The maximum length of a string of consecutive primes is 2: that is, the primes 2, 3. This is easily proven, as no even number other than 2 is prime.
In contrast, consider the set of numbers which are a product of exactly two primes (they don't need to be distinct). This set begins 4, 6, 9, 10, 12 ... It goes on to include the numbers 33, 34, 35 - a run of three consecutive integers. Is this the longest consecutive run of such numbers?
My conjecture: that each set of numbers which are the product of exactly n primes contains one and only one run of consecutive numbers which is n+1 long.
No idea whether this is true! Thoughts, anyone?
The limits each time are defined by the powers of $2$. So a consecutive set of numbers which each have exactly two prime factors (not necessarily distinct) will be bounded by the need to avoid multiples of $2^4=4$ (given that including $4$ itself is not an option).
And a consecutive set of numbers which have exactly three prime factors (not necessarily distinct) will be bounded by the need to avoid multiples of $2^3=8$. So we can possibly expect a run of $7$ such numbers, and the smallest such run starts at $211763$: $$\begin{array}{c|c} n &f_1 & f_2 & f_3 \\ \hline 211673 & 7 & 11 & 2749 \\ 211674 & 2 & 3 & 35279 \\ 211675 & 5 & 5 & 8467 \\ 211676 & 2 & 2 & 52919 \\ 211677 & 3 & 37 & 1907 \\ 211678 & 2 & 109 & 971 \\ 211679 & 13 & 19 & 857 \\ \end{array}$$