NCERT Solutions · Class 11 Chemistry Hydrocarbons

Chlorination of methane is a free-radical chain reaction, and ethane appears because one of the possible termination steps couples two methyl radicals together.Methane reacts with chlorine only when energy is supplied (UV light or heat around $\displaystyle 570$–$\displaystyle 770$ K), and the reaction does not stop cleanly at methyl chloride. It runs as a three-stage free-radical chain: initiation, propagation, and termination. Ethane's formation only makes sense once you see all three stages, because ethane comes from the termination step, not from the main propagation cycle that makes \(\displaystyle \text{CH}_3\text{Cl} \).Step $\displaystyle 1$ — Initiation: chlorine splits into atomsLight or heat supplies enough energy to break the weak \(\displaystyle \text{Cl–Cl} \) bond homolytically (one electron goes to each atom, giving two neutral atoms each with an unpaired electron — a free radical): \[\text{Cl}_2 \xrightarrow{h\nu} 2\,\text{Cl}^{\bullet} \] This equation is already balanced: $\displaystyle 2$ chlorine atoms in \(\displaystyle \text{Cl}_2\) become $\displaystyle 2$ chlorine atoms as $\displaystyle 2$ \(\displaystyle \text{Cl}^\bullet\) radicals — nothing is created or destroyed, only the bond is broken.Step $\displaystyle 2$ — Propagation: the chain that makes methyl chlorideA chlorine radical is extremely reactive and pulls a hydrogen atom off methane, leaving a methyl radical: \[\text{CH}_4 + \text{Cl}^{\bullet} \longrightarrow \text{CH}_3^{\bullet} + \text{HCl} \] Check the balance: left side has $\displaystyle 1$ C, $\displaystyle 4$ H, $\displaystyle 1$ Cl; right side has $\displaystyle 1$ C, $\displaystyle 3$ H (on the radical) + $\displaystyle 1$ H (on HCl) = $\displaystyle 4$ H, and $\displaystyle 1$ Cl. It balances.The methyl radical \(\displaystyle \text{CH}_3^{\bullet}\) then attacks a fresh \(\displaystyle \text{Cl}_2\) molecule: \[\text{CH}_3^{\bullet} + \text{Cl}_2 \longrightarrow \text{CH}_3\text{Cl} + \text{Cl}^{\bullet} \] Balance check: left side $\displaystyle 1$ C, $\displaystyle 3$ H, $\displaystyle 2$ Cl; right side \(\displaystyle \text{CH}_3\text{Cl}\) has $\displaystyle 1$ C, $\displaystyle 3$ H, $\displaystyle 1$ Cl, plus the freed \(\displaystyle \text{Cl}^\bullet\) makes the second Cl. It balances.This second step regenerates a \(\displaystyle \text{Cl}^{\bullet}\) radical, which goes back into the first propagation step and pulls another hydrogen off another methane molecule. This is why it is called a chain reaction: one initiation event can trigger thousands of these two-step cycles, each one turning a methane molecule into a molecule of methyl chloride (named: chloromethane, formula \(\displaystyle \text{CH}_3\text{Cl}\)).The step people usually skip over is realising that the chain regenerates its own radical — that is what makes a small amount of light or heat enough to convert a large amount of methane, and it is also why the reaction does not simply stop after one substitution.Step $\displaystyle 3$ — Termination: this is where ethane comes fromA chain reaction ends when two radicals meet and combine, using up both unpaired electrons to form a new covalent bond, without regenerating a radical. Three termination combinations are possible, since two kinds of radical (\(\displaystyle \text{Cl}^{\bullet}\) and \(\displaystyle \text{CH}_3^{\bullet}\)) are both present in the reaction mixture: \[\text{Cl}^{\bullet} + \text{Cl}^{\bullet} \longrightarrow \text{Cl}_2 \] \[\text{CH}_3^{\bullet} + \text{Cl}^{\bullet} \longrightarrow \text{CH}_3\text{Cl} \] \[\text{CH}_3^{\bullet} + \text{CH}_3^{\bullet} \longrightarrow \text{CH}_3-\text{CH}_3 \] Each of these is already balanced — a termination step is simply the pairing of two radicals' unpaired electrons into one new bond, so the atoms on each side match automatically ($\displaystyle 2$ C + $\displaystyle 6$ H on both sides of the last equation).The last of these three is the one that matters here: when two methyl radicals happen to collide with each other instead of with a chlorine molecule, they combine to give ethane — a hydrocarbon, formula \(\displaystyle \text{CH}_3-\text{CH}_3\), whose carbon atoms are joined directly by the new bond formed from the two radicals' unpaired electrons. Since methyl radicals are present throughout the reaction (produced continuously in the propagation step), some of them are bound to meet each other rather than a \(\displaystyle \text{Cl}_2\) molecule, especially once the chlorine concentration starts to drop. Ethane is therefore not a product of any deliberate substitution — it is a side product of the chain-termination stage of the free-radical mechanism.Answer: Chlorination of methane proceeds by a free-radical chain mechanism: \(\displaystyle \text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^{\bullet}\) (initiation); \(\displaystyle \text{CH}_4 + \text{Cl}^{\bullet} \rightarrow \text{CH}_3^{\bullet} + \text{HCl}\) and \(\displaystyle \text{CH}_3^{\bullet} + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^{\bullet}\) (propagation). Ethane forms in the termination step, when two methyl radicals collide and combine directly: \(\displaystyle \text{CH}_3^{\bullet} + \text{CH}_3^{\bullet} \rightarrow \text{CH}_3-\text{CH}_3\) (ethane).

