Chlorination of methane proceeds by a free-radical chain mechanism. Chlorine radicals abstract hydrogen from methane to form methyl radicals: CH4 + Cl· → CH3· + HCl. Two methyl radicals may combine in a termination step: CH3· + CH3· → CH3CH3. This accounts for ethane formation.
Ethane forms by coupling of methyl free radicals.
For C4H8 with one double bond, possible open-chain alkenes are CH2=CHCH2CH3, CH3CH=CHCH3 and CH2=C(CH3)2. For C5H8 with one triple bond, possible open-chain alkynes are HC≡CCH2CH2CH3, CH3C≡CCH2CH3 and HC≡CCH(CH3)2.
(a) But-1-ene, but-2-ene and 2-methylpropene. (b) Pent-1-yne, pent-2-yne and 3-methylbut-1-yne.
Ozonolysis cleaves the C=C bond and converts each alkene carbon into a carbonyl carbon. Pent-2-ene gives ethanal and propanal. 3,4-Dimethylhept-3-ene gives CH3COCH2CH3 and CH3COCH2CH2CH3. 2-Ethylbut-1-ene gives methanal from the terminal carbon and pentan-3-one from the substituted carbon. 1-Phenylbut-1-ene gives benzaldehyde and propanal.
(i) Ethanal and propanal. (ii) Butan-2-one and pentan-2-one. (iii) Methanal and pentan-3-one. (iv) Benzaldehyde and propanal.
Rejoin the carbonyl carbons of ethanal (CH3CHO) and pentan-3-one (CH3CH2COCH2CH3) by replacing C=O bonds with a C=C bond. This gives CH3CH=C(CH2CH3)2. The longest chain containing the double bond has five carbons with an ethyl substituent at C-3, so the name is 3-ethylpent-2-ene.
A is CH3CH=C(CH2CH3)2, named 3-ethylpent-2-ene.
The aldehyde of molar mass 44 u is ethanal, CH3CHO. Since ozonolysis gives two moles of ethanal, the alkene must be symmetrical: CH3CH=CHCH3. This alkene has three C-C sigma bonds, eight C-H sigma bonds and one π bond. Its IUPAC name is but-2-ene.
But-2-ene.
Propanal contributes an alkene carbon bearing H and ethyl, while pentan-3-one contributes an alkene carbon bearing two ethyl groups. Rejoining the carbonyl carbons gives CH3CH2CH=C(CH2CH3)2, whose IUPAC name is 3-ethylhex-3-ene.
CH3CH2CH=C(CH2CH3)2.
Complete combustion converts all carbon to CO2 and hydrogen to H2O. Balancing C, H and then O gives the equations listed.
(i) 2C4H10 + 13O2 → 8CO2 + 10H2O; (ii) 2C5H10 + 15O2 → 10CO2 + 10H2O; (iii) 2C6H10 + 17O2 → 12CO2 + 10H2O; (iv) C7H8 + 9O2 → 7CO2 + 4H2O.
Benzene is planar, cyclic and fully conjugated with six π electrons, satisfying Huckel's 4n+2 rule for n = 1. The π electrons are delocalised over the whole ring, giving resonance stabilisation. Therefore benzene is much more stable than a normal triene with three isolated double bonds.
Because benzene is aromatic and has delocalised π electrons.
For aromaticity, the system must form a ring, all ring atoms must have p orbitals allowing continuous overlap, the ring must be planar enough for delocalisation, and the delocalised π electron count must satisfy Huckel's rule, 4n+2 where n is 0, 1, 2 and so on.
It must be cyclic, planar, fully conjugated and contain (4n+2) π electrons.
(i) Benzene + Br2/FeBr3 → bromobenzene; nitration with HNO3/H2SO4 gives mainly o- and p-nitrobromobenzene, from which the para isomer is separated. (ii) Benzene → nitrobenzene by nitration; NO2 is meta-directing, so chlorination with Cl2/FeCl3 gives m-nitrochlorobenzene. (iii) Benzene + CH3Cl/AlCl3 → toluene; nitration gives chiefly o- and p-nitrotoluene, from which the para isomer is separated. (iv) Benzene + CH3COCl/AlCl3 → acetophenone.
(i) Brominate, then nitrate and isolate para product. (ii) Nitrate, then chlorinate. (iii) Methylate, then nitrate and isolate para product. (iv) Friedel-Crafts acylation with acetyl chloride.
