Question Description
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.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
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.