IUPAC naming always works the same way: find the longest carbon chain (or the ring, for an aromatic compound) that carries the double/triple bonds, number it so the multiple bonds get the lowest possible locants, then name every branch as a prefix at its own locant. Get the numbering direction wrong and you get a name that is chemically the same molecule but not the accepted IUPAC name — so the working below always checks both directions before choosing.(a) \(\displaystyle \mathrm{CH_3-CH=C(CH_3)_2}\)Write out every carbon: \(\displaystyle \mathrm{CH_3(C1)-CH(C2)=C(C3)(CH_3)(CH_3)}\). The carbon \(\displaystyle \mathrm{C_{3}}\) carries a double bond and two methyl groups — one of those methyls has to be treated as the continuation of the main chain (it's the longest chain rule: a $\displaystyle 4$-carbon chain through the double bond, not a $\displaystyle 3$-carbon one with a stray branch).So the parent chain is but-$\displaystyle 2$-ene, with the remaining methyl group as a substituent on the double-bond carbon. Two ways to number it: The double-bond locant is tied at $\displaystyle 2$ either way, so the tie-breaker is the substituent locant — $\displaystyle 2$ beats 3. That is the step people skip, jumping to whichever numbering they wrote the formula in.(b) \(\displaystyle \mathrm{CH_2=CH-C\equiv C-CH_3}\)Five carbons in a straight line, no branching — number from either end and compare the locant set for the two multiple bonds together, not one bond at a time. \[\text{from the left: double bond at 1, triple bond at 3} \;\Rightarrow\; \{1,3\} \] \[\text{from the right: triple bond at 2, double bond at 4} \;\Rightarrow\; \{2,4\} \] Comparing term by term, \(\displaystyle 1 < 2\), so \(\displaystyle \{1,3\}\) wins regardless of which bond type "should" get priority. Suffix order is always "-ene" before "-yne" in the name (never the reverse), giving pent-$\displaystyle 1$-en-$\displaystyle 3$-yne.(c) \(\displaystyle \mathrm{CH_2=CH-CH=CH_2}\)A four-carbon chain, double bond at each end. Numbering from either end gives the same locant set \(\displaystyle \{1,3\}\) by symmetry, so there's nothing to choose — buta-$\displaystyle 1,3$-diene.(d) A benzene ring carrying the chain \(\displaystyle \mathrm{-CH_2-CH_2-CH=CH_2}\), i.e. \(\displaystyle \mathrm{C_6H_5-CH_2-CH_2-CH=CH_2}\)Here the double bond sits on the chain, not the ring, so the chain — not the ring — is the parent (the ring becomes the substituent "phenyl"). Numbering the four-carbon chain from the double-bond end (to give the double bond the lower locant, as always) puts the phenyl group on \(\displaystyle \mathrm{C_{4}}\): \[\underbrace{\mathrm{CH_2}}_{C1}=\underbrace{\mathrm{CH}}_{C2}-\underbrace{\mathrm{CH_2}}_{C3}-\underbrace{\mathrm{CH_2}}_{C4}-\mathrm{C_6H_5} \] giving $\displaystyle 4$-phenylbut-$\displaystyle 1$-ene. The trap here is reflexively naming the ring as the parent because it's the "bigger" piece of the drawing — the parent is chosen by which fragment carries the point of unsaturation and gives it the lowest locant, not by carbon count of the ring versus the chain.(e) A benzene ring carrying \(\displaystyle \mathrm{-OH}\) and \(\displaystyle \mathrm{-CH_3}\) on adjacent (ortho) ring carbons\(\displaystyle \mathrm{-OH}\) on a benzene ring is named with the retained parent name "phenol", and the hydroxyl carbon is always \(\displaystyle \mathrm{C_{1}}\) by definition of that name — it isn't something you number to minimize. The methyl group sits on the very next ring carbon, so it is C2. That gives $\displaystyle 2$-methylphenol (its everyday name is o-cresol, "o-" short for ortho, but the systematic answer is $\displaystyle 2$-methylphenol).(f) \(\displaystyle \mathrm{CH_3(CH_2)_4-CH(CH_2)_3CH_3}\) with a \(\displaystyle \mathrm{-CH_2-CH(CH_3)_2}\) branch hanging off the middle \(\displaystyle \mathrm{CH}\)First find the longest chain — don't just accept the one drawn horizontally. Counting along the drawn line: \(\displaystyle \mathrm{CH_3(1)-CH_2(2)-CH_2(3)-CH_2(4)-CH_2(5)-CH(6)-CH_2(7)-CH_2(8)-CH_2(9)-CH_3(10)}\) — ten carbons, a decane. Could routing through the branch instead give something longer? The branch is \(\displaystyle \mathrm{-CH_2-CH(CH_3)_2}\), only $\displaystyle 4$ carbons long from the attachment point; going into it from either side of the main chain tops out at $\displaystyle 8$–$\displaystyle 9$ carbons, shorter than 10. So the horizontal line drawn in the question really is the longest chain — checking it is the point of the exercise, not a formality.The branch itself, read from its point of attachment, is \(\displaystyle \mathrm{-CH_2-CH(CH_3)-CH_3}\) with a methyl also sitting on that second carbon: a propyl chain with a methyl on \(\displaystyle \mathrm{C_{2}}\), i.e. "$\displaystyle 2$-methylpropyl" (the everyday name is isobutyl).Numbering the ten-carbon chain to give this one substituent the lower locant: from the branch's near end it sits at \(\displaystyle \mathrm{C_{6}}\), from the far end at \(\displaystyle 10-6+1=5\). Five is lower, so number from that end: $\displaystyle 5$-($\displaystyle 2$-methylpropyl)decane.(g) \(\displaystyle \mathrm{CH_3-CH=CH-CH_2-CH=CH-CH(C_2H_5)-CH_2-CH=CH_2}\)Ten carbons, three double bonds, one ethyl branch on the seventh carbon (counting from the left \(\displaystyle \mathrm{CH_3}\)). Label the carbons left to right \(\displaystyle L1\) to \(\displaystyle L10\): double bonds sit at \(\displaystyle L2\)=\(\displaystyle L3\), \(\displaystyle L5\)=\(\displaystyle L6\), \(\displaystyle L9\)=\(\displaystyle L10\), and the ethyl group is on \(\displaystyle L7\).Compare the two numbering directions by the full set of double-bond locants, since three double bonds must all be counted together before you even look at the branch: \[\text{left-to-right: } \{2,5,9\} \qquad \text{right-to-left: } \{1,5,8\} \] At the first point of difference, \(\displaystyle 1<2\), so numbering from the right wins. Under that numbering the double bonds fall at $\displaystyle 1$, $\displaystyle 5$, $\displaystyle 8$, and the carbon that was \(\displaystyle L7\) becomes position $\displaystyle 4$ (it is now the fourth carbon from that end), carrying the ethyl group.This gives $\displaystyle 4$-ethyldeca-$\displaystyle 1,5,8$-triene. The trap is stopping after comparing just one double bond's locant instead of the whole set — a numbering that wins on the first bond can still lose once all three are compared.Answer: (a) $\displaystyle 2$-methylbut-$\displaystyle 2$-ene (b) pent-$\displaystyle 1$-en-$\displaystyle 3$-yne (c) buta-$\displaystyle 1,3$-diene (d) $\displaystyle 4$-phenylbut-$\displaystyle 1$-ene (e) $\displaystyle 2$-methylphenol (o-cresol) (f) $\displaystyle 5$-($\displaystyle 2$-methylpropyl)decane (g) $\displaystyle 4$-ethyldeca-$\displaystyle 1,5,8$-triene

