Draw a structure for an amine that exhibits a molecular ion at m+ = 87 and that produces fragments at m/z = 72 and m/z = 58.

             The amine is N-methylbutan-2-amine and the structure is drawn below.

Explanation:

                      In order to find out the correct structure for said data we will first try out to find the molecular formula by applying rule of thirteen.

Rule of Thirteen:

First divide the parent peak value by 13 as,

                = 87 ÷ 13

                = 6.6923

Now, multiply 13 by 6,

                = 13 × 6 (here, 6 specifies number of carbon atoms)

                = 78

Now subtract 78 from 87,

                = 87 - 78

                = 9

Add 9 into 11,

                = 9 + 6

                = 15 (hydrogen atoms)

So, the rough formula we have is,

                                                       C₆H₁₅

Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,

                C₆H₁₅   -------N-------->    C₅H₁₃N

Hence, the molecular formula is C₅H₁₃N.

The correct structural formula is being predicted by using the stability of the fragments formed by the cleavage of the parent molecule. Therefore, as shown below, following secondary (stable) carbocations are formed as the major fragments.

the answer for sodium is 0.186

the answer for rubidium is 0.251