Propanoic acid to Ethane
1. To convert propanoic acid to ethane, we perform a reduction reaction using a strong reducing agent like LiAlH‚„. The steps are: 1. Reduction of Propanoic acid: Propanoic acid is reduced by lithium aluminum hydride (LiAlH‚„) to ethanol. \[ \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \] 2. Reduction of Ethanol to Ethane: Ethanol is then reduced further by a reducing agent like zinc in the presence of hydrochloric acid (Zn/HCl) to ethane. \[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Zn/HCl}} \text{CH}_3\text{CH}_3 \] Thus, Propanoic acid is first converted to ethanol and then ethanol to ethane.
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]
Acidified KMnO_4 oxidizes sulphite to:
Which element is a strong reducing agent in +2 oxidation state and why?