Q 13-13-23 P - Organic Chemistry (Mcmurry)

Question

We saw in Section 9-3 that addition of H-Br to a terminal alkyne leads to the Markovnikov addition product, with the Br bonding to the more highly substituted carbon. How could you use ¹³C NMR to identify the product of the addition of 1 equivalent of H-Br to 1-hexyne?

Step-by-Step Solution

Answer

The two possible products are easy to distinguish by using ¹³C NMR 2-bromo-1-hexene.

13 C NMR Spectroscopy

Carbon-13 nuclear magnetic resonance is the application of nuclear magnetic resonance spectroscopy to carbon. It is analogous to proton NMR and allows the identification of carbon atom in an organic molecule just as proton NMR identifies Hydrogen atoms.

DEPT 13 C NMR spectra

A modern technique called Distortionless Enhancement by Polarization Transfer, better known as DEPT ¹³C NMR spectra has been developed to distinguish between groups. It shows four spectra of the same compound each spectra provides different information.

Broadband decoupled Spectrum: It shows signal for all type of carbon atoms.

DEPT-45 spectrum: It shows signals for all carbon that are covalently bonded to hydrogens $({\mathrm{CH}}_3, {\mathrm{CH}}_2 \mathrm{and} \mathrm{CH})$

DEPT-90 spectrum: It shows signals for all carbon that are covalently bonded to one hydrogen (CH).

DEPT-135 spectrum: It shows signals for all carbon that are covalently bonded to hydrogens,$({\mathrm{CH}}_3, {\mathrm{CH}}_2 \mathrm{and} \mathrm{CH})$ but the phase of the signal will be different, depending on whether the number of attached hydrogen to each atom are odd or even. Signal arising from $({\mathrm{CH}}_3, \mathrm{CH})$odd groups will give positive peak and signal arising from $\left({\mathrm{CH}}_2\right)$even group will give negative peak.

Markovnikov’s addition rule

According to Markovnikov’s rule when a protic acid HX is added to an unsymmetric alkene, the acidic hydrogen attaches itself to the carbon having a greater number of hydrogen substituents whereas halide group attach itself to the carbon atom which has a greater number of alkyl substituents.

For example,            

Identify the product by 13 C NMR

                            Products of given addition reaction

The two possible products are easy to distinguish by using ¹³C NMR 2-bromo-1-hexene, the actual product formed, show no peaks in DEPT-90 ¹³C NMR spectrum because it has no CH carbons. The other possible product, 1-bromo-1-hexene, shows two peaks in its DEPT-90 ¹³C NMR spectrum.