NCERT Solutions · Class 12 Chemistry Haloalkanes and Haloarenes

Ten separate conversions, each one built from the same small toolkit — free-radical halogenation, nucleophilic substitution, elimination (Zaitsev), and the Wurtz/Fittig carbon-carbon coupling — applied in the right order.(i) Ethanol to but-$\displaystyle 1$-yne — a four-carbon terminal alkyne has to be assembled, not found inside a two-carbon starting material, so ethanol is converted into two different pieces that are then joined. One portion of ethanol, \(\displaystyle CH_3CH_2OH \), is turned into the alkylating agent, ethyl bromide: \[CH_3CH_2OH \xrightarrow[\Delta]{PBr_3} CH_3CH_2Br \] A second portion is carried all the way to ethyne by first eliminating once to a double bond, adding bromine across it, then eliminating twice more (excess alcoholic KOH removes both hydrogen halide equivalents): \[CH_3CH_2OH \xrightarrow[\Delta]{HBr} CH_3CH_2Br \xrightarrow[\Delta]{alc.\,KOH} CH_2\!=\!CH_2 \xrightarrow{Br_2} BrCH_2-CH_2Br \xrightarrow[\Delta]{alc.\,KOH\,(excess)} HC\!\equiv\!CH \] Sodium amide, a much stronger base than hydroxide, is needed to pull off the one hydrogen sitting on the sp carbon of ethyne (it is unusually acidic for a C–H bond, but still far weaker an acid than water, so alcoholic KOH cannot deprotonate it). The resulting acetylide is then alkylated with the ethyl bromide from the first portion: \[HC\!\equiv\!CH \xrightarrow{NaNH_2} HC\!\equiv\!C^{-}Na^{+} \xrightarrow{CH_3CH_2Br} HC\!\equiv\!C-CH_2-CH_3 \;+\; NaBr \] The product is but-$\displaystyle 1$-yne, \(\displaystyle HC\!\equiv\!C-CH_2CH_3 \).(ii) Ethane to bromoethene — a vinylic bromide is reached from an alkane by first over-brominating to a $\displaystyle 1,2$-dihalide, then eliminating only one equivalent of HBr. \[CH_3CH_3 \xrightarrow[hv]{Br_2} CH_3CH_2Br \xrightarrow[hv]{Br_2} BrCH_2-CH_2Br \xrightarrow[\Delta]{alc.\,KOH} CH_2\!=\!CHBr \] Stopping the dehydrohalogenation at a single equivalent (rather than the excess used in part (i)) is what leaves one C–Br bond intact instead of driving through to the alkyne. The product is bromoethene (vinyl bromide), \(\displaystyle CH_2\!=\!CHBr \).(iii) Propene to $\displaystyle 1$-nitropropane — the bromine has to land on the terminal carbon (anti-Markovnikov), and the nitro group has to form through nitrogen, not oxygen. Hydrogen bromide adds to propene, \(\displaystyle CH_3-CH\!=\!CH_2 \), in the presence of a peroxide (the Kharasch/free-radical pathway), so the bromine atom — not the proton — attacks first and goes to the less hindered terminal carbon: \[CH_3-CH\!=\!CH_2 \xrightarrow[peroxide]{HBr} CH_3CH_2CH_2Br \] Heating $\displaystyle 1$-bromopropane with silver nitrite (never sodium nitrite) then substitutes the halide: \[CH_3CH_2CH_2Br \xrightarrow{AgNO_2} CH_3CH_2CH_2NO_2 \] The nitrite ion is ambident — it can bond through either O or N. Silver nitrite is covalent, and the more covalent Ag–O bond it already has favours substitution through the nitrogen end, giving the nitroalkane as the major product; the ionic sodium salt would instead favour the alkyl nitrite ester through oxygen. The product is $\displaystyle 1$-nitropropane, \(\displaystyle CH_3CH_2CH_2NO_2 \).(iv) Toluene to benzyl alcohol — the methyl hydrogens, not the ring, are the ones that react, because chlorination here is run as a free-radical (photochemical) reaction rather than an electrophilic aromatic substitution. \[C_6H_5-CH_3 \xrightarrow[hv,\;no\ catalyst]{Cl_2} C_6H_5-CH_2Cl \xrightarrow{aq.\,NaOH} C_6H_5-CH_2OH \] Light with no Lewis-acid catalyst directs chlorine to the weaker benzylic C–H bond instead of the ring; hydroxide then displaces the chloride by nucleophilic substitution. The product is benzyl alcohol, \(\displaystyle C_6H_5CH_2OH \).(v) Propene to propyne — the same vicinal-dihalide-then-double-elimination route as part (i), starting one carbon shorter. \[CH_3-CH\!=\!CH_2 \xrightarrow{Br_2} CH_3-CHBr-CH_2Br \xrightarrow[\Delta]{alc.\,KOH\,(excess)} CH_3-C\!\equiv\!CH \] The first equivalent of base removes one HBr to give a bromopropene; a second equivalent (hence "excess," at higher temperature) removes the second HBr to install the triple bond. The product is propyne, \(\displaystyle CH_3-C\!\equiv\!CH \).(vi) Ethanol to ethyl fluoride — fluorine cannot be installed straight from the alcohol or from HF; it goes in last, by halogen exchange on the already-formed bromide (a Swarts reaction). \[CH_3CH_2OH \xrightarrow[\Delta]{HBr} CH_3CH_2Br \xrightarrow[dry]{AgF} CH_3CH_2F \] Silver fluoride is used, not aqueous HF, because the very strong, insoluble Ag–Br bond that forms is what pulls the halogen-exchange equilibrium over to the alkyl fluoride; direct nucleophilic attack by fluoride on carbon is otherwise too slow to be useful. The product is ethyl fluoride (fluoroethane), \(\displaystyle CH_3CH_2F \).(vii) Bromomethane to propanone — three one-carbon or acid units are stitched together and then decarboxylated as a pair, since methyl bromide itself supplies only one carbon. Bromomethane first becomes a two-carbon nitrile by nucleophilic substitution (cyanide attacks through carbon, not nitrogen, because carbon is the better nucleophilic centre in the cyanide ion for this reaction): \[CH_3Br \xrightarrow{KCN} CH_3-C\!\equiv\!N \] The nitrile is hydrolysed to the carboxylic acid: \[CH_3CN \xrightarrow{H_3O^{+}} CH_3COOH \] Two acid molecules are then combined as the calcium salt and dry-distilled, a classic ketonic decarboxylation that loses one carbon as carbonate and links the other two acyl fragments through a shared carbonyl: \[2\,CH_3COOH \xrightarrow{Ca(OH)_2} (CH_3COO)_2Ca \xrightarrow[dry\ distillation]{\Delta} CH_3-CO-CH_3 \;+\; CaCO_3 \] The product is propanone (acetone), \(\displaystyle CH_3COCH_3 \).(viii) But-$\displaystyle 1$-ene to but-$\displaystyle 2$-ene — this is an isomerisation, done by adding HX to move the halogen inward and then eliminating it back out toward the more substituted, more stable alkene (Zaitsev's rule). \[CH_2\!=\!CH-CH_2CH_3 \xrightarrow{HBr} CH_3-CHBr-CH_2CH_3 \xrightarrow[\Delta]{alc.\,KOH} CH_3-CH\!=\!CH-CH_3 \] Markovnikov addition of HBr puts the bromine on \(\displaystyle \mathrm{C_{2}}\) (the internal carbon); elimination can then only reform a double bond between C2–C3, since \(\displaystyle \mathrm{C_{1}}\) no longer carries the leaving group. Zaitsev's rule (the more substituted, more stable alkene predominates) reinforces the same outcome. The product is but-$\displaystyle 2$-ene, \(\displaystyle CH_3-CH\!=\!CH-CH_3 \).(ix) $\displaystyle 1$-Chlorobutane to n-octane — two identical four-carbon halides are joined end to end by a Wurtz reaction, which is why using a single alkyl halide (not a mixture) gives one clean product. \[2\,CH_3CH_2CH_2CH_2Cl \xrightarrow[dry\ ether]{2\,Na} CH_3CH_2CH_2CH_2-CH_2CH_2CH_2CH_3 \;+\; 2\,NaCl \] Sodium inserts between two molecules of the halide and couples their carbon skeletons; because both halide molecules are the same $\displaystyle 1$-chlorobutane, the eight-carbon chain that results is a single, unbranched product rather than a mixture. The product is n-octane, \(\displaystyle CH_3(CH_2)_6CH_3 \).