Question

Using the included 'H NMR spectra, identify which compound is the starting material and which is the product by drawing the Lewis structures directly on the corresponding spectrum. Then designate which peaks correspond to which protons on your Lewis structure (designate using (a), (b), etc.). -3.592 -3.578 -3.567 2.356 2.349 0.50 1.535 - 1.518 1500 -1480 1464 - 1.381 1.362 100 343 0.99 -1.325 -1.306 -1.289 -0.903 0.00 4.884- -0.867 1511 0.00 1.0

synthesis of 1-bromobutane

Answer 1

The Product formed here is 1-Bromobutane and from the ¹H NMR Signals, given below it can be confirmed

ABCD HHHH H-C-C-CC-Br HHAH Signal Proton A (Triplet) 0.8 B (Multiplet) 1.35 C(Multiplet) D (Triplet)

There is only 5 signals in the ¹H NMR spectra given and four signals are assigned to product. From this we can understand that, the starting material have similar structure and hydrogen atoms in similar environment (Magnetic field). From these assumptions and a broad peak at 2.3 ppm (Characteristic peak of -OH - Broad and varies from 0.5 ppm to 4.5 ppm) it can be concluded that the starting material is 1-Butanol. The ¹H NMR signals of 1-Butanol is given below.

ABCDE HHHH H-ċ-ċ-ċ-O-H HHH Proton A (Triplet) Signal 0.8 B (Multiplet) C(Multiplet) D (Triplet) E (Singlet)

(The signals of all protons in both starting material won't be same. There will be a slight difference. But in the ¹H NMR spectra of mixture of starting material and product an overlapped spectra is obtained because only slight variation in signal is observed.)