Question

Construct a simulated 1H NMR spectrum for 2-chloropropane by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.

It has signals at 1.5 and about 3.8 ppm

Answer 1

Answer:

1.5 ppm (6H, doublet) and 3.8 ppm (1H, septet)

Explanation:

2-chloropropane is a symmetrical molecule along the second carbon that has chlorine. Essentially, the protons of the two methyl groups on both sides of the vertical symmetry plane are equivalent producing a peak at a lower ppm value of 1.5 ppm for a total of 6 protons (3 protons in each methyl group). This is simply because these protons are distant from the strongly electronegative chlorine atom and are, therefore, more shielded.

The middle carbon already has 3 bonds and only contains 1 proton having a high ppm value of 3.8 ppm due to being attached to the strongly deshielding (electron-withdrawing) chlorine directly.

The methyl protons would exhibit a doublet, as they have only one proton neighbour at the second carbon (according to the n + 1 rule, n neighbors produce n + 1 splitting). The proton at the second carbon would split into septet, as it has a total of 6 neighboring protons. That said, we have two peaks:

1.5 ppm (6H, doublet) and 3.8 ppm (1H, septet).

The diagram is attached below: