Alkynes and benzene

MDCAT Chemistry · Chemistry · Hydrocarbons. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Alkynes (C_nH_2n−2) have a C≡C bond and undergo electrophilic addition twice. A terminal alkyne has an H atom on an sp carbon. An sp orbital has 50% s character, so it holds the bonding electrons close to carbon and this H can leave as H⁺. Terminal alkynes are therefore weak acids; alkenes and alkanes are not.

Benzene, C₆H₆, is a flat ring of six sp² carbons. The six π electrons are delocalised over the whole ring, so all C–C bonds are equal and the ring is very stable.

To keep this stability, benzene prefers electrophilic substitution to addition: an electrophile replaces one H atom.

Rules to remember

Terminal alkynes give a white precipitate with ammoniacal AgNO₃ and a reddish-brown precipitate with ammoniacal Cu₂Cl₂; they also react with Na or NaNH₂
Hydration of an alkyne (dilute H₂SO₄, HgSO₄): ethyne gives ethanal, all other alkynes give ketones
Benzene: all C–C bonds 139 pm (between C–C and C=C), bond angle 120°, resonance energy about 150 kJ mol⁻¹
Nitration: concentrated HNO₃ + concentrated H₂SO₄, electrophile NO₂⁺. Sulphonation: fuming H₂SO₄, electrophile SO₃
Halogenation: Cl₂ or Br₂ with FeCl₃, FeBr₃ or AlCl₃. Friedel–Crafts: RCl or RCOCl with anhydrous AlCl₃ (a Lewis acid that makes R⁺ or RCO⁺)
–CH₃ and –OH activate the ring and direct to positions 2 and 4; –NO₂ deactivates the ring and directs to position 3
Faraday discovered benzene in 1825; Kekulé proposed the ring structure in 1865. Toluene = methylbenzene, cumene = (1-methylethyl)benzene

Common mistake

Students let benzene decolourise bromine water like an alkene. Benzene does not add bromine in the dark; it needs a catalyst such as FeBr₃ and then substitution takes place, with HBr as the other product.

Worked example

1.3 g of ethyne, C₂H₂, reacts completely with excess sodium to form disodium ethynide. What mass of hydrogen is formed? (H = 1, C = 12)

Equation: HC≡CH + 2Na → NaC≡CNa + H₂, so 1 mol of ethyne gives 1 mol of H₂.
Moles of ethyne = 1.3 ÷ 26 = 0.050 mol, so moles of H₂ = 0.050 mol.
Mass of H₂ = 0.050 × 2 = 0.10 g.

Answer: 0.10 g of hydrogen

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