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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 concerning product(s) formed when Y is treated with excess of HCI is
- a)A pair of enantiomers is formed in equal amount
- b)Two pairs of diastereomers are forme
- c)Only a meso-dichloride is formed
- d)Only one pair of diastereomers is formed
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Passage IIAn organic compound X (C7H11Br) shows optical isomerism as w...
Diastereomers are always chiral, and always different from one another. Diastereomers occur in
pairs, and each has
two chiral centers. In the original classic diastereomer, the chirality of one of them would be ( for example ) “ R, S “, and the other would be “ R, R
Most Upvoted Answer
Passage IIAn organic compound X (C7H11Br) shows optical isomerism as w...
Product Formation when Y is Treated with Excess of HCl
Two pairs of diastereomers are formed when Y is treated with excess of HCl.
Explanation:
Y is a cyclic compound with seven carbon atoms, and it can exist in cis and trans forms due to the presence of a double bond. When Y is treated with excess of HCl, the double bond is converted into a single bond, resulting in the formation of a cyclic compound with two chlorine atoms.
However, the formation of diastereomers is possible due to the presence of a chiral center in the molecule. The chiral center is formed due to the replacement of one of the carbon atoms in the ring with a chlorine atom.
The two pairs of diastereomers are formed due to the presence of two chlorine atoms in the molecule. The two pairs of diastereomers are formed due to the different arrangements of the two chlorine atoms in the molecule.
The two pairs of diastereomers differ in their physical and chemical properties, such as melting point, boiling point, and reactivity. They also have different optical activities, which means that they rotate the plane of polarized light in different directions. Therefore, they can be separated by techniques such as chromatography.
Conclusion:
When Y is treated with excess of HCl, two pairs of diastereomers are formed due to the presence of a chiral center in the molecule. The two pairs of diastereomers differ in their physical and chemical properties and can be separated by chromatography.
Two pairs of diastereomers are formed when Y is treated with excess of HCl.
Explanation:
Y is a cyclic compound with seven carbon atoms, and it can exist in cis and trans forms due to the presence of a double bond. When Y is treated with excess of HCl, the double bond is converted into a single bond, resulting in the formation of a cyclic compound with two chlorine atoms.
However, the formation of diastereomers is possible due to the presence of a chiral center in the molecule. The chiral center is formed due to the replacement of one of the carbon atoms in the ring with a chlorine atom.
The two pairs of diastereomers are formed due to the presence of two chlorine atoms in the molecule. The two pairs of diastereomers are formed due to the different arrangements of the two chlorine atoms in the molecule.
The two pairs of diastereomers differ in their physical and chemical properties, such as melting point, boiling point, and reactivity. They also have different optical activities, which means that they rotate the plane of polarized light in different directions. Therefore, they can be separated by techniques such as chromatography.
Conclusion:
When Y is treated with excess of HCl, two pairs of diastereomers are formed due to the presence of a chiral center in the molecule. The two pairs of diastereomers differ in their physical and chemical properties and can be separated by chromatography.
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Passage IIAn organic compound X (C7H11Br) shows optical isomerism as w...
B 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. 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 concerning product(s) formed when Y is treated with excess of HCI isa)A pair of enantiomers is formed in equal amountb)Two pairs of diastereomers are formec)Only a meso-dichloride is formedd)Only one pair of diastereomers is formedCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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