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The Cannizaro Reaction

a An organic ...

Question

a) An organic compound (A) of molecular formula $C_{7} H_{6} O$ is not reduced by fehling's solution but will undergo cannizaro reaction. Compound (A) reacts with aniline to give compound (B).
compound (A) also reacts with $C l_{2}$ in the absence of catalyst to give compound (C). Identify A,B and C and explain the reactions.
b) The $K_{a}$ of propanoic acid is $1.34 \times 10^{- 5}$ . What is the $p H$ of a solution containing 0.5M propanoic acid and 0.5M sodium propanoate?

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Solution

a)
i) An organic compound (A) is not reduced by fehling solution but undergoes cannizaro reaction means (A) is benzaldehyde
$C_{6} H_{5} C H O + C_{6} H_{5} C H O N a O H \to C_{6} H_{5} C O O N a + C_{6} H_{5} C H_{2} O H$
ii) Compound (A) benzaldehyde reacts with aniline to compound (B) called schiff's base.
$C_{6} H_{5} C H = O + C_{6} H_{5} N H_{2} \to C_{6} H_{5} C H = N C_{6} H_{5}$
iii) Benzaldehyde reacts with $C l_{2}$ in the absence of catalyst to give (C) benzoyl chloride.
$C_{6} H_{5} C H O + C l_{2} ⟶ C_{6} H_{5} C (C l) = O + H C l$
A is Benzaldehyde $C_{6} H_{5} C H O$
B is Schiff's base $C_{6} H_{5} C H = N C_{6} H_{5}$
C is Benzoyl chloride
(b)
The $p H$ of an acidic buffer solution is given by the equation.
$p H = P_{k_{a}} + log \frac{[s a l t]}{[a c i d]}$
$p H = 1.34 \times 10^{- 5}$ and $log \frac{[0.5]}{[0.5]} = log 1 = 0$
$∴ p H = P_{k_{a}} = - log 1.34 \times 10^{- 5}$
$= - [5.01271] = - log \frac{0.5 m}{0.5 m} = 1$
$= - [- 4.8729] = 4.8729$

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