Why does bash behave differently, when it is called as sh?

Question

I have an ubuntu machine with default shell set to bash and both ways to the binary in $PATH:

$ which bash
/bin/bash
$ which sh
/bin/sh
$ ll /bin/sh
lrwxrwxrwx 1 root root 4 Mar  6  2013 /bin/sh -> bash*

But when I try to call a script that uses the inline file descriptor (that only bash can handle, but not sh) both calls behave differently:

$ . ./inline-pipe
reached
$ bash ./inline-pipe
reached
$ sh ./inline-pipe
./inline-pipe: line 6: syntax error near unexpected token `<'
./inline-pipe: line 6: `done < <(echo "reached")'

The example-script I am referring to looks like that

#!/bin/sh
while read line; do
if [[ "$line" == "reached" ]]; then echo "reached"; fi
done < <(echo "reached")

the real one is a little bit longer:

#!/bin/sh
declare -A elements
while read line
do
    for ele in $(echo $line | grep -o "[a-z]*:[^ ]*")
    do
        id=$(echo $ele | cut -d ":" -f 1)
        elements["$id"]=$(echo $ele | cut -d ":" -f 2)
    done
done < <(adb devices -l)
echo ${elements[*]}

Answer 1

When bash is invoked as sh, it (mostly) restricts itself to features found in the POSIX standard. Process substitution is not one of those features, hence the error.