Problem 18

C₈H₁₄
8 - 14/2 + 1 = 2 degrees of unsaturation.
Look for any 2 of pi bonds or aliphatic rings.

The bands at 3000-2850 indicate C-H alkane stretches. There really aren't many other bands in the spectrum to indicate functional groups. The compound is an alkyne; we would expect to see a carbon-carbon triple bond stretch at 2260-2100, however, this is a weak band at best and often does not show up on IR.

This is the structure. See if you can assign the peaks on your own.

The integral values for A, B, and C need to be multiplied by 2 to get a total of 14 protons in the molecule. By convention, integral values on an NMR spectrum are reported as the lowest common multiple.