Open In App

Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 14: Q62E (page 599)

Draw the idealized NMR spectra for the following compounds.Assume that the five hydrogen atoms in the benzene ring are equivalent with no spin-spin coupling.

Short Answer

Expert verified

The NMR spectra for the compounds are

Step by step solution

Definition of NMR spectra

Nuclear magnetic resonance (NMR) spectroscopy is a technique used in spectroscopy for observing local magnetic fields around the nuclei of atoms.

Determination of NMR spectra of the compounds

The NMR spectra for the (a) is :

The NMR spectra for (b) is:

The NMR spectra for ( c) is:

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

In exercise 71 in chapter 13 the lewis structure for benzene was drawn. Using one of the lewis structures estimate $∆ H_{f}^{\circ}$ for $C_{6} H_{6} (g)$ . Using bond energies and given standard enthalpies of formation of C(g) is 717kJ/mol. The experimental _{$∆ H_{f}^{\circ}$}for $C_{6} H_{6}$ is 83 kJ/mol. Explain the discrepancy between the experimental value and the calculated $∆ H_{f}^{\circ}$ value of $C_{6} H_{6} (g)$ .

Vitamin B₆is an organic compound whose deficiency in the human body can cause apathy, irritability, and an increased susceptibility to infections. Below is an incomplete Lewis structure, for vitamin B₆ . Complete the Lewis structure and answer the following questions. [Hint: Vitamin B₆canbe classified as an organic compound (a compound based on carbon atoms). The majority of Lewis structures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero.

a. How many σ bonds and π bonds exist in vitamin B₆ ?

b.Give approximate values for the bond angles marked‘a’through‘g'in the structure.

c. How many carbon atoms are SP²hybridized?

d. How many carbon, oxygen, and nitrogen atoms are $s p^{3}$ hybridized?

e. Does vitamin B₆ exhibit delocalized π bonding? Explain.

Use the MO model to explain the bonding in BeH₂. When constructing the MO energy-level diagram, assume that the Be 1s electrons are not involved in bond formation.

The diatomic molecule OH exist in the gas phase. OH plays an important role in combustion reaction and is a reactive oxidizing agent in polluted air. The bond length and bond energy have been measured to be 97.06 pm and 424.7 kJ/mol, respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that the MOs result from the overlap of p_z orbital from oxygen and 1s orbital of hydrogen. (The O-H bond lies along the z-axis)

a. Draw a picture of the sigma bonding and antibonding molecular orbitals in OH.

b. Which of the two MOs has the greater hydrogen 1s character?

c. Can the 2px orbital of oxygen form MOs with 1s orbital of hydrogen? Explain.

d. Knowing that only the 2p orbitals of oxygen interact significantly with the 1s orbital of hydrogen, complete the MO diagram for OH. Place the correct number of electrons in the energy level.

e. Estimate the bond order for OH

f. Predict whether the bond order of OH⁺ is greater than, lesser than, or the same as that of OH+. Explain.

Which of the following statements concerning SO₂ is (are) true?

a. The central sulfur atom is sp² hybridized.

b. One of the sulfur–oxygen bonds is longer than the other(s).

c. The bond angles about the central sulfur atom are about 120 degrees.

d. There are two bonds in SO₂.

e. There are no resonance structures for SO₂.

See all solutions

Recommended explanations on Chemistry Textbooks

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

Draw the idealized NMR spectra for the following compounds.Assume that the five hydrogen atoms in the benzene ring are equivalent with no spin-spin coupling.

Short Answer

Step by step solution

Definition of NMR spectra

Determination of NMR spectra of the compounds

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Physical Chemistry

Chemical Reactions

Organic Chemistry

Chemical Analysis

Kinetics

Study anywhere. Anytime. Across all devices.

Company

Product

Help