(x) Benzene to biphenyl — the aryl-aryl analogue of the Wurtz reaction (called the Fittig reaction) couples two aromatic rings once each carries a halogen. Benzene is first brominated on the ring using a Lewis-acid catalyst (electrophilic aromatic substitution): \[C_6H_6 \xrightarrow{Br_2,\;FeBr_3} C_6H_5Br \] Two molecules of bromobenzene are then coupled with sodium in dry ether: \[2\,C_6H_5Br \xrightarrow[dry\ ether]{2\,Na} C_6H_5-C_6H_5 \;+\; 2\,NaBr \] This aryl-aryl coupling is specifically the Fittig reaction; the same coupling between one aryl and one alkyl halide is instead called the Wurtz-Fittig reaction. The product is biphenyl, \(\displaystyle C_6H_5-C_6H_5 \).Answer: (i) but-$\displaystyle 1$-yne, \(\displaystyle HC\equiv C-CH_2CH_3\) — via ethyl bromide + sodium acetylide (from ethanol → ethene → $\displaystyle 1,2$-dibromoethane → ethyne). (ii) bromoethene, \(\displaystyle CH_2=CHBr\) — via double bromination then single dehydrohalogenation. (iii) $\displaystyle 1$-nitropropane, \(\displaystyle CH_3CH_2CH_2NO_2\) — via anti-Markovnikov HBr addition then AgNO2. (iv) benzyl alcohol, \(\displaystyle C_6H_5CH_2OH\) — via side-chain (benzylic) chlorination then hydrolysis. (v) propyne, \(\displaystyle CH_3C\equiv CH\) — via $\displaystyle 1,2$-dibromopropane and double dehydrohalogenation. (vi) ethyl fluoride, \(\displaystyle CH_3CH_2F\) — via ethyl bromide + AgF (Swarts reaction). (vii) propanone, \(\displaystyle CH_3COCH_3\) — via methyl cyanide → acetic acid → calcium acetate → dry distillation. (viii) but-$\displaystyle 2$-ene, \(\displaystyle CH_3CH=CHCH_3\) — via Markovnikov HBr addition then Zaitsev elimination. (ix) n-octane, \(\displaystyle CH_3(CH_2)_6CH_3\) — via the Wurtz reaction. (x) biphenyl, \(\displaystyle C_6H_5-C_6H_5\) — via bromobenzene and the Fittig reaction.

Resonance, hydrogen bonding, and moisture explain three different pieces of halide chemistry -- an aryl C-Cl bond picks up double-bond character it never had in an alkyl halide, a polar C-X bond still cannot buy its way into water's hydrogen-bonded network, and a Grignard carbon is basic enough to seize a proton from the very first water molecule it meets.(i) Dipole moment: chlorobenzene versus cyclohexyl chlorideIn cyclohexyl chloride (\(\displaystyle \text{C}_6\text{H}_{11}\text{Cl}\), a cyclohexane ring with Cl on an ordinary sp3 carbon) the C-Cl bond is a simple, isolated σ-bond. Chlorine is more electronegative than carbon, so it pulls the bonding electron pair toward itself with the full strength of that electronegativity difference, and nothing on the ring opposes it. The measured dipole moment is large, about \(\displaystyle 2.15\ \text{D}\).In chlorobenzene (\(\displaystyle \text{C}_6\text{H}_5\text{Cl}\)) the carbon carrying chlorine is part of the aromatic ring and is sp2 hybridized. Chlorine has three lone pairs; one of them sits in a p-orbital that is parallel to, and overlaps with, the ring's \(\displaystyle \pi\) system. That overlap lets the lone pair delocalize into the ring, giving resonance contributors in which chlorine carries a formal positive charge and the ortho and para ring carbons carry a formal negative charge:\(\displaystyle \text{C}_6\text{H}_5-\overset{\displaystyle \cdot\cdot}{\text{Cl}} \longleftrightarrow {}^{+}\text{Cl}=\text{C}_6\text{H}_4{}^{-}\) (and the equivalent para form)This \(\displaystyle p\pi\)-\(\displaystyle \pi\) conjugation does two things: it shortens the C-Cl bond (about \(\displaystyle 169\ \text{pm}\) in chlorobenzene against \(\displaystyle 177\text{-}179\ \text{pm}\) for an ordinary sp3 C-Cl bond, i.e. it gains partial double-bond character), and it pushes electron density from chlorine back onto the ring -- which is exactly the opposite direction to the plain inductive pull that dominates in cyclohexyl chloride. The bond's ionic (dipolar) character is therefore partly cancelled by this back-donation. Net result: chlorobenzene's dipole moment, about \(\displaystyle 1.69\ \text{D}\), is lower than cyclohexyl chloride's, because chlorobenzene's C-Cl bond is behaving partly like a double bond and giving some electron density back, while cyclohexyl chloride's C-Cl bond has no such resonance escape and stays a full, uncompensated dipole.(ii) Why polar alkyl halides do not mix with waterAn alkyl halide such as \(\displaystyle \text{CH}_3-\text{CH}_2-\text{Cl}\) (chloroethane) is polar -- the C-Cl bond has a real dipole because Cl is more electronegative than C. Polarity alone, though, is not enough to guarantee mixing with water; what decides miscibility is whether the energy released by forming new solute-solvent interactions can pay back the energy needed to break the solvent's own interactions.Water molecules are held together by an extensive three-dimensional network of hydrogen bonds (O-H\(\displaystyle \cdots\)O), and breaking part of that network to make room for a solute costs a large amount of energy. To get that energy back, the solute has to form comparably strong interactions with water in return -- normally by donating or accepting hydrogen bonds itself (an O-H or N-H group can do this).An alkyl halide has no O-H or N-H bond. All it can offer water is a weak dipole-dipole interaction (and the halogen's lone pairs are, at best, poor hydrogen-bond acceptors). The energy released by these weak alkyl-halide-water interactions is far smaller than the energy required to break water's own hydrogen bonds. Since the process would cost more energy than it returns, it is not thermodynamically favourable, and the alkyl halide separates out as its own layer rather than dissolving -- alkyl halides are polar but immiscible with water.(iii) Why Grignard reagents must be made under anhydrous conditionsA Grignard reagent, general formula \(\displaystyle \text{R-Mg-X}\) (R = alkyl or aryl group, X = Cl, Br or I), is made by treating the corresponding halide \(\displaystyle \text{R-X}\) with magnesium turnings in dry ether. Because magnesium is far more electropositive than carbon, the C-Mg bond is strongly polarized toward carbon, so the alkyl/aryl carbon carries substantial negative (carbanion-like) character. That makes \(\displaystyle \text{R-Mg-X}\) simultaneously a strong base and a strong nucleophile -- which is exactly why it is useful for building new C-C bonds, and exactly why it cannot tolerate a proton source.Water is a proton donor, even though it is a weak acid, and the carbanion-like carbon of the Grignard reagent reacts with it instantly:\(\displaystyle \text{R-Mg-X} + \text{H}_2\text{O} \longrightarrow \text{R-H} + \text{Mg(OH)X}\)Here the C-Mg bond breaks, the carbon picks up a proton from water, and the Grignard reagent is destroyed -- converted into the plain alkane R-H (or arene, if R is aryl) and a basic magnesium halide/hydroxide salt, with no C-C bond ever formed. Even trace moisture in the glassware, the solvent, or the atmosphere is enough to quench the reagent this way. That is why Grignard reagents are prepared and used with rigorously dried ether, dried apparatus, and often under an inert, moisture-free atmosphere -- any water present reacts preferentially and irreversibly before the Grignard reagent can do its intended chemistry.Answer: (i) Chlorobenzene's C-Cl bond gains partial double-bond character from resonance (lone-pair donation from Cl into the aromatic ring), which shortens the bond and opposes its dipole, giving chlorobenzene a lower dipole moment (\(\displaystyle \approx 1.69\ \text{D}\)) than cyclohexyl chloride (\(\displaystyle \approx 2.15\ \text{D}\)), whose sp3 C-Cl bond has no such resonance and shows the full inductive dipole. (ii) Alkyl halides are polar but cannot form hydrogen bonds with water, so the weak interactions they can offer water do not release enough energy to compensate for breaking water's hydrogen-bonded network, making them immiscible with water. (iii) Grignard reagents (R-Mg-X) have a strongly nucleophilic, carbanion-like carbon that reacts instantly with water (\(\displaystyle \text{R-Mg-X} + \text{H}_2\text{O} \rightarrow \text{R-H} + \text{Mg(OH)X}\)), destroying the reagent, so they must be prepared and handled under strictly anhydrous conditions.