The molecular formula alone fixes how much unsaturation must be present — after that, every isomer is just a different place to put the double or triple bond on a different carbon skeleton.For a hydrocarbon \(\displaystyle \text{C}_c\text{H}_h \), the degree of unsaturation is \[\text{DoU} = \frac{2c+2-h}{2} \] where \(\displaystyle c\) is the number of carbon atoms and \(\displaystyle h\) the number of hydrogen atoms; each ring or double bond uses up $\displaystyle 1$ unit, and each triple bond uses up $\displaystyle 2$ units (a triple bond is "worth" two double bonds' worth of missing hydrogens).Checking this against what each part already tells you: (a) \(\displaystyle \text{C}_4\text{H}_8 \), one double bondOnly two carbon skeletons exist for four carbons: the straight chain and the one with a single methyl branch. Slide the double bond along each, keeping the numbering rule that the point of unsaturation gets the lowest possible locant.Straight chain: The step that is easy to miss here: in but-$\displaystyle 1$-ene, one end of the double bond is \(\displaystyle \mathrm{=\text{CH}_2}\), which carries two identical H's, so there is only one way to arrange it in space. In but-$\displaystyle 2$-ene, by contrast, each doubly-bonded carbon carries two different groups — one H and one \(\displaystyle \text{CH}_3\) and a C=C bond cannot rotate. That gives two genuinely different molecules from the one structural formula: \(\displaystyle \text{CH}_3\) groups on the same side is cis-but-$\displaystyle 2$-ene, and on opposite sides is trans-but-$\displaystyle 2$-ene. Writing "but-$\displaystyle 2$-ene" without noting this hides two of your isomers.Branched chain: the only place a double bond can sit on a four-carbon chain with one methyl branch is between the branch carbon and a terminal carbon: \(\displaystyle \text{CH}_2\text{=C(CH}_3\text{)-CH}_3 \) — $\displaystyle 2$-methylprop-$\displaystyle 1$-ene (also called isobutylene). Here the branch carbon carries two \(\displaystyle \text{CH}_3\) groups that are identical to each other, so there is no cis/trans version of this one.That is every possibility — a double bond one carbon further along the branched skeleton would just be relabelling the same molecule from the other end, and a branch anywhere else would either exceed four carbons or duplicate the straight chain.So \(\displaystyle \text{C}_4\text{H}_8 \) with one double bond gives four isomers: (Quick self-check: each formula has $\displaystyle 4$ carbons and, counting every H, adds up to $\displaystyle 8$ — matching \(\displaystyle \text{C}_4\text{H}_8 \).)(b) \(\displaystyle \text{C}_5\text{H}_8 \), one triple bondThe rule that controls this one: a carbon at either end of a C≡C bond is sp-hybridised and linear, so it has room for only one other atom or group besides its triple-bond partner. A terminal alkyne carbon uses that one slot for an H; an internal alkyne carbon uses it to continue the chain. Either way, a triple-bonded carbon can never itself be a branch point — that is the step people get wrong, trying to hang a methyl group directly off a \(\displaystyle \text{C}\equiv\text{C}\) carbon, which its geometry simply does not allow. Branches can only sit on the ordinary (sp3) carbons elsewhere in the chain.Straight chain ($\displaystyle 5$ carbons): (Triple bond at C3–C4 is pent-$\displaystyle 2$-yne again, renumbered from the other end; at C4–C5 it is pent-$\displaystyle 1$-yne again — so the straight chain gives only these two distinct compounds.)Branched chain ($\displaystyle 4$-carbon main chain, one methyl branch): the triple bond must sit at one end (C1–C2), because a triple bond at C2–C3 of a four-carbon chain — \(\displaystyle \text{CH}_3\text{-C}{\equiv}\text{C-CH}_3 \), but-$\displaystyle 2$-yne — has only terminal \(\displaystyle \text{CH}_3\) carbons left to branch from, and putting a methyl on either of those just extends the chain into pent-$\displaystyle 2$-yne, which is already counted, not a new isomer. With the triple bond at C1–C2, the only sp3 carbon with a spare hydrogen to replace is \(\displaystyle \mathrm{C_{3}}\): \(\displaystyle \text{HC}{\equiv}\text{C-CH(CH}_3\text{)-CH}_3 \) — $\displaystyle 3$-methylbut-$\displaystyle 1$-yne. The chain is numbered from the triple-bond end because the point of unsaturation always gets the lower locant before substituents are considered, so it is "but-$\displaystyle 1$-yne," not "but-$\displaystyle 3$-yne."So \(\displaystyle \text{C}_5\text{H}_8 \) with one triple bond gives three isomers (no cis/trans possibilities here — a linear triple bond has no "sides" to be on): (Self-check: each has $\displaystyle 5$ carbons, and the hydrogens on every formula total $\displaystyle 8$ — matching \(\displaystyle \text{C}_5\text{H}_8 \).)**Answer: (a) \(\displaystyle \text{C}_4\text{H}_8 \) with one double bond gives but-$\displaystyle 1$-ene \(\displaystyle \text{CH}_2\text{=CH-CH}_2\text{-CH}_3 \), cis- and trans-but-$\displaystyle 2$-ene \(\displaystyle \text{CH}_3\text{-CH=CH-CH}_3 \), and $\displaystyle 2$-methylprop-$\displaystyle 1$-ene \(\displaystyle \text{CH}_2\text{=C(CH}_3\text{)-CH}_3 \). (b) \(\displaystyle \text{C}_5\text{H}_8 \) with one triple bond gives pent-$\displaystyle 1$-yne \(\displaystyle \text{HC}{\equiv}\text{C-CH}_2\text{-CH}_2\text{-CH}_3 \), pent-$\displaystyle 2$-yne \(\displaystyle \text{CH}_3\text{-C}{\equiv}\text{C-CH}_2\text{-CH}_3 \), and $\displaystyle 3$-methylbut-$\displaystyle 1$-yne \(\displaystyle \text{HC}{\equiv}\text{C-CH(CH}_3\text{)-CH}_3 \).

