Question

Question: (A true story.) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp 100–105 °C), marked only “STUDENT PREP.” She ran a quick mass spectrum, which is printed below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, “I know what it is.”

(a) What compound is the “student prep”? Any uncertainty in the structure?

(b) Suggest structures for the fragments at 136, 107, and 93. Why is the base peak (at m/z 57) so strong?

Short answer

Answer

a. The compound student prep is 1-bromobutane.

b. The fragmentation pattern is:

Step-by-step solution

Mass spectrum

A graph between the intensity (abundance) and $\frac{m}{z}$ (mass to charge ratio) values of the ions is known as the mass spectrum.

Explanation for part (a)

The unknown compound will be 1-bromobutane since there are two equal peaks as seen for M⁺and M⁺² in the mass spectrum.

Explanation for part (b)

The fragmentation pattern is shown below:

The base peak at is formed by leaving bromine (Br) , hence it is strong.