Each of these four halogen compounds is used for a specific reason tied to one property of the C–X bond — thermal/chemical stability, density, or the ease with which it releases a reactive fragment. Name the compound, then the property, then the use.Freon $\displaystyle 12$ is dichlorodifluoromethane, \(\displaystyle \text{CCl}_2\text{F}_2 \) — a carbon atom carrying two chlorine and two fluorine atoms, one of the chlorofluorocarbons (CFCs). DDT is dichlorodiphenyltrichloroethane, \(\displaystyle (4\text{-ClC}_6\text{H}_4)_2\text{CH-CCl}_3 \) — a central \(\displaystyle \text{CH} \) carbon bonded to a \(\displaystyle \text{CCl}_3 \) group and to two para-chlorophenyl rings. Carbon tetrachloride is tetrachloromethane, \(\displaystyle \text{CCl}_4 \) — a carbon atom bonded to four chlorine atoms, with no C–H bond at all. Iodoform is triiodomethane, \(\displaystyle \text{CHI}_3 \) — one hydrogen and three iodine atoms on a single carbon. Answer: Freon $\displaystyle 12$ (\(\displaystyle \text{CCl}_2\text{F}_2 \)) is used as a refrigerant and aerosol propellant; DDT is used as an insecticide against disease-carrying insects and agricultural pests (now banned/restricted for non-biodegradability); carbon tetrachloride (\(\displaystyle \text{CCl}_4 \)) is used as a solvent for fats/oils/resins, formerly in fire extinguishers and dry cleaning; iodoform (\(\displaystyle \text{CHI}_3 \)) was used as an antiseptic because it slowly liberates free iodine, though this use has now been discontinued in favour of odour-free iodine formulations.

A primary carbon has no steric shield on its back side, so the cyanide ion attacks directly opposite the leaving group in one concerted step — this is \(\displaystyle S_N2 \), not \(\displaystyle S_N1 \).Name every species first.\(\displaystyle n\text{-BuBr} \) is n-butyl bromide, i.e. $\displaystyle 1$-bromobutane: \(\displaystyle CH_3-CH_2-CH_2-CH_2-Br \). The carbon bearing bromine (\(\displaystyle C_1 \)) carries the leaving group \(\displaystyle Br^- \) and is attached to only one other carbon plus two hydrogens — a primary carbon, wide open on the side away from bromine.\(\displaystyle KCN \) is potassium cyanide. In solution it dissociates completely: \(\displaystyle KCN \rightarrow K^+ + CN^- \). The cyanide ion \(\displaystyle CN^- \) is the nucleophile, and it is an ambident nucleophile — it can attack through either the carbon or the nitrogen end, because negative charge is delocalised over both atoms:\[{}^{-}:C\equiv N: \;\longleftrightarrow\; :C\equiv N:^{-} \]\(\displaystyle n\text{-BuCN} \), the product, is n-butyl cyanide, systematically pentanenitrile: \(\displaystyle CH_3-CH_2-CH_2-CH_2-C\equiv N \). The fact that the product is named as a cyanide (nitrile, C attached) and not an isocyanide (N attached) tells you which end of \(\displaystyle CN^- \) actually bonds — the carbon end. That happens because the carbon end of \(\displaystyle CN^- \) is the better nucleophile (it holds more electron density available for bonding) and because the new \(\displaystyle C-C \) bond formed is stronger than a \(\displaystyle C-N \) bond would be; this is a kinetically controlled outcome, not a thermodynamic one.Now the mechanism itself, step by step.Step $\displaystyle 1$ — approach. Since \(\displaystyle C_1 \) of n-BuBr is primary, the carbon end of \(\displaystyle CN^- \) can approach \(\displaystyle C_1 \) from the side directly opposite the \(\displaystyle C-Br \) bond (a $\displaystyle 180$° backside approach), unobstructed by the small \(\displaystyle -CH_2CH_2CH_2CH_3\) chain sitting on the other side.Step $\displaystyle 2$ — concerted bond formation and bond breaking (single transition state). As the carbon of \(\displaystyle CN^- \) begins forming a new \(\displaystyle C-C \) bond to \(\displaystyle C_1 \), the \(\displaystyle C_1-Br \) bond simultaneously begins to weaken and stretch — the two events are not sequential, they happen together in one transition state. \(\displaystyle C_1 \) passes through a five-coordinate, trigonal-bipyramidal transition state in which it is partially bonded to both the incoming \(\displaystyle CN^- \) carbon and the departing \(\displaystyle Br^- \), with the three original substituents (two H atoms and the propyl chain) splayed out in the equatorial plane:\[NC^{\delta-}\cdots C_1 \cdots Br^{\delta-} \]Step $\displaystyle 3$ — departure of the leaving group. The transition state collapses as the \(\displaystyle C_1-CN \) bond finishes forming and the \(\displaystyle C_1-Br \) bond finishes breaking, releasing \(\displaystyle Br^- \) (which pairs with \(\displaystyle K^+ \) already in solution to give \(\displaystyle KBr \)) and leaving pentanenitrile behind. Because the nucleophile enters as the leaving group exits from the opposite face, configuration at \(\displaystyle C_1 \) is inverted (Walden inversion) — though here \(\displaystyle C_1 \) bears two identical hydrogens, so it is not a stereocentre and this inversion produces no observable change in optical activity; it is still mechanistically an inversion.Overall equation:\[CH_3CH_2CH_2CH_2-Br + K^+CN^- \longrightarrow CH_3CH_2CH_2CH_2-C\equiv N + KBr \]This is a single-step, bimolecular process, so the rate law is first order in each reactant and second order overall:\[\text{Rate} = k[n\text{-BuBr}][CN^-] \]— exactly the kinetic signature of \(\displaystyle S_N2 \), consistent with the primary substrate reacting by backside attack rather than through a carbocation.Answer: The reaction proceeds by an \(\displaystyle S_N2 \) mechanism — the carbon end of the ambident nucleophile \(\displaystyle CN^- \) attacks \(\displaystyle C_1 \) of n-butyl bromide from the side opposite to \(\displaystyle Br^- \) in one concerted step (backside attack through a five-coordinate transition state, with inversion at \(\displaystyle C_1 \)), displacing \(\displaystyle Br^- \) directly and giving n-butyl cyanide (pentanenitrile), \(\displaystyle CH_3CH_2CH_2CH_2CN \), plus \(\displaystyle KBr \); rate \(\displaystyle = k[n\text{-BuBr}][CN^-] \).

