\!CH_3C_6H_4COOMgBr} \] The carbon introduced from carbon dioxide becomes the carboxyl carbon. \(\textbf{Third stage — Acidification:}\) Acidic hydrolysis protonates the carboxylate: \[ \mathrm{4\!-\!CH_3C_6H_4COOMgBr \xrightarrow{H_3O^+} 4\!-\!CH_3C_6H_4COOH} \] The product is 4-methylbenzoic acid. \(\textbf{Distinguish the possible structural changes:}\) This route does not oxidise the original methyl group. Instead, the carbon dioxide-derived carboxyl group becomes attached at the aromatic carbon originally bearing bromine. The methyl group remains at the para position relative to the newly introduced carboxyl group. \(\textbf{Integrated conclusion:}\) The reaction sequence combines formation of an arylmagnesium halide with carbon dioxide carboxylation. It demonstrates that the position of an aryl halogen can determine the position at which a new carbon-containing functional group is introduced. Thus, the final product is 4-methylbenzoic acid.

Explanation: \(\textbf{Identify the coupling principle:}\) The Wurtz reaction involves formation of carbon–carbon bonds using suitable organic halides and sodium metal in dry ether. When a molecule contains halogen atoms at both ends of a carbon chain, intramolecular coupling can join the terminal carbon atoms and form a ring under suitable conditions. The ring size depends on the number of carbon atoms in the original chain. \(\textbf{Reaction P — Three-carbon chain:}\) 1,3-Dibromopropane has the structure: \[ \mathrm{BrCH_2CH_2CH_2Br} \] Its two bromine-bearing terminal carbons are separated by one methylene carbon. The intended intramolecular coupling joins these terminal carbons through a new carbon–carbon bond: \[ \mathrm{BrCH_2CH_2CH_2Br+2Na \xrightarrow{dry\ ether} C_3H_6+2NaBr} \] The resulting three-carbon ring is cyclopropane. \(\textbf{Reaction Q — Four-carbon chain:}\) 1,4-Dibromobutane has the structure: \[ \mathrm{BrCH_2CH_2CH_2CH_2Br} \] The two terminal carbons are joined under the specified intramolecular coupling conditions: \[ \mathrm{Br(CH_2)_4Br+2Na \xrightarrow{dry\ ether} C_4H_8+2NaBr} \] The product is cyclobutane. \(\textbf{Track the carbon skeletons:}\) Neither reaction removes carbon atoms from the starting chain. Ring formation adds a bond between carbon atoms that were already present in the molecule. Therefore, the number of carbon atoms in each ring equals the number of carbon atoms in its starting dihalide. \(\textbf{Distinguish ring closure from ordinary coupling:}\) Intermolecular coupling would connect carbon chains belonging to different molecules and could lead to higher molecular mass products. The question explicitly specifies conditions favouring intramolecular coupling, so the corresponding ring-closure products are selected. \(\textbf{Chemical conclusion:}\) The three-carbon dihalide gives a three-membered ring, while the four-carbon dihalide gives a four-membered ring. Thus, the products are cyclopropane and cyclobutane, respectively.

Explanation: \(\textbf{Identify the functional arrangement:}\) Both compounds contain chlorine directly attached to a benzene ring. The chlorine atom occupies the para position relative to the alkyl side chain. The difference between P and Q is the structure of the carbon directly attached to the aromatic ring. \(\textbf{Analyse compound P:}\) P is 4-chlorotoluene. Its methyl carbon is benzylic because it is directly attached to the aromatic ring. This carbon contains three hydrogen atoms. Strong oxidation with alkaline potassium permanganate can convert a suitable benzylic alkyl side chain containing a benzylic hydrogen into a carboxyl group. Thus: \[ \mathrm{4\!-\!ClC_6H_4CH_3 \xrightarrow[\text{then }H^+]{KMnO_4/OH^-,\ \Delta} 4\!-\!ClC_6H_4COOH} \] The product is 4-chlorobenzoic acid. \(\textbf{Analyse compound Q:}\) Q contains a tert-butyl group attached directly to the aromatic ring. The benzylic carbon is bonded to the benzene ring and three methyl groups. It therefore possesses no benzylic hydrogen. The ordinary side-chain oxidation that converts suitable alkylbenzenes into aromatic carboxylic acids depends on the presence of at least one benzylic hydrogen. Consequently, Q does not undergo the corresponding ordinary oxidation to 4-chlorobenzoic acid under the stated reaction model. \(\textbf{Track the aromatic chlorine:}\) The carbon–chlorine bond in both molecules remains attached to the aromatic ring during the proposed oxidation. Its position relative to the side chain is retained. Therefore, the acid formed from P contains chlorine at the para position. \(\textbf{Resolve the misconception:}\) The total number of hydrogen atoms in an alkyl group is not the decisive condition. Hydrogen must be present on the carbon directly attached to the aromatic ring for the usual benzylic oxidation. P satisfies this requirement, whereas Q does not. Thus, P gives 4-chlorobenzoic acid, while Q resists the corresponding side-chain oxidation.

