Hydrocarbons — Practice

Question 1

Predict the major monobromination product of $2,3$-dimethylbutane and explain why chlorination would be far less selective.

Accepted Answer

The molecule has two tertiary hydrogens and twelve primary hydrogens.

For bromine the relative rate is $1600:1$, so tertiary attack outweighs primary by $\dfrac{2	imes1600}{12	imes1}=267$ to $1$.

The major product is $2$-bromo-$2,3$-dimethylbutane, essentially exclusively.

For chlorine the ratio is only $5:1$, so primary attack outweighs tertiary by $\dfrac{12}{2\times5}$, giving a mixture in which the primary product actually dominates.

Question 2

Draw both chair conformations of cis-$1,3$-dimethylcyclohexane and identify the more stable, explaining why the trans isomer behaves differently.

Accepted Answer

For the **cis** $1,3$ isomer, one chair places both methyls equatorial and the other places both axial. The diequatorial conformation is far more stable and dominates.

For the **trans** $1,3$ isomer, every conformation has one methyl axial and one equatorial, so neither chair is favoured and the isomer is inherently less stable.

Cis-$1,3$ is therefore the more stable of the two isomers.

This reverses the pattern for $1,2$ and $1,4$ substitution, where trans is the isomer that achieves the diequatorial arrangement.

Question 3

Give reagents to convert $2$-methylbut-$2$-ene into (a) $2$-methylbutan-$2$-ol and (b) $3$-methylbutan-$2$-ol.

Accepted Answer

**(a)** The hydroxyl is on the more substituted carbon, so this is the Markovnikov product. Use oxymercuration with mercuric acetate in water, then sodium borohydride.

Dilute acid would also work here, since no rearrangement is possible from an already tertiary cation.

**(b)** The hydroxyl is on the less substituted carbon, so this is anti-Markovnikov. Use borane in tetrahydrofuran followed by alkaline hydrogen peroxide.

Hydroboration also adds syn, which fixes the relative stereochemistry of the two new stereocentres.

Question 4

Treatment of trans-but-$2$-ene with bromine gives a single achiral product, while cis-but-$2$-ene gives a racemic mixture. Explain.

Accepted Answer

Bromine adds **anti** through a bromonium ion in both cases.

Anti addition to the trans alkene gives the **meso** dibromide, which has an internal mirror plane and is achiral.

Anti addition to the cis alkene gives a pair of enantiomers, formed in equal amounts because the bromonium ion can form on either face.

The result is stereospecific: each starting geometry gives one definite stereochemical outcome, which is strong evidence for the bridged intermediate.

Question 5

Give the product of treating cyclohexene with (a) a peracid followed by aqueous acid and (b) cold dilute alkaline permanganate.

Accepted Answer

**(a)** The peracid gives an epoxide, which aqueous acid opens by backside attack of water. The two hydroxyls end up on opposite faces, giving **trans**-cyclohexane-$1,2$-diol as a racemic mixture.

**(b)** Permanganate forms a cyclic manganate ester delivering both oxygens from the same face. Hydrolysis gives **cis**-cyclohexane-$1,2$-diol, which is meso.

Both routes give a diol from the same alkene, with opposite stereochemistry.

The difference lies entirely in whether the oxygen is delivered by the ring or attacks it from outside.

Question 6

Explain why but-$1$-yne reacts with sodium amide but but-$2$-yne does not.

Accepted Answer

Acidity here requires a hydrogen attached **directly to an $sp$ carbon**, so that the resulting carbanion occupies an $sp$ orbital with fifty per cent $s$ character.

But-$1$-yne is terminal and has such a hydrogen, with $pK_a$ about $25$.

But-$2$-yne is internal, and its hydrogens are on $sp^{3}$ carbons with $pK_a$ around $50$, far beyond the reach of amide.

This difference is the basis of a simple test: only terminal alkynes give a precipitate with ammoniacal silver nitrate or cuprous chloride.

