NCERT Solutions · Class 11 Chemistry Hydrocarbons

Hydrogenation just adds one \(\displaystyle H_2\) molecule across the \(\displaystyle C=C\) double bond — so to reverse it, look at every C–C single bond in the saturated skeleton and ask "could this have been the double bond that got saturated?"First write out $\displaystyle 2$-methylbutane and number its carbons so nothing gets lost:\[\underset{C_1}{CH_3}-\underset{C_2}{CH}(-\underset{C_b}{CH_3})-\underset{C_3}{CH_2}-\underset{C_4}{CH_3} \]It is a $\displaystyle 4$-carbon chain (butane) carrying one methyl branch on \(\displaystyle C_2\). There are exactly four C–C single bonds where a double bond could sit: \(\displaystyle C_1\!-\!C_2\), \(\displaystyle C_2\!-\!C_b\), \(\displaystyle C_2\!-\!C_3\), and \(\displaystyle C_3\!-\!C_4\). Each alkene has the molecular formula \(\displaystyle C_5H_{10}\) — one degree of unsaturation, one \(\displaystyle C=C\) — so each equation below balances simply because adding one \(\displaystyle H_2\) turns \(\displaystyle C_5H_{10}\) into \(\displaystyle C_5H_{12}\): carbon count stays at $\displaystyle 5$, and hydrogen goes \(\displaystyle 10+2=12\), matching $\displaystyle 2$-methylbutane exactly.Double bond at \(\displaystyle C_1\!-\!C_2\) (or, equivalently, at \(\displaystyle C_2\!-\!C_b\)). \(\displaystyle C_1\) and \(\displaystyle C_b\) are both plain methyl groups hanging off \(\displaystyle C_2\), so moving the double bond from one to the other gives the same molecule — this is the step people miss, and it's why the count of distinct alkenes is $\displaystyle 3$, not 4. Removing one H from \(\displaystyle C_2\) and one H from \(\displaystyle C_1\) (or from \(\displaystyle C_b\)) gives:\[CH_2=C(CH_3)-CH_2-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH(CH_3)-CH_2-CH_3 \]This alkene is $\displaystyle 2$-methylbut-$\displaystyle 1$-ene.Double bond at \(\displaystyle C_2\!-\!C_3\). Remove one H from \(\displaystyle C_2\) and one H from \(\displaystyle C_3\):\[CH_3-CH=C(CH_3)-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH(CH_3)-CH_2-CH_3 \]Both carbons of this double bond carry a \(\displaystyle CH_3\) on the same side (\(\displaystyle C_2\) has \(\displaystyle C_1\)'s methyl and \(\displaystyle C_b\)'s methyl, which are identical), so there is no cis/trans pair to worry about — only one compound exists here. This is $\displaystyle 2$-methylbut-$\displaystyle 2$-ene.Double bond at \(\displaystyle C_3\!-\!C_4\). Remove one H from \(\displaystyle C_3\) and one H from \(\displaystyle C_4\):\[CH_2=CH-CH(CH_3)-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH_2-CH(CH_3)-CH_3 \](written from the other end, this is the same target molecule, \(\displaystyle CH_3-CH(CH_3)-CH_2-CH_3\)). Since \(\displaystyle C_4\) becomes a terminal \(\displaystyle =CH_2\) with two identical H's, again no cis/trans isomer arises. This is $\displaystyle 3$-methylbut-$\displaystyle 1$-ene.Checking that nothing was missed: those four bonds are the only C–C single bonds in the skeleton, and the first two collapse into one product by symmetry, so three is the complete list.Answer: There are exactly three such alkenes — $\displaystyle 2$-methylbut-$\displaystyle 1$-ene, \(\displaystyle CH_2=C(CH_3)CH_2CH_3\); $\displaystyle 2$-methylbut-$\displaystyle 2$-ene, \(\displaystyle CH_3CH=C(CH_3)CH_3\); and $\displaystyle 3$-methylbut-$\displaystyle 1$-ene, \(\displaystyle CH_2=CHCH(CH_3)CH_3\) — each of which adds one molecule of \(\displaystyle H_2\) (Ni/Pt/Pd catalyst) to give $\displaystyle 2$-methylbutane, \(\displaystyle CH_3CH(CH_3)CH_2CH_3\).

