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

The standard 13CNMR spectrum of phenyl propanoate is shown here. Predict the appearance of the DEPT-90 and DEPT-135 spectra.

Short Answer

Expert verified

The full carbon spectrum of phenyl propanoate is presented below:

DEPT-90 will show only the methine carbons and all other peaks disappear.

The DEPT-135 will show only the methyl and methine peaks pointed upwards and the methylene peaks pointed downwards.

Step by step solution

01

DEPT carbon NMR:

Distortionless Enhancement by Polarization Transfer (DEPT) is a double resonance pulse program that transfers polarization from an excited nucleus to another. There are three DEPT experiments and they differ only in the “APT portion” of the pulse program, meaning, that they differ only in magnitude of the final 1H tip angle (X = 45, 90, 130). DEPT-45 leaves all resonances with a positive phase, DEPT-90 only shows methine carbon and DEPT-135 shows methine/methyl with a positive phase and methylene with a negative phase.

02

Application of DEPT-90 and DEPT-135 spectra for phenyl propionate:

The full carbon NMR spectrum for phenyl propanoate is presented below with chemical shifts marked with carbon atoms indicated by alphabets.

DEPT-90 will show only the methine carbons and all other peaks disappear. DEPT-90 only gives methine peaks. The carbon atoms which are methine carbons have been displayed in DEPT-90 spectrum with alphabet numbering that which carbon corresponds to value of chemical shift.

The DEPT-135 will show only the methyl and methine peaks pointed upwards and the methylene peaks pointed downwards. DEPT-135 gives signals of all protonated carbons and signals of quaternary carbons are absent in all DEPT spectra. The chemical shift values have been indicated for the carbons in the spectra.

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.

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

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.

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?

Predict the chemical shifts of the protons in the following compounds

If the imaginary replacement of either of two protons forms enantiomers, then those protons are said to be enantiotopic.The NMR is not a chiral probe, and it cannot distinguish between enantiotopic protons. They are seen to be “equivalent by NMR”.

  1. Use the imaginary replacement technique to show that the two allylic protons (those on) of allyl bromide are enantiotopic.
  2. Similarly, show that the two HCprotons in cyclobutanol are enantiotopic.
  3. What other protons in cyclobutanol are enantiotopic?

Hexamethylbenzene undergoes radical bromination with N-bromosuccinimide to give one monobrominated productand four dibrominated products. These products are easily separated by GC-MS, but the dibrominated products are difficult to distinguish by their mass spectra. Draw the monobrominated product and the four dibrominated products and explain howNMR would easily distinguish among these compounds.

See all solutions

Recommended explanations on Chemistry Textbooks

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