Solution:

Single numbers give sums $1$ through $5$, complements give sums $10$ through $14$, and all five numbers give $15$. So an arrangement is good exactly when consecutive blocks can make sums $6$ and $7$.

If sum $6$ is impossible, then $1$ is not adjacent to $5$. By rotating and reflecting, write the arrangement as $1bc5e$. The adjacent pair $bc$ cannot be $\{2,3\}$ or $\{2,4\}$, since $1+2+3=6$ and $2+4=6$. Thus $e=2$, and avoiding the consecutive block $2,1,3$ forces the bad arrangement $14352$.

If sum $7$ is impossible, then $2$ is not adjacent to $5$. Similarly write the arrangement as $2bc5e$. Now $bc$ cannot be $\{3,4\}$ or $\{1,4\}$, so $e=4$. To avoid the consecutive block $4,2,1$, the remaining order must be $b=3,\ c=1$, giving $23154$.

These two arrangements are indeed bad, one missing sum $6$ and the other missing sum $7$. Hence there are $2$ bad arrangements.

Thus, the correct answer is B .