Ozonolysis snaps the C=C double bond in half and caps each broken end with a =O; whether that end turns into an aldehyde or a ketone depends only on how many H atoms sat on that particular alkene carbon.The reaction happens in two stages. First, ozone adds across the double bond to give a cyclic ozonide (all three oxygen atoms of \(\displaystyle \mathrm{O_3} \) are incorporated, so no atoms are lost): \[\mathrm{R_2C{=}CR'_2 + O_3 \rightarrow ozonide\ (R_2C(O{-}O{-}O)CR'_2)} \] Second, the ozonide is decomposed with zinc dust and water. Zinc reduces the peroxide-type intermediate, splitting the O–O–O bridge and handing one oxygen atom to each carbon as a C=O: \[\mathrm{ozonide + Zn + H_2O \rightarrow (carbonyl_1) + (carbonyl_2) + Zn(OH)_2} \] In every equation below, the carbons and hydrogens of the ozonide reappear unchanged, split between the two carbonyl fragments — that is how each equation is balanced; the extra oxygen and the water's two hydrogens end up in \(\displaystyle \mathrm{Zn(OH)_2} \).The step people skip: it must be Zn dust + water, not plain water. Hydrolysing the ozonide with water alone releases hydrogen peroxide, which is a strong enough oxidant to push an aldehyde on to a carboxylic acid. Zn is added specifically to consume that peroxide (Zn gets oxidised to \(\displaystyle \mathrm{Zn(OH)_2} \) instead), so the aldehyde survives as an aldehyde.The rule for reading off the products: look at what is attached to each of the two doubly-bonded carbons. (i) Pent-$\displaystyle 2$-ene, \(\displaystyle \mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_3} \) (numbering C1–C5, double bond \(\displaystyle \mathrm{C_{2}}\)=C3). \[\mathrm{C_5H_{10}O_3\ (ozonide) + Zn + H_2O \rightarrow CH_3CHO + CH_3CH_2CHO + Zn(OH)_2} \] Check: carbons $\displaystyle 2$ + $\displaystyle 3$ = $\displaystyle 5$ ✓; hydrogens ($\displaystyle 10$ + $\displaystyle 2$ on the left) = ($\displaystyle 4$ + $\displaystyle 6$ + $\displaystyle 2$) on the right = $\displaystyle 12$ ✓; oxygens ($\displaystyle 3$ + $\displaystyle 1$) = ($\displaystyle 1$ + $\displaystyle 1$ + $\displaystyle 2$) = $\displaystyle 4$ ✓.Products: ethanal (acetaldehyde), \(\displaystyle \mathrm{CH_3CHO} \), and propanal (propionaldehyde), \(\displaystyle \mathrm{CH_3CH_2CHO} \).(ii) $\displaystyle 3,4$-Dimethylhept-$\displaystyle 3$-ene. The parent chain is hept-$\displaystyle 3$-ene, C1–C7 with the double bond at \(\displaystyle \mathrm{C_{3}}\)=\(\displaystyle \mathrm{C_{4}}\), and a methyl branch sits on each of those two carbons: \(\displaystyle \mathrm{CH_3{-}CH_2{-}C(CH_3){=}C(CH_3){-}CH_2{-}CH_2{-}CH_3} \). \[\mathrm{C_9H_{18}O_3\ (ozonide) + Zn + H_2O \rightarrow CH_3COCH_2CH_3 + CH_3COCH_2CH_2CH_3 + Zn(OH)_2} \] Check: carbons $\displaystyle 4$ + $\displaystyle 5$ = $\displaystyle 9$ ✓; hydrogens ($\displaystyle 18$ + $\displaystyle 2$) = ($\displaystyle 8$ + $\displaystyle 10$ + $\displaystyle 2$) = $\displaystyle 20$ ✓; oxygens ($\displaystyle 3$ + $\displaystyle 1$) = ($\displaystyle 1$ + $\displaystyle 1$ + $\displaystyle 2$) = $\displaystyle 4$ ✓.Products: butan-$\displaystyle 2$-one (methyl ethyl ketone), \(\displaystyle \mathrm{CH_3COCH_2CH_3} \), and pentan-$\displaystyle 2$-one (methyl propyl ketone), \(\displaystyle \mathrm{CH_3COCH_2CH_2CH_3} \).(iii) $\displaystyle 2$-Ethylbut-$\displaystyle 1$-ene. But-$\displaystyle 1$-ene is \(\displaystyle \mathrm{CH_2{=}CH{-}CH_2{-}CH_3} \) (C1=C2–C3–C4); putting an ethyl group on \(\displaystyle \mathrm{C_{2}}\) in place of its one H gives \(\displaystyle \mathrm{CH_2{=}C(CH_2CH_3){-}CH_2{-}CH_3} \) — \(\displaystyle \mathrm{C_{2}}\) now bears two identical ethyl groups (the substituent, and the C3–C4 chain). \[\mathrm{C_6H_{12}O_3\ (ozonide) + Zn + H_2O \rightarrow HCHO + CH_3CH_2COCH_2CH_3 + Zn(OH)_2} \] Check: carbons $\displaystyle 1$ + $\displaystyle 5$ = $\displaystyle 6$ ✓; hydrogens ($\displaystyle 12$ + $\displaystyle 2$) = ($\displaystyle 2$ + $\displaystyle 10$ + $\displaystyle 2$) = $\displaystyle 14$ ✓; oxygens ($\displaystyle 3$ + $\displaystyle 1$) = ($\displaystyle 1$ + $\displaystyle 1$ + $\displaystyle 2$) = $\displaystyle 4$ ✓.Products: methanal (formaldehyde), \(\displaystyle \mathrm{HCHO} \), and pentan-$\displaystyle 3$-one (diethyl ketone), \(\displaystyle \mathrm{CH_3CH_2COCH_2CH_3} \).(iv) $\displaystyle 1$-Phenylbut-$\displaystyle 1$-ene. But-$\displaystyle 1$-ene's \(\displaystyle \mathrm{C_{1}}\) (normally \(\displaystyle \mathrm{=CH_2} \)) has one of its H's replaced by a phenyl group: \(\displaystyle \mathrm{C_6H_5{-}CH{=}CH{-}CH_2{-}CH_3} \) (double bond \(\displaystyle \mathrm{C_{1}}\)=C2). \[\mathrm{C_{10}H_{12}O_3\ (ozonide) + Zn + H_2O \rightarrow C_6H_5CHO + CH_3CH_2CHO + Zn(OH)_2} \] Check: carbons $\displaystyle 7$ + $\displaystyle 3$ = $\displaystyle 10$ ✓; hydrogens ($\displaystyle 12$ + $\displaystyle 2$) = ($\displaystyle 6$ + $\displaystyle 6$ + $\displaystyle 2$) = $\displaystyle 14$ ✓; oxygens ($\displaystyle 3$ + $\displaystyle 1$) = ($\displaystyle 1$ + $\displaystyle 1$ + $\displaystyle 2$) = $\displaystyle 4$ ✓.Products: benzaldehyde (the retained IUPAC name; systematically benzenecarbaldehyde), \(\displaystyle \mathrm{C_6H_5CHO} \), and propanal (propionaldehyde), \(\displaystyle \mathrm{CH_3CH_2CHO} \).Answer: (i) ethanal ( \(\displaystyle \mathrm{CH_3CHO} \) ) and propanal ( \(\displaystyle \mathrm{CH_3CH_2CHO} \) ). (ii) butan-$\displaystyle 2$-one ( \(\displaystyle \mathrm{CH_3COCH_2CH_3} \) ) and pentan-$\displaystyle 2$-one ( \(\displaystyle \mathrm{CH_3COCH_2CH_2CH_3} \) ). (iii) methanal ( \(\displaystyle \mathrm{HCHO} \) ) and pentan-$\displaystyle 3$-one ( \(\displaystyle \mathrm{CH_3CH_2COCH_2CH_3} \) ). (iv) benzaldehyde ( \(\displaystyle \mathrm{C_6H_5CHO} \) ) and propanal ( \(\displaystyle \mathrm{CH_3CH_2CHO} \) ).

