(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 CHEMISTRYPKGDRL+MLCRL MDL
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- The following 'H NMR spectrum was taken with a 750 MHz spectrometer: 1.0 0.5 0.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 ' 2.0 1.0 0.0 (ppm) What is the difference Av in the frequency of RF ac Δν ac radiation absorbed by the a and c protons? (Note: it's not equal to the difference in chemical shifts.) Round your answer to 2 significant digits, and be sure it has an appropriate unit symbol. = O O a will shift left, c will shift right. O a will shift right, c will shift left. a and c will both shift left, with more space between them. Suppose a new spectrum is taken with a 500 MHz spectrometer. What will be true about this new spectrum? O a and c will both shift left, with less space between them. O a and c will both shift right, with more space between them. O a and c will both shift right, with less space between them. Which protons have the largest energy gap between spin up and spin down states? O None of the above. ○ a Ob Explanation Check C Ar B 2025 McGraw Hill LLC. All Rights Reserved.…arrow_forwardWhat mass of Na2CO3 must you add to 125g of water to prepare 0.200 m Na2CO3? Calculate mole fraction of Na2CO3, mass percent, and molarity of the resulting solution. MM (g/mol): Na2CO3 105.99; water 18.02. Final solution density is 1.04 g/mL.arrow_forward(ME EX2) Prblms Can you please explain problems to me in detail, step by step? Thank you so much! If needed color code them for me.arrow_forward
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