Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidative Cleavage
Oxidative cleavage is a process where molecules are split into smaller fragments with the help of oxidizing agents. It is a chemical reaction that typically breaks carbon-carbon (C-C) bonds, converting them into carbon-oxygen bonds (like carbonyl groups or carboxylic acids). This process is particularly useful in organic chemistry for transforming complex molecules into simpler, more functional groups.

In the given exercise, oxidative cleavage involves the breaking of the carbon-carbon triple bond in alkynes. The presence of strong oxidizing agents like \( ext{NaIO}_4\) (sodium periodate) and \( ext{KMnO}_4\) (potassium permanganate) helps in this transformation. These reagents work together to facilitate the cleavage of the alkyne into two separate molecules.

Oxidative cleavage is an important reaction because it often leads to the formation of functional groups such as carboxylic acids, which can participate in further chemical synthesis or serve as endpoints in analysis.
Alkyne Reactions
Alkynes are hydrocarbons characterized by a carbon-carbon triple bond (C≡C). This triple bond makes alkynes highly reactive and versatile in organic reactions. The reaction of alkynes with oxidizing agents like \(\text{NaIO}_4\) and \(\text{KMnO}_4\) involves significant transformations.

Here's how the oxidative cleavage of alkynes typically proceeds:
  • The triple bond is broken through the addition of oxygen from the oxidizing agents.
  • The cleavage results in the formation of two new molecules.
  • These molecules are usually oxidized further into carboxylic acids.
In the exercise, the compound \(\text{CH}_3-\text{C}\equiv\text{CH}_2-\text{CH}_3\) undergoes oxidative cleavage, leading to the creation of acetic acid (\(\text{CH}_3\text{COOH}\)) and formic acid (\(\text{CH}_2\text{COOH}\)). This exemplifies alkyne reactions' influence in producing vital organic molecules through oxidative methods.
Carboxylic Acid Formation
Carboxylic acids are a class of organic compounds identified by the presence of at least one carboxyl group (\(-COOH\)). The formation of these compounds is a crucial result of oxidative cleavage reactions.

In oxidative cleavage of alkynes, the triple bond is transformed into two carboxyl groups. Understanding the process:
  • Upon oxidative cleavage, electrons from the triple bond are redistributed to form new bonds with oxygen.
  • This results in the creation of carboxylic acids, which have a distinct -OH (hydroxyl) group attached to the carbonyl carbon.
  • These newly formed carboxylic acids are often more reactive and can engage in various subsequent chemical reactions.
In our specific example, the alkyne compound gives rise to acetic acid and formic acid following oxidative cleavage, illustrating how valuable carboxylic acids are in organic chemistry for forming new compounds and materials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Each question in this section has four suggested answers out of which ONE OR MORE answers will be correct. Which of the following statements is correct? (a) Cyclopropane will react with chlorine or bromine at room temperature, in diffused light to give a dihalogen derivative. (b) A hydrocarbon of molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6}\) should contain always a triple bond. (c) The trans isomer of an alkene is generally less polar than the cis isomer. (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) is an example of cumulated diene.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) product The product formed is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO} \mathrm{CH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO} \mathrm{CO} \mathrm{CH}_{3}\) (d)

An organic compound (A) \(\mathrm{C}_{5} \mathrm{H}_{8}\) does not react with Tollens reagent but reacts with \(\mathrm{H}_{2}\) in the presence of Lindlar catalyst to give (B). Aqueous chlorine water reacts with (B) to give (C) which when reacted with hot alcoholic KOH gives (D). The compound (D) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\) (c) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d)

(A) and (B) are respectively (a) But-2-yne and but-2-ene (b) But-2-ene and but-2-yne (c) Butane and but-2-ene (d) But-2-ene and butane

A compound that decolourizes alkaline \(\mathrm{KMnO}_{4}\) and gives a red precipitate with ammoniacal cuprous chloride is (a) 1 -butyne (b) 1 -pentene (c) 2 -butyne (d) 2-pentyne

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free