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Chapter 14: Q.21558-14-38P (page 562)

Question: Using a 300 MHz NMR instrument:

a. How many Hz downfield from TMS is a signal at 2.5 ppm?

b. If a signal comes at 1200 Hz downfield from TMS, at what ppm does it occur?

c. If two signals are separated by 2 ppm, how many Hz does this correspond to?

Short Answer

Expert verified

Answer

a. 750 Hz

b. 4 ppm

c. 600 Hz

Step by step solution

01

Chemical shift

The chemical shift indicates the resonant frequency of a standard compound. The δ scale is used for measuring chemical shifts.

02

Upfield and downfield

  • An upfield signal has low energy and ppm value.
  • A downfield signal has high energy and ppm value.
03

Identifying the chemical shift 

The chemical shift (in ppm) on the scale can be using the given formula:

chemical shift (in ppm on δscale)

=observedchemicalshift(inHZ)dowmfieldfromTMSvofNMRspectrometer(inMHZ)

a. The chemical shift is 2.5 ppm.

of NMR is 300 MHz.

The observed chemical shift (in Hz) that is downfield from TMS can be found as shown:

chemical shift (in ppm on δscale) =Observedchemicalshift(inHZ)downfieldfromTMSvofNMRspectrometer(inMHZ)

2.5ppm=xHZ300MHZ

x=750 HZ

b. The observed chemical shift in Hz is 1200 Hz.

of NMR is 300 MHz.

The chemical shift (in ppm) can be found as shown:

2.5ppm=xHZ300MHZ

x =4 ppm

c. The chemical shift is 2 ppm.

of NMR is 300 MHz.

The observed chemical shift (in HZ) that is downfield from TMS can be found as shown:

2.5ppm=xHZ300MHZ

x=600HZ

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Most popular questions from this chapter

Propose a structure for a compound of molecular formula C7H14O2 with an IR absorption at 1740 cm-1 and the following 1HNMR data:

AbsorptionppmRelativeareasinglet1.29triplet1.33quartet4.12

How many signals are present in the 1HNMR spectrum for each molecule? What splitting is observed in each signal?

a.

b.

c.

Identify products A and B from the given 1HNMR data.

a. Treatment of CH2=CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1HNMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2H) ppm. What is the structure of A?

b.Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1HNMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?

Compound A exhibits two signals in its 1H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the absorbing signals is 2:3. Compound B exhibits two signals in its 1H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to dimethyl succinate and which compound corresponds to ethylene diacetate?

Propose a structure consistent with each set of data.

a. Compound J: molecular ion at 72; IR peak at 1710cm-1 ; 1H -NMR data (ppm) at 1.0 (triplet, 3 H), 2.1 (singlet, 3 H), and 2.4 (quartet, 2 H)

b. Compound K: molecular ion at 88; IR peak at 3600–3200 ;1H-NMR data (ppm) at 0.9 (triplet, 3 H), 1.2 (singlet, 6 H), 1.5 (quartet, 2 H), and 1.6 (singlet, 1 H)

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