Ozonolysis snips the C=C double bond exactly in half, and each of the two carbons that were joined by it turns into the carbon of its own C=O group — so to find the alkene, you run that cleavage backwards, starting from the two carbonyl products you were given.The general reaction is\[\text{R}_1\text{R}_2\text{C}=\text{CR}_3\text{R}_4 \;\xrightarrow{\text{(i) O}_3}\; \text{ozonide} \;\xrightarrow{\text{(ii) Zn, H}_2\text{O}}\; \text{R}_1\text{R}_2\text{C}=\text{O} \;+\; \text{O}=\text{CR}_3\text{R}_4 \]Step (i) is simple addition of ozone across the double bond, giving a cyclic ozonide. Step (ii) is where people trip up: splitting the ozonide also releases hydrogen peroxide alongside the two carbonyl compounds, and if that peroxide is left in solution it will oxidise an aldehyde product straight on to a carboxylic acid. Using zinc dust with the water destroys the peroxide as it forms, so the aldehyde/ketone comes out clean — this is why the reagent is always written as Zn/H₂O (or Zn/AcOH), never water alone.Reading the two carbonyl carbons off the productsEthanal, \(\displaystyle \text{CH}_3-\text{CHO} \): its carbonyl carbon carries one \(\displaystyle \text{CH}_3\) group and one H.Pentan-$\displaystyle 3$-one, \(\displaystyle \text{CH}_3\text{CH}_2-\text{CO}-\text{CH}_2\text{CH}_3 \): its carbonyl carbon carries two \(\displaystyle \text{C}_2\text{H}_5\) (ethyl) groups and no hydrogen at all. That absence of H on the carbonyl carbon is exactly what makes it a ketone rather than an aldehyde — carry that fact backwards and this carbon of the alkene must have had two alkyl substituents, not one alkyl group and one H.Rebuilding the alkeneReverse the cleavage: delete the oxygen from each carbonyl carbon and join the two carbons with a double bond in its place.\[\underbrace{\text{CH}_3-\text{CH}}_{\text{from ethanal}}\;=\;\underbrace{\text{C}(\text{C}_2\text{H}_5)_2}_{\text{from pentan-3-one}} \]So alkene A is \(\displaystyle \text{CH}_3-\text{CH}=\text{C}(\text{CH}_2\text{CH}_3)_2 \). In words: take a chain of five carbons with a double bond between the second and third carbon; the third carbon (the far end of that double bond) also carries a second ethyl group hanging off it, in addition to the one that continues the main chain.Checking this by balancing the forward ozonolysis of ACounting atoms in A: \(\displaystyle \text{CH}_3\) ($\displaystyle 3$ H) + \(\displaystyle \text{CH}=\) ($\displaystyle 1$ H) + \(\displaystyle \text{C}\) with two \(\displaystyle \text{C}_2\text{H}_5\) groups ($\displaystyle 0$ H on this carbon itself, $\displaystyle 10$ H on the two ethyls) gives \(\displaystyle \text{C}_7\text{H}_{14}\) — one double bond, so \(\displaystyle \text{C}_n\text{H}_{2n}\) with \(\displaystyle n=7\), as it should be for an open-chain monoene.Step (i), addition of ozone — balanced automatically because addition puts every atom from both reactants into one product, nothing lost:\[\text{C}_7\text{H}_{14}\ (\text{A}) + \text{O}_3 \;\longrightarrow\; \text{C}_7\text{H}_{14}\text{O}_3\ (\text{ozonide}) \]Step (ii), reductive cleavage with zinc:\[\text{C}_7\text{H}_{14}\text{O}_3\ (\text{ozonide}) + \text{Zn} \;\longrightarrow\; \text{CH}_3\text{CHO} + (\text{C}_2\text{H}_5)_2\text{C}=\text{O} + \text{ZnO} \]Check the count on both sides: left has C\(\displaystyle _7\)H\(\displaystyle _{14}\)O\(\displaystyle _3\) plus one Zn. On the right, ethanal is C\(\displaystyle _2\)H\(\displaystyle _4\)O and pentan-$\displaystyle 3$-one is C\(\displaystyle _5\)H\(\displaystyle _{10}\)O, which together already account for C\(\displaystyle _7\)H\(\displaystyle _{14}\)O\(\displaystyle _2\); the ozonide's third oxygen — the one that doesn't end up in either carbonyl group — is exactly the one that leaves as ZnO. Both sides now read C = $\displaystyle 7$, H = $\displaystyle 14$, O = $\displaystyle 3$, Zn = $\displaystyle 1$, so the equation is balanced, and the two products are precisely ethanal and pentan-$\displaystyle 3$-one, matching the question.Naming APick the longest chain that runs through the double bond: using one of the two ethyl groups on the right-hand carbon to extend the chain gives a five-carbon chain, \(\displaystyle \text{C}_1(\text{CH}_3)-\text{C}_2(\text{CH})=\text{C}_3(\text{C})-\text{C}_4(\text{CH}_2)-\text{C}_5(\text{CH}_3)\), with the other ethyl group left over as a branch on \(\displaystyle \text{C}_3\).Number that chain to give the double bond the lower locant. Numbering from the left puts the double bond at $\displaystyle 2$–$\displaystyle 3$ (locant $\displaystyle 2$); numbering from the right would put it at $\displaystyle 3$–$\displaystyle 4$ (locant $\displaystyle 3$). Lower wins, so numbering runs left to right, and the ethyl branch sits at C3.That makes the parent name pent-$\displaystyle 2$-ene with a $\displaystyle 3$-ethyl substituent: $\displaystyle 3$-ethylpent-$\displaystyle 2$-ene. No cis/trans (E/Z) label is needed, because \(\displaystyle \mathrm{C_{3}}\) — one of the two double-bond carbons — carries two identical ethyl groups, so there's no distinct "other side" at that end of the double bond to compare against.Answer: A is \(\displaystyle \text{CH}_3-\text{CH}=\text{C}(\text{C}_2\text{H}_5)_2 \), i.e. a pentene chain carrying an extra ethyl group on \(\displaystyle \mathrm{C_{3}}\); its IUPAC name is $\displaystyle 3$-ethylpent-$\displaystyle 2$-ene. Ozonolysis (O₃ then Zn/H₂O) cleaves the \(\displaystyle \mathrm{C_{2}}\)=\(\displaystyle \mathrm{C_{3}}\) bond to give \(\displaystyle \text{CH}_3-\text{CH}=\text{O}\) (ethanal) from the \(\displaystyle \mathrm{C_{2}}\) end and \(\displaystyle (\text{C}_2\text{H}_5)_2\text{C}=\text{O}\) (pentan-$\displaystyle 3$-one) from the \(\displaystyle \mathrm{C_{3}}\) end.

