"Hydrocarbons" — Practice

Question 1

Give the major product of propene with hydrogen bromide, and with hydrogen bromide in the presence of benzoyl peroxide, explaining both.

Accepted Answer

Without peroxide the mechanism is electrophilic addition: the proton adds to the terminal carbon because that generates a secondary carbocation rather than a primary one, and bromide then captures it, giving **2-bromopropane**. With peroxide the mechanism becomes a radical chain: a bromine radical adds first, again to the terminal carbon, because that leaves a secondary radical rather than a primary one, and hydrogen is delivered second, giving **1-bromopropane**. Both products come from choosing the more stable intermediate; only the order of addition changed.

Question 2

Why does the peroxide effect operate with hydrogen bromide but not with hydrogen chloride or hydrogen iodide?

Accepted Answer

Because the radical chain requires two propagation steps to be energetically favourable, and only bromine satisfies both. For HCl the H-Cl bond is too strong, so the step in which a carbon radical abstracts hydrogen is endothermic and the chain stalls. For HI the I-I bond is so weak that iodine radicals recombine rather than adding to the alkene. Bromine sits in the narrow range where both steps proceed, so only HBr shows anti-Markovnikov addition.

Question 3

An alkene of formula $\mathrm{C_5H_{10}}$ gives propanone and ethanal on ozonolysis with reductive workup. Identify the alkene.

Accepted Answer

Ozonolysis cleaves the double bond and places an oxygen on each of the two carbons that formerly shared it. To reverse it, remove the oxygen from each carbonyl carbon and join those two carbons with a double bond. Propanone contributes a carbon bearing two methyls, and ethanal a carbon bearing a hydrogen and a methyl, so the alkene is **2-methylbut-2-ene**, $\mathrm{(CH_3)_2C{=}CHCH_3}$, which checks as $\mathrm{C_5H_{10}}$. Had an oxidative workup been used instead, ethanoic acid would have appeared in place of ethanal while propanone was unchanged.

Question 4

Give the product of hydrating but-1-yne with dilute sulphuric acid and mercury(II) sulphate, and explain why the immediate addition product is not isolated.

Accepted Answer

Water adds across the triple bond following Markovnikov's rule, placing the hydroxyl on the internal carbon and giving an enol, but-1-en-2-ol. Enols are unstable relative to their carbonyl tautomers because a $\mathrm{C{=}O}$ bond is considerably stronger than a $\mathrm{C{=}C}$ one, so the enol immediately tautomerises by proton migration. The isolated product is **butan-2-one**, a methyl ketone. Every terminal alkyne except ethyne gives a methyl ketone this way, while ethyne gives ethanal.

Question 5

How would you distinguish between ethane, ethene and ethyne using simple chemical tests?

Accepted Answer

First add bromine water to each. Ethane leaves it orange, while ethene and ethyne both decolourise it because each has a pi bond available for addition. Then add ammoniacal silver nitrate to the two that reacted. Ethyne gives a white precipitate of silver acetylide, because its terminal hydrogen sits on an electronegative sp carbon and is weakly acidic, while ethene gives nothing. Two tests separate all three, and Baeyer's reagent could replace bromine water in the first step.

Question 6

Explain why bromination of 2-methylpropane gives almost entirely the tertiary product while chlorination gives a mixture.

Accepted Answer

Both proceed by radical chains in which a halogen atom abstracts hydrogen to form an alkyl radical, and abstracting the tertiary hydrogen gives the most stable radical. What differs is where the transition state sits. Abstraction by bromine is endothermic, so the transition state comes late with the radical almost fully formed, and its stability is fully felt in the barrier: the per-hydrogen preference for tertiary over primary is about 1600 to 1. Abstraction by chlorine is exothermic, so the transition state is early and the radical has barely begun to form, giving a preference of only about 5 to 1 and hence a mixture. Greater reactivity means lower selectivity.

Question 7

Why is the Wurtz reaction unsuitable for preparing an alkane with an odd number of carbon atoms, and why does it fail with tertiary halides?

