Chapter 9: Hydrocarbons (Hydrocarbons)

Chlorination is a free-radical chain reaction. In propagation, methyl radicals are formed: $\mathrm{CH}{}_4+\mathrm{Cl}{}^{\bullet }\to \mathrm{CH}{}_3{}^{\bullet }+\mathrm{HCl}$. In the termination step two methyl radicals can combine: $\mathrm{CH}{}_3{}^{\bullet }+\mathrm{CH}{}_3{}^{\bullet }\to \mathrm{CH}{}_3-\mathrm{CH}{}_3$. So a little ethane is always formed as a by-product, which supports the free-radical mechanism.

Two CH₃• radicals combine in the termination step.

(e) OH gets C1 (it is the principal group). (f) The longest chain has 10 carbons, branch on C5. (g) Ten-carbon chain numbered from the CH₂= end: double bonds at 1, 5, 8 and ethyl on C4.

(a) 2-Methylbut-2-ene (b) pent-1-en-3-yne (c) buta-1,3-diene (d) 4-phenylbut-1-ene (e) 2-methylphenol (f) 5-(2-methylpropyl)decane (g) 4-ethyldeca-1,5,8-triene

(a) CH₂=CH–CH₂–CH₃ but-1-ene; CH₃–CH=CH–CH₃ but-2-ene (cis and trans); CH₂=C(CH₃)–CH₃ 2-methylpropene.

(b) HC≡C–CH₂–CH₂–CH₃ pent-1-yne; CH₃–C≡C–CH₂–CH₃ pent-2-yne; HC≡C–CH(CH₃)–CH₃ 3-methylbut-1-yne.

(a) But-1-ene, cis/trans-but-2-ene, 2-methylpropene (b) pent-1-yne, pent-2-yne, 3-methylbut-1-yne

Each C=C becomes two C=O.

(i) CH₃CH=CHCH₂CH₃ → CH₃CHO + CH₃CH₂CHO. (ii) CH₃CH₂C(CH₃)=C(CH₃)CH₂CH₂CH₃ → CH₃CH₂COCH₃ + CH₃COCH₂CH₂CH₃. (iii) CH₂=C(C₂H₅)₂ → HCHO + (C₂H₅)₂CO. (iv) C₆H₅CH=CHCH₂CH₃ → C₆H₅CHO + CH₃CH₂CHO.

(i) Ethanal + propanal (ii) butan-2-one + pentan-2-one (iii) methanal + pentan-3-one (iv) benzaldehyde + propanal

Join the carbonyl carbons of CH₃CHO and (C₂H₅)₂CO by a double bond: CH₃CH=C(C₂H₅)₂.

CH₃CH=C(CH₂CH₃)₂, 3-ethylpent-2-ene

Three C–C σ bonds and eight H: C₄H₈. The aldehyde of molar mass 44 is CH₃CHO; two moles of it come from a symmetrical alkene, CH₃CH=CHCH₃.

But-2-ene

CH₃CH₂CHO + (C₂H₅)₂CO: join the carbonyl carbons: CH₃CH₂CH=C(CH₂CH₃)₂.

CH₃CH₂CH=C(C₂H₅)₂, 3-ethylhex-3-ene (NCERT key writes 4-ethylhex-3-ene, the same compound numbered from the other end)

(i) $\mathrm{C}{}_4\mathrm{H}{}_{10}+\frac{13}{2}\mathrm{O}{}_2\to 4\mathrm{CO}{}_2+5\mathrm{H}{}_2\mathrm{O}$ (ii) $\mathrm{C}{}_5\mathrm{H}{}_{10}+\frac{15}{2}\mathrm{O}{}_2\to 5\mathrm{CO}{}_2+5\mathrm{H}{}_2\mathrm{O}$ (iii) $\mathrm{C}{}_6\mathrm{H}{}_{10}+\frac{17}{2}\mathrm{O}{}_2\to 6\mathrm{CO}{}_2+5\mathrm{H}{}_2\mathrm{O}$ (iv) $\mathrm{C}{}_7\mathrm{H}{}_8+9\mathrm{O}{}_2\to 7\mathrm{CO}{}_2+4\mathrm{H}{}_2\mathrm{O}$. All are exothermic.