Reactivity toward an electrophile \(\displaystyle E^+\) tracks how much electron density is sitting on the ring — a substituent that pushes electron density in speeds up attack, a substituent that pulls electron density out slows it down.Electrophilic aromatic substitution starts with \(\displaystyle E^+\) attacking the \(\displaystyle \pi\)-cloud of the ring. Whatever raises the electron density of that \(\displaystyle \pi\)-cloud above benzene's own level makes the ring a better nucleophile toward \(\displaystyle E^+\) (faster reaction); whatever lowers it makes the ring a poorer nucleophile (slower reaction). So to rank a set of substituted benzenes, look at every substituent already on the ring and ask whether it is a net electron-donor or a net electron-withdrawer, and how many of each are present.(a) Chlorobenzene, \(\displaystyle p\)-nitrochlorobenzene, $\displaystyle 2,4$-dinitrochlorobenzeneThese three are: \(\displaystyle \text{C}_6\text{H}_5\text{Cl}\) (one Cl on the ring), \(\displaystyle \text{C}_6\text{H}_4(\text{Cl})(\text{NO}_2)\) (one Cl and one \(\displaystyle \text{NO}_2\), para to each other), and \(\displaystyle \text{C}_6\text{H}_3(\text{Cl})(\text{NO}_2)_2\) (one Cl and two \(\displaystyle \text{NO}_2\) groups, at the $\displaystyle 2$- and $\displaystyle 4$-positions).Chlorine carries a lone pair that it can donate into the ring by resonance (this is exactly why Cl is an ortho/para director — that donation puts extra electron density, and so extra stability for the intermediate cation, at the ortho and para carbons). At the same time, chlorine is more electronegative than carbon and pulls electron density out of the ring through the sigma framework (its inductive, \(\displaystyle -I\), effect). For chlorobenzene the inductive pull slightly outweighs the resonance donation, so the ring ends up a little less reactive than plain benzene — but chlorine is still by far the mildest deactivator of the three substituents in this question.A short aside, because this is the step most people get backwards: "ortho/para-directing" and "activating" are not the same claim. Chlorine directs to ortho/para (resonance argument, about where the intermediate is most stable) while still net deactivating the ring relative to benzene (inductive argument, about how fast overall). Both are true of the same atom at once.\(\displaystyle \text{NO}_2\), by contrast, has no lone pair to donate. Its nitrogen is doubly bonded to one oxygen and carries a formal positive charge balanced by the other, so it withdraws ring electron density both by induction (\(\displaystyle -I\), through the sigma bonds) and by resonance (\(\displaystyle -M\): the ring's own \(\displaystyle \pi\) electrons can delocalize onto the electronegative oxygens, most strongly draining density from the ortho and para carbons — which is why \(\displaystyle \text{NO}_2\) is a meta director and a strong deactivator, not a mild one like Cl).So each successive compound adds a strongly-deactivating \(\displaystyle \text{NO}_2\) group on top of the same mildly-deactivating Cl: Each added \(\displaystyle \text{NO}_2\) drains more electron density from the ring, so each added \(\displaystyle \text{NO}_2\) makes the ring a slower target for \(\displaystyle E^+\):\[\text{Chlorobenzene} \;>\; p\text{-Nitrochlorobenzene} \;>\; 