CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317599
Assertion : Chiral molecules do not have any element of symmetry. Reason : \(\mathrm{sp}^{3}\) hybridised carbon atom having four types of substituents is asymmetric and is called chiral.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Chiral molecules contain asymmetric carbon and molecules exhibiting chirality do not have any element of symmetry. Carbon atom with four different groups is called chiral carbon. So , the option (2) is correct.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317600
Optical isomerism arises from the presence of
1 a centre of symmetry
2 a line of symmetry
3 an asymmetric carbon atom
4 All of the above
Explanation:
A compound could be optically active only when it contains at least one asymmetric carbon atom or a chiral centre.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317601
Identify the chiral molecular from the following:
1 2-Bromo-2-methylbutane
2 2-Bromo-3-methylbutane
3 3-Bromopentane
4 2-Bromopropane
Explanation:
\(2^{\text {nd }}\) carbon atom is chiral.
MHTCET - 2021
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317602
Total number of optically active forms in molecules with ' \({\text{n}}\) ' number of asymmetric \(\mathrm{C}\) - atoms and which are not divisible into two equal halves is
Total number of optically active forms in molecules with ' \(\mathrm{n}\) ' number of asymmetric \(\mathrm{C}\) atoms and which are not divisible into two equal halves is \(2^{\mathrm{n}}\). When molecules are divisible into two equal halves where, \(\mathrm{n}=\) even number, then number of optically active forms is \(2^{\mathrm{n}-1}\). If \(\mathrm{n}=\) odd number then number, of optically active forms is \({{\text{2}}^{\left( {\frac{{{\text{n - 1}}}}{{\text{2}}}} \right)}}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317599
Assertion : Chiral molecules do not have any element of symmetry. Reason : \(\mathrm{sp}^{3}\) hybridised carbon atom having four types of substituents is asymmetric and is called chiral.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Chiral molecules contain asymmetric carbon and molecules exhibiting chirality do not have any element of symmetry. Carbon atom with four different groups is called chiral carbon. So , the option (2) is correct.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317600
Optical isomerism arises from the presence of
1 a centre of symmetry
2 a line of symmetry
3 an asymmetric carbon atom
4 All of the above
Explanation:
A compound could be optically active only when it contains at least one asymmetric carbon atom or a chiral centre.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317601
Identify the chiral molecular from the following:
1 2-Bromo-2-methylbutane
2 2-Bromo-3-methylbutane
3 3-Bromopentane
4 2-Bromopropane
Explanation:
\(2^{\text {nd }}\) carbon atom is chiral.
MHTCET - 2021
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317602
Total number of optically active forms in molecules with ' \({\text{n}}\) ' number of asymmetric \(\mathrm{C}\) - atoms and which are not divisible into two equal halves is
Total number of optically active forms in molecules with ' \(\mathrm{n}\) ' number of asymmetric \(\mathrm{C}\) atoms and which are not divisible into two equal halves is \(2^{\mathrm{n}}\). When molecules are divisible into two equal halves where, \(\mathrm{n}=\) even number, then number of optically active forms is \(2^{\mathrm{n}-1}\). If \(\mathrm{n}=\) odd number then number, of optically active forms is \({{\text{2}}^{\left( {\frac{{{\text{n - 1}}}}{{\text{2}}}} \right)}}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317599
Assertion : Chiral molecules do not have any element of symmetry. Reason : \(\mathrm{sp}^{3}\) hybridised carbon atom having four types of substituents is asymmetric and is called chiral.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Chiral molecules contain asymmetric carbon and molecules exhibiting chirality do not have any element of symmetry. Carbon atom with four different groups is called chiral carbon. So , the option (2) is correct.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317600
Optical isomerism arises from the presence of
1 a centre of symmetry
2 a line of symmetry
3 an asymmetric carbon atom
4 All of the above
Explanation:
A compound could be optically active only when it contains at least one asymmetric carbon atom or a chiral centre.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317601
Identify the chiral molecular from the following:
1 2-Bromo-2-methylbutane
2 2-Bromo-3-methylbutane
3 3-Bromopentane
4 2-Bromopropane
Explanation:
\(2^{\text {nd }}\) carbon atom is chiral.
MHTCET - 2021
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317602
Total number of optically active forms in molecules with ' \({\text{n}}\) ' number of asymmetric \(\mathrm{C}\) - atoms and which are not divisible into two equal halves is
Total number of optically active forms in molecules with ' \(\mathrm{n}\) ' number of asymmetric \(\mathrm{C}\) atoms and which are not divisible into two equal halves is \(2^{\mathrm{n}}\). When molecules are divisible into two equal halves where, \(\mathrm{n}=\) even number, then number of optically active forms is \(2^{\mathrm{n}-1}\). If \(\mathrm{n}=\) odd number then number, of optically active forms is \({{\text{2}}^{\left( {\frac{{{\text{n - 1}}}}{{\text{2}}}} \right)}}\)
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CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317599
Assertion : Chiral molecules do not have any element of symmetry. Reason : \(\mathrm{sp}^{3}\) hybridised carbon atom having four types of substituents is asymmetric and is called chiral.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Chiral molecules contain asymmetric carbon and molecules exhibiting chirality do not have any element of symmetry. Carbon atom with four different groups is called chiral carbon. So , the option (2) is correct.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317600
Optical isomerism arises from the presence of
1 a centre of symmetry
2 a line of symmetry
3 an asymmetric carbon atom
4 All of the above
Explanation:
A compound could be optically active only when it contains at least one asymmetric carbon atom or a chiral centre.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317601
Identify the chiral molecular from the following:
1 2-Bromo-2-methylbutane
2 2-Bromo-3-methylbutane
3 3-Bromopentane
4 2-Bromopropane
Explanation:
\(2^{\text {nd }}\) carbon atom is chiral.
MHTCET - 2021
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317602
Total number of optically active forms in molecules with ' \({\text{n}}\) ' number of asymmetric \(\mathrm{C}\) - atoms and which are not divisible into two equal halves is
Total number of optically active forms in molecules with ' \(\mathrm{n}\) ' number of asymmetric \(\mathrm{C}\) atoms and which are not divisible into two equal halves is \(2^{\mathrm{n}}\). When molecules are divisible into two equal halves where, \(\mathrm{n}=\) even number, then number of optically active forms is \(2^{\mathrm{n}-1}\). If \(\mathrm{n}=\) odd number then number, of optically active forms is \({{\text{2}}^{\left( {\frac{{{\text{n - 1}}}}{{\text{2}}}} \right)}}\)
