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Passage II
An organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4 gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. C on further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E (C8H8O2) as one of the product.
Q.
The statement that is true regarding A to D is
- a)If oxygen of A is labelled by 18O, D will retain 18O
- b)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ring
- c)A can show both enantiomerism and diastereomerism
- d)B can show both enantiomerism as well as diastereomerism
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Passage IIAn organic compound A(C9H10O) is chiral and does not evolve ...
The compound A shown above satisfy the given criteria.
If oxygen of A is labelled as O18, it will be part of phenolic —OH, hence retained in D.
Most Upvoted Answer
Passage IIAn organic compound A(C9H10O) is chiral and does not evolve ...
Understanding the Organic Compound A to D Transformation
The transformation of the organic compound A (C9H10O) to D involves several important reactions and characteristics. Let's break down the key points related to the correct statement (option A).
Key Characteristics of Compound A
- Chirality: Compound A is chiral, meaning it has a non-superimposable mirror image, which allows for the existence of enantiomers.
- Hydrolysis: Upon hydrolysis with dilute H2SO4, A yields B (C9H12O2). Since B does not give a yellow precipitate with iodine, it suggests that B does not contain a methyl ketone or an active methylene group, eliminating potential enantiomeric forms.
Labeling with 18O
- Oxygen Labeling: If the oxygen of A is labeled with 18O, it will be retained in the product D after all transformations. This is due to the fact that the oxygen in the molecule remains intact through the transformations, meaning the labeled oxygen will not be lost or exchanged during the reactions.
Transformation Steps
- Excess Conc. HBr Reaction: Treatment of A with excess conc. HBr produces C (C9H11OBr). This indicates that the reaction introduces a bromine atom without altering the core structure significantly.
- Formation of D: After treating C with sodium ethoxide and followed by acidification, D, which is an isomer of A, is formed. The transformation does not involve the loss of the labeled oxygen.
Conclusion
- Retention of 18O: Therefore, option A is correct because the 18O label on compound A will be retained in compound D after the sequence of reactions, confirming the integrity of the functional group throughout the transformations.
This understanding of the mechanism and transformations supports the conclusion that option A is indeed the true statement regarding the relationship between compounds A and D.
The transformation of the organic compound A (C9H10O) to D involves several important reactions and characteristics. Let's break down the key points related to the correct statement (option A).
Key Characteristics of Compound A
- Chirality: Compound A is chiral, meaning it has a non-superimposable mirror image, which allows for the existence of enantiomers.
- Hydrolysis: Upon hydrolysis with dilute H2SO4, A yields B (C9H12O2). Since B does not give a yellow precipitate with iodine, it suggests that B does not contain a methyl ketone or an active methylene group, eliminating potential enantiomeric forms.
Labeling with 18O
- Oxygen Labeling: If the oxygen of A is labeled with 18O, it will be retained in the product D after all transformations. This is due to the fact that the oxygen in the molecule remains intact through the transformations, meaning the labeled oxygen will not be lost or exchanged during the reactions.
Transformation Steps
- Excess Conc. HBr Reaction: Treatment of A with excess conc. HBr produces C (C9H11OBr). This indicates that the reaction introduces a bromine atom without altering the core structure significantly.
- Formation of D: After treating C with sodium ethoxide and followed by acidification, D, which is an isomer of A, is formed. The transformation does not involve the loss of the labeled oxygen.
Conclusion
- Retention of 18O: Therefore, option A is correct because the 18O label on compound A will be retained in compound D after the sequence of reactions, confirming the integrity of the functional group throughout the transformations.
This understanding of the mechanism and transformations supports the conclusion that option A is indeed the true statement regarding the relationship between compounds A and D.
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Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer?
Question Description
Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer?.
Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Passage IIAn organic compound A(C9H10O) is chiral and does not evolve any gas on treatment with Na-metal but on hydrolysis with dil. H2SO4gives B (C9H12O2) which on further treatment with alkaline solution of iodine does not give an yellow precipitate. Also A on treatment with excess of conc. HBr gives C (C9H11OBr) as the major product. Con further treatment with C2H5ONa/C2H5OH followed by acidification of product gives D (an isomer of A). D on ozonolysis followed by work-up with dimethyl sulphide gives E(C8H8O2) as one of the product.Q.The statement that is true regarding A to D isa)If oxygen of A is labelled by 18O, D will retain 18Ob)If A is hydrolysed with H2O18/H+, Swill have 18O on phenyl ringc)A can show both enantiomerism and diastereomerismd)B can show both enantiomerism as well as diastereomerismCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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