As written above

cis: CH₃ and C₃H₇ on the same side of the C=C (the two H atoms together on the other side); trans: on opposite sides.

cis-Hex-2-ene has the higher boiling point: its bond dipoles do not cancel, so it is more polar with stronger dipole–dipole attractions; in the trans form they nearly cancel.

cis-Hex-2-ene (more polar)

The π electrons are not localised in three double bonds: the six p orbitals overlap all round the ring, giving a delocalised π cloud (benzene is a resonance hybrid of the two Kekulé structures). Delocalisation lowers the energy (resonance energy); benzene is planar, cyclic and has 6 = (4n + 2) π electrons: aromatic. Hence it prefers substitution, which keeps the ring intact, to addition.

Delocalised π electrons (resonance, aromaticity).

(i) Planar, cyclic structure. (ii) Complete delocalisation of π electrons round the ring (every ring atom sp², with a p orbital). (iii) (4n + 2) π electrons, n = 0, 1, 2, … (Hückel's rule).

Planar, cyclic, fully conjugated, with (4n + 2) π electrons.

(i) The ring contains an sp³ (CH₂) carbon, so delocalisation round the ring is not complete.

(ii) Cyclopentadiene has an sp³ CH₂ and only 4 π electrons (not 4n + 2).

(iii) Cyclooctatetraene has 8 π electrons (4n, n = 2) and is non-planar (tub-shaped), so its π electrons are not delocalised.

(i), (ii) contain an sp³ ring carbon; (ii) and (iii) also lack (4n + 2) π electrons, and (iii) is non-planar.

(i) Brominate first (Br₂/FeBr₃) → bromobenzene; then nitrate (conc. HNO₃ + H₂SO₄): Br is o/p directing → o- and p-nitrobromobenzene; separate the para isomer.

(ii) Nitrate first → nitrobenzene; then chlorinate (Cl₂/FeCl₃): NO₂ is m-directing → m-nitrochlorobenzene.

(iii) Friedel–Crafts methylation (CH₃Cl/anhyd. AlCl₃) → toluene; nitrate → o- and p-nitrotoluene; separate the para isomer.

(iv) Friedel–Crafts acylation: $\mathrm{C}{}_6\mathrm{H}{}_6+\mathrm{CH}{}_3\mathrm{COCl}\stackrel{\text{anhyd. AlCl}{}_3}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COCH}{}_3+\mathrm{HCl}$.

The order of substitution follows the directing effect of the first group.

1° carbons (bonded to one C): the five CH₃ groups, 15 H. 2° carbons: the two CH₂ groups, 4 H. 3° carbon: the CH, 1 H. (The C(CH₃)₂ carbon is 4°, with no H.)

1°: 5 C, 15 H; 2°: 2 C, 4 H; 3°: 1 C, 1 H

Branching lowers the boiling point. A branched molecule is more compact (nearly spherical), with a smaller surface area, so the van der Waals forces between molecules are weaker: pentane 309 K, 2-methylbutane 301 K, 2,2-dimethylpropane 282.5 K.

Boiling point decreases with branching.

Without peroxide (ionic, Markovnikov): H⁺ adds to the CH₂ end, giving the more stable 2° carbocation CH₃C⁺HCH₃; Br⁻ attacks it → 2-bromopropane.

