1H NMR: Novice Level, Spectrum 2

Formula: C3H7I

Answer: Isopropyl Iodide (2-iodopropane)

Chemical Shift Assignments:

1H NMR: δ 1.90 (d, 6H), 4.32 (septuplet, 1H)

The two methyl groups are enantiotopic and necessarily equivalent. The 13C spectrum shows two peaks: 31.3 and 21.2 ppm. Which singlet is the methyl carbon? Click here. The six protons at δ 1.90 split the methine proton at δ 4.32 into a septuplet. Conversely, the methine proton splits the methyl groups into a doublet. There is a weak doublet at δ1.13 and an even weaker one at δ 1.23. Blow-up this region. What do you think the signal at δ 1.13 might be? How might this sample of isopropyl iodide been made? Try the next spectrum. Return to Menu.

iodine=violet