Question

Which substance can oxidize ethanol to ethanoic acid?

• A. Alkaline potassium permanganate
• B. Sodium hydroxide
• C. Dilute hydrochloric acid
• D. Sodium ethoxide

Hint:

Remember that oxidizing alcohols typically involves the use of oxidizing agents like potassium permanganate (\(KMnO_4\)), potassium dichromate (\(K_2Cr_2O_7\)), or chromium trioxide (\(CrO_3\)). The strength of the oxidizing agent and the reaction conditions can determine whether the alcohol is oxidized to an aldehyde/ketone or further to a carboxylic acid.

Answer 1

The Correct Option is A

Step 1: Understand the oxidation of ethanol.
Ethanol is a primary alcohol. Oxidation of primary alcohols can proceed in two stages: first to an aldehyde (ethanal in this case), and further oxidation leads to a carboxylic acid (ethanoic acid).
Step 2: Analyze the properties of the given substances.

• 
  Alkaline potassium permanganate (\(KMnO_4\)):
  Potassium permanganate is a strong oxidizing agent. In an alkaline medium, it is capable of oxidizing primary alcohols all the way to carboxylic acids. The purple \(MnO_4^-\) ion is reduced to the green \(MnO_4^{2-}\) ion (and further to \(MnO_2\) depending on conditions).
• 
  Sodium hydroxide (\(NaOH\)):
  Sodium hydroxide is a strong base. It can react with alcohols to form alkoxides, but it is not an oxidizing agent and will not oxidize ethanol.
• 
  Dilute hydrochloric acid (\(HCl\)):
  Hydrochloric acid is a strong acid. It can react with alcohols in a dehydration reaction under specific conditions (high temperature, presence of a catalyst) to form alkenes or ethers, but it does not oxidize alcohols.
• 
  Sodium ethoxide (\(CH_3CH_2ONa\)):
  Sodium ethoxide is the conjugate base of ethanol and a strong base. It is not an oxidizing agent and will not oxidize ethanol.

Step 3: Identify the oxidizing agent capable of converting ethanol to ethanoic acid.
Alkaline potassium permanganate is a well-known strong oxidizing agent that can oxidize primary alcohols like ethanol to carboxylic acids like ethanoic acid. The reaction proceeds as follows: \[ CH_3CH_2OH \xrightarrow{[O]} CH_3CHO \xrightarrow{[O]} CH_3COOH \] where \([O]\) represents oxidation by alkaline \(KMnO_4\).