2,4\text{-Dinitrochlorobenzene} \](b) Toluene, \(\displaystyle p\)-nitrotoluene, \(\displaystyle p\)-dinitrobenzeneThese are: \(\displaystyle \text{C}_6\text{H}_5\text{CH}_3\) (one \(\displaystyle \text{CH}_3\) on the ring), \(\displaystyle \text{C}_6\text{H}_4(\text{CH}_3)(\text{NO}_2)\) (one \(\displaystyle \text{CH}_3\) and one \(\displaystyle \text{NO}_2\), para to each other, written in the question as \(\displaystyle p\text{-H}_3\text{C–C}_6\text{H}_4\text{–NO}_2\)), and \(\displaystyle \text{C}_6\text{H}_4(\text{NO}_2)_2\) (two \(\displaystyle \text{NO}_2\) groups, para to each other, written as \(\displaystyle p\text{-O}_2\text{N–C}_6\text{H}_4\text{–NO}_2\)).The methyl group has no lone pair either, but its \(\displaystyle \text{C–H}\) sigma electrons can overlap with the ring's \(\displaystyle \pi\) system (hyperconjugation), and this pushes electron density into the ring rather than pulling it out. So \(\displaystyle \text{CH}_3\) is a genuine electron donor — it makes toluene's ring more electron-rich than benzene's, hence more reactive toward \(\displaystyle E^+\), and it directs ortho/para for the same reason (extra density, and extra cation stability, at those positions).Now compare the three rings directly: \[\text{Toluene} \;>\; p\text{-Nitrotoluene} \;>\; p\text{-Dinitrobenzene} \]The two lists run on the same logic: count the electron-donors, count the electron-withdrawers (and how strong each one is), and the ring with the most net electron density reacts fastest with \(\displaystyle E^+\).Answer: (a) Chlorobenzene > \(\displaystyle p\)-Nitrochlorobenzene > $\displaystyle 2,4$-Dinitrochlorobenzene; (b) Toluene > \(\displaystyle p\)-Nitrotoluene > \(\displaystyle p\)-Dinitrobenzene — in both series, reactivity falls as strongly electron-withdrawing \(\displaystyle \text{NO}_2\) groups are added, since each one drains electron density from the ring and makes it a slower target for the electrophile.

Nitration is an electrophilic attack on the ring, so whichever ring has the most electron density reacts fastest.How the electrophile is made and what it doesNitration replaces a ring hydrogen with a nitro group, \(\displaystyle -NO_2\). The actual attacking species is not nitric acid itself but the nitronium ion, \(\displaystyle NO_2^+\), generated when concentrated nitric acid is protonated by concentrated sulfuric acid:\[HNO_3 + 2\,H_2SO_4 \rightarrow NO_2^{+} + H_3O^{+} + 2\,HSO_4^{-} \]Check the balance atom by atom: left side has \(\displaystyle H = 1 + 2(2) = 5\), \(\displaystyle N = 1\), \(\displaystyle O = 3 + 2(4) = 11\), \(\displaystyle S = 2\); right side has \(\displaystyle H = 0 + 3 + 2(1) = 5\), \(\displaystyle N = 1\), \(\displaystyle O = 2 + 1 + 2(4) = 11\), \(\displaystyle S = 2\). Every atom matches, and the charges balance too: the left side is neutral, and the right side is \(\displaystyle (+1) + (+1) + 2(-1) = 0\).Once \(\displaystyle NO_2^{+}\) is formed, it attacks a ring carbon. This is the slow, rate-determining step — the ring's electrons attack the electrophile and form a positively charged intermediate (the arenium ion, or sigma complex), which then loses \(\displaystyle \mathrm{H^{+}}\) to restore the aromatic ring and give the nitro compound:\[C_6H_6 + NO_2^{+} \rightarrow C_6H_5NO_2 + H^{+} \]This is already balanced: \(\displaystyle C = 6\) and \(\displaystyle