Here is a solution. I'm afraid I used a computer to help find it. Some explanation is below.
2184 2 3 7 13
2198 2 7
2196 2 3
2190 2 3 5
2200 2 5 11
17 numbers, from 2184 to 2200. Each row shows which "small" prime numbers divide the number on that row, and you can see that each pair of consecutive rows has a prime number in common.
To find this without requiring too much human brute force or computer cleverness, I
defined a set of numbers to be "plausible" if each shares a common factor with at least one other, and no more than two share a common factor with only one other,
a condition it's easy to check by computer, and then
conducted an exhaustive search for the shortest plausible string of consecutive numbers below 5000.
This might have yielded a non-solution, in which case I'd have continued the search. That would have happened if, e.g.,
you could arrange the numbers into a "chain" and some "rings", but there was no way to connect these together into a single "chain".
Is this solution unique?
This set of numbers is unique. (An earlier version of this answer had a sketch of how to prove it, with a remark to the effect that the details are fiddly enough to put me off doing it properly. Peter Taylor, in comments below, correctly observes that the details are much less fiddly than I'd thought.) Suppose you have a "working" set of numbers, and it includes a prime number $p$. This must be adjacent to at least one other number, which must therefore be a multiple of $p$. Therefore your range must extend at least as far as $2p$. But in the range $(p,2p)$ there is at least one prime number, by Bertrand's postulate. So if you have a prime number then you have a larger prime number; hence, any solution must include no prime numbers. The longest gap between prime numbers below 5000 has length 33; so brute force up to size 33 suffices. I've run my program further than that and found no viable sets of numbers other than the one above.
the exact sequence is not unique; e.g., we could swap 2187 and 2193.