bash - Syntax error: EOF in backquote substitution -

i'm getting error syntax error: eof in backquote substitution , don't have faintest idea why. mind taking quick look?

#! /bin/sh # chkconfig 345 85 60 # description: startup script produtcrawler-router # processname: producrawler-router  name=productcrawler-router dir=/etc/productcrawler/services exec=router.py pid_file=/var/run/productcrawler.pid iexe=/etc/init.d/productcrawler-router run_as=root  ### begin init info # provides:          productcrawler-router # required-start:    $remote_fs $syslog # required-stop:     $remote_fs $syslog # default-start:     5 # default-stop:      0 1 2 3 6 # description:       starts productcrawler-router service ### end init info  if [ ! -f $dir/$exec ]         echo "$dir/$exec not found."         exit fi  case "$1" in   start)         echo -n "starting $name"         cd $ldir         start-stop-daemon -d $dir --start --background --pidfile $pid_file --make-pidfile --exec $exec --quiet         echo "$name running."         ;;   stop)         echo -n "stopping $name"         kill -term `cat $pid_file         rm $pid_file         echo "$name."         ;;   force-reload|restart)         $0 stop         $0 start         ;;   *)         echo "use: /etc/init.d/$name {start|stop|restart|force-reload}"         exit 1         ;; esac 

the syntax coloring didn't give away?

 stop)        echo -n "stopping $name"        kill -term `cat $pid_file 

it's backtick , end of kill line.