Avoid bash to interpret \ in a double quoted string

For bash's builtin echo command to output a given string verbatim followed by a newline character, you need:

# switch from PWB/USG/XPG/SYSV-style of echo to BSD/Unix-V8-style of echo
# where -n/-e options are recognised and backslash sequences not enabled by
# default
shopt -u xpg_echo

# Use -n (skip adding a newline) with $'\n' (add a newline by hand) to make
# sure the contents of `$VAR` is not treated  as an option if it starts with -
echo -n "$VAR"$'\n'

Or:

# disable POSIX mode so options are recognised even if xpg_echo is also on:
set +o posix

# use -E to disable escape processing, and we use -n (skip adding a newline)
# with $'\n' (add a newline by hand) to make sure the contents of `$VAR` is not
# treated  as an option if it starts with -
echo -En "$VAR"$'\n'

Those are specific to the bash shell, you'd need different approaches for other shells/echos, and note that some implementations won't let you output arbitrary strings.

But best here is to use printf instead which is the standard command for that:

printf '%s\n' "$VAR"

See Why is printf better than echo? for details.

use 4 backslashes instead of 2