H = 6\) on both sides once you count the \(\displaystyle H^+\) released, \(\displaystyle N = 1\), \(\displaystyle O = 2\), and the charge goes from \(\displaystyle +1\) on the left to \(\displaystyle +1\) on the right (the proton). The product, \(\displaystyle C_6H_5NO_2\), is nitrobenzene.Because the slow step is the electrophile attacking the ring, anything that makes the ring more electron-rich lowers the energy of that step and speeds up nitration; anything that makes the ring electron-poor does the opposite. This is the whole question.Toluene — the methyl group pushes electrons inToluene is \(\displaystyle C_6H_5CH_3\), benzene with one \(\displaystyle H\) replaced by a methyl group, \(\displaystyle -CH_3\). The \(\displaystyle C-H\) bonds of that methyl group can align with the ring's \(\displaystyle \pi\) system and donate a little electron density into it — a no-bond delocalization called hyperconjugation — with a small additional push from the inductive (\(\displaystyle +I\)) effect of the alkyl group. This raises the electron density of the ring, especially at the carbons ortho and para to the methyl group, so the arenium-ion intermediate formed when \(\displaystyle NO_2^+\) attacks there is extra stabilized. That is why toluene nitrates markedly faster than benzene itself.m-Dinitrobenzene — two nitro groups pull electrons outm-Dinitrobenzene, \(\displaystyle C_6H_4(NO_2)_2\), already carries two \(\displaystyle -NO_2\) groups. The nitro group is strongly electron-withdrawing in two ways at once: inductively (\(\displaystyle -I\)), because the electronegative \(\displaystyle N\) and \(\displaystyle O\) atoms pull electron density through the sigma bonds, and by resonance (\(\displaystyle -M\)), because the nitrogen can draw a ring electron pair into the \(\displaystyle N=O\) system, leaving the ring carbons ortho and para to it short of electrons. With two such groups already deactivating the ring, its electron density is far below that of plain benzene, so the arenium-ion intermediate for a third nitration is much less stable and forms much more slowly. (This is the step people get backwards: it is tempting to think a ring that already has nitro groups on it should accept another one just as easily — it does not, because a nitro group is a deactivator, not an activator, for further substitution.)Putting the three in orderRanking by how easily each accepts a further nitro group, from fastest to slowest:\[\text{toluene} > \text{benzene} > \text{m-dinitrobenzene} \]Toluene's electron-donating methyl group speeds up attack by \(\displaystyle NO_2^+\); benzene is the unactivated, undeactivated reference; and m-dinitrobenzene's two electron-withdrawing nitro groups strongly resist further attack by the same electrophile.Answer: Toluene undergoes nitration most easily, because its \(\displaystyle -CH_3\) group donates electron density into the ring (by hyperconjugation and the \(\displaystyle +I\) effect), stabilizing the arenium-ion intermediate formed by attack of \(\displaystyle NO_2^{+}\); benzene is intermediate; m-dinitrobenzene is hardest to nitrate because its two \(\displaystyle -NO_2\) groups withdraw electron density from the ring by both the \(\displaystyle -I\) and \(\displaystyle -M\) effects, so the order of ease of nitration is toluene > benzene > m-dinitrobenzene.