Question 7

Starting from ethyne, outline syntheses of cis-hex-$3$-ene and trans-hex-$3$-ene.

Accepted Answer

Deprotonate ethyne with sodium amide and alkylate with bromoethane to give but-$1$-yne.

Deprotonate again and alkylate with a second bromoethane to give hex-$3$-yne.

For the **cis** alkene, hydrogenate over Lindlar's catalyst, which delivers both hydrogens from the same face.

For the **trans** alkene, reduce with sodium in liquid ammonia, which proceeds through a vinyl radical anion adopting the less crowded trans geometry.

Question 8

Give the products of hydrating pent-$1$-yne and pent-$2$-yne with mercuric sulphate in dilute acid, and comment.

Accepted Answer

Pent-$1$-yne: the hydroxyl adds to carbon two, giving an enol that tautomerises to **pentan-$2$-one**.

Pent-$2$-yne is internal and unsymmetrical, so Markovnikov offers no clear preference. Both pentan-$2$-one and pentan-$3$-one are formed.

A **mixture** results, which is why acid hydration is synthetically useful only for terminal or symmetrical alkynes.

Hydroboration of pent-$1$-yne would instead give pentanal, the aldehyde.

Question 9

Explain why nitration of chlorobenzene is slower than that of benzene yet still gives mainly ortho and para products.

Accepted Answer

Chlorine is strongly electronegative and withdraws electron density from the whole ring by induction, destabilising the positively charged arenium ion. Every position is therefore deactivated and the overall rate falls.

Chlorine also has lone pairs that can be donated by resonance, and resonance donation reaches only the ortho and para positions.

Those two positions are therefore deactivated **least**, and substitution occurs there.

Rate is governed by the inductive effect and orientation by the resonance effect, and the two point in different directions without contradiction.

Question 10

Explain why Friedel-Crafts reactions cannot be carried out on nitrobenzene or on aniline, giving a different reason for each.

Accepted Answer

**Nitrobenzene:** the nitro group withdraws electron density so strongly that the ring is around a hundred thousand times less reactive than benzene. It cannot attack the weakly electrophilic acylium or alkyl cation at all.

**Aniline:** the problem is the catalyst rather than the ring. The nitrogen lone pair coordinates to aluminium chloride, forming a complex in which the nitrogen carries a positive charge.

That positively charged group is now strongly deactivating and meta directing, so the ring behaves like nitrobenzene.

The standard workround is to acetylate the amine first, moderating its donation before carrying out the substitution.

Question 11

An alkene $	ext{C}_7	ext{H}_{14}$ gives only butanone and propanal on reductive ozonolysis. Identify it and give the products of oxidative ozonolysis.

Accepted Answer

Butanone supplies a $	ext{CH}_3	ext{CH}_2\left(	ext{CH}_3ight)	ext{C}$ fragment; propanal supplies $	ext{CH}_3	ext{CH}_2	ext{CH}$.

Rejoining gives $	ext{CH}_3	ext{CH}_2\left(	ext{CH}_3ight)	ext{C}=	ext{CHCH}_2	ext{CH}_3$, that is $3$-methylhex-$3$-ene, formula $	ext{C}_7	ext{H}_{14}$.

Under oxidative work-up the butanone fragment is unchanged, since its carbon carries no hydrogen.

The propanal fragment is oxidised further to **propanoic acid**.

Question 12

Suggest a synthesis of $p$-nitroethylbenzene from benzene, and explain the order of steps.

Accepted Answer

Acylate benzene with acetyl chloride and aluminium chloride to give acetophenone, then reduce the carbonyl by the Clemmensen method to give ethylbenzene.

Nitrate the ethylbenzene; the ethyl group is activating and ortho-para directing, so the para product predominates.

The order cannot be reversed. Nitrating first would deactivate the ring so severely that no Friedel-Crafts reaction could follow.

Alkylating directly with chloroethane would also work here, since an ethyl cation cannot rearrange, but acylation and reduction is the general route for longer chains.

Practice — Hydrocarbons

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