Predict the direction of the following equilibrium and the approximate value of Keq.
$R_{2} C = O$ + $HCl$ $⇌$ $R_{2} C = O^{+} H$ + $C l^{-}$ ( $p K_{a}$ : $HCl$ ≈ -6, protonated ketone ≈ -7)

Question

Predict the direction of the following equilibrium and the approximate value of Keq. R 2 ​ C = O + HCl ⇌ R 2 ​ C = O + H + C l − ( p K a ​ : HCl ≈ -6, protonated ketone ≈ -7)

TestHub · Accepted Answer

The acid on the left is HCl ( p K a ​ ≈ -6) and the acid on the right is the protonated ketone ( p K a ​ ≈ -7). The protonated ketone is the slightly stronger acid, so the equilibrium lies slightly to the left : in practice only a small fraction of the ketone is protonated by HCl . Keq ≈ 10^(-7-(-6)) = 1 0 − 1 ⋅

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