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The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/are
- a)I has perpendicular pi planes while II has parallel pi planes
- b)Both gives the same product on adding 1.0 equivalent at HBr of -80° C
- c)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperature
- d)Both have same length of C = C double bonds
Correct answer is option 'A,C'. Can you explain this answer?
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Verified Answer
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene,...
The correct answers are options A & C
Adjacent hybrid orbitals are perpendicular to each other irrespective of the type of the hybridization. In 1,2-butadiene, adjacent 2× bonds are perpendicular to each other while in 1,3-butadiene, 2× bond and
1× bonds are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
Adjacent hybrid orbitals are perpendicular to each other irrespective of the type of the hybridization. In 1,2-butadiene, adjacent 2× bonds are perpendicular to each other while in 1,3-butadiene, 2× bond and
1× bonds are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
Most Upvoted Answer
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene,...
Explanation:
Statement I: 1,2-butadiene(l) has perpendicular pi planes.
Statement II: 1,3-butadiene has parallel pi planes.
Statement A: I has perpendicular pi planes while II has parallel pi planes
This statement is correct. In 1,2-butadiene(l), the two double bonds are present in a perpendicular arrangement with respect to each other. This is due to the presence of a sigma bond between the two carbon atoms in the middle. On the other hand, in 1,3-butadiene, the double bonds are present in a parallel arrangement with respect to each other.
Statement B: Both give the same product on adding 1.0 equivalent of HBr at -80°C
This statement is incorrect. When 1,2-butadiene(l) reacts with 1.0 equivalent of HBr at -80°C, it gives 2,2-dibromobutane as the product. On the other hand, when 1,3-butadiene reacts with 1.0 equivalent of HBr at -80°C, it gives 2,3-dibromobutane as the product. Therefore, the products obtained are different for the two compounds.
Statement C: I gives 2,2-dibromobutane while II gives 2,3-dibromobutane, when reacted with excess of HBr at very low temperature
This statement is correct. When 1,2-butadiene(l) reacts with excess HBr at very low temperature, it gives 2,2-dibromobutane as the major product. On the other hand, when 1,3-butadiene reacts with excess HBr at very low temperature, it gives 2,3-dibromobutane as the major product. This is due to the difference in the arrangement of the double bonds in the two compounds.
Statement D: Both have the same length of C=C double bonds
This statement is incorrect. The C=C double bonds in 1,2-butadiene(l) are shorter than the C=C double bonds in 1,3-butadiene. This is because in 1,2-butadiene(l), the double bonds are conjugated, which leads to increased bond strength and shorter bond length. On the other hand, in 1,3-butadiene, the double bonds are not conjugated, resulting in longer bond length.
Therefore, the correct statements are A) I has perpendicular pi planes while II has parallel pi planes and C) I gives 2,2-dibromobutane while II gives 2,3-dibromobutane when reacted with excess of HBr at very low temperature.
Statement I: 1,2-butadiene(l) has perpendicular pi planes.
Statement II: 1,3-butadiene has parallel pi planes.
Statement A: I has perpendicular pi planes while II has parallel pi planes
This statement is correct. In 1,2-butadiene(l), the two double bonds are present in a perpendicular arrangement with respect to each other. This is due to the presence of a sigma bond between the two carbon atoms in the middle. On the other hand, in 1,3-butadiene, the double bonds are present in a parallel arrangement with respect to each other.
Statement B: Both give the same product on adding 1.0 equivalent of HBr at -80°C
This statement is incorrect. When 1,2-butadiene(l) reacts with 1.0 equivalent of HBr at -80°C, it gives 2,2-dibromobutane as the product. On the other hand, when 1,3-butadiene reacts with 1.0 equivalent of HBr at -80°C, it gives 2,3-dibromobutane as the product. Therefore, the products obtained are different for the two compounds.
Statement C: I gives 2,2-dibromobutane while II gives 2,3-dibromobutane, when reacted with excess of HBr at very low temperature
This statement is correct. When 1,2-butadiene(l) reacts with excess HBr at very low temperature, it gives 2,2-dibromobutane as the major product. On the other hand, when 1,3-butadiene reacts with excess HBr at very low temperature, it gives 2,3-dibromobutane as the major product. This is due to the difference in the arrangement of the double bonds in the two compounds.
Statement D: Both have the same length of C=C double bonds
This statement is incorrect. The C=C double bonds in 1,2-butadiene(l) are shorter than the C=C double bonds in 1,3-butadiene. This is because in 1,2-butadiene(l), the double bonds are conjugated, which leads to increased bond strength and shorter bond length. On the other hand, in 1,3-butadiene, the double bonds are not conjugated, resulting in longer bond length.
Therefore, the correct statements are A) I has perpendicular pi planes while II has parallel pi planes and C) I gives 2,2-dibromobutane while II gives 2,3-dibromobutane when reacted with excess of HBr at very low temperature.
Community Answer
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene,...
Adjacent hybrid orbitals are perpendicular to each other irrespective of the type of the hybridization. In 1,2-butadiene, adjacent 2× bonds are perpendicular to each other while in 1,3-butadiene, 2× bond and
1× bond are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
I hope it helps. If u still have any doubt, u can discuss with me.
1× bond are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
I hope it helps. If u still have any doubt, u can discuss with me.
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The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer?
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The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. 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 The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. 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 The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer?.
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. 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 The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. 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 The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer?.
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The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer?, a detailed solution for The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer? has been provided alongside types of The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/area)I has perpendicular pi planes while II has parallel pi planesb)Both gives the same product on adding 1.0 equivalent at HBr of -80° Cc)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperatured)Both have same length of C = C double bondsCorrect answer is option 'A,C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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