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Chapter 13: Q25P (page 688)

Draw the expected broadband-decoupled 13C NMR spectra of the following compounds. Use Figure 13-41 (page 686) to estimate the chemical shift.

Short Answer

Expert verified

In this molecule, only three carbons is present which are chemically equivalent. Hence three carbon peak is formed with different chemical shift.

Step by step solution

01

Broadband-decoupled 13C-NMR spectra

The 13C NMR spectra tell us about the carbon skeleton of organic molecules. The 13C-NMR spectra gives two main pieces of information:

  1. The number of signals shows equivalent and different carbons or sets of carbons in molecules.
  2. The splitting of a also shows, how many hydrogens are attached to each carbon atom.

All spin-spin interaction is suppressed (decoupled) in NMR spectroscopy approach, resulting in singlets for all signals.

02

Estimation of 13C-spectra of compound (a)

In this molecule total four carbon is present in the skeleton but only two peak is formed because two-carbon set (a, b)are equivalent.

Chemical shift:2C (75.5), 2C(125.9).

03

Estimation 13C-spectra of Compound (b)

In this molecule, total si carbon is present but only three13C-NMR peak formed because in this organic skeleton three carbons set is present which is chemically equivalent.

Chemical shift:2C (22.4), 2C(24.9), 2C (126.0)

04

Estimation 13C-spectra of Compound (c)

In this molecule, five different carbons is present hence five peak is formed in the spectrum.

Chemical shift:1C (17.4), 1C(26.4), 1C (124.4), 1C(144.6), 1C (196.5).

05

Estimation 13C-spectra of Compound (c)

Chemical shift:1C (137.6), 1C(138.5), 1C (191.1).

In this molecule, only three carbons is present, which are chemically equivalent. Hence three carbon peak is formed with different chemical shifts.

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Most popular questions from this chapter

Question:Using a 60-MHz spectrometer, a chemist observes the following absorption: doublet, J = 7 Hz, at 4.00

(a) What would the chemical shift () be in the 500-MHz spectrum?

(b) What would the splitting value J be in the 500-MHz spectrum?

(c) How many hertz from the TMS peak is this absorption in the 60-MHz spectrum? In the 500-MHz spectrum?

When 2-chloro-2-methylbutane is treated with a variety of strong bases, the products always seem to contain two isomers (A and B) of formula. When sodium hydroxide is used as base, isomer A predominates. When potassium tert-butoxide is used as the base, isomer B predominates. TheandNMR spectra of A and B are given below.

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