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(a)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Hybridization: hybridization is mixing of valence atomic orbitals to get equivalent (equally energetic) hybridized orbitals. Sp3 hybridized orbital has four orbitals. Sp3 hybridized carbon atom in a compound has 4 sigma bonds.
Identify: the sp3 hybridized carbon atoms in the cyclopentadiene.
(b)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Most acidic proton of a compound depends upon the stability of conjugate base formed after the deprotonation.
Identify: most acidic proton presented in cyclopentadiene.
(c)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Resonance structures: Resonance structures are electrically equivalent Lewis structures. Formal charges on these structures are same, these structures represented through a double headed arrow between the structures. Resonance structures are resulted during the delocalization of pi-bonding electrons and lone pair of electrons in the molecule.
To draw: resonance structures of cyclopentadienyl anion.
(d)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Hybridization: hybridization is mixing of valence atomic orbitals to get equivalent (equally energetic) hybridized orbitals. Geometry can be predicted according to the hybridization of the molecule.
To find: the number sp3- hybridized carbon atoms present in cyclopentadienyl anion.
(e)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Geometry is predicted according to the hybridization of the molecule. Hybridization: hybridization is mixing of valence atomic orbitals to get equivalent (equally energetic) hybridized orbitals.
To find: geometry of cyclopentadienyl anion.
(f)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
In bond-line structure, bonds between other than hydrogen atoms and lone pairs presented are shown. Number of hydrogen atoms attached to a carbon atom depends upon the number of other than hydrogen atoms bonded and lone pairs presented to that carbon atom. Covalence of carbon is four. So hydrogen atoms are presented accordingly if the counting of carbon atom’s bonds with other than hydrogen atom and lone pair of electrons are less than four.
To find: number of hydrogen atoms presented in the cyclopentadienyl anion.
(g)
Interpretation:
sp3 hybridized carbon atoms, most acidic proton, sp3 hybridized carbon atoms of conjugate base, geometry of conjugate base, number of hydrogen atoms in the conjugate base, number of lone pairs in the conjugate base and the resonance structure of conjugate base of cyclopentadiene are needed to find out.
Concept introduction:
Lone pairs of the molecules are electron pairs which do not participate in the sigma bonding of the molecule. It can participate in the delocalization of molecule.
To find: the lone pairs present in the cyclopentadienyl anion.
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Chapter 3 Solutions
Organic Chemistry, 3e WileyPLUS Registration Card + Loose-leaf Print Companion
- NH2 1. CH3–MgCl 2. H3O+ ? As the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C - C bond as its major product: If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new C - C bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new C - C bond. Х ☐: Carrow_forwardPredict the major products of this organic reaction. If there will be no major products, check the box under the drawing area instead. No reaction. : + Х è OH K Cr O 2 27 2 4' 2 Click and drag to start drawing a structure.arrow_forwardLaminar compounds are characterized by havinga) a high value of the internal surface of the solid.b) a high adsorption potential.arrow_forward
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