[bash] BASH Syntax error near unexpected token 'done'

There's a way you can get this problem without having mixed newline problems (at least, in my shell, which is GNU bash v4.3.30):

#!/bin/bash
# foo.sh

function foo() {
    echo "I am quoting a thing `$1' inside a function."
}

while [ "$input" != "y" ]; do
    read -p "Hit `y' to continue: " -n 1 input
    echo
done

foo "What could possibly go wrong?"

$ ./foo.sh
./foo.sh: line 11: syntax error near unexpected token `done'
./foo.sh: line 11: `done'

This is because bash expands backticks inside double-quoted strings (see the bash manual on quoting and command substitution), and before finding a matching backtick, will interpret any additional double quotes as part of the command substitution:

$ echo "Command substitution happens inside double-quoted strings: `ls`"
Command substitution happens inside double-quoted strings: foo.sh
$ echo "..even with double quotes: `grep -E "^foo|wrong" foo.sh`"
..even with double quotes: foo "What could possibly go wrong?"

You can get around this by escaping the backticks in your string with a backslash, or by using a single-quoted string.

I'm not really sure why this only gives the one error message, but I think it has to do with the function definition:

#!/bin/bash
# a.sh

function a() {
    echo "Thing's `quoted'"
}
a
while true; do
    echo "Other `quote'"
done

#!/bin/bash
# b.sh

echo "Thing's `quoted'"
while true; do
    echo "Other `quote'"
done

$ ./a.sh
./a.sh: line 10: syntax error near unexpected token `done'
./a.sh: line 10: `done'
$ ./b.sh
./b.sh: command substitution: line 6: unexpected EOF while looking for matching `''
./b.sh: command substitution: line 9: syntax error: unexpected end of file
Thing's quote'
./b.sh: line 7: syntax error near unexpected token `done'
./b.sh: line 7: `done'