Ethylation of benzene is Friedel–Crafts alkylation, and the catalyst's real job is generating a carbocation — any strong enough Lewis acid can do that, not just \(\displaystyle \mathrm{AlCl_3} \).In this reaction, benzene, \(\displaystyle \mathrm{C_6H_6} \) (six \(\displaystyle \mathrm{CH}\) units joined in the aromatic ring), is treated with ethyl chloride, \(\displaystyle \mathrm{C_2H_5Cl} \) (chloroethane, the alkylating agent supplying the ethyl group), in the presence of a catalyst. Anhydrous \(\displaystyle \mathrm{AlCl_3} \) is the catalyst usually named in the textbook, but its role is purely that of a Lewis acid — a species with an empty orbital that can accept the lone pair on the chlorine of \(\displaystyle \mathrm{C_2H_5Cl} \) and pull the \(\displaystyle \mathrm{Cl} \) away as a chloride complex. Once that happens, the ethyl group is left behind as an ethyl carbocation, \(\displaystyle \mathrm{CH_3CH_2^+} \), which is the actual electrophile that attacks the ring.Because the job is just "accept a chloride ion strongly enough to ionise the C–Cl bond," several other metal halides do it equally well. The one most commonly used as a substitute, and the one NCERT accepts as the answer, is anhydrous ferric chloride, \(\displaystyle \mathrm{FeCl_3} \) (iron(III) chloride — the iron centre has the same kind of electron-deficient, empty-orbital character as aluminium in \(\displaystyle \mathrm{AlCl_3} \)). Anhydrous \(\displaystyle \mathrm{BF_3} \) (boron trifluoride), anhydrous \(\displaystyle \mathrm{SnCl_4} \) (tin(IV) chloride), and anhydrous \(\displaystyle \mathrm{ZnCl_2} \) (zinc chloride) work the same way and are also acceptable answers.The word "anhydrous" is doing real work here, not decoration. If the catalyst is hydrated, water molecules occupy the empty orbital on the metal first, satisfying its Lewis acidity before it ever meets the alkyl halide — the catalyst is effectively dead. This is a step people skip: naming the right metal halide is not enough; it must be moisture-free to ionise the \(\displaystyle \mathrm{C-Cl} \) bond at all.With \(\displaystyle \mathrm{FeCl_3} \) as the catalyst, the overall ethylation is\[\mathrm{C_6H_6 + C_2H_5Cl \xrightarrow{\text{anhydrous } FeCl_3} C_6H_5CH_2CH_3 + HCl} \]Balancing check: count atoms on each side — left side has \(\displaystyle \mathrm{C}: 6+2=8 \), \(\displaystyle \mathrm{H}: 6+5=11 \), \(\displaystyle \mathrm{Cl}: 1 \); right side, ethylbenzene \(\displaystyle \mathrm{C_6H_5CH_2CH_3} \) contributes \(\displaystyle \mathrm{C_8H_{10}} \) and the by-product \(\displaystyle \mathrm{HCl} \) contributes one more \(\displaystyle \mathrm{H} \) and one \(\displaystyle \mathrm{Cl} \), giving \(\displaystyle \mathrm{C}: 8 \), \(\displaystyle \mathrm{H}: 10+1=11 \), \(\displaystyle \mathrm{Cl}: 1 \). Both sides match with all coefficients equal to $\displaystyle 1$, so the equation is already balanced as written.Mechanistically: \(\displaystyle \mathrm{FeCl_3} \) coordinates to the chlorine of \(\displaystyle \mathrm{C_2H_5Cl} \), generating the ethyl carbocation \(\displaystyle \mathrm{CH_3CH_2^+} \) and the complex ion \(\displaystyle \mathrm{FeCl_4^-} \). The carbocation is attacked by the \(\displaystyle \pi \) electrons of benzene to give an arenium (sigma-complex) intermediate, which then loses a proton to reform the aromatic ring — giving ethylbenzene, \(\displaystyle \mathrm{C_6H_5CH_2CH_3} \) (benzene with one hydrogen replaced by an ethyl group). The lost proton combines with \(\displaystyle \mathrm{FeCl_4^-} \) to release \(\displaystyle \mathrm{HCl} \) and regenerate \(\displaystyle \mathrm{FeCl_3} \), so the catalyst is not consumed overall.**Answer: Anhydrous ferric chloride, \(\displaystyle \mathrm{FeCl_3} \), can be used in place of anhydrous \(\displaystyle \mathrm{AlCl_3} \) as the Lewis acid catalyst for ethylation of benzene (anhydrous \(\displaystyle \mathrm{BF_3} \), \(\displaystyle \mathrm{SnCl_4} \), or \(\displaystyle \mathrm{ZnCl_2} \) work the same way); it ionises \(\displaystyle \mathrm{C_2H_5Cl} \) into the electrophilic ethyl carbocation that attacks benzene, giving \(\displaystyle \mathrm{C_6H_6 + C_2H_5Cl \xrightarrow{FeCl_3} C_6H_5CH_2CH_3 + HCl} \).

