Example 2.26 provides an NMR spectrum and asks for the common name of the compound with molecular formula C8H10O2. The spectrum shows two signals, a 2H singlet and a 3H singlet.
The solution explains that the only way to get two types of hydrogens on a disubstituted benzene is to have two equal substituents on benzene para to one another, which eliminates any compounds with ortho substituents.
I looked this up in an intro O-chem textbook (L. G. Wade, 6th edition, page 573), and the spectrum for 1,2-dimethylbenzene shows only two peaks even though there are three different types of protons. Wade is NOT saying that there are two types of hydrogens on 1,2-dimethylbenzene -- there are three -- but that when looking at the NMR spectrum for 1,2-dimethylbenzene there are only two signals, which normally would be interpreted as evidence for the presence of two types of protons. Wade uses the term "accidentally equivalent" to describe protons that are not chemically equivalent but happen to absorb at the same chemical shift.
With "accidentally equivalent" protons in mind, how would we be able to choose between answer choice C (para-methoxy anisole) and answer choice D (ortho-methoxy anisole) if we are only provided with the NMR spectrum (we are not explicitly told that there are only two types of protons)?
Is TBR ignoring the concept of "accidentally equivalent" protons?
Also, why is the downfield singlet in the spectrum a 2H singlet if there are 4 chemically equivalent hydrogens on the benzene ring of para-methoxy anisole?
Any help with this would be greatly appreciated!
Thanks!
Edit: if you don't have TBR O-chem, how would you describe (in terms of integration, splitting, relative shift values) the expected proton NMR spectrum of para-methoxy anisole? How could it be differentiated from the expected proton NMR spectrum of ortho-methoxy anisole?
The solution explains that the only way to get two types of hydrogens on a disubstituted benzene is to have two equal substituents on benzene para to one another, which eliminates any compounds with ortho substituents.
I looked this up in an intro O-chem textbook (L. G. Wade, 6th edition, page 573), and the spectrum for 1,2-dimethylbenzene shows only two peaks even though there are three different types of protons. Wade is NOT saying that there are two types of hydrogens on 1,2-dimethylbenzene -- there are three -- but that when looking at the NMR spectrum for 1,2-dimethylbenzene there are only two signals, which normally would be interpreted as evidence for the presence of two types of protons. Wade uses the term "accidentally equivalent" to describe protons that are not chemically equivalent but happen to absorb at the same chemical shift.
With "accidentally equivalent" protons in mind, how would we be able to choose between answer choice C (para-methoxy anisole) and answer choice D (ortho-methoxy anisole) if we are only provided with the NMR spectrum (we are not explicitly told that there are only two types of protons)?
Is TBR ignoring the concept of "accidentally equivalent" protons?
Also, why is the downfield singlet in the spectrum a 2H singlet if there are 4 chemically equivalent hydrogens on the benzene ring of para-methoxy anisole?
Any help with this would be greatly appreciated!
Thanks!
Edit: if you don't have TBR O-chem, how would you describe (in terms of integration, splitting, relative shift values) the expected proton NMR spectrum of para-methoxy anisole? How could it be differentiated from the expected proton NMR spectrum of ortho-methoxy anisole?
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