NMR: Novice Level, Spectrum 12

Formula: C3H7Br

Answer: 1-Bromopropane (n-propyl bromide)

Chemical Shift Assignments: δ 1.01(t, 3H), 1.87 (sextuplet. 2H), 3.37 (t, 2H)

 

This sample of 1-bromopropane was prepared by the reaction of NaBr and H2SO4 in the presence of 1-propanol followed by distillation. The experiment was conducted by Jessica Chen in the undergraduate laboratory course (Chem 222La) during the fall of 2000. The triplet at δ 1.01 is the methyl group and the low field triplet δ 3.37 (t, 2H) is adjacent to the heteroatom. These five hydrogens are equally coupled (J = ~ 7 Hz) to the methylene group at δ 1.87 to produce the sextuplet. For a comparison of the 1H NMR spectra of n-propyl bromide and isopropyl bromide, click here.

The 13C spectrum displays the three carbon atoms at 36.06, 26.47, and 13.17 ppm. The signals are at the expected region of the spectrum. For a comparison of the 13C NMR spectra of n-propyl bromide and isopropyl bromide, click here. Return to Menu.

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