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Passage II
An organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.
Q. The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 is
- a)Only a pair of enantiomers are forme
- b)Only a pair of diastereomers are forme
- c)Two pairs of enantiomers are forme
- d)A meso product is forme
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Passage IIAn organic compound X (C7H11Br) shows optical isomerism as w...
Explanation:
To understand why a meso product is formed when compound X reacts with HBr in the presence of H2O2, we need to examine the reaction mechanism and the nature of the reactants involved.
Reaction Mechanism:
The reaction of compound X (C7H11Br) with HBr in the presence of H2O2 proceeds through a free radical mechanism. The bromine radical generated from HBr reacts with the compound X, resulting in the formation of a bromine atom radical and a brominated intermediate. This intermediate can undergo further reactions to form various products.
Optical Isomerism:
Compound X is optically active, indicating the presence of a chiral center in its structure. When it reacts with HBr in the absence of a peroxide, it forms a pair of diastereomers, both of which are optically active. This suggests that the reaction proceeds through a stereospecific pathway, resulting in the formation of two different diastereomers.
Decolourization of Bromine Water:
The decolourization of bromine water by compound X suggests that it is capable of undergoing an addition reaction with bromine. This reaction occurs through the electrophilic addition of bromine to the double bond present in compound X.
Formation of Meso Product:
When compound X reacts with HBr in the presence of H2O2, a meso product is formed. A meso compound is a type of stereoisomer that contains chiral centers but is not optically active due to internal symmetry. In this case, the formation of a meso product suggests the presence of an internal plane of symmetry within the molecule.
Conclusion:
In summary, when compound X (C7H11Br) reacts with HBr in the presence of H2O2, a meso product is formed. This indicates the presence of an internal plane of symmetry within the molecule, which makes it optically inactive despite having chiral centers. The reaction proceeds through a stereospecific pathway, forming a pair of diastereomers when HBr is added in the absence of a peroxide. The decolourization of bromine water suggests that compound X is capable of undergoing an addition reaction with bromine.
To understand why a meso product is formed when compound X reacts with HBr in the presence of H2O2, we need to examine the reaction mechanism and the nature of the reactants involved.
Reaction Mechanism:
The reaction of compound X (C7H11Br) with HBr in the presence of H2O2 proceeds through a free radical mechanism. The bromine radical generated from HBr reacts with the compound X, resulting in the formation of a bromine atom radical and a brominated intermediate. This intermediate can undergo further reactions to form various products.
Optical Isomerism:
Compound X is optically active, indicating the presence of a chiral center in its structure. When it reacts with HBr in the absence of a peroxide, it forms a pair of diastereomers, both of which are optically active. This suggests that the reaction proceeds through a stereospecific pathway, resulting in the formation of two different diastereomers.
Decolourization of Bromine Water:
The decolourization of bromine water by compound X suggests that it is capable of undergoing an addition reaction with bromine. This reaction occurs through the electrophilic addition of bromine to the double bond present in compound X.
Formation of Meso Product:
When compound X reacts with HBr in the presence of H2O2, a meso product is formed. A meso compound is a type of stereoisomer that contains chiral centers but is not optically active due to internal symmetry. In this case, the formation of a meso product suggests the presence of an internal plane of symmetry within the molecule.
Conclusion:
In summary, when compound X (C7H11Br) reacts with HBr in the presence of H2O2, a meso product is formed. This indicates the presence of an internal plane of symmetry within the molecule, which makes it optically inactive despite having chiral centers. The reaction proceeds through a stereospecific pathway, forming a pair of diastereomers when HBr is added in the absence of a peroxide. The decolourization of bromine water suggests that compound X is capable of undergoing an addition reaction with bromine.
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Passage IIAn organic compound X (C7H11Br) shows optical isomerism as w...
D is correct
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Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer?
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Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer?.
Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Here you can find the meaning of Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Passage IIAn organic compound X (C7H11Br) shows optical isomerism as well as decolourises brown colour of bromine water solution. X on treatment with HBr in the absence of a peroxide forms a pair of diastereomers, both of them are optically active. Also, X with C2H5ONa in C2H5OH gives a single product Y (C7H10). Y on treatment with ozone followed by reduction with (CH3)2S gives 1, 3,-cyclopentanedione as one product.Q.The correct statement regarding produces) formed by the reaction of X with HBr in th e presence of H2O2 isa)Only a pair of enantiomers are formeb)Only a pair of diastereomers are formec)Two pairs of enantiomers are formed)A meso product is formeCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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