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.

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Hexamethylbenzene undergoes radical bromination with N-bromosuccinimide to give one monobrominated product (C12H17Br) and four dibrominated products (C12H16Br2). 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 how 13C NMR would easily distinguish among these compounds .

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