Question

Which of the following statement is wrong? (1) \Lambdalkencs decolouriscs alkalinc \(\mathrm{KMnO}_{4}\) solution but does not give precipitate with silver nitrate (2) Oxidation of alkencs with cold and dilute alkaline \(\mathrm{KMnO}_{4}\) gives 1,2 -diols (glycols) (3) IIydroxylation of alkenes with alkaline permanganate is always syn- addition (4) During oxidation of alkenes by alkaline permanganate, the oxidation state of Mn changes from 17 to 16

Short answer

Statement 4 is wrong.

Step-by-step solution

- Analyze Statement 1

Statement 1 says that alkenes decolourize alkaline KMnO4 solution but do not give a precipitate with silver nitrate. This is generally correct because alkenes react with alkaline KMnO4 (potassium permanganate) to form diols, resulting in decolorization.

- Analyze Statement 2

Statement 2 states that the oxidation of alkenes with cold and dilute alkaline KMnO4 gives 1,2-diols (glycols). This is correct as this reaction is known as syn-dihydroxylation.

- Analyze Statement 3

Statement 3 mentions that hydroxylation of alkenes with alkaline permanganate is always syn-addition. This is a correct statement as syn-addition refers to the simultaneous addition of the hydroxy groups on the same side of the double bond.

- Analyze Statement 4

Statement 4 claims that during the oxidation of alkenes by alkaline permanganate, the oxidation state of Mn changes from 17 to 16. This is incorrect. Manganese in KMnO4 starts with an oxidation state of +7 and reduces to +4, not to 16.

Key concepts

KMnO4 Reaction with Alkenes

Potassium permanganate (KMnO4) is a powerful oxidizing agent, and it reacts readily with alkenes. This reaction is often used to test for the presence of a double bond in organic compounds. When an alkene is added to an alkaline KMnO4 solution, the solution changes color, usually from purple to brown or colorless.

This color change happens because KMnO4 oxidizes the alkene, forming a diol (a molecule with two alcohol groups). The reaction can be represented as:
Alkene + KMnO4 (cold, dilute, alkaline) → Diol (glycol) + MnO2 (manganese dioxide)

Keep in mind this reaction does not produce a precipitate with silver nitrate, highlighting the specific reactivity of KMnO4 with alkenes.
- Potassium permanganate (KMnO4) is a purple compound.
- It turns brown or colorless when it reacts with alkenes.
- This reaction forms diols, which are compounds with two -OH groups.

Syn-Dihydroxylation

Syn-dihydroxylation refers to the addition of two hydroxyl groups (-OH) to the same side of a double bond in an alkene. This process is a form of addition reaction, which KMnO4 facilitates.

When you add cold, dilute alkaline KMnO4 to an alkene, the hydrogen groups attach simultaneously to the same face of the double bond. This type of addition is called 'syn' because both -OH groups are added to the same side. The final product is a 1,2-diol (also known as glycol).

The steps involved in syn-dihydroxylation are:
1. The alkene double bond reacts with KMnO4.
2. Two hydroxyl groups are added to the same side of the double bond.
3. A 1,2-diol is formed.

This reaction is highly useful in organic chemistry for transforming alkenes into diols, providing a way to further study and manipulate these molecules.

Oxidation States of Manganese

Understanding the oxidation states of manganese is essential in explaining its reactivity. In potassium permanganate (KMnO4), manganese is in its highest oxidation state, +7.

During the oxidation of alkenes, the oxidation state of manganese changes. Specifically, in the reaction with cold, dilute alkaline KMnO4, the Mn+7 state is reduced to Mn+4. This reduction is responsible for the color change from purple to brown or colorless.

Important notes about manganese oxidation states:
- In KMnO4, manganese starts with an oxidation state of +7.
- When oxidizing alkenes, manganese is reduced to an oxidation state of +4.
- It is incorrect to state that manganese changes oxidation state from 17 to 16; the proper range is from +7 to +4.

Recognizing these oxidation state changes helps clarify why and how the reaction between KMnO4 and alkenes results in the observed color changes and products.