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Mobile App Chapter 14: Q77P (page 527)
Cyclohex-2-enone has two protons on its carbon-carbon double bond (labeled \({{\rm{H}}_{\rm{a}}}\) and \({{\rm{H}}_{\rm{b}}}\) and two protons on the carbon adjacent to the double bond (labeled \({{\rm{H}}_{\rm{c}}}\)).
(a) If \({{\rm{J}}_{{\rm{ab}}}}\) = 11 Hz and \({{\rm{J}}_{{\rm{bc}}}}\) = 4 Hz, sketch the splitting pattern observed for each proton on the \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)hybridized carbons.
(b) Despite the fact that \({{\rm{H}}_{\rm{a}}}\) is located adjacent to an electron-withdrawing C=O, its absorption occurs up-field from the signal due to \({{\rm{H}}_{\rm{b}}}\) (6.0 vs. 7.0 ppm). Offer an explanation.
Short Answer
Expert verified
(a) Splitting pattern for \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized carbon:Doublet
Splitting pattern for proton \({{\rm{H}}_{\rm{a}}}\): Doublet
Splitting pattern for proton \({{\rm{H}}_{\rm{b}}}\): triplet of doublet
(b) \({{\rm{H}}_{\rm{b}}}\) signal comes at the downfield region because of the carbon-carbon double bond conjugation with a carbonyl group.
Step by step solution
01
Spin-spin coupling
The magnetic interaction between the spins of neighboring, non-equivalent NMR-active nuclei causes splitting of the NMR spectrum which is known as spin-spin coupling.
The splitting pattern is due to the equivalent H-atom at the nearby nuclei.
02
Splitting pattern of protons
(a) Splitting pattern for proton \({{\rm{H}}_{\rm{a}}}\): Proton \({{\rm{H}}_{\rm{a}}}\) is surrounded by a single proton \({{\rm{H}}_{\rm{b}}}\). Hence it couples with \({{\rm{H}}_{\rm{b}}}\) to give a doublet of 11Hz.
Splitting pattern for proton \({{\rm{H}}_{\rm{a}}}\)
(b) Splitting pattern for proton\({{\rm{H}}_{\rm{b}}}\): Protons\({{\rm{H}}_{\rm{b}}}\)are surrounded by two kinds of proton which include one proton\({{\rm{H}}_{\rm{a}}}\)and two protons\({{\rm{H}}_{\rm{c}}}\).
\({{\rm{H}}_{\rm{b}}}\)couples with\({{\rm{H}}_{\rm{a}}}\)to give a doublet of 11Hz, then each signal couples with one\({{\rm{H}}_{\rm{c}}}\)which further splits each signal into 2 signals with coupling constant 4Hz.
Finally, each signal couples with a second\({{\rm{H}}_{\rm{c}}}\)and furthersplits each signal into 2 signals with a coupling constant 4Hz.
Therefore, \({{\rm{H}}_{\rm{b}}}\) protons give a triplet of the doublet.
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