Question

Identify the product and it's quantity obtained when 9.2 g ethanol is treated with acidified potassium dichromate under ideal conditions?

Options

• Ethanal, 3.2 g

• Ethanoic acid, 3.2 g

• Ethanoic acid, 12.0 g

• Ethanal, 4.6 g

Answer 1

Ethanoic acid, 12.0 g

Explanation:

\[\begin{array}{cc}
\phantom{................................................}\ce{O}\\
\phantom{................................................}||\\
\ce{\underset{\text{Ethanol}}{H3C - CH2OH} ->[K2Cr2O7/HNO3][(O)] \underset{\text{Acetaldehyde}}{H3C - CHO} -> \underset{\text{Ethanoic acid}}{H3C - C - OH}}
\end{array}\]

M. W. = 46 g
Mass = 9.2 g

M. W. = 60 g
Mass = ?

46 g of Ethanol ≡ 60 g of Ethanoic acid

∴ 9.2 g of Ethanol ≡ `(60 xx 9.2)/46` = 12.0 g of Ethanoic acid

By using K₂Cr₂O₇, primary alcohol is first oxidised to aldehyde and then carboxylic acid. Oxidation does not stop at aldehyde stage.