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Chapter 14: Q23P (page 527)

Question: How many peaks are observed in the \({}^1H\) NMR signal for each proton shown in red in palau'amine, the complex chapter-opening molecule?

Short Answer

Expert verified

\({H_a}\): doublet

\({H_b}\): doublet

\({H_c}\): singlet

\({H_d}\): doublet

Step by step solution

01

The (n+1) rule

The (n +1) rule is used to predict the multiplicity of\({}^{\bf{1}}{\bf{H}}\)NMR peaks.

n represents the number of hydrogen atoms present on the carbon atoms adjacent to the carbon to which the desired hydrogen is connected.

02

Peaks observed in the \({}^{\bf{1}}{\bf{H}}\) NMR spectrum of palau'amine

The number of peaks observed for the indicated hydrogen atoms is as follows:

\({H_a}\): \(1\; + \;1\; = \;2\) (doublet).

\({H_b}\): \(1\; + \;1\; = \;2\) (doublet)

\({H_c}\): singlet (as it is connected to a nitrogen atom).

\({H_d}\): \(1\; + \;1\; = \;2\) (doublet)

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Most popular questions from this chapter

Identify A and B, isomers of molecular formula C3H4Cl2 , from the given 1HNMR data: Compound A exhibits signals at 1.75 (doublet, 3H,J = 6.9 Hz) and 5.89 (quartet, 1H, J = 6.9 Hz) ppm. Compound B exhibits signals at 4.16 (singlet, 2 H), 5.42 (doublet, 1H, J = 1.9 Hz), and 5.59 (doublet, 1H, J = 1.9 Hz) ppm.

Question. When 2-bromo-3,3-dimethylbutane is treated with K+- OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4 , the major product U has the same molecular formula. Given the following -NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm 1 H NMR of U: 1.60 (singlet) ppm.

The 1H NMR spectrum of CH3OH recorded on a 500 MHz NMR spectrometer consists of two signals, one due to the CH3 protons at 1715 Hz and one due to the OH proton at 1830 Hz, both measured downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) Do the CH3 protons absorb upfield or downfield from the OH proton?

Question:Reaction of unknown A with HCl forms chlorohydrin B as the major product. A shows no absorptions in its IR spectrum at 1700 cm-1 or 3600-3200 cm-1 , and gives the following 1H NMR data: 1.4 (doublet, 3 H), 3.0 (quartet of doublets, 1 H), 3.5 (doublet, 1 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.2 (doublet, 2 H) ppm.

(a) Propose a structure for A, including stereochemistry.

(b) Explain why B is the major product in this reaction.

Question: How could you use chemical shift and integration data in 1H NMR spectroscopy to distinguish between CH3OCH2CH2OCH3and CH3OCH2OCH3? The 1H NMR spectrum of each compound contains only singlets.
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