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 isomer 13C NMR spectrum, including the DEPT technique, would allow you to distinguish.

Short Answer

Expert verified

b)

Step by step solution

01

Step-1. Explanation of part (a) :

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.
In part (a), there are six peaks observed in carbon-NMR and carbons have been labelled in alphabets in structure to depict different types of carbons which gives rise to peaks. One peak is observed in DEPT-90 as only one methine carbon is present. Structure has four methylene groups; thus, four peaks appear downward in DEPT-135 and one peak each of methine and methyl carbon appears upwards in DEPT-135

Carbon-NMR, DEPT-90 and DEPT-135 peaks

02

Step-2. Explanation of part (b):

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.

In part (a), there are six peaks observed in carbon-NMR and carbons have been labelled in alphabets in structure to depict different types of carbons which gives rise to peaks. One peak is observed in DEPT-90 as only one methine carbon is present. Structure has four methylene groups; thus, four peaks appear downward in DEPT-135 and one peak each of methine and methyl carbon appears upwards in DEPT-135

Carbon-NMR, DEPT-90 and DEPT-135 peaks

03

Step-3. Explanation of part (c) :

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.
In part (c), there are five peaks observed in carbon-NMR and carbons have been labelled in alphabets in structure to depict different types of carbons which gives rise to peaks. One peak is observed in DEPT-90 as only one methine carbon is present. Structure has two methylene groups; thus, two peaks appear downward in DEPT-135 and one peak each of methine and methyl carbon appears upwards in DEPT-135.

Carbon NMR, DEPT-90, and DEPT-135 peaks

04

Step-4. Explanation of part (d):

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.
In part (d), there are five peaks observed in carbon-NMR and carbons have been labelled in alphabets in structure to depict different types of carbons which gives rise to peaks. Two peaks are observed in DEPT-90 as two methine carbons are present. Structure has one methylene group; thus, one peak appears downward in DEPT-135 and two peaks each of methine and methyl carbon appears upwards in DEPT-135

Carbon NMR peaks, DEPT-90, and DEPT-135 peaks

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

Question: An unknown compound has the molecular formula C9H11Br. Its proton NMR spectrum shows the following absorptions: singlet, d7.1, integral 44 mm singlet, d2.3, integral 130 mm singlet, d2.2, integral 67 mm Propose a structure for this compound.

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.

Each of these four structures has molecular formula . Match the structure with its characteristic proton NMR signals. (Not all of the signals are listed in each case.)

  1. Sharpsinglet atδ8.0 andtriplet atδ4.0
  2. Sharpsinglet atδ2.0 andquartet atδ4.1
  3. Sharpsinglet atδ3.7 andquartet atδ2.3
  4. Broadsinglet atδ11.5 andtriplet atδ2.3

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.

  1. Determine the structures of isomers A and B.
  2. Explain why A is the major product when using sodium hydroxide as the base and why B is the major product when using potassium tert-butoxide as the base.

Draw a splitting tree, similar to Figures 13-32 and 13-33, for proton Hc in styrene. What is the chemical shift of proton Hc?

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