Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Q1P (page 650)

Question:In a 300-MHz spectrometer, the protons in bromomethane absorb at a position 660 Hz downfield from TMS.

(a) What is the chemical shift of these protons?

(b) What is the chemical shift of the bromomethane protons in a 60-MHz spectrometer?

(c) How many hertz downfield from TMS would they absorb at 60 MHz?

Short Answer

Expert verified

Answer

  1. The chemical shift value is 2.20 ppm.
  2. The chemical shift value is 2.20 ppm.
  3. The answer is 132 Hz.

Step by step solution

01

Chemical shift

The chemical shift is the fraction and its formula is as shown:

Chemical shift=shift downfieldHzspectrometer frequencyMHz

02

Subpart (a)

The spectrometer frequency given is 300 MHz.

Downfield shift in bromomethane is 660 Hz.

The value of chemical shift can be calculated as shown below:

Chemical shift=shift downfieldHzspectrometer frequencyMHz=660 Hz300 MHz=2.20 ppm

03

Subpart (b)

The chemical shift value of bromomethane remains the same and it is 2.20 ppm.

04

Subpart (c)

The frequency shift is calculated as shown below:

frequency shift=chemical shift×spectrometer frequency=2.20 ppm×60 MHz=132  Hz

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Phenyl Grignard reagent adds to 2-methylpropanal to give the secondary alcohol shown. The proton NMR of 2-methylpropanal shows the two methyl groups as equivalent (one doublet at δ 1.1), yet the product alcohol, a racemic mixture, shows two different 3H doublets, one at δ 0.75 and one around δ 1.0

(a)Draw a Newmann projection of the product along the C1-C2 axis.
(b)Explain why the two methyl groups have different NMR chemical shifts. What is the term applied to protons such as these?

Different types of protons and carbons in alkanes tend to absorb at similar chemical shifts, making structure determination difficult. Explain how the among the following four isomerNMR spectrum, including the DEPT technique, would allow you to distinguish.

A laboratory student was converting cyclohexanol to cyclohexyl bromide by using one equivalent of sodium bromide in a large excess of concentrated sulfuric acid. The major product she recovered was not cyclohexyl bromide, but a compound of formula C6H10that gave the following 13CNMR spectrum:

  1. Propose a structure for this product.
  2. Assign the peaks in the 13CNMR spectrum to the carbon atoms in the structure.
  3. Suggest modifications in the reaction to obtain a better yield of cyclohexyl bromide.

The following spectra are taken from a compound that is an important starting material for organic synthesis. Determine the structure, first by considering each spectrum individually, and then by considering all the spectra together. Assign peaks to show that your proposed structure accounts for all the major features of each spectrum. DEPT information is given in blue on the carbon NMR

A compound(C10H1202)whose spectrum is shown here was isolated from a reaction mixture containing 2-phenylethanol and acetic acid.

  1. Propose a structure for this compound.
  2. Assign peaks to show which protons give rise to which signals in the spectrum.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free