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Hex-4-ene-2-ol on treatment with PCC gives 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is
- a)2-pentene
- b)proponaldehyde
- c)2-butene
- d)4-methylpent-2-ene
Correct answer is option 'C'. Can you explain this answer?
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Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium h...
Hex-4-ene-2-ol undergoes a series of reactions to form compound C. Let's analyze each step of the reaction:
1. Treatment with PCC:
Hex-4-ene-2-ol reacts with pyridinium chlorochromate (PCC) to form compound A. PCC is an oxidizing agent that converts primary alcohols to aldehydes. In this case, the alcohol group (-OH) in hex-4-ene-2-ol is oxidized to an aldehyde group (-CHO) to form compound A.
2. Reaction with sodium hypoiodite:
Compound A reacts with sodium hypoiodite (NaIO) to form compound B. Sodium hypoiodite is a mild oxidizing agent that converts aldehydes to carboxylic acids. In this reaction, the aldehyde group (-CHO) in compound A is oxidized to a carboxylic acid group (-COOH) to form compound B.
3. Heating with soda lime:
Compound B is further heated with soda lime (a mixture of sodium hydroxide and calcium oxide) to form compound C. This reaction is known as the decarboxylation reaction, where the carboxylic acid group (-COOH) in compound B is removed as carbon dioxide (CO2), resulting in the formation of compound C.
Therefore, compound C is 2-butene, as it is formed by the removal of the carboxylic acid group (-COOH) from compound B.
In summary:
- Hex-4-ene-2-ol (starting compound)
- Treatment with PCC: Hex-4-ene-2-ol -> A (oxidation of alcohol to aldehyde)
- Reaction with sodium hypoiodite: A -> B (oxidation of aldehyde to carboxylic acid)
- Heating with soda lime: B -> C (decarboxylation, removal of carboxylic acid group)
Hence, the compound C formed is 2-butene, which corresponds to option C.
1. Treatment with PCC:
Hex-4-ene-2-ol reacts with pyridinium chlorochromate (PCC) to form compound A. PCC is an oxidizing agent that converts primary alcohols to aldehydes. In this case, the alcohol group (-OH) in hex-4-ene-2-ol is oxidized to an aldehyde group (-CHO) to form compound A.
2. Reaction with sodium hypoiodite:
Compound A reacts with sodium hypoiodite (NaIO) to form compound B. Sodium hypoiodite is a mild oxidizing agent that converts aldehydes to carboxylic acids. In this reaction, the aldehyde group (-CHO) in compound A is oxidized to a carboxylic acid group (-COOH) to form compound B.
3. Heating with soda lime:
Compound B is further heated with soda lime (a mixture of sodium hydroxide and calcium oxide) to form compound C. This reaction is known as the decarboxylation reaction, where the carboxylic acid group (-COOH) in compound B is removed as carbon dioxide (CO2), resulting in the formation of compound C.
Therefore, compound C is 2-butene, as it is formed by the removal of the carboxylic acid group (-COOH) from compound B.
In summary:
- Hex-4-ene-2-ol (starting compound)
- Treatment with PCC: Hex-4-ene-2-ol -> A (oxidation of alcohol to aldehyde)
- Reaction with sodium hypoiodite: A -> B (oxidation of aldehyde to carboxylic acid)
- Heating with soda lime: B -> C (decarboxylation, removal of carboxylic acid group)
Hence, the compound C formed is 2-butene, which corresponds to option C.
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Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer?
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Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer?.
Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hex-4-ene-2-ol on treatment with PCC gives A on reaction with sodium hypoiodite gives B, which on further heating with soda lime gives C. The compound C isa)2-penteneb)proponaldehydec)2-butened)4-methylpent-2-eneCorrect answer is option 'C'. Can you explain this answer?.
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