What kind of attack does the current brokenness of SHA-1 allow? I understand that the following is becoming feasible, or already is:

Find any 2 data (d1 and d2), for which SHA1(d1) = SHA1(d2)


However, it is not entirely clear to me if there is evidence of the feasibility of:

Find d2 for a specific d1, such that SHA1(d1) = SHA1(d2)


My difficulty in understanding the available literature is that I typically see the attack referred to as "seeking collisions" rather than "seeking a collision"; implying that what is being identified is two data that happen to share a SHA-1, rather than finding a datum which shares the same SHA-1 as a specific target datum.

EDIT: My question is partially redundant to, though more specific in purpose than, this question Second pre-image resistance vs Collision resistance
That's true. There should not be any publicly known attack on SHA-1 that allows for a given $d1$ to find a $d2$ such that $h(d1) = h(d2)$.

SHA-1 is vulnerable  for finding a pair of $d1$ and $d2$ such that $h(d1) = h(d2)$.

The same, however, applies to MD5, which is unusable for SSL/TLS certificates and a successful attack has been performed. So, this kind of collision can allow (under some circumstances) more that you might think.

The notable difference between SHA-1 and MD5 attacks is the cost of such attacks. For MD5, the attack is very fast. For SHA-1, the attack would be very expensive, but for some groups (e.g. governments) feasible.

(There might be also some more differences that I am not aware of. However, both of them use the Merkle–Damgård construction, which implies some common security characteristics.)