Ozonolysis snips the C=C bond and puts an oxygen on each of the two carbons that were joined by it — so to get back to the alkene, undo exactly that: strip the oxygen off each carbonyl carbon and weld the two carbons together with a double bond.When an alkene is treated with ozone and the ozonide formed is then split apart with zinc dust and water — written as \(\displaystyle \text{O}_3\) followed by \(\displaystyle \text{Zn}/\text{H}_2\text{O}\) — each of the two alkene carbons turns into a carbonyl carbon, \(\displaystyle \text{C=O}\). Nothing else attached to that carbon changes: every H and every alkyl group that was already sitting on an alkene carbon is still sitting on it once it becomes a carbonyl carbon. So the substituents on each product's carbonyl carbon tell you exactly what was on that alkene carbon.Read the substituents off each productPropanal, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{CHO}\) (an aldehyde; common name propionaldehyde), has its carbonyl carbon attached to only two things besides the oxygen: one hydrogen atom and one ethyl group, \(\displaystyle -\text{C}_2\text{H}_5\).Pentan-$\displaystyle 3$-one, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3\) (a ketone, carbonyl on carbon $\displaystyle 3$; common name diethyl ketone), has its carbonyl carbon attached to two ethyl groups, \(\displaystyle -\text{C}_2\text{H}_5\) and \(\displaystyle -\text{C}_2\text{H}_5\) — never an H, because a ketone's carbonyl carbon can't carry one.The step people rush is gluing the two molecules together at their oxygens. What actually happens is the opposite: the oxygen leaves each carbonyl carbon, and it is those two now-oxygen-free carbons that become the two ends of the new C=C bond. Everything else each carbon was already holding — the lone H on propanal's carbon, both ethyls on pentan-$\displaystyle 3$-one's carbon — comes across unchanged.Rebuild the alkeneDelete the oxygen from propanal's carbonyl carbon (leaving it holding H and \(\displaystyle -\text{C}_2\text{H}_5\)) and from pentan-$\displaystyle 3$-one's carbonyl carbon (leaving it holding two \(\displaystyle -\text{C}_2\text{H}_5\) groups), then join the two bare carbons with a double bond:\[\text{CH}_3-\text{CH}_2-\text{CH}=\text{C}(\text{C}_2\text{H}_5)_2 \]That is the structural (condensed) formula of the alkene: an ethyl group and an H on the left-hand double-bond carbon (inherited from propanal), two ethyl groups on the right-hand one (inherited from pentan-$\displaystyle 3$-one).Name it and check the atom countCarbons: propanal supplies $\displaystyle 3$, pentan-$\displaystyle 3$-one supplies $\displaystyle 5$, so the alkene has \(\displaystyle 3+5=8\) carbons — molecular formula \(\displaystyle \text{C}_8\text{H}_{16}\). Hydrogens check out the same way: propanal (\(\displaystyle \text{C}_3\text{H}_6\text{O}\)) has $\displaystyle 6$ H, pentan-$\displaystyle 3$-one (\(\displaystyle \text{C}_5\text{H}_{10}\text{O}\)) has $\displaystyle 10$ H, and \(\displaystyle 6+10=16\), matching \(\displaystyle \text{C}_8\text{H}_{16}\) (an alkene follows \(\displaystyle \text{C}_n\text{H}_{2n}\), and \(\displaystyle 2\times8=16\)).To name it, trace the longest chain through the double bond. Starting from propanal's end, \(\displaystyle \text{CH}_3-\text{CH}_2-\text{CH}=\), the chain reaches the right-hand carbon, which carries two ethyl branches — but a chain can only continue through one of them, not fork into both. Following just one ethyl group onward gives a chain of $\displaystyle 6$ carbons (hex-), with the double bond falling between carbons $\displaystyle 3$ and $\displaystyle 4$ either way you number this particular chain — so it comes out as a "$\displaystyle 3$-ene" from both ends, a tie. The tie is broken by the leftover ethyl group (the one the chain didn't absorb): numbering from the end that puts it on carbon $\displaystyle 3$ rather than carbon $\displaystyle 4$ gives it the lower locant. That makes the name $\displaystyle 3$-ethylhex-$\displaystyle 3$-ene.The trap here is stopping at something like "ethylpentene" because only five carbons look chain-like at first glance — the two ethyl groups on that one carbon are two separate two-carbon branches, and only one of them can be folded into the main chain. The other always has to be reported separately, as the "$\displaystyle 3$-ethyl" prefix.Check it by running ozonolysis forwardOzonolysis happens in two steps: ozone adds across the double bond to give a cyclic ozonide, and zinc with water then cleaves that ozonide into the two carbonyl products.\[\text{C}_8\text{H}_{16} + \text{O}_3 \longrightarrow \text{C}_8\text{H}_{16}\text{O}_3 \]This needs no coefficients to balance: all $\displaystyle 8$ carbons and $\displaystyle 16$ hydrogens of the alkene pass into the ozonide untouched, and all $\displaystyle 3$ oxygens of \(\displaystyle \text{O}_3\) end up in it too — $\displaystyle 8$ C, $\displaystyle 16$ H, $\displaystyle 3$ O on both sides.\[\text{C}_8\text{H}_{16}\text{O}_3 + \text{Zn} \longrightarrow \underbrace{\text{C}_3\text{H}_6\text{O}}_{\text{propanal}} \ + \ \underbrace{\text{C}_5\text{H}_{10}\text{O}}_{\text{pentan-3-one}} \ + \ \text{ZnO} \]Balance check, atom by atom: carbon, \(\displaystyle 3+5=8\), matching the $\displaystyle 8$ on the left; hydrogen, \(\displaystyle 6+10=16\), matching the $\displaystyle 16$ on the left; oxygen, the ozonide's $\displaystyle 3$ oxygens split one to propanal, one to pentan-$\displaystyle 3$-one, and one taken up by zinc as \(\displaystyle \text{ZnO}\), so \(\displaystyle 1+1+1=3\), matching the $\displaystyle 3$ on the left; zinc, one atom on each side. Every atom lands somewhere on the right, and the products it lands as are exactly propanal and pentan-$\displaystyle 3$-one — confirming the alkene.Answer: \(\displaystyle \text{CH}_3-\text{CH}_2-\text{CH}=\text{C}(\text{C}_2\text{H}_5)_2\), i.e. $\displaystyle 3$-ethylhex-$\displaystyle 3$-ene (\(\displaystyle \text{C}_8\text{H}_{16}\)) — one double-bond carbon carrying an H and an ethyl group (giving propanal on ozonolysis), the other carrying two ethyl groups (giving pentan-$\displaystyle 3$-one).