An \(\displaystyle \mathrm{SN_{2}}\) attack comes from directly behind the C–Br bond, so anything that crowds that backside approach — a bulky group on the carbon bearing the halogen, or even bulky groups one carbon further out — slows the reaction, regardless of whether the carbon is called primary, secondary, or tertiary.In an \(\displaystyle \mathrm{SN_{2}}\) (substitution, nucleophilic, bimolecular) displacement, the nucleophile attacks the carbon bearing the leaving group from the side opposite the C–Br bond, passing through a five-coordinate transition state before bromide leaves. The rate-determining step is this single concerted attack, so the reaction is fastest when that backside path is open and slows as alkyl groups pile up around the reacting carbon — first from groups on the carbon itself, then from groups on the carbon next door.(i) $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 1$-Bromopentane, $\displaystyle 2$-BromopentaneNaming the three halides by the class of carbon carrying bromine: A primary carbon leaves the backside almost unobstructed; a secondary carbon has one extra alkyl group in the way; a tertiary carbon has three alkyl groups crowding the approach, and in practice a tertiary halide does not go by \(\displaystyle \mathrm{SN_{2}}\) at all (it goes by \(\displaystyle \mathrm{SN_{1}}\) instead). So reactivity toward \(\displaystyle \mathrm{SN_{2}}\) falls straight from primary to secondary to tertiary:\[\text{1-Bromopentane} > \text{2-Bromopentane} > \text{2-Bromo-2-methylbutane} \](ii) $\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 2$-Bromo-$\displaystyle 3$-methylbutaneAgain classing each by the carbon bearing bromine: The same primary-beats-secondary-beats-tertiary logic applies, since the branching in the first compound is too far from the reacting carbon to add real steric hindrance there:\[\text{1-Bromo-3-methylbutane} > \text{2-Bromo-3-methylbutane} > \text{2-Bromo-2-methylbutane} \](iii) $\displaystyle 1$-Bromobutane, $\displaystyle 1$-Bromo-$\displaystyle 2,2$-dimethylpropane, $\displaystyle 1$-Bromo-$\displaystyle 2$-methylbutane, $\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutaneHere every bromine sits on a primary carbon, so the class of that carbon cannot distinguish them — the deciding factor is how much alkyl bulk sits on the carbon right next door (the beta carbon), since that bulk still crowds the nucleophile's backside path even though it is not directly on the reacting carbon: Because the branching in the beta position sits right next to the carbon undergoing attack, it hinders the backside approach almost as effectively as branching on the reacting carbon itself, while branching one carbon further away (the gamma position) hinders it much less. Ranking from the open, unbranched chain down to the doubly-branched neopentyl case:\[\text{1-Bromobutane} > \text{1-Bromo-3-methylbutane} > \text{1-Bromo-2-methylbutane} > \text{1-Bromo-2,2-dimethylpropane} \]Answer: (i) $\displaystyle 1$-Bromopentane > $\displaystyle 2$-Bromopentane > $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane; (ii) $\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutane > $\displaystyle 2$-Bromo-$\displaystyle 3$-methylbutane > $\displaystyle 2$-Bromo-$\displaystyle 2$-methylbutane; (iii) $\displaystyle 1$-Bromobutane > $\displaystyle 1$-Bromo-$\displaystyle 3$-methylbutane > $\displaystyle 1$-Bromo-$\displaystyle 2$-methylbutane > $\displaystyle 1$-Bromo-$\displaystyle 2,2$-dimethylpropane — in each set, \(\displaystyle \mathrm{SN_{2}}\) reactivity falls as alkyl bulk builds up at or next to the carbon bearing bromine, since that bulk blocks the nucleophile's backside attack.

A carbocation sitting on a carbon attached to two benzene rings is far more stable than one attached to only one ring — so the compound whose leaving group sits between two phenyl groups ionises fastest, and it is \(\displaystyle C_6H_5CHClC_6H_5\) that is hydrolysed more easily by aqueous KOH.Name both halides first.\(\displaystyle C_6H_5CH_2Cl\) is benzyl chloride — IUPAC name (chloromethyl)benzene. Here the carbon bearing chlorine carries one phenyl group and two hydrogens: a primary benzylic halide.\(\displaystyle C_6H_5CHClC_6H_5\) is benzhydryl chloride — IUPAC name chlorodiphenylmethane. Here the carbon bearing chlorine carries two phenyl groups and one hydrogen: a secondary benzylic halide, with chlorine flanked on both sides by an aromatic ring.Both carbons are benzylic (directly attached to an aromatic ring), so aqueous KOH does not displace chlorine by a clean back-side attack (\(\displaystyle S_N2\)); instead hydrolysis goes through the \(\displaystyle S_N1\) pathway, because the ring can donate electron density into an empty p-orbital once the carbocation forms. In \(\displaystyle S_N1\), the slow, rate-determining step is loss of the leaving group to generate the carbocation — so whichever substrate gives the more stable carbocation ionises faster, and hydrolyses faster.Step $\displaystyle 1$ — ionisation (rate-determining) for benzyl chloride: \[C_6H_5-CH_2-Cl \longrightarrow C_6H_5-CH_2^{+} + Cl^{-} \] The benzylic cation \(\displaystyle C_6H_5CH_2^{+}\) is stabilised by resonance: the empty orbital on the benzylic carbon overlaps with the ring's \(\displaystyle \pi\) system, so the positive charge is delocalised onto the ortho and para carbons of that one ring. This gives the cation, in total, four resonance contributors (the original plus three ring-delocalised forms).Step $\displaystyle 1$ — ionisation for benzhydryl chloride: \[C_6H_5-CHCl-C_6H_5 \longrightarrow C_6H_5-CH^{+}-C_6H_5 + Cl^{-} \] The diphenylmethyl (benzhydryl) cation \(\displaystyle C_6H_5CH^{+}C_6H_5\) has the same empty orbital, but now it is flanked by two rings, and each ring independently delocalises the positive charge onto its own ortho and para carbons. Counting the resonance forms from each ring separately (plus the original structure) gives roughly seven contributors instead of four, and the positive charge is spread over a much larger volume of the molecule.Spreading a charge over more atoms always lowers a cation's energy more than spreading it over fewer atoms, so the benzhydryl cation is markedly more stable than the benzyl cation. Because the rate of the \(\displaystyle S_N1\) ionisation step tracks carbocation stability, the C–Cl bond in \(\displaystyle C_6H_5CHClC_6H_5\) breaks heterolytically faster than the one in \(\displaystyle C_6H_5CH_2Cl\) — this, not any difference in how the nucleophile attacks, is the step that decides which substrate reacts faster.Step $\displaystyle 2$ — fast capture of the (planar, sp²) carbocation by hydroxide/water, common to both: \[C_6H_5CH_2^{+} + OH^{-} \longrightarrow C_6H_5CH_2OH \text{ (benzyl alcohol)} \] \[C_6H_5CH^{+}C_6H_5 + OH^{-} \longrightarrow C_6H_5CH(OH)C_6H_5 \text{ (benzhydrol, diphenylmethanol)} \]So overall, with aqueous KOH: \[C_6H_5CH_2Cl + KOH(aq) \longrightarrow C_6H_5CH_2OH + KCl \quad \text{(slower)} \] \[C_6H_5CHClC_6H_5 + KOH(aq) \longrightarrow C_6H_5CH(OH)C_6H_5 + KCl \quad \text{(faster)} \]The one thing students often get backwards here: it is not "more substituted carbon = slower" as in simple alkyl halides — for benzylic/allylic systems the extra substituent is a second phenyl ring, and a second ring means a second path for delocalising the positive charge, which speeds up ionisation rather than slowing it by sterics.Answer: \(\displaystyle C_6H_5CHClC_6H_5\) (benzhydryl chloride) is hydrolysed more easily than \(\displaystyle C_6H_5CH_2Cl\) (benzyl chloride) by aqueous KOH, because its carbocation intermediate is stabilised by resonance with two benzene rings instead of one, making that carbocation more stable and its \(\displaystyle S_N1\) ionisation step faster.

