Understanding the Reaction Sequence

The given reaction involves a cyclic alkene undergoing ozonolysis (reaction with A) followed by a specific treatment (B) to form a dicarboxylic acid. Let's analyze step by step.

Step 1: Identify Reaction A (Ozonolysis)

The first step converts the alkene into carbonyl compounds. Ozonolysis is used to cleave alkenes at the double bond. For an alkene, ozonolysis with O₃ followed by Zn/H₂O gives aldehydes or ketones. Here, the product after A is not shown, but the final product is a dicarboxylic acid, indicating that the intermediate carbonyls are aldehydes (which can be oxidized to acids). Thus, A is O₃, Zn/H₂O.

Reaction: $\mathrm{Alkene}+O_3\stackrel{\mathrm{Zn}/H_{2}O}{\to }\mathrm{Carbonyl}\mathrm{compounds}$

Step 2: Identify Reaction B (Oxidation to Dicarboxylic Acid)

The intermediate from ozonolysis is likely an aldehyde (since it forms a carboxylic acid upon oxidation). To convert aldehyde to carboxylic acid, we need an oxidizing agent. KMnO₄ (under acidic or basic conditions) is commonly used for this oxidation. NaOH(alc)/I₂ is used for iodoform test or oxidation of methyl ketones, not for direct aldehyde oxidation to acid. Thus, B is KMnO₄.

Reaction: $\mathrm{Aldehyde}\stackrel{\mathrm{KMnO}4}{\to }\mathrm{Carboxylic}\mathrm{acid}$

Final Answer

Therefore, A is O₃, Zn/H₂O and B is KMnO₄. The correct option is: O₃, Zn/H₂O and KMnO₄.

Related Topics & Formulae

• Ozonolysis: Cleavage of alkenes with O₃ and Zn/H₂O to form carbonyl compounds. If double bond is terminal, it gives aldehyde.
• Oxidation of Aldehydes: Aldehydes are oxidized to carboxylic acids using KMnO₄/H⁺ or K₂Cr₂O₇/H⁺.
• NaOH(alc)/I₂: Used for iodoform test; oxidizes methyl ketones (CH₃C=O) to carboxylic acids with one carbon less.

Key Equations:

Ozonolysis: $R-\mathrm{CH}=\mathrm{CH}-R^{\text{'}}\stackrel{1.O_3,2.\mathrm{Zn}/H_{2}O}{\to }R-\mathrm{CHO}+R^{\text{'}}-\mathrm{CHO}$

Oxidation: $R-\mathrm{CHO}+\left[O\right]\stackrel{\mathrm{KMnO}4}{\to }R-\mathrm{COOH}$