Complete combustion turns every carbon atom of a hydrocarbon into carbon dioxide, \(\displaystyle \mathrm{CO_2}\), and every hydrogen atom into water, \(\displaystyle \mathrm{H_2O}\); oxygen, \(\displaystyle \mathrm{O_2}\), is the atom to balance last because it is split between both products.The order that keeps this fast: fix the \(\displaystyle \mathrm{CO_2}\) coefficient from the carbon count, fix the \(\displaystyle \mathrm{H_2O}\) coefficient from the hydrogen count, then add up how many oxygen atoms those two totals demand and supply that many via \(\displaystyle \mathrm{O_2}\) — doubling every coefficient at the end if the oxygen count comes out odd, since a finished equation cannot carry half a molecule.(i) Butane. Butane is the four-carbon alkane, \(\displaystyle \mathrm{C_4H_{10}}\) (alkanes follow \(\displaystyle \mathrm{C_nH_{2n+2}}\), here \(\displaystyle n=4\)): $\displaystyle 4$ carbons and $\displaystyle 10$ hydrogens per molecule. \[\mathrm{C_4H_{10}} + \mathrm{O_2} \rightarrow 4\,\mathrm{CO_2} + 5\,\mathrm{H_2O} \] Oxygen on the right: \(\displaystyle 4\times2\) (from \(\displaystyle \mathrm{CO_2}\)) \(\displaystyle +\ 5\times1\) (from \(\displaystyle \mathrm{H_2O}\)) \(\displaystyle =13\) atoms, i.e. \(\displaystyle \tfrac{13}{2}\,\mathrm{O_2}\). A fractional \(\displaystyle \mathrm{O_2}\) is not a valid final coefficient, so every coefficient is doubled: \[2\,\mathrm{C_4H_{10}} + 13\,\mathrm{O_2} \rightarrow 8\,\mathrm{CO_2} + 10\,\mathrm{H_2O} \] Check: C, \(\displaystyle 2\times4=8\); H, \(\displaystyle 2\times10=20=10\times2\); O, \(\displaystyle 13\times2=26=8\times2+10\times1\). All three match.(ii) Pentene. Pentene is a five-carbon alkene, \(\displaystyle \mathrm{C_5H_{10}}\) (alkenes follow \(\displaystyle \mathrm{C_nH_{2n}}\), here \(\displaystyle n=5\)). An aside: whether the double bond sits at \(\displaystyle \mathrm{C_{1}}\), \(\displaystyle \mathrm{C_{2}}\), or elsewhere (pent-$\displaystyle 1$-ene, pent-$\displaystyle 2$-ene, …), the molecular formula — and so the combustion equation — is exactly the same, because combustion counts only atoms, never their arrangement. \[\mathrm{C_5H_{10}} + \mathrm{O_2} \rightarrow 5\,\mathrm{CO_2} + 5\,\mathrm{H_2O} \] Oxygen needed: \(\displaystyle 5\times2+5\times1=15\) atoms \(\displaystyle =\tfrac{15}{2}\,\mathrm{O_2}\). Doubling to clear the fraction: \[2\,\mathrm{C_5H_{10}} + 15\,\mathrm{O_2} \rightarrow 10\,\mathrm{CO_2} + 10\,\mathrm{H_2O} \] Check: C, \(\displaystyle 10=10\); H, \(\displaystyle 20=20\); O, \(\displaystyle 30=20+10\).(iii) Hexyne. Hexyne is a six-carbon alkyne, \(\displaystyle \mathrm{C_6H_{10}}\) (alkynes follow \(\displaystyle \mathrm{C_nH_{2n-2}}\), here \(\displaystyle n=6\)). This is the step people slip on: a triple bond removes two more hydrogens than a double bond does for the same chain length, so the alkyne count is \(\displaystyle 2n-2\), not \(\displaystyle 2n\). \[\mathrm{C_6H_{10}} + \mathrm{O_2} \rightarrow 6\,\mathrm{CO_2} + 5\,\mathrm{H_2O} \] Oxygen needed: \(\displaystyle 6\times2+5\times1=17\) atoms \(\displaystyle =\tfrac{17}{2}\,\mathrm{O_2}\). Doubling: \[2\,\mathrm{C_6H_{10}} + 17\,\mathrm{O_2} \rightarrow 12\,\mathrm{CO_2} + 10\,\mathrm{H_2O} \] Check: C, \(\displaystyle 12=12\); H, \(\displaystyle 20=20\); O, \(\displaystyle 34=24+10\).(iv) Toluene. Toluene (methylbenzene) is the benzene ring with one ring hydrogen replaced by a methyl group, \(\displaystyle \mathrm{-CH_3}\): $\displaystyle 6$ ring carbons plus $\displaystyle 1$ methyl carbon gives \(\displaystyle \mathrm{C_7H_8}\) ($\displaystyle 5$ remaining ring hydrogens plus $\displaystyle 3$ methyl hydrogens \(\displaystyle =8\)). \[\mathrm{C_7H_8} + \mathrm{O_2} \rightarrow 7\,\mathrm{CO_2} + 4\,\mathrm{H_2O} \] Oxygen needed: \(\displaystyle 7\times2+4\times1=18\) atoms \(\displaystyle =9\,\mathrm{O_2}\) — this one lands on a whole number directly, so no doubling is needed: \[\mathrm{C_7H_8} + 9\,\mathrm{O_2} \rightarrow 7\,\mathrm{CO_2} + 4\,\mathrm{H_2O} \] Check: C, \(\displaystyle 7=7\); H, \(\displaystyle 8=8\); O, \(\displaystyle 18=14+4\).**Answer: (i) \(\displaystyle 2\,\mathrm{C_4H_{10}} + 13\,\mathrm{O_2} \rightarrow 8\,\mathrm{CO_2} + 10\,\mathrm{H_2O}\); (ii) \(\displaystyle 2\,\mathrm{C_5H_{10}} + 15\,\mathrm{O_2} \rightarrow 10\,\mathrm{CO_2} + 10\,\mathrm{H_2O}\); (iii) \(\displaystyle 2\,\mathrm{C_6H_{10}} + 17\,\mathrm{O_2} \rightarrow 12\,\mathrm{CO_2} + 10\,\mathrm{H_2O}\); (iv) \(\displaystyle \mathrm{C_7H_8} + 9\,\mathrm{O_2} \rightarrow 7\,\mathrm{CO_2} + 4\,\mathrm{H_2O}\)

A cis–trans pair exists here because each double‑bond carbon carries one H and one different alkyl group — the two ends are "unsymmetrical," which is exactly the condition needed for geometrical isomerism.Setting up the skeletonHex‑$\displaystyle 2$‑ene is \(\displaystyle \mathrm{CH_3-CH=CH-CH_2-CH_2-CH_3} \), numbered C‑$\displaystyle 1$ to C‑$\displaystyle 6$, with the double bond fixed between C‑$\displaystyle 2$ and C‑$\displaystyle 3$ (that is what "$\displaystyle 2$‑ene" means — no rotation is possible about this bond, which is exactly why cis/trans forms can be separated at all).Look at what each double‑bond carbon is carrying: Because C‑$\displaystyle 2$ and C‑$\displaystyle 3$ each hold two different groups (H versus an alkyl chain), the substituents can be locked either on the same side of the double bond or on opposite sides — two distinct, non‑interconvertible molecules.The two structures, describedcis-hex-$\displaystyle 2$-ene: hold the \(\displaystyle \mathrm{C2=C3} \) double bond as the fixed reference. The methyl group on C‑$\displaystyle 2$ and the propyl group (\(\displaystyle \mathrm{CH_2CH_2CH_3} \)) on C‑$\displaystyle 3$ sit on the same side of the double bond; the two H atoms then sit together on the other side. So the two alkyl chains — the "bulky" parts of the molecule — are bunched on one face of the planar C=C system.trans-hex-$\displaystyle 2$-ene: the methyl group on C‑$\displaystyle 2$ and the propyl group on C‑$\displaystyle 3$ sit on opposite sides of the double bond — each alkyl group is diagonally across from an H on the other carbon, not from the other alkyl group. The molecule is now the more "spread out," symmetric-looking arrangement.(A common slip here: don't confuse this with the E/Z system needing a full CIP priority comparison — but note that in this case cis and Z agree, and trans and E agree, because on both carbons the alkyl group is automatically the higher-priority group over H.)Which one boils higher, and whycis-Hex-$\displaystyle 2$-ene has the higher boiling point.A molecule's boiling point is set by how strong the attractions between its molecules are, not by how "neat" or symmetric its shape looks. Each \(\displaystyle \mathrm{C-CH_3} \) and \(\displaystyle \mathrm{C-C_3H_7} \) bond carries a small dipole (alkyl carbon is very slightly electron-donating relative to the alkene carbon). Since boiling requires supplying enough energy to overcome all the intermolecular attractions holding the liquid together, the isomer with the extra dipole–dipole pull needs more energy (a higher temperature) to vaporize. That is the cis isomer.(Don't let this trick you when melting points come up instead — there, the flatter, more symmetric trans isomer usually packs into a crystal lattice more efficiently and so tends to have the higher melting point. Boiling point and melting point are governed by different things: liquid-phase dipole attraction versus solid-state packing.)Answer: cis-Hex-$\displaystyle 2$-ene has \(\displaystyle \mathrm{CH_3} \) (on C‑$\displaystyle 2$) and \(\displaystyle \mathrm{CH_2CH_2CH_3} \) (on C‑$\displaystyle 3$) on the same side of the \(\displaystyle \mathrm{C2=C3} \) double bond (H's together on the other side); trans-hex-$\displaystyle 2$-ene has these two alkyl groups on opposite sides. The cis isomer has the higher boiling point, because its unsymmetrical (same-side) arrangement leaves it with a net dipole moment, giving it dipole–dipole attractions in addition to van der Waals forces, whereas in the trans isomer the bond dipoles cancel by symmetry, leaving only the weaker van der Waals forces to overcome on boiling.