A primary carbon is attached to one carbon, secondary to two, tertiary to three and quaternary to four. The five methyl carbons are primary and carry 3 H each. The two -CH2- carbons are secondary and carry 2 H each. The -CH(CH3)2 carbon is tertiary and carries 1 H. The central C attached to two methyl groups and two chain carbons is quaternary and carries no hydrogen.
There are five 1° carbons, each bonded to 3 H; two 2° carbons, each bonded to 2 H; one 3° carbon, bonded to 1 H. The central C in C(CH3)2 is 4° and has no H.
Branching makes an alkane molecule more compact and decreases its surface area. Weaker London dispersion forces then operate between molecules, so the boiling point decreases as branching increases among isomeric alkanes.
Branching lowers the boiling point.
Without peroxide, propene is protonated to form the more stable secondary carbocation, CH3-CH+-CH3, which is attacked by Br- to give 2-bromopropane. In the presence of benzoyl peroxide, Br radicals form. Br· adds to propene so that the more stable secondary carbon radical is produced: Br· + CH3CH=CH2 → CH3-CH·-CH2Br. This radical abstracts H from HBr to give CH3CH2CH2Br and regenerate Br·. Hence the product is 1-bromopropane.
Without peroxide, Markovnikov electrophilic addition gives 2-bromopropane. With peroxide, free-radical anti-Markovnikov addition gives 1-bromopropane.
Acidity of C-H bonds increases with the s-character of the carbon orbital holding the bonded electron pair: sp > sp2 > sp3. Ethyne has sp C-H bonds, benzene has sp2 C-H bonds and n-hexane has sp3 C-H bonds. Therefore the decreasing acidity order is ethyne, benzene, n-hexane.
Ethyne > benzene > n-hexane.
The π electron cloud of benzene attracts electrophiles, so electrophilic attack occurs readily. After substitution, loss of H+ restores aromaticity. Nucleophiles are repelled by the electron-rich ring, and nucleophilic substitution would require difficult disruption of aromatic stabilisation or a suitable leaving group/electron-withdrawing activation.
Benzene is electron-rich and aromatic; electrophilic substitution preserves aromaticity, while nucleophilic substitution is difficult.
(i) 3HC≡CH → C6H6 on passing ethyne through a red-hot iron tube. (ii) CH2=CH2 + Br2 → BrCH2CH2Br; on treatment with alcoholic KOH, the vicinal dibromide gives ethyne, which trimerises to benzene. (iii) n-Hexane undergoes dehydrocyclisation/aromatisation over catalysts such as Cr2O3/V2O5/Mo2O3 at high temperature to give benzene.
(i) Trimerise ethyne. (ii) Convert ethene to ethyne, then trimerise. (iii) Aromatise hexane.
Hydrogenation keeps the carbon skeleton unchanged and converts C=C to C-C. For the 2-methylbutane skeleton, placing one double bond at non-equivalent adjacent carbon positions gives CH2=C(CH3)CH2CH3, CH2=CHCH(CH3)CH3 and (CH3)2C=CHCH3.
2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene.
Electron-withdrawing nitro groups deactivate the ring toward electrophilic substitution; more nitro groups mean lower reactivity. Alkyl groups activate the ring, so toluene is most reactive in (b). A nitro group reduces reactivity, and two nitro groups reduce it still further.
(a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (b) Toluene > p-nitrotoluene > p-dinitrobenzene.
The methyl group in toluene activates the benzene ring through +I effect and hyperconjugation, increasing electron density and making electrophilic attack easier. Benzene is less reactive than toluene. Nitro groups are strongly deactivating, so m-dinitrobenzene is least reactive.
Toluene undergoes nitration most easily.
Friedel-Crafts alkylation needs a Lewis acid to generate the electrophile from an alkyl halide. Besides anhydrous AlCl3, Lewis acids such as FeCl3, BF3 or SnCl4 can be used; FeCl3 is one suitable example.
Anhydrous ferric chloride, FeCl3, can be used.
Wurtz reaction couples alkyl halides with sodium. A single alkyl halide gives an even-carbon symmetrical alkane. To prepare an odd-carbon alkane such as propane, one would use CH3Br and C2H5Br, but coupling gives ethane (CH3-CH3), propane (CH3-CH2CH3) and butane (C2H5-C2H5). Because the desired product is mixed with other alkanes, Wurtz reaction is not preferred for odd-carbon alkanes.
Because preparing an odd-carbon alkane requires two different alkyl halides, which gives a mixture of products.