We count the $4$ vertices adjacent to the $1$. Place the $1$ on top. The numbers on the side of the die, going in one direction, are $2, 3, 5, 4$. We know this because: A) $6$ must be on the bottom so it's not there, and B) Of the remaining numbers $2, 3, 4, 5$, the pairs are $2 + 5 = 3 + 4 = 7$, so they can't touch each other. So, the sum of our four products on top is $1 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$ Now we look at the bottom face of the die, a $6$. We get the same order of adjacent faces: $2, 3, 5, 4$. We go counterclockwise with $1$, and clockwise with $6$ (or vice versa - I don't have a die with me), but the direction you go is irrelevant. So, our four products on bottom are $6 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$ Adding those gives $7 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$, which simplifies to $7 \cdot 49$, or $\boxed{343}$.