Melting point is a solid-state property — it is decided by how tightly the molecules can stack into a crystal lattice, not just by how strong the pull between any one pair of molecules is. That packing question turns on molecular shape and symmetry, and that is exactly where the three dichlorobenzenes differ.All three isomers — $\displaystyle 1,2$-dichlorobenzene (ortho), $\displaystyle 1,3$-dichlorobenzene (meta) and $\displaystyle 1,4$-dichlorobenzene (para) — share the same molecular formula \(\displaystyle \text{C}_6\text{H}_4\text{Cl}_2 \) and the same molar mass. If the strength of intermolecular attraction alone fixed the melting point, the three should melt at similar temperatures. They do not: p-dichlorobenzene melts around $\displaystyle 53$ °C, while o-dichlorobenzene melts around −$\displaystyle 17$ °C and m-dichlorobenzene around −$\displaystyle 25$ °C — a gap of roughly $\displaystyle 70$–$\displaystyle 80$ °C between the para isomer and its two siblings.The reason is shape, not force. In $\displaystyle 1,4$-dichlorobenzene the two chlorine atoms sit on directly opposite carbons of the ring (C1 and C4). This places every substituent on an axis of symmetry running through the ring, so the molecule as a whole is symmetric — it has the same "footprint" looked at from either chlorine end. A molecule shaped like this stacks the way a symmetric brick stacks: layer upon layer with no wasted space, every molecule surrounded by neighbours making close, uniform contact with it. That close, regular packing lets a very large number of van der Waals contacts form per mole in the solid, and it is the sum of all those contacts — not any single strong interaction — that has to be overcome to melt the crystal.In $\displaystyle 1,2$-dichlorobenzene the chlorines are on adjacent carbons, and in $\displaystyle 1,3$-dichlorobenzene they are one carbon apart; in both, the substituents break the symmetry of the ring and give the molecule an irregular, "bent" outline. Irregularly shaped molecules cannot tile a lattice as efficiently — they leave gaps, and the crystal that results is looser, with fewer molecule-to-molecule contacts per mole. A looser lattice takes less thermal energy to break apart, so both isomers melt at much lower temperatures than the para isomer, even though the chemical bonding within each molecule is essentially identical.There is a genuine paradox worth naming, because it is what makes this question instructive rather than obvious. p-Dichlorobenzene is actually the least polar of the three. Each C–Cl bond carries a dipole pointing from carbon toward the more electronegative chlorine. In the para isomer these two bond dipoles point in exactly opposite directions along the same axis, so they cancel: \[\vec{\mu}_{\text{C–Cl (C1)}} + \vec{\mu}_{\text{C–Cl (C4)}} = 0 \] and the molecule has zero net dipole moment. In the ortho isomer the two C–Cl dipoles are about $\displaystyle 60$° apart and add to a large resultant; in the meta isomer they are about $\displaystyle 120$° apart and add to a smaller but still nonzero resultant. So by dipole moment alone, one would predict the order o- > m- > p- for melting point, since dipole–dipole attraction is a real intermolecular force and the para isomer has none of it. The observed order is the reverse for the para isomer: p- is far above both o- and m-.This is the point of the question: for a solid, how efficiently the molecules can pack outweighs how strongly any one pair of them attracts. The many extra van der Waals contacts that the symmetric, close-packing para molecule gains in the crystal add up to more cohesive energy than the single dipole–dipole interaction available to the bulkier, poorly packing ortho and meta molecules — even though, molecule for molecule, o- and m-dichlorobenzene are the more polar species.The liquid state confirms this reading, because it removes the packing advantage. Once melted, molecules tumble freely and lattice geometry no longer matters; here dipole–dipole forces are free to dominate, and boiling point (not melting point) tracks polarity instead of symmetry. Consistently, o-dichlorobenzene has the highest boiling point of the three (about $\displaystyle 180$ °C), with the meta and para isomers close together and lower (roughly $\displaystyle 173$ °C and $\displaystyle 174$ °C) — essentially the reverse of the melting-point ranking. The fact that the ranking flips between the solid and the liquid property is itself the evidence that packing symmetry, not intermolecular force strength, is what controls the melting point.Answer: p-Dichlorobenzene ($\displaystyle 1,4$-dichlorobenzene) has a much higher melting point (≈ $\displaystyle 53$ °C) than o-dichlorobenzene ($\displaystyle 1,2$-dichlorobenzene, ≈ −$\displaystyle 17$ °C) or m-dichlorobenzene ($\displaystyle 1,3$-dichlorobenzene, ≈ −$\displaystyle 25$ °C) because its molecule is symmetrical — the two chlorines sit directly opposite each other on the ring — which lets p-dichlorobenzene molecules pack closely and efficiently into the crystal lattice, maximizing van der Waals contacts between molecules. The unsymmetrical o- and m-isomers cannot pack as tightly, so their crystals are held together more loosely and melt at much lower temperatures, despite o- and m-dichlorobenzene actually being the more polar molecules (p-dichlorobenzene's symmetry makes its net dipole moment zero).

Each of these twenty conversions is a short synthetic route: work out which bond has to break and which has to form, then pick the reagent that does exactly that — nothing more.(i) Propene to propan-$\displaystyle 1$-ol . Markovnikov addition of water to \(\displaystyle CH_3-CH=CH_2\) would put \(\displaystyle OH\) on the middle carbon (propan-$\displaystyle 2$-ol), so the anti-Markovnikov route is needed: hydroboration–oxidation. Diborane, \(\displaystyle B_2H_6\), adds boron to the less hindered terminal carbon of the double bond (steric and electronic effects keep boron off the more crowded carbon), giving a trialkylborane. Oxidative work-up with alkaline hydrogen peroxide, \(\displaystyle H_2O_2/OH^-\), replaces that boron by \(\displaystyle OH\) with the same regiochemistry. Net result: \(\displaystyle CH_3-CH_2-CH_2-OH\), propan-$\displaystyle 1$-ol.(ii) Ethanol to but-$\displaystyle 1$-yne. The chain has to grow from two carbons to four, so ethanol is used twice — once to build the alkyne skeleton and once to build the alkylating agent. Dehydrate ethanol with conc. \(\displaystyle H_2SO_4\) at $\displaystyle 443$ K to ethene, \(\displaystyle CH_2=CH_2\) ; add \(\displaystyle Br_2\) to get $\displaystyle 1,2$-dibromoethane, \(\displaystyle CH_2Br-CH_2Br\); then double dehydrohalogenation with excess alcoholic \(\displaystyle KOH\) removes both \(\displaystyle HBr\) molecules to give ethyne, \(\displaystyle CH\equiv CH\). Treating ethyne with sodamide, \(\displaystyle NaNH_2\) (a strong enough base to remove the acidic terminal alkyne proton), gives sodium acetylide, \(\displaystyle HC\equiv C^{-}Na^{+}\). Separately, ethanol is converted to ethyl bromide, \(\displaystyle C_2H_5Br\), with \(\displaystyle HBr\). The acetylide carbanion then displaces bromide from ethyl bromide in an \(\displaystyle S_N2\) alkylation: \(\displaystyle HC\equiv C^{-}Na^{+} + C_2H_5Br \rightarrow HC\equiv C-CH_2-CH_3\), which is but-$\displaystyle 1$-yne.(iii) $\displaystyle 1$-Bromopropane to $\displaystyle 2$-bromopropane. The halogen has to move from the end carbon to the middle one, so the route goes through the alkene. Alcoholic \(\displaystyle KOH\) eliminates \(\displaystyle HBr\) from \(\displaystyle CH_3-CH_2-CH_2Br\) to give propene, \(\displaystyle CH_3-CH=CH_2\). Adding \(\displaystyle HBr\) back on now follows Markovnikov's rule — the proton adds to the carbon that already has more hydrogens, and \(\displaystyle Br\) goes to the carbon that gives the more stable (secondary) carbocation — placing bromine on the middle carbon: \(\displaystyle CH_3-CHBr-CH_3\), $\displaystyle 2$-bromopropane.(iv) Toluene to benzyl alcohol . The methyl group's benzylic hydrogens are attacked by chlorine radicals under photochemical conditions, \(\displaystyle Cl_2/h\nu\) (side-chain, not ring, chlorination, since ring substitution needs a Lewis-acid catalyst instead), giving benzyl chloride , \(\displaystyle C_6H_5-CH_2Cl\). Aqueous \(\displaystyle NaOH\) then does an \(\displaystyle S_N2\) hydrolysis at that primary benzylic carbon — hydroxide attacks the carbon bearing chlorine, displacing chloride — to give benzyl alcohol, \(\displaystyle C_6H_5-CH_2OH\).