[CentOS] Booting CentOS 6 with Grub from CentOS 7

I had CentOS 6 installed on an MBR hard disk. I converted the disk to
GPT (with UEFI) and installed CentOS 7 on a new partition.

CentOS 7 itself seems to work properly.

Grub from CentOS 7 automatically created an entry for CentOS 6, but I
get an error if I try to boot CentOS 6.

The menu entry is:

menuentry 'CentOS release 6.10 (Final) (on /dev/sda7)' --class
gnu-linux --class gnu --class os $menuentry_id_option
'osprober-gnulinux-simple-92dcc576-4e61-41f3-ae42-d06b3aec01ae' {
        insmod part_gpt
        insmod ext2
        set root='hd0,gpt7'
        if [ x$feature_platform_search_hint = xy ]; then
          search --no-floppy --fs-uuid --set=root --hint-bios=hd0,gpt7
--hint-efi=hd0,gpt7 --hint-baremetal=ahci0,gpt7 
92dcc576-4e61-41f3-ae42-d06b3aec01ae
        else
          search --no-floppy --fs-uuid --set=root
92dcc576-4e61-41f3-ae42-d06b3aec01ae
        fi
        linux /boot/vmlinuz-2.6.32-754.3.5.el6.centