The compound \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}_{2}-\mathrm{CH}_{3}\) when treated with \(\mathrm{NaIO}_{4}\) in presence of \(\mathrm{KMnO}_{4}\) gives

Short Answer

Expert verified
Answer: The products formed are acetic acid (CH₃-COOH), formic acid (CH₂COOH), sodium manganate (NaMnO₄), potassium iodate (KIO₄), and water (H₂O).

Step by step solution

01

Identify the functional groups

The given compound is CH₃-C≡CH₂-CH₃, which contains an alkyne functional group (C≡C) at the second and third carbon atoms.
02

Reaction of alkynes with NaIO₄ and KMnO₄

When an alkyne reacts with NaIO₄ in the presence of KMnO₄, the alkyne undergoes oxidative cleavage. In this reaction, the triple bond (C≡C) gets broken, and the two fragments formed are oxidized to obtain carboxylic acid groups (-COOH).
03

Apply the oxidative cleavage to the given compound

CH₃-C≡CH₂-CH₃ undergoes oxidative cleavage at the alkyne bond, resulting in two fragments: 1. CH₃-COOH (acetic acid or ethanoic acid) 2. CH₂COOH (formic acid or methanoic acid)
04

Write the final balanced equation

The final balanced equation for the reaction of CH₃-C≡CH₂-CH₃ with NaIO₄ in the presence of KMnO₄ can be written as: CH₃-C≡CH₂-CH₃ + 2 NaIO₄ + 2 KMnO₄ → CH₃-COOH + CH₂COOH + 2 NaMnO₄ + 2 KIO₄ + 2 H₂O

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Most popular questions from this chapter

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