Accepted Answer

Because the reaction couples two alkyl fragments, so the product always has an even number of carbons when a single halide is used. Producing an odd-numbered chain would require two different halides, and coupling is indiscriminate, so three products form in comparable amounts and separation is impractical. With tertiary halides the strongly basic conditions promote elimination rather than coupling, so alkenes are obtained instead of the desired alkane.

Question 8

Compare the staggered and eclipsed conformations of ethane and the anti and gauche conformations of butane.

Accepted Answer

In ethane the staggered conformation places the hydrogens as far apart as possible and lies about 12.5 kJ mol$^{-1}$ below the eclipsed form. In butane the rotation moves two methyl groups rather than hydrogens, so a further distinction appears among the staggered forms: the anti conformation with the methyls 180 degrees apart is most stable, while the gauche forms at 60 degrees sit about 3.8 kJ mol$^{-1}$ higher because of the bulk of the two groups. The two eclipsed maxima also differ, methyl against methyl at about $+19$ and methyl against hydrogen at about $+16$: a torsional cost plus a steric cost, added. All the barriers are small enough that rotation is rapid and conformers cannot be separated.

Question 9

Classify benzene, cyclobutadiene and cyclooctatetraene as aromatic, antiaromatic or non-aromatic, with reasons.

Accepted Answer

Benzene is cyclic, planar and fully conjugated with six pi electrons, fitting $4n+2$ with $n=1$, so it is aromatic. Cyclobutadiene is cyclic, planar and conjugated with four pi electrons, fitting $4n$, so it is antiaromatic and so destabilised it can only be isolated at very low temperature. Cyclooctatetraene has eight pi electrons but its more immediate disqualification is that it is not planar, adopting a tub shape in which the p orbitals cannot overlap continuously, so it is non-aromatic rather than antiaromatic. Note that the rule counts pi electrons and ignores charge, so the cyclopentadienyl anion and cycloheptatrienyl cation are both aromatic at six.

Question 10

Predict the major mononitration products of toluene and of nitrobenzene, and explain the difference in both position and rate.

Accepted Answer

Toluene carries a methyl group, which donates electron density by the inductive effect and by hyperconjugation, so it activates the ring and the reaction is faster than for benzene. It directs to ortho and para, giving 2-nitrotoluene and 4-nitrotoluene, because attack there produces an arenium ion whose positive charge can sit next to the donating methyl. Nitrobenzene carries a strongly withdrawing nitro group, which deactivates the ring so harsher conditions are needed, and directs meta, giving 1,3-dinitrobenzene, since meta attack keeps the positive charge away from the nitro group.

Question 11

Explain why chlorobenzene is less reactive than benzene towards nitration yet still gives ortho and para products.

Accepted Answer

Chlorine exerts two opposing effects. Its high electronegativity withdraws electron density through the sigma bond, and because this operates at every position it lowers the overall ring density and slows electrophilic attack, so chlorobenzene nitrates about 30 times more slowly than benzene. Its lone pairs, however, can be donated into the ring by resonance, and that donation places extra density specifically at ortho and para. So induction controls the **rate** while resonance controls the **position**, which is why halogens are the only common deactivating groups that direct ortho and para.

Question 12

Why can propylbenzene not be made cleanly by Friedel-Crafts alkylation of benzene with 1-chloropropane, and what is the better route?

Accepted Answer

The reaction generates a primary propyl carbocation, which rearranges by a hydride shift to the more stable secondary isopropyl cation before it can attack the ring, so the major product is isopropylbenzene. There is a second problem even without rearrangement: the alkyl group installed is activating, so the product reacts faster than benzene itself and polyalkylation follows. The better route is Friedel-Crafts acylation with propanoyl chloride, since the acylium ion $\mathrm{RCO^+}$ is resonance-stabilised and does not rearrange while the acyl group deactivates the ring, followed by Clemmensen or Wolff-Kishner reduction of the ketone to $\mathrm{CH_2}$.

Practice — "Hydrocarbons"

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