Question

1 -Propanol and 2-propanol can be distinguished by

• A. oxidation with alkaline $KMnO_{4}$ followed by reaction with Fehling solution
• B. oxidation with acidic dichromate followed by reaction with Fehling solution
• C. oxidation by heating with copper followed by reaction with Fehling solution
• D. oxidation with concentrated $H_{2}SO_{4}$ followed by reaction with Fehling solution

Answer 1

The Correct Option is C

(a)$CH_{3}CH_{2} CH_{2}OH $ ${ ->[{alk. KMnO_{4}}][{or\, K_{2}Cr_{2}O_{7} /H+}] }$ $CH_{3}CH_{2}CHO$
$\rightarrow CH_{3}CH_{2}COOH $ ${ ->[Fehling \, sol.] }$ No reaction
$\left(b\right) CH_{3}CHOHCH_{3}$ ${ ->[{alk. \, KMnO_{4}}][{or\, K_{2}Cr_{2}O_{7} / H+}] }$ $CH_{3}COCH_{3}$
$\to CH_{3}COOH$ ${ ->[Feh \, sol.] }$ No reaction
$\left(C\right)CH_{3}CH_{2}CH_{2}OH $ ${ ->[Cu] }$ $CH_{3}CH_{2}CHO$ ${ ->[{Fehling}][{Solution}] }$ $Cu_{2}O ?\quad_{\left(red\right)}$
$CH_{3}CHOHCH_{3} $ ${ ->[Cu] }$ $CH_{3}COCH_{3}$ ${ ->[{Fehling}][{Solution}] }$ No reaction
$\left(d\right) CH_{3} CH_{2}CH_{2}OH or\,$ $ CH_{3} CHOHCH_{3} $ ${ ->[ Conc.\, H_{2}SO_{4}] }$ $CH_{3} CH=CH_{2}$

Answer 2

The correct option is (C): “oxidation by heating with copper followed by reaction with Fehling solution”
1-Propanol \(\underset{\underset{−H_{2}}{\longrightarrow}}{\frac{Cu}{573K}} \text{ Propanol}\)  \(\text{(Give positive test with Fehling solution)}\)
2-Propanol \(\underset{\underset{−H_{2}}{\longrightarrow}}{\frac{Cu}{573K}} \text{ Propanone}\)
Oxidation with K₂Cr₂O₇H⁺ or KMnO₄ will give propanoic acid in the case of 1-propanol which can not be tested with Fehling solution. Reaction with conc. H₂SO₄ will lead to dehydration. Therefore, no oxidation is possible.

Answer 3

 It can be distinguished with the help of fehling’s solution. 1-propanol is a primary alcohol and can be converted into propanal by oxidation. There will be a colour change that will also indicate the positive test for reducing sugar. 

In case of 2-propanol, it is a secondary alcohol and therefore does not  have an aldehyde group that can be oxidised with the Fehling’s solution.