Class 11 Exam > Class 11 Questions > A compound X (C5HgBr) is optically active. X ...
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A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/are
- a)is 1-bromp-1ynethyl cyclobutane
- b)X is 1-bromo-2-imethyl cyclobutane
- c)Y is a chiral hydrocarbon
- d)Z is an oxoaldebyde
Correct answer is option 'B,D'. Can you explain this answer?
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A compound X (C5HgBr) is optically active. X on treatment with KOH in ...
Reagent. Identify the compound X, Y, and Z.
Compound X is optically active, which means it contains a chiral center. Since it is given as C5HgBr, it suggests that it is a brominated compound with a chiral center.
On treatment with KOH in ethanol, compound X (C5HgBr) gives compound Y (C5H8) as the major product. This reaction suggests that compound X undergoes elimination of HBr to form compound Y. The elimination of HBr indicates the presence of a leaving group on compound X.
Ozonolysis of compound Y followed by Zn-hydrolysis gives compound Z (C5H8O2). This reaction suggests that compound Y contains a double bond that is cleaved by ozone, forming two carbonyl groups in compound Z.
Compound Z forms a yellow precipitate with NaOH/l2, indicating the presence of an aldehyde or ketone functional group. It also reduces Tollen's reagent, suggesting the presence of an aldehyde group.
Based on the given information, we can propose the following structures for the compounds:
Compound X: C5HgBr (brominated compound with a chiral center)
Compound Y: C5H8 (formed by elimination of HBr from X)
Compound Z: C5H8O2 (formed by ozonolysis and Zn-hydrolysis of Y)
Without further information, it is difficult to determine the exact structures of X, Y, and Z. Additional information, such as the stereochemistry of X or the position of the double bond in Y, would be needed to determine their structures more precisely.
Compound X is optically active, which means it contains a chiral center. Since it is given as C5HgBr, it suggests that it is a brominated compound with a chiral center.
On treatment with KOH in ethanol, compound X (C5HgBr) gives compound Y (C5H8) as the major product. This reaction suggests that compound X undergoes elimination of HBr to form compound Y. The elimination of HBr indicates the presence of a leaving group on compound X.
Ozonolysis of compound Y followed by Zn-hydrolysis gives compound Z (C5H8O2). This reaction suggests that compound Y contains a double bond that is cleaved by ozone, forming two carbonyl groups in compound Z.
Compound Z forms a yellow precipitate with NaOH/l2, indicating the presence of an aldehyde or ketone functional group. It also reduces Tollen's reagent, suggesting the presence of an aldehyde group.
Based on the given information, we can propose the following structures for the compounds:
Compound X: C5HgBr (brominated compound with a chiral center)
Compound Y: C5H8 (formed by elimination of HBr from X)
Compound Z: C5H8O2 (formed by ozonolysis and Zn-hydrolysis of Y)
Without further information, it is difficult to determine the exact structures of X, Y, and Z. Additional information, such as the stereochemistry of X or the position of the double bond in Y, would be needed to determine their structures more precisely.
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A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer?
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A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,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 A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,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 A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer?.
A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,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 A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,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 A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer?.
Solutions for A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,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 A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer?, a detailed solution for A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer? has been provided alongside types of A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A compound X (C5HgBr) is optically active. X on treatment with KOH in ethanol gives Y (C5H8) as major product. Ozonolysis of Y followed by Zn-hydrolysis gives Z (C5H8O2) . Z forms yellow precipitate with NaO H/l2 as well as it reduces Tollen’s reagent. Z on treatment with Zn (Hg)/HCI gives pentane. The correct statement(s) regarding X, Y and Z is/area)is 1-bromp-1ynethyl cyclobutaneb)X is 1-bromo-2-imethyl cyclobutanec)Y is a chiral hydrocarbond)Z is an oxoaldebydeCorrect answer is option 'B,D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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