Pair up $\{1..2n\}$ that the sums of each pair are different primes.

Pair up $$\{1..2n\}$$ that the sums of each pair are different primes. I found 9 examples:

$$\{(1,2)\},$$

$$\{(1,2),(3,4)\},$$

$$\{(1,2),(3,4),(5,6)\},$$

$$\{(1,4),(2,5),(3,8),(6,7)\},\{(2,3),(1,6),(4,7),(5,8)\},$$

$$\{(1,4),(2,5),(3,8),(6,7),(9,10)\},\{(2,3),(1,6),(4,7),(5,8),(9,10)\},$$

$$\{(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)\},\{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)\}.$$

Is there another example like this?

There are no more. There are none for $$n=7$$. We would need the sums to be seven distinct odd primes below $$27$$. The most the sum of these sums can be is $$5+7+11+13+17+19+23=95$$ but the sum of all the numbers up to $$14$$ is $$105$$.

For $$n=8$$ we would need the sum of eight primes below $$31$$ to be $$120$$. We can only do this with $$3,5,11,13,17,19,23,29$$. We need $$1+2$$ to get $$3$$ but cannot get $$5$$.

For $$n=9$$ we need all the primes up to $$31$$ except $$2,5$$, but then $$1+2=3,3+4=7,5+5=11$$ and we are stuck for $$13$$.

For $$n=10$$ we need all the primes up to $$37$$ except $$2,5$$ and the $$n=9$$ proof works.

For $$n=11$$ we need all the primes up to $$43$$ except $$2,7,11$$ or $$2,5,13$$. We can do $$21+22=43$$ but cannot get $$41$$.

For $$n=12$$ the sum of all the primes up to $$43$$ is $$271$$ while the sum of the numbers up to $$24$$ is $$276$$

Now going from $$n$$ to $$n+1$$ the sum of the numbers increases by $$4n+3$$ while the sum of primes increases by $$4n+1, 4n+3,$$ or $$8n+4$$. As less than $$\frac {2\cdot 3 \cdot 5}{3 \cdot 5 \cdot 7} \lt \frac 12$$ are primes, the sum of primes will never catch up. $$n=13$$ doesn't add any new primes because the odd numbers added are $$49,51$$ and again we miss at $$n=19$$. Each intervening $$n$$ has only added one prime.