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Exercises · 9.18

Q.Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

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Acidity increases with the stability of the conjugate base, which depends on the hybridisation of the carbon bearing the negative charge. The order is: ethyne > benzene > n-hexane.

The acidity of a hydrocarbon is determined by how readily it donates a proton (HX+\ce{H+}), which in turn depends on how stable the resulting conjugate base (the carbanion) is. The more stable the anion, the more acidic the parent compound.

The key insight here is that s-character of the carbon orbital holding the lone pair in the conjugate base dictates stability. Electrons in orbitals with higher s-character are held closer to the nucleus and are more stable. This is because s-orbitals are spherical and centreed on the nucleus, while p-orbitals have lobes extending away from it.

Let's examine each compound:

  1. Ethyne (HC≡CH\ce{HC≡CH}): The terminal carbon is spsp-hybridised (50% s-character). When ethyne loses a proton, the conjugate base is HC≡CX−\ce{HC≡C^-}, where the negative charge resides in an spsp orbital. This high s-character means the electrons are held tightly, making the anion very stable. Ethyne is therefore the most acidic of the three, with pKa≈25\text{p}K_a \approx 25.

  2. Benzene (CX6HX6\ce{C_6H_6}): The carbon atoms are sp2sp^2-hybridised (33% s-character). Removing a proton from benzene gives the phenyl anion CX6HX5X−\ce{C_6H_5^-}, where the negative charge is in an sp2sp^2 orbital. This is less stable than the ethynide ion but more stable than an sp3sp^3 carbanion. Benzene has pKa≈43\text{p}K_a \approx 43.

  3. n-Hexane (CHX3(CHX2)X4CHX3\ce{CH_3(CH_2)_4CH_3}): All carbons are sp3sp^3-hybridised (25% s-character). The conjugate base would be an alkyl anion with the negative charge in an sp3sp^3 orbital. This is the least stable arrangement because the electrons are held furthest from the nucleus. n-Hexane is extremely weakly acidic, with pKa≈50\text{p}K_a \approx 50.

Acidity∝s-character of C in conjugate base\text{Acidity} \propto \text{s-character of C in conjugate base}

sp (50%)>sp2 (33%)>sp3 (25%)sp \, (50\%) > sp^2 \, (33\%) > sp^3 \, (25\%) …

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