Number the friends $1$ through $6$ in original seating order; we count orderings of $1, \ldots, 6$ in which no two consecutive integers are adjacent. Apply inclusion-exclusion over which of the five pairs $\{i, i+1\}$ are forced to sit together. If a chosen set of $k$ pairs forms $r$ maximal runs of consecutive integers, gluing each run into a block (orderable ascending or descending) gives $2^r (6 - k)!$ seatings containing all chosen adjacencies.

Tallying by $k:$ for $k = 0,$ $720.$ For $k = 1,$ five sets, each $2 \cdot 120,$ total $1200.$ For $k = 2,$ four sets form one run $(2 \cdot 24)$ and six form two runs $(4 \cdot 24),$ total $768.$ For $k = 3,$ three sets form one run $(2 \cdot 6),$ six form two runs $(4 \cdot 6),$ one forms three runs $(8 \cdot 6),$ total $228.$ For $k = 4,$ two sets form one run $(2 \cdot 2)$ and three form two runs $(4 \cdot 2),$ total $32.$ For $k = 5,$ one set, total $2.$

The count is $720 - 1200 + 768$ $- 228 + 32 - 2 = 90.$