Concept explainers
a) C8H?O2, has two rings and one double bond.
Interpretation:
The number of hydrogens present in C8H?O2 which has two rings and one double bond.
Concept introduction:
Each ring or a double bond in a molecule corresponds to a loss of two hydrogens from the formula of
To determine:
The number of hydrogens present in C8H?O2 having two rings and one double bond.
b) C7H?N has two double bonds.
Interpretation:
The number of hydrogens present in C7H?N which has two double bonds.
Concept introduction:
Each ring or a double bond in a molecule corresponds to a loss of two hydrogens from the formula of alkane. Hence two hydrogens are to be added for each to arrive at the molecular formula of the parent alkane. Further, If the molecular formula of a compound contains halogens, oxygen and/or nitrogen, then the number of halogens is to be subtracted from the number of hydrogens, the number of oxygens to be ignored and number of the nitrogens is to be added to the number of hydrogens, in arriving the number of hydrogens.
To determine:
The number of hydrogens present in C7H?N having two double bonds.
c) C9H?NO, has one ring and three double bonds.
Interpretation:
The number of hydrogens present in C9H?NO which has one ring and three double bonds.
Concept introduction:
Each ring or a double bond in a molecule corresponds to a loss of two hydrogens from the formula of alkane. Hence two hydrogens are to be added for each to arrive at the molecular formula of the parent alkane. Further, If the molecular formula of a compound contains halogens, oxygen and/or nitrogen, then the number of halogens is to be subtracted from the number of hydrogens, the number of oxygens to be ignored and number of the nitrogens is to be added to the number of hydrogens, in arriving the number of hydrogens.
To determine:
The number of hydrogens present in C9H?NO having one ring and three double bonds.
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Chapter 7 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
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- 5. The existence of compounds of the noble gases was once a great surprise and stimulated a great deal of theoretical work. Label the molecular orbital diagram for XeF (include atom chemical symbol, atomic orbitals, and molecular orbitals) and deduce its ground state electron configuration. Is XeF likely to have a shorter bond length than XeF+? Bond Order XeF XeF+arrow_forward6. Draw the molecular orbital diagram shown to determine which of the following is paramagnetic. B22+ B22+, B2, C22, B22 and N22+ Molecular Orbital Diagram B2 C22- B22- N22+ Which molecule is paramagnetic?arrow_forward3. Put the following species in order of increasing bond length by using molecular orbital diagrams and calculating their bond orders: F2, F2, F2+ Molecular Orbital Diagram F2 F2 F2+ Bond Order Shortest bond: Longest bondarrow_forward
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
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