Question

7.23 Th e condensation reaction of acetone, (CH3)2CO (propanone), in aqueous solution is catalyzed by bases, B, which react reversibly with acetone to form the carbanion C3H5O−. Th e carbanion then reacts with a molecule of acetone to give the product. A simplifi ed version of the mechanism is (1) AH + B → BH+ + A− (2) A− + BH+ → AH + B (3) A− + HA → product where AH stands for acetone and A− its carbanion. Use the steadystate approximation to fi nd the concentration of the carbanion and derive the rate equation for the formation of the product.

Answer 1

Answer:

$\frac{d[product]}{dt}=K_{3}[A^{-}][HA]$

Explanation:

The multiple steps are:

1) $AH+B--->BH^{+}+A^{-}$

2) $BH^{+}+A^{-} --->AH + B$

3) $A^{-}+HA-->Products$

For carbanion (A⁻) the rate law will be

$\frac{d[A^{-}]}{dt}=K_{1}[AH][B] - K_{2}[A^{-}][BH^{+}]-K_{3}[A^{-}][HA]$...(4)

As here B is in intermediate

so we may apply Steady state approximation (SSA) on it.

$\frac{d[B]}{dt}=-K_{1}[AH][B] +K_{2}[A^{-}][BH^{+}]=0$

$[B]=\frac{K_{2}[A^{-}][BH^{+}]}{K_{1}[AH]}$ ....(5)

Put this in above expression (4)

$\frac{d[A^{-}]}{dt}=K_{1}[AH][\frac{K_{2}[A^{-}][BH^{+}]}{K_{1}[AH]} - K_{2}[A^{-}][BH^{+}]-K_{3}[A^{-}][HA]$

Thus

$\frac{d[A^{-}]}{dt}=-K_{3}[A^{-}][HA]$

This is rate equation for formation of product

$\frac{d[product]}{dt}=K_{3}[A^{-}][HA]$