With peroxide (free radical, anti-Markovnikov): $(\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COO}){}_2\to 2\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COO}{}^{\bullet }\to 2\mathrm{C}{}_6\mathrm{H}{}_5{}^{\bullet }+2\mathrm{CO}{}_2$; $\mathrm{C}{}_6\mathrm{H}{}_5{}^{\bullet }+\mathrm{HBr}\to \mathrm{C}{}_6\mathrm{H}{}_6+\mathrm{Br}{}^{\bullet }$; Br• adds to the terminal CH₂ to give the more stable 2° radical CH₃C•HCH₂Br, which takes H from HBr → CH₃CH₂CH₂Br + Br• (chain continues). Product: 1-bromopropane.

Ionic path via 2° carbocation (Markovnikov) vs free-radical path via 2° radical (peroxide effect).

Kekulé structure I (double bond between the two CH₃-bearing carbons) would give 2 mol methylglyoxal (CH₃COCHO) + 1 mol glyoxal (OHC–CHO); structure II would give 1 mol dimethylglyoxal (CH₃COCOCH₃) + 2 mol glyoxal.

In fact all three (glyoxal, methylglyoxal and dimethylglyoxal) are obtained. No single Kekulé structure can give all three, so benzene is a resonance hybrid of the two Kekulé structures.

Glyoxal, methylglyoxal and dimethylglyoxal; both Kekulé structures contribute.

The H is attached to an sp (50% s), sp² (33%) and sp³ (25%) carbon respectively. More s character holds the bonding pair closer to carbon (more electronegative carbon), so H⁺ leaves more easily.

Ethyne > benzene > n-hexane

The ring has six delocalised π electrons above and below its plane: it is electron rich, so it attracts electron-deficient reagents (electrophiles) and repels electron-rich ones (nucleophiles). Substitution (rather than addition) keeps the stable aromatic sextet.

The electron-rich π cloud attracts electrophiles and repels nucleophiles.

(i) Cyclic polymerisation: $3\mathrm{HC}\equiv \mathrm{CH}\stackrel{\text{red-hot Fe tube, 873 K}}{\to }\mathrm{C}{}_6\mathrm{H}{}_6$.

(ii) $\mathrm{CH}{}_2=\mathrm{CH}{}_2+\mathrm{Br}{}_2\to \mathrm{CH}{}_2\mathrm{BrCH}{}_2\mathrm{Br}$; alc. KOH → CH₂=CHBr; NaNH₂ → HC≡CH; red-hot iron tube → C₆H₆.

(iii) Aromatisation: $\mathrm{CH}{}_3(\mathrm{CH}{}_2){}_4\mathrm{CH}{}_3\to \mathrm{C}{}_6\mathrm{H}{}_6+4\mathrm{H}{}_2$ (Cr₂O₃/V₂O₅/Mo₂O₃ on alumina, 773 K, 10–20 atm).

(i) Polymerisation (ii) via ethyne (iii) aromatisation

The C=C can be placed anywhere in the 2-methylbutane skeleton.

CH₂=C(CH₃)CH₂CH₃ (2-methylbut-1-ene); (CH₃)₂C=CHCH₃ (2-methylbut-2-ene); (CH₃)₂CHCH=CH₂ (3-methylbut-1-ene)

Electron-withdrawing NO₂ groups deactivate the ring; CH₃ activates it.

(a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene (b) toluene > p-nitrotoluene > p-dinitrobenzene

The CH₃ group releases electrons (+I and hyperconjugation), increasing the electron density of the ring; m-dinitrobenzene, with two strongly withdrawing groups, reacts least readily.

Toluene

It accepts Cl⁻ from C₂H₅Cl to form the electrophile C₂H₅⁺.

Anhydrous FeCl₃ (also BF₃, SnCl₄ or ZnCl₂)

An alkane with an odd number of carbons needs two different alkyl halides, and then three alkanes form, which are hard to separate. E.g. for heptane: CH₃(CH₂)₂Br + CH₃(CH₂)₃Br + Na gives heptane (C₃ + C₄) together with hexane (C₃ + C₃) and octane (C₄ + C₄).

A mixture of three alkanes is formed.