Explanation: \(\textbf{First transformation — Alcohol to alkyl bromide:}\) Propan-1-ol contains a three-carbon chain with a terminal hydroxyl group. A suitable brominating reagent replaces this hydroxyl group by bromine: \[ \mathrm{CH_3CH_2CH_2OH\xrightarrow{HBr}CH_3CH_2CH_2Br} \] The resulting compound is 1-bromopropane. \(\textbf{Second transformation — Ester formation:}\) Silver ethanoate supplies an ethanoate group capable of replacing bromide in a suitable primary alkyl halide. The reaction is: \[ \mathrm{CH_3CH_2CH_2Br+CH_3COOAg \rightarrow CH_3COOCH_2CH_2CH_3+AgBr} \] The product P contains the ester functional group \(\mathrm{-COO-}\). Its correct name is propyl ethanoate. \(\textbf{Identify the carbon fragments:}\) The propyl portion of the ester originates from the original propan-1-ol. The ethanoate portion originates from silver ethanoate. The ester therefore contains five carbon atoms in total, but these carbon atoms belong to two distinguishable structural fragments. \(\textbf{Third transformation — Ester hydrolysis:}\) Complete acid-catalysed hydrolysis breaks the ester linkage to regenerate the corresponding carboxylic acid and alcohol: \[ \mathrm{CH_3COOCH_2CH_2CH_3+H_2O \xrightarrow{H^+} CH_3COOH+CH_3CH_2CH_2OH} \] The products are ethanoic acid and propan-1-ol. \(\textbf{Verify the acid–alcohol assignment:}\) The carbonyl-containing portion of the ester determines the carboxylic acid formed during hydrolysis. The alkoxy-containing portion determines the alcohol. Reversing these assignments would produce an incorrect description of the starting ester. \(\textbf{Integrated conclusion:}\) The sequence combines alcohol-to-halide conversion, carboxylate substitution and ester hydrolysis. The key reasoning involves tracking which carbon fragment originates from each reagent and identifying the functional group formed at every stage. Thus, P is propyl ethanoate, and hydrolysis gives ethanoic acid and propan-1-ol.

Explanation: \(\textbf{Identify the alpha carbon:}\) 1-Bromo-1-methylcyclohexane contains bromine and a methyl group attached to the same ring carbon. This ring carbon is the alpha carbon for beta-elimination. The neighbouring ring carbons and the attached methyl carbon can provide beta-hydrogens. \(\textbf{Elimination involving a neighbouring ring carbon:}\) Removal of a beta-hydrogen from either adjacent ring carbon produces a carbon–carbon double bond within the cyclohexane ring. Because the two adjacent ring directions are equivalent in the starting structure, they give the same constitutional alkene. The product is 1-methylcyclohexene. \(\textbf{Determine its substitution level:}\) The endocyclic double bond has three carbon-containing substituent connections across its two alkene carbons. It is therefore a trisubstituted alkene. \(\textbf{Elimination involving the methyl group:}\) A beta-hydrogen can also be removed from the methyl group attached to the bromine-bearing ring carbon. This gives a double bond between the ring carbon and the exocyclic carbon: \[ \mathrm{Ring{-}C(Br)(CH_3)\rightarrow Ring{-}C(=CH_2)} \] The product is methylenecyclohexane. Its exocyclic double bond is disubstituted because the terminal methylene carbon carries no additional carbon substituent. \(\textbf{Apply the major-product tendency:}\) Under the specified ordinary Zaitsev preference, the more substituted alkene is generally favoured. 1-Methylcyclohexene is trisubstituted, whereas methylenecyclohexane is disubstituted. Consequently, 1-methylcyclohexene is the principal constitutional product. \(\textbf{Check the structural inference:}\) The two products differ in whether the double bond is located inside or outside the ring. Both retain the same total number of carbon atoms as the starting bromide. The existence of an exocyclic elimination pathway must not be overlooked merely because the substrate is cyclic. However, the presence of that pathway does not establish that its alkene is the major product. Thus, both P and Q are possible, with P favoured under the stated conditions.