(Solved): please help with 1-6  Why is Fourier transform NMR spectroscopy preferred over continuous wave ...

Why is Fourier transform NMR spectroscopy preferred over continuous wave as a technique for \( { }^{13} \mathrm{C} \) NMR? is commonly used as an internal reference in NMR spectroscopy; its signal is assigned \( \delta=0 \) in \( { }^{1} \mathrm{H} \) and \( { }^{13} \mathrm{C} \) NMR spectroscopy. \( \mathrm{H}-\mathrm{H} \) coupling is observed in the \( { }^{1} \mathrm{H} \) NMR spectrum of \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{3} \) and the signal from the methylene \( \mathrm{H}^{\prime} \) 's does not appear as a single peak. How does this signal appear? Explain in detail what is occurring. An unknown compound, \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{Br}_{2} \), gave the following proton NMR data: Singlet at \( 1.97 \mathrm{ppm}(6 \mathrm{H}) \) Singlet at \( 3.89 \mathrm{ppm}(2 \mathrm{H}) \) What is the compound? An unknown compound, \( \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Cl}_{3} \), gave the following proton NMR data: Doublet at \( 1.70 \mathrm{ppm}(3 \mathrm{H} \) ) Multiplet at \( 4.32 \mathrm{ppm} \) (1H) Doublet at \( 5.85 \mathrm{ppm}(1 \mathrm{H}) \) What is the structure of the compound? An unknown compound, \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \), gave the following proton NMR data: Triplet at \( 1.13 \mathrm{ppm} \) Quartet at \( 3.38 \mathrm{ppm} \) What is the structure of the compound?