CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317153
Empirical formula of compound having molar mass 58 is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Number of structural isomers possible, are
1 1
2 2
3 3
4 4
Explanation:
Empirical formula is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Empirical formula mass \(=29 \mathrm{~g} / \mathrm{mol}\) Molar mass \(=58 \mathrm{~g} / \mathrm{mol}\) \(\eta=\dfrac{\text { molar mass }}{\text { empirical mass formula }}=\dfrac{58}{29}=2\) Molecular formula \(=\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}=\mathrm{C}_{4} \mathrm{H}_{10}\) The molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{10}\) and the isomers are two (chain isomers) in number.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317154 show which type of isomerism?
1 Positional isomerism
2 Chain isomerism
3 Functional group isomerism
4 None of these
Explanation:
They are not isomers as they don't have same molecular formula.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317155
Number of acyclic isomers of the compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) is
1 4
2 6
3 5
4 7
Explanation:
The acyclic isomers of compound having molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) are as,
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317156
The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by
1 R and E
2 S and E
3 \(\mathrm{R}\) and \(\mathrm{Z}\)
4 \(\mathrm{S}\) and \(\mathrm{Z}\)
Explanation:
In the above compound, the configuration about the chiral carbon \(\left(\mathrm{C}^{*}\right)\) is \(\mathrm{R}\). Since, the groups with higher priority marked as(1) are on the same side of the double bond and those with lower priority are on the other side,thus the geometry about the \(\mathrm{C}=\mathrm{C}\) double bond is \(\mathrm{Z}\).
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317153
Empirical formula of compound having molar mass 58 is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Number of structural isomers possible, are
1 1
2 2
3 3
4 4
Explanation:
Empirical formula is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Empirical formula mass \(=29 \mathrm{~g} / \mathrm{mol}\) Molar mass \(=58 \mathrm{~g} / \mathrm{mol}\) \(\eta=\dfrac{\text { molar mass }}{\text { empirical mass formula }}=\dfrac{58}{29}=2\) Molecular formula \(=\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}=\mathrm{C}_{4} \mathrm{H}_{10}\) The molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{10}\) and the isomers are two (chain isomers) in number.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317154 show which type of isomerism?
1 Positional isomerism
2 Chain isomerism
3 Functional group isomerism
4 None of these
Explanation:
They are not isomers as they don't have same molecular formula.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317155
Number of acyclic isomers of the compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) is
1 4
2 6
3 5
4 7
Explanation:
The acyclic isomers of compound having molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) are as,
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317156
The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by
1 R and E
2 S and E
3 \(\mathrm{R}\) and \(\mathrm{Z}\)
4 \(\mathrm{S}\) and \(\mathrm{Z}\)
Explanation:
In the above compound, the configuration about the chiral carbon \(\left(\mathrm{C}^{*}\right)\) is \(\mathrm{R}\). Since, the groups with higher priority marked as(1) are on the same side of the double bond and those with lower priority are on the other side,thus the geometry about the \(\mathrm{C}=\mathrm{C}\) double bond is \(\mathrm{Z}\).
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317153
Empirical formula of compound having molar mass 58 is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Number of structural isomers possible, are
1 1
2 2
3 3
4 4
Explanation:
Empirical formula is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Empirical formula mass \(=29 \mathrm{~g} / \mathrm{mol}\) Molar mass \(=58 \mathrm{~g} / \mathrm{mol}\) \(\eta=\dfrac{\text { molar mass }}{\text { empirical mass formula }}=\dfrac{58}{29}=2\) Molecular formula \(=\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}=\mathrm{C}_{4} \mathrm{H}_{10}\) The molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{10}\) and the isomers are two (chain isomers) in number.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317154 show which type of isomerism?
1 Positional isomerism
2 Chain isomerism
3 Functional group isomerism
4 None of these
Explanation:
They are not isomers as they don't have same molecular formula.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317155
Number of acyclic isomers of the compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) is
1 4
2 6
3 5
4 7
Explanation:
The acyclic isomers of compound having molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) are as,
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317156
The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by
1 R and E
2 S and E
3 \(\mathrm{R}\) and \(\mathrm{Z}\)
4 \(\mathrm{S}\) and \(\mathrm{Z}\)
Explanation:
In the above compound, the configuration about the chiral carbon \(\left(\mathrm{C}^{*}\right)\) is \(\mathrm{R}\). Since, the groups with higher priority marked as(1) are on the same side of the double bond and those with lower priority are on the other side,thus the geometry about the \(\mathrm{C}=\mathrm{C}\) double bond is \(\mathrm{Z}\).
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317153
Empirical formula of compound having molar mass 58 is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Number of structural isomers possible, are
1 1
2 2
3 3
4 4
Explanation:
Empirical formula is \(\mathrm{C}_{2} \mathrm{H}_{5}\). Empirical formula mass \(=29 \mathrm{~g} / \mathrm{mol}\) Molar mass \(=58 \mathrm{~g} / \mathrm{mol}\) \(\eta=\dfrac{\text { molar mass }}{\text { empirical mass formula }}=\dfrac{58}{29}=2\) Molecular formula \(=\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}=\mathrm{C}_{4} \mathrm{H}_{10}\) The molecular formula is \(\mathrm{C}_{4} \mathrm{H}_{10}\) and the isomers are two (chain isomers) in number.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317154 show which type of isomerism?
1 Positional isomerism
2 Chain isomerism
3 Functional group isomerism
4 None of these
Explanation:
They are not isomers as they don't have same molecular formula.
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317155
Number of acyclic isomers of the compound having the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) is
1 4
2 6
3 5
4 7
Explanation:
The acyclic isomers of compound having molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) are as,
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317156
The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by
1 R and E
2 S and E
3 \(\mathrm{R}\) and \(\mathrm{Z}\)
4 \(\mathrm{S}\) and \(\mathrm{Z}\)
Explanation:
In the above compound, the configuration about the chiral carbon \(\left(\mathrm{C}^{*}\right)\) is \(\mathrm{R}\). Since, the groups with higher priority marked as(1) are on the same side of the double bond and those with lower priority are on the other side,thus the geometry about the \(\mathrm{C}=\mathrm{C}\) double bond is \(\mathrm{Z}\).