The Wurtz reaction always joins two identical alkyl groups, so it can only build alkanes with an even number of carbons cleanly — an odd-carbon alkane forces you to mix two different halides, and that mixture reacts three ways at once.The Wurtz reaction is: \[2\,R\text{–}X + 2\,Na \xrightarrow{\text{dry ether}} R\text{–}R + 2\,NaX \] Here \(\displaystyle R\) is an alkyl group, \(\displaystyle X\) is a halogen (usually Br or Cl), and the two \(\displaystyle R\text{–}X\) molecules that come together must be identical — sodium removes the halogen from each and the two carbon fragments join to give a single symmetrical alkane \(\displaystyle R\text{–}R\), which always has an even number of carbon atoms (two identical halves always add to an even total).The step people get wrong: to make an odd-carbon alkane you cannot use one halide alone — you must react two different alkyl halides together, say bromomethane, \(\displaystyle CH_3Br\), and bromoethane, \(\displaystyle C_2H_5Br\), aiming for propane, \(\displaystyle C_3H_8\). But sodium does not know to pick one \(\displaystyle CH_3Br\) and one \(\displaystyle C_2H_5Br\) every time. All three possible pairings happen at once, statistically:Self-coupling of bromomethane gives ethane: \[2\,CH_3Br + 2\,Na \longrightarrow CH_3\text{–}CH_3 + 2\,NaBr \] (balance check: left side has \(\displaystyle 2\,C,\ 6\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_2H_6\) plus \(\displaystyle 2\,NaBr\))Self-coupling of bromoethane gives n-butane: \[2\,C_2H_5Br + 2\,Na \longrightarrow C_2H_5\text{–}C_2H_5 + 2\,NaBr \] (balance check: left side has \(\displaystyle 4\,C,\ 10\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_4H_{10}\) plus \(\displaystyle 2\,NaBr\))Cross-coupling, the one you actually wanted, gives propane: \[CH_3Br + C_2H_5Br + 2\,Na \longrightarrow CH_3\text{–}C_2H_5 + 2\,NaBr \] (balance check: left side has \(\displaystyle 3\,C,\ 8\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_3H_8\) plus \(\displaystyle 2\,NaBr\))So the flask ends up with a mixture of three alkane gases — ethane, propane, and n-butane — along with sodium bromide. Since these three hydrocarbons are all low-boiling gases with boiling points close together, separating pure propane from the ethane and butane formed alongside it is impractical, and the yield of the alkane you actually wanted is low. That is why the Wurtz reaction is a good route to even-carbon alkanes (where only one alkyl halide, reacting with itself, is needed) but a poor one for odd-carbon alkanes (where a second, unwanted, symmetrical coupling is unavoidable alongside the cross product).Answer: The Wurtz reaction couples two alkyl halide molecules into one symmetrical alkane \(\displaystyle R\text{–}R\), so a single halide always gives an even-carbon product. To reach an odd-carbon alkane like propane, two different halides — \(\displaystyle CH_3Br\) and \(\displaystyle C_2H_5Br\) — must be used together, but sodium then couples them in all three possible ways at once: \(\displaystyle 2CH_3Br+2Na\rightarrow C_2H_6+2NaBr\) (ethane), \(\displaystyle 2C_2H_5Br+2Na\rightarrow C_4H_{10}+2NaBr\) (n-butane), and the wanted \(\displaystyle CH_3Br+C_2H_5Br+2Na\rightarrow C_3H_8+2NaBr\) (propane). The three gaseous alkanes formed are hard to separate from each other, so the method gives a poor, impure yield of the odd-carbon alkane — which is why Wurtz synthesis is not preferred for odd-carbon-number alkanes.