Benzene's "three double bonds" are a bookkeeping device, not physical reality — every carbon–carbon bond in benzene is identical, because the electrons that would make three separate double bonds are spread out (delocalized) over all six carbons at once.Step $\displaystyle 1$: Why one Kekulé structure is not the real molecule.If benzene, \(\displaystyle \text{C}_6\text{H}_6 \), really had three isolated double bonds fixed at positions $\displaystyle 1$-$\displaystyle 2$, $\displaystyle 3$-$\displaystyle 4$, $\displaystyle 5$-$\displaystyle 6$ (a "cyclohexatriene"), it should show two different bond lengths around the ring:\[\text{C–C single bond} \approx 154\text{ pm}, \qquad \text{C=C double bond} \approx 133\text{ pm} \]Experimentally this is not what's seen. Every one of the six C–C bonds in benzene measures the same, \(\displaystyle 139\text{ pm} \) — a value sitting almost exactly between a single and a double bond. A single fixed arrangement of alternating double and single bonds cannot produce six equal bond lengths, so benzene cannot be that one structure.Step $\displaystyle 2$: Resonance — benzene is a hybrid of two (and more) contributing structures.You can draw the double bonds in the ring two equally valid ways:Structure I: double bonds between C1–C2, C3–C4, C5–C6 Structure II: double bonds between C2–C3, C4–C5, C6–C1Neither structure by itself is correct. The true molecule is a resonance hybrid of these (and other, minor) contributing structures — the actual π electrons are not sitting fixed between any one pair of carbons; they are delocalized as a continuous cloud above and below the plane of the ring, shared equally by all six carbon atoms. That is why all six bonds come out identical, at $\displaystyle 139$ pm: each C–C bond is really "one-and-a-half" bonds' worth of electron density, not alternately single and double.This is the point people get wrong: resonance structures are not different molecules flipping back and forth — only one real molecule exists, and it is the average (hybrid) of the drawn structures, not a mixture of them.Step $\displaystyle 3$: Delocalization lowers the energy — the resonance (delocalization) energy.Spreading the π electrons over the whole ring, instead of confining them between fixed pairs of carbons, lowers the total energy of the molecule compared to the hypothetical, non-delocalized cyclohexatriene. This extra stability can be measured using enthalpies of hydrogenation.Hydrogenating one isolated double bond (as in cyclohexene) releases about\[\Delta H = -119.5 \text{ kJ mol}^{-1} \]If benzene really behaved as three independent double bonds, hydrogenating all three (adding \(\displaystyle 3\text{H}_2 \) to give cyclohexane) should release three times as much heat:\[\Delta H_{\text{calculated}} = 3 \times (-119.5) = -358.5 \text{ kJ mol}^{-1} \]But the experimentally measured enthalpy of hydrogenation of benzene to cyclohexane,\[\text{C}_6\text{H}_6 + 3\text{H}_2 \rightarrow \text{C}_6\text{H}_{12}, \qquad \Delta H_{\text{observed}} = -208.4 \text{ kJ mol}^{-1} \]releases far less heat than predicted. Less heat given out means benzene started out at a lower energy (more stable) than the hypothetical cyclohexatriene. The gap between the two values is the stabilization gained purely from delocalizing the π electrons:\[\Delta H_{\text{calculated}} - \Delta H_{\text{observed}} = (-358.5) - (-208.4) = -150.1 \text{ kJ mol}^{-1} \]So benzene is roughly \(\displaystyle 150 \text{ kJ mol}^{-1} \) more stable than a structure with three localized, independent double bonds would be. This ~$\displaystyle 150$ kJ/mol is called the resonance energy (or delocalization energy) of benzene.Step $\displaystyle 4$: Why this shows up as "extraordinary" stability.Because this large amount of energy would have to be put back in to localize the electrons again, benzene resists the addition reactions that normal alkenes undergo readily (which would break up the delocalized system and destroy that $\displaystyle 150$ kJ/mol of stabilization). Instead it prefers electrophilic substitution reactions, which keep the delocalized ring intact — a direct chemical consequence of the extra stability calculated above.**Answer: Benzene is not a molecule with three fixed, alternating double and single bonds; it is a resonance hybrid in which all six π electrons are delocalized equally over all six carbon atoms. This is confirmed structurally, since all six C–C bonds measure the same $\displaystyle 139$ pm (between a single bond's $\displaystyle 154$ pm and a double bond's $\displaystyle 133$ pm), and energetically, since the observed hydrogenation enthalpy of benzene (\(\displaystyle -208.4 \text{ kJ mol}^{-1} \)) is about \(\displaystyle 150 \text{ kJ mol}^{-1} \) less negative than the value calculated for three isolated double bonds (\(\displaystyle -358.5 \text{ kJ mol}^{-1} \)). This $\displaystyle 150$ kJ/mol resonance (delocalization) energy is what makes benzene extraordinarily stable and makes it undergo substitution rather than the addition reactions typical of alkenes.