Problem 4

C₆H₁₄O
Rule 2, omit O, gives C₆H₁₄
6 - 14/2 + 1 = 0 degrees of unsaturation.
No pi bonds or rings.

The broad band at 3350 indicates O-H stretch, probably an alcohol. The bands at 3000-2850 indicate C-H alkane stretches. The bands from 1320-1000 indicate C-O stretch, consistent with an alcohol.

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

As noted, C and D are not equivalent even though they're on the same carbon.