(v) Benzene to $\displaystyle 4$-bromonitrobenzene. The order of the two substitutions is the whole trick: the nitro group is a meta director, so nitrating first and brominating second would put bromine meta to the nitro group, never para. Instead, brominate first: \(\displaystyle Br_2/FeBr_3\) gives bromobenzene , \(\displaystyle C_6H_5Br\). Bromine is a weak deactivator but an ortho/para director (its lone pair still donates into the ring by resonance even though its electronegativity withdraws electron density inductively), so nitrating bromobenzene with \(\displaystyle HNO_3/H_2SO_4\) gives a mixture of ortho- and para-bromonitrobenzene, dominated by the less hindered para isomer. Fractional crystallisation (the para isomer has a much higher melting point) separates out $\displaystyle 4$-bromonitrobenzene.(vi) Benzyl alcohol to $\displaystyle 2$-phenylethanoic acid. One extra carbon is needed, so a cyanide has to be installed and then hydrolysed. Thionyl chloride, \(\displaystyle SOCl_2\), converts benzyl alcohol to benzyl chloride , \(\displaystyle C_6H_5-CH_2Cl\) (chosen over \(\displaystyle HCl/ZnCl_2\) because it leaves only gaseous by-products and avoids side reactions). Alcoholic \(\displaystyle KCN\) then substitutes chloride by cyanide at this primary benzylic carbon (\(\displaystyle S_N2\)): the carbon count rises by one, giving phenylacetonitrile, \(\displaystyle C_6H_5-CH_2-CN\). Acidic hydrolysis (\(\displaystyle H_3O^+\), heat) converts the nitrile carbon to a carboxyl carbon, releasing ammonium ion and giving phenylacetic acid, \(\displaystyle C_6H_5-CH_2-COOH\), which is $\displaystyle 2$-phenylethanoic acid.(vii) Ethanol to propanenitrile. Again the chain must gain one carbon, from the cyanide carbon. Ethanol reacts with \(\displaystyle HBr\) (or \(\displaystyle PBr_3\)) to give ethyl bromide, \(\displaystyle CH_3-CH_2-Br\). Alcoholic \(\displaystyle KCN\) displaces bromide by an \(\displaystyle S_N2\) mechanism, and the incoming \(\displaystyle -C\equiv N\) carbon becomes the third carbon of the chain: \(\displaystyle CH_3-CH_2-CN\), propanenitrile.(viii) Aniline to chlorobenzene . The amino nitrogen cannot simply be swapped for chlorine directly, so it is first converted to a leaving group that chlorine can replace. Treating aniline, \(\displaystyle C_6H_5-NH_2\), with \(\displaystyle NaNO_2/HCl\) at $\displaystyle 273$–$\displaystyle 278$ K (diazotisation, kept cold so the diazonium salt does not decompose) gives benzenediazonium chloride, \(\displaystyle C_6H_5-N_2^{+}Cl^{-}\). Warming this with cuprous chloride and \(\displaystyle HCl\), \(\displaystyle Cu_2Cl_2/HCl\) — the Sandmeyer reaction — replaces the \(\displaystyle -N_2^{+}\) group by chlorine with loss of nitrogen gas, giving chlorobenzene, \(\displaystyle C_6H_5-Cl\).(ix) $\displaystyle 2$-Chlorobutane to $\displaystyle 3,4$-dimethylhexane. This is a Wurtz coupling: two alkyl halide molecules join at the carbon that carried the halogen when treated with sodium metal in dry ether, \(\displaystyle 2R-Cl + 2Na \rightarrow R-R + 2NaCl\). Here \(\displaystyle R\) is the sec-butyl group, \(\displaystyle -CH(CH_3)(CH_2CH_3)\), from $\displaystyle 2$-chlorobutane, \(\displaystyle CH_3-CHCl-CH_2-CH_3\). Coupling two of these radicals at the carbon that bore chlorine gives \(\displaystyle CH_3-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3\). Numbering the longest chain (six carbons, an ethyl group on each side of the new bond) shows a methyl branch on carbon $\displaystyle 3$ and carbon $\displaystyle 4$: $\displaystyle 3,4$-dimethylhexane.(x) $\displaystyle 2$-Methyl-$\displaystyle 1$-propene to $\displaystyle 2$-chloro-$\displaystyle 2$-methylpropane. Adding \(\displaystyle HCl\) across \(\displaystyle (CH_3)_2C=CH_2\) follows Markovnikov's rule: the proton adds to the \(\displaystyle =CH_2\) end (which already carries two hydrogens), generating the more stable tertiary carbocation on the trisubstituted carbon, and chloride then attacks that carbocation. The product is \(\displaystyle (CH_3)_3C-Cl\), $\displaystyle 2$-chloro-$\displaystyle 2$-methylpropane (tert-butyl chloride).(xi) Ethyl chloride to propanoic acid . Alcoholic \(\displaystyle KCN\) substitutes chloride in \(\displaystyle CH_3-CH_2-Cl\) by cyanide (\(\displaystyle S_N2\)), adding one carbon to give propanenitrile, \(\displaystyle CH_3-CH_2-CN\). Acidic hydrolysis (dilute \(\displaystyle H_2SO_4\) or \(\displaystyle H_3O^+\), heat) proceeds through the amide, \(\displaystyle CH_3CH_2CONH_2\), to the carboxylic acid, releasing ammonium ion: \(\displaystyle CH_3-CH_2-COOH\), propanoic acid.(xii) But-$\displaystyle 1$-ene to n-butyl iodide. Ordinary \(\displaystyle HBr\) addition to \(\displaystyle CH_2=CH-CH_2-CH_3\) would follow Markovnikov's rule and put bromine on the internal carbon, so the peroxide effect (Kharasch effect) is used instead: in the presence of peroxides, \(\displaystyle HBr\) adds by a free-radical chain mechanism in which the bromine atom, not \(\displaystyle H^+\), attacks first and adds to the terminal, less hindered carbon (giving the more stable secondary radical at the other carbon), so the halogen ends up terminal. This gives $\displaystyle 1$-bromobutane, \(\displaystyle CH_3-CH_2-CH_2-CH_2-Br\). Treating this with sodium iodide in dry acetone, \(\displaystyle NaI/\text{acetone}\) (the Finkelstein reaction), swaps bromide for iodide — the reaction is driven forward because \(\displaystyle NaBr\) is insoluble in acetone and precipitates out — giving n-butyl iodide, \(\displaystyle CH_3CH_2CH_2CH_2I\).(xiii) $\displaystyle 2$-Chloropropane to $\displaystyle 1$-propanol . As in (i), anti-Markovnikov hydration is needed, so the halide is first eliminated to the alkene and then hydroborated. Alcoholic \(\displaystyle KOH\) removes \(\displaystyle HCl\) from \(\displaystyle CH_3-CHCl-CH_3\) to give propene, \(\displaystyle CH_3-CH=CH_2\). Hydroboration with \(\displaystyle B_2H_6\) followed by oxidation with \(\displaystyle H_2O_2/OH^-\) places \(\displaystyle OH\) on the terminal carbon, giving \(\displaystyle CH_3-CH_2-CH_2-OH\), $\displaystyle 1$-propanol.(xiv) Isopropyl alcohol to iodoform. This is the haloform reaction: iodine in aqueous \(\displaystyle NaOH\) generates sodium hypoiodite in situ, which first oxidises the secondary alcohol's \(\displaystyle -CH(OH)-\) carbon (this is the step people forget — the alcohol must first become a methyl ketone before haloform chemistry can happen) to a carbonyl, effectively giving acetone , \(\displaystyle CH_3-CO-CH_3\), as an intermediate. The three acidic hydrogens on the methyl group next to that carbonyl are then progressively replaced by iodine, and hydroxide finally cleaves the resulting \(\displaystyle CI_3-CO-CH_3\) at the carbon-carbon bond next to the carbonyl (nucleophilic acyl substitution), releasing triiodomethane and the carboxylate: \(\displaystyle CH_3-CHOH-CH_3 + 4I_2 + 6NaOH \rightarrow CHI_3\downarrow + CH_3COONa + 5NaI + 5H_2O\). The yellow precipitate is iodoform, \(\displaystyle CHI_3\).(xv) Chlorobenzene to p-nitrophenol . Chlorine is an ortho/para director, so nitrating chlorobenzene with \(\displaystyle HNO_3/H_2SO_4\) gives mainly a mixture of ortho- and para-chloronitrobenzene; the para isomer is separated by fractional distillation. In p-chloronitrobenzene, the nitro group's strong electron withdrawal from the para position makes the ring carbon bearing chlorine susceptible to nucleophilic aromatic substitution (unlike unactivated chlorobenzene, which needs far harsher conditions) — aqueous \(\displaystyle NaOH\) under heat and pressure displaces chloride, and after acidification the product is p-nitrophenol, \(\displaystyle 4-O_2N-C_6H_4-OH\).(xvi) $\displaystyle 2$-Bromopropane to $\displaystyle 1$-bromopropane. Alcoholic \(\displaystyle KOH\) eliminates \(\displaystyle HBr\) from \(\displaystyle CH_3-CHBr-CH_3\) to give propene, \(\displaystyle CH_3-CH=CH_2\). Adding \(\displaystyle HBr\) back under the peroxide effect (free-radical addition, bromine atom attacking first and adding to the terminal carbon) reverses the original regiochemistry and gives $\displaystyle 1$-bromopropane, \(\displaystyle CH_3-CH_2-CH_2-Br\).(xvii) Chloroethane to butane. A straightforward Wurtz coupling: sodium metal in dry ether couples two ethyl groups, \(\displaystyle 2CH_3-CH_2-Cl + 2Na \rightarrow CH_3-CH_2-CH_2-CH_3 + 2NaCl\), giving butane.(xviii) Benzene to diphenyl. This looks like a Wurtz reaction but is not — coupling two aryl halides with sodium is specifically called the Fittig reaction (the Wurtz–Fittig name is reserved for coupling one aryl halide with one alkyl halide). Chlorinating benzene, \(\displaystyle Cl_2/FeCl_3\), gives chlorobenzene , \(\displaystyle C_6H_5-Cl\). Treating this with sodium metal in dry ether, \(\displaystyle 2C_6H_5Cl + 2Na \rightarrow C_6H_5-C_6H_5 + 2NaCl\), joins the two rings directly to give diphenyl (biphenyl), \(\displaystyle C_6H_5-C_6H_5\).(xix) tert-Butyl bromide to isobutyl bromide . The halogen has to move off a carbon with no hydrogens onto one with two, so, as in (iii) and (xvi), the route goes through the alkene. A tertiary halide eliminates readily with alcoholic \(\displaystyle KOH\): \(\displaystyle (CH_3)_3C-Br \rightarrow (CH_3)_2C=CH_2\) ($\displaystyle 2$-methylpropene). Adding \(\displaystyle HBr\) under the peroxide effect (anti-Markovnikov, free-radical addition) puts bromine on the terminal, less substituted carbon instead of the tertiary one: \(\displaystyle (CH_3)_2CH-CH_2-Br\), isobutyl bromide ($\displaystyle 1$-bromo-$\displaystyle 2$-methylpropane).(xx) Aniline to phenyl isocyanide . This is the carbylamine (isocyanide) reaction, a good confirmatory test for a primary amine: heating aniline with chloroform and alcoholic \(\displaystyle KOH\), \(\displaystyle CHCl_3/alc.\,KOH,\Delta\), generates dichlorocarbene in situ, which the amine nitrogen's lone pair attacks; after loss of the two chlorines as \(\displaystyle KCl\) and a proton, the nitrogen ends up doubly bonded to a terminal carbon. Product: \(\displaystyle C_6H_5-NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5-NC + 3KCl + 3H_2O\), phenyl isocyanide.**Answer: (i) propan-$\displaystyle 1$-ol, \(\displaystyle CH_3CH_2CH_2OH\) (ii) but-$\displaystyle 1$-yne, \(\displaystyle HC\equiv C-CH_2CH_3\) (iii) $\displaystyle 2$-bromopropane, \(\displaystyle CH_3CHBrCH_3\) (iv) benzyl alcohol, \(\displaystyle C_6H_5CH_2OH\) (v) $\displaystyle 4$-bromonitrobenzene (vi) $\displaystyle 2$-phenylethanoic acid, \(\displaystyle C_6H_5CH_2COOH\) (vii) propanenitrile, \(\displaystyle CH_3CH_2CN\) (viii) chlorobenzene, \(\displaystyle C_6H_5Cl\) (ix) $\displaystyle 3,4$-dimethylhexane (x) $\displaystyle 2$-chloro-$\displaystyle 2$-methylpropane, \(\displaystyle (CH_3)_3CCl\) (xi) propanoic acid, \(\displaystyle CH_3CH_2COOH\) (xii) n-butyl iodide, \(\displaystyle CH_3CH_2CH_2CH_2I\) (xiii) $\displaystyle 1$-propanol, \(\displaystyle CH_3CH_2CH_2OH\) (xiv) iodoform, \(\displaystyle CHI_3\) (xv) p-nitrophenol (xvi) $\displaystyle 1$-bromopropane, \(\displaystyle CH_3CH_2CH_2Br\) (xvii) butane, \(\displaystyle CH_3CH_2CH_2CH_3\) (xviii) diphenyl, \(\displaystyle C_6H_5-C_6H_5\) (xix) isobutyl bromide, \(\displaystyle (CH_3)_2CHCH_2Br\) (xx) phenyl isocyanide, \(\displaystyle C_6H_5NC\).

Aqueous and alcoholic KOH hand the alkyl chloride the same base–nucleophile, OH⁻ (or its ethanol‑born cousin, ethoxide), but drop it into two very different solvents — and it is the solvent, not the halide, that decides whether that species attacks the carbon (substitution) or plucks off a hydrogen (elimination).Step $\displaystyle 1$ — what "aqueous KOH" and "alcoholic KOH" actually contain. KOH is fully ionic; dissolved in water it exists as K⁺ and OH⁻, and each OH⁻ is wrapped in a shell of hydrogen-bonded water molecules (it is heavily solvated). Dissolved instead in ethanol, KOH sets up the equilibrium KOH + \(\displaystyle \mathrm{C_{2}H_{5}OH}\) ⇌ \(\displaystyle \mathrm{C_{2}H_{5}O}\)⁻ (ethoxide ion) + \(\displaystyle \mathrm{H_{2}O}\), so "alcoholic KOH" is really a solution containing OH⁻ and the ethoxide ion \(\displaystyle \mathrm{C_{2}H_{5}O}\)⁻, sitting in a solvent that is a far poorer solvator of small anions than water is. This difference in solvation is the whole story.Step $\displaystyle 2$ — aqueous KOH: the anion behaves as a nucleophile, so you get substitution. Because water solvates OH⁻ so effectively, the ion approaches the alkyl chloride from the side opposite the leaving group and attacks the electrophilic carbon that carries the chlorine — this is nucleophilic substitution (SN2 for a primary or secondary carbon; \(\displaystyle \mathrm{SN_{1}}\), through a carbocation, for a tertiary one). The C–Cl bond breaks heterolytically, chlorine leaves as Cl⁻, and the new C–OH bond forms at that same carbon. Take $\displaystyle 2$-chloropropane, CH3-CHCl-CH3, as the working example:CH3-CHCl-CH3 + KOH(aq) → CH3-CH(OH)-CH3 + KClThe product is propan-$\displaystyle 2$-ol (isopropanol): the halogen-bearing carbon is untouched in its connectivity — only Cl is swapped for OH.Step $\displaystyle 3$ — alcoholic KOH: the anion behaves as a base, so you get elimination. In ethanol there is very little water left to stabilize an anion sitting on oxygen, so both OH⁻ and, more importantly, the bulky ethoxide ion \(\displaystyle \mathrm{C_{2}H_{5}O}\)⁻ are relatively "naked" and hungry to grab a proton rather than to squeeze past a crowded carbon and displace chloride. Instead of attacking the carbon bearing Cl, the base abstracts a hydrogen from the carbon next to it (the β-carbon). This is a concerted, one-step E2 process: as the C–H bond on the β-carbon breaks and its electron pair swings in to form a new π bond, the C–Cl bond on the α-carbon breaks at the same time and Cl⁻ departs — so a proton leaves from one carbon and a chloride ion leaves from the adjacent carbon, and a carbon–carbon double bond is created between them (β-elimination / dehydrohalogenation). For the same substrate:CH3-CHCl-CH3 + KOH(alc) → CH3-CH=\(\displaystyle \mathrm{CH_{2}}\) + KCl + \(\displaystyle \mathrm{H_{2}O}\)The product is propene, an alkene, not an alcohol.Step $\displaystyle 4$ — why the base "chooses" elimination once it is in alcohol. Two effects push the same direction. First, a poorly solvated base is a stronger, more aggressive base (less of its reactivity has been "used up" satisfying hydrogen bonds to solvent), and a stronger base preferentially removes a proton rather than performs the slower backside attack on carbon. Second, the β-hydrogens sit on the outside of the molecule and are sterically far more accessible than the crowded α-carbon that already bears the bulky halogen — so a big, base-hungry anion like ethoxide reaches the hydrogen much more easily than it reaches the carbon. Aqueous OH⁻, well solvated and smaller in effective reactivity, does not have this bias and simply substitutes instead.A step students often get backwards: it is easy to assume "aqueous solvent = more reactive = elimination," but it runs the other way — water's strong solvation of OH⁻ suppresses its basicity and lets its nucleophilicity dominate (→ substitution), while alcohol's weaker solvation lets basicity dominate (→ elimination). Also remember that when there is more than one type of β-hydrogen (as in $\displaystyle 2$-chlorobutane, say), Zaitsev's rule applies: the alcoholic-KOH elimination gives mainly the more substituted, more stable alkene, not the alkene formed by removing the "easiest" hydrogen.Answer: Aqueous KOH provides a well-solvated OH⁻ ion that acts as a nucleophile, so it substitutes the halogen by \(\displaystyle \mathrm{SN_{1}}\)/\(\displaystyle \mathrm{SN_{2}}\) attack on the α-carbon, giving an alcohol (e.g. CH3-CHCl-CH3 + KOH(aq) → CH3-CH(OH)-CH3 + KCl, propan-$\displaystyle 2$-ol). Alcoholic KOH generates a poorly solvated, more basic anion (OH⁻/\(\displaystyle \mathrm{C_{2}H_{5}O}\)⁻) that instead abstracts a β-hydrogen in a concerted E2 process, expelling the halide from the adjacent carbon and forming a C=C bond, so the major product is an alkene (e.g. CH3-CHCl-CH3 + KOH(alc) → CH3-CH=\(\displaystyle \mathrm{CH_{2}}\) + KCl + \(\displaystyle \mathrm{H_{2}O}\), propene) — substitution in water, elimination in alcohol, because the solvent controls whether the base attacks carbon or hydrogen.

Six different fates for six different halides — three kinds of substitution, one elimination, one Grignard formation and one radical coupling — because "same functional group" does not mean "same reaction."(i) n-Butyl chloride + alcoholic KOHAlcoholic KOH is loaded with ethoxide, and ethoxide behaves as a base, not a nucleophile — with a primary halide it pulls off a β-hydrogen instead of displacing the leaving group.n-Butyl chloride is $\displaystyle 1$-chlorobutane, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{Cl} \) . In alcoholic KOH, the ethoxide ion abstracts a proton from the carbon next to the one bearing chlorine (C-$\displaystyle 2$, the β-carbon) at the same time as the C–Cl bond breaks and chloride ion leaves — a concerted, one-step β-elimination (E2). A new \(\displaystyle \pi \) bond forms between C-$\displaystyle 1$ and C-2. Because the chlorine sits on the very end of a straight chain, C-$\displaystyle 1$ has only one neighbouring carbon that carries hydrogens, so there is no Saytzeff/Hofmann choice to make — only one alkene is geometrically possible.Product: but-$\displaystyle 1$-ene, \(\displaystyle \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3 \).(ii) Bromobenzene + Mg, dry etherMagnesium metal inserts straight into the aryl carbon–bromine bond — dry ether is not incidental, it is what keeps the product from being destroyed the instant it forms.Bromobenzene, \(\displaystyle \text{C}_6\text{H}_5-\text{Br} \) , is stirred with magnesium turnings in ether that has been rigorously dried, because a Grignard reagent reacts instantly (and irreversibly) with any trace of water to regenerate the starting hydrocarbon. At the metal surface, magnesium is oxidised from the $\displaystyle 0$ to the +$\displaystyle 2$ state: one pair of electrons forms a new carbon–magnesium bond and the other forms the magnesium–bromine bond, so the aryl and the bromine end up on the same magnesium atom rather than being separated.Product: phenylmagnesium bromide, \(\displaystyle \text{C}_6\text{H}_5-\text{MgBr} \) — a Grignard reagent, ether-solvated and never isolated dry.(iii) Chlorobenzene subjected to hydrolysisChlorobenzene resists hydrolysis because the chlorine's lone pair is delocalised into the ring — the C–Cl bond gets shorter and stronger, and the ring carbon becomes a poor target for a nucleophile, not a good one.In chlorobenzene, \(\displaystyle \text{C}_6\text{H}_5-\text{Cl} \) , a lone pair on chlorine conjugates with the aromatic \(\displaystyle \pi \) system, giving the C–Cl bond partial double-bond character. Two consequences follow: the bond is shorter and stronger than an ordinary sp³ C–Cl bond (harder to break), and the ring carbon carrying chlorine, being sp² and fed electron density by that same resonance, is a poor electrophile — it does not attract an incoming \(\displaystyle \text{OH}^- \) the way an alkyl halide's carbon does. Because of this, ordinary aqueous hydrolysis (dilute NaOH, room conditions) does essentially nothing to chlorobenzene.Only under forcing industrial conditions — chlorobenzene heated with aqueous NaOH at $\displaystyle 623$ K under about $\displaystyle 300$ atmospheres pressure (the Dow process) — does nucleophilic substitution occur, giving sodium phenoxide, \(\displaystyle \text{C}_6\text{H}_5-\text{O}^-\text{Na}^+ \); acidifying this afterwards liberates phenol.Product: no reaction under ordinary hydrolysis; under Dow's high-temperature, high-pressure conditions, phenol, \(\displaystyle \text{C}_6\text{H}_5-\text{OH} \) (isolated after acidifying the sodium phenoxide intermediate).(iv) Ethyl chloride + aqueous KOHAqueous KOH supplies a strong, unhindered hydroxide ion that substitutes the chlorine outright instead of removing a proton — this is the direct contrast with part (i).Ethyl chloride, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{Cl} \), is a primary halide with an accessible back side opposite the C–Cl bond. Hydroxide ion attacks that carbon from the side directly opposite the leaving chlorine, and as the new C–O bond forms the C–Cl bond breaks, expelling chloride ion in a single concerted step \(\displaystyle \mathrm{(SN_{2})}\) through a five-coordinate transition state. In water, \(\displaystyle \text{OH}^- \) behaves overwhelmingly as a nucleophile rather than a base — the opposite balance from the ethoxide-in-ethanol system of part (i) — which is why substitution wins here while elimination won there.Product: ethanol, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{OH} \) .(v) Methyl bromide + sodium, dry etherSodium metal welds two alkyl halide molecules together end to end, doubling the carbon count — this is the Wurtz reaction.Two molecules of methyl bromide, \(\displaystyle \text{CH}_3-\text{Br} \), react with two atoms of sodium in dry ether. Each sodium atom donates an electron into a C–Br bond, ejecting bromide ion and leaving a methyl fragment; the two methyl fragments then combine, forming a new carbon–carbon bond, while the two sodium atoms end up paired with the two bromide ions as NaBr.\[2\,\text{CH}_3\text{Br} + 2\,\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3-\text{CH}_3 + 2\,\text{NaBr} \]Product: ethane, \(\displaystyle \text{CH}_3-\text{CH}_3 \) .(vi) Methyl chloride + KCNCyanide has two ends that could attack, carbon and nitrogen, and with KCN it is the carbon end that bonds — not because carbon is more electronegative (it isn't), but because the C–C bond that results is more stable than the C–N bond nitrogen-attack would give.Potassium cyanide is essentially fully ionic in solution, releasing a "free" cyanide ion, \(\displaystyle \text{C}\!\equiv\!\text{N}^- \), which is an ambident nucleophile — it can bond through either its carbon or its nitrogen. Methyl chloride, \(\displaystyle \text{CH}_3-\text{Cl} \), undergoes the same back-side \(\displaystyle \mathrm{SN_{2}}\) attack described in part (iv): the nucleophile approaches the carbon opposite the chlorine, and as the new bond forms the C–Cl bond breaks, releasing chloride ion. With ionic KCN, attack occurs preferentially through the carbon atom of cyanide, because the resulting C–C bond is stronger and the product more stable than the isomeric C–N-bonded product would be — this is the reason KCN gives predominantly the nitrile rather than the isonitrile (the opposite happens with covalent AgCN, which reacts through nitrogen to give the isocyanide, but that is not the reagent here).Product: methyl cyanide (ethanenitrile / acetonitrile), \(\displaystyle \text{CH}_3-\text{C}\!\equiv\!\text{N} \).Answer: (i) but-$\displaystyle 1$-ene, \(\displaystyle \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3 \), by E2 elimination; (ii) phenylmagnesium bromide, \(\displaystyle \text{C}_6\text{H}_5\text{MgBr} \); (iii) no reaction under ordinary hydrolysis — phenol, \(\displaystyle \text{C}_6\text{H}_5\text{OH} \), forms only via sodium phenoxide under Dow's process ($\displaystyle 623$ K, ~$\displaystyle 300$ atm); (iv) ethanol, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH} \), by \(\displaystyle \mathrm{SN_{2}}\) substitution; (v) ethane, \(\displaystyle \text{CH}_3-\text{CH}_3 \), by the Wurtz reaction; (vi) methyl cyanide (ethanenitrile), \(\displaystyle \text{CH}_3\text{CN} \), by \(\displaystyle \mathrm{SN_{2}}\) attack through the carbon of \(\displaystyle \text{CN}^- \).