We check each:

A 30 = 13+17 = 2+11+17 = ?+?+?+?  

Let's show why we cannot have 4 groups of different primes adding up to 30.

One of the 4 groups cannot be 2, because 2 is the only even prime, and three
more odd primes added to 2 would given an odd number, not 30. 

The smallest 4 odd primes have sum 3+5+7+11=26

If we use 13 instead of 11, we have sum 3+5+7+13=28

If we use 17 instead of 11, we have sum 3+5+7+17=32, too much.

We must have a group of 3, because if we left it out, the four smallest
odd primes without it would have sum 5+7+11+13=36, too much.

We must have a group of 5, because if we left it out, the four smallest 
odd primes without it would have sum 3+7+11+13=34, too much.

We must have a group of 7, because if we left it out, the four smallest 
odd primes without ie would have sum 3+5+11+13=32, too much.

We must have a group of 11, because if we left it out, the four smallest 
odd primes would have sum 3+5+7+13=28, too much.

So Jon cannot have exactly 30 coins. 

B 45 = 2+43 = 5+17+23 = 2+11+13+19, so Jon could have 45 coins.

C 60 = 29+31 = 2+17+41 = 11+13+17+19, so Jon could have 60 coins.

D 75 = 2+73 = 13+19+43 = 2+19+23+31, so Jon could have 75 coins.

E 90 = 43+47 = 2+41+47 = 17+19+23+31, so Jon could have 90 coins.

Of the choices listed, the only number of coins which Jon could not possibly 
have is 30 coins.

Edwin