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Passage I
An unknown organic compound A is determined to have molecular formula C6H12O and passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.
Q.
The correct statement regarding C is
- a)Reduction of a pure enantiomer of C with NaBH4 gives achiral product
- b)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixture
- c)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomers
- d)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diols
Correct answer is option 'C'. Can you explain this answer?
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Passage IAn unknown organic compound A is determined to have molecular...
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Passage IAn unknown organic compound A is determined to have molecular...
Reduction of C with NaBH4:
- When a pure enantiomer of compound C is reduced with NaBH4, it results in the formation of a pair of diastereomers.
- This is because the reduction of a chiral compound like C with a chiral reducing agent like NaBH4 leads to the formation of diastereomers due to the different spatial arrangements of the substituents around the newly formed chiral center.
Diastereomers Formation:
- Diastereomers are stereoisomers that are not mirror images of each other and have different physical and chemical properties.
- In this case, the reduction of the optically active compound C with NaBH4 results in the formation of two diastereomers due to the different orientations of the substituents around the chiral center created during the reduction process.
Explanation:
- The positive iodoform test obtained from compound C indicates the presence of a methyl ketone group.
- The treatment of a pure enantiomer of compound C with excess CH3MgBr followed by acid hydrolysis would lead to the formation of a racemic mixture of diols.
- This is because the reaction with excess CH3MgBr would result in the formation of a mixture of two diastereomeric alcohols, which upon acid hydrolysis, would give a racemic mixture of diols due to the presence of a chiral center and the formation of diastereomers in the reaction.
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Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer?
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Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer?.
Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Here you can find the meaning of Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Passage IAn unknown organic compound A is determined to have molecular formula C6H12Oand passing it through chiral column does not separate it into enantiomers. A does not react with Br2 nor with cold, dilute alkaline KMnO4. Heating A with concentrated H2SO4 gives product B (C6H10) which can be separated into enantiomers. Ozonolysis of a single enantiomer of B produces C, an optically active keto-aldehyde of formula C6H10O2 which gives positive iodoform test.Q.The correct statement regarding C isa)Reduction of a pure enantiomer of C with NaBH4 gives achiral productb)Reduction of a pure enantiomer of C with NaBH4 gives racemic mixturec)Reduction of a pure enantiomer of C with NaBH4 gives a pair of diastereomersd)Treatment of a pure enantiomer of C with excess of CH3MgBr followed by acid hydrolysis gives racemic mixture of diolsCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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