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CBSE Class 10 Science 2020 Delhi Set 3

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Question : 6 of 10
Marks: +1, -0
(a) Draw the structures for (i) ethanol, (ii) ethanoic acid.
(b) Why is the conversion of ethanol to ethanoic acid considered an oxidation reaction? Write the oxidising agent used in the reaction involved.
Solution:  
(a) Structures for ethanol and ethanoic acid are as follows:
(i) H−C∣H∣H−C∣H∣H−O−HEthanol\underset{\text{Ethanol}}{ \mathrm{H}- \underset{ \mathrm{H} }{ \underset{ \vert }{ \overset{ \mathrm{H} }{ \overset{ \vert }{ \mathrm{C} } } } } - \underset{ \mathrm{H} }{ \underset{ \vert }{ \overset{ \mathrm{H} }{ \overset{ \vert }{ \mathrm{C} } } } } - \mathrm{O}-\mathrm{H} }
(ii) H−C∣H∣H−C∥OEthanoic acid\underset{\text{Ethanoic acid}}{ \mathrm{H}- \underset{ \mathrm{H} }{ \underset{ \vert }{ \overset{ \mathrm{H} }{ \overset{ \vert }{ \mathrm{C} } } } } - \overset{ \mathrm{O} }{ \overset{ \parallel }{ \mathrm{C} } } }
(b) Conversion of ethanol to ethanoic acid is considered as an oxidation reaction because oxygen is added to ethanol to convert it to ethanoic acid.
CH3−CH2OHEthanol→ (or) Acidified  K2Cr2O7+ Heat Alc. KMO4+HeatCH3COOHEthanoic acid\underset{\text{Ethanol}}{ \mathrm{CH}_3-\mathrm{CH}_2\mathrm{OH} } \xrightarrow[\text{ (or) Acidified }\,\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + \text{ Heat }]{\text{Alc.}\,\mathrm{KMO}_4 + \text{Heat}} \underset{\text{Ethanoic acid}}{\mathrm{CH}_3\mathrm{COOH}}
In the above reaction alkaline KMnO4\mathrm{KMnO}_4 or acidified K2Cr2O7\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 adds oxygen to ethanol hence they are the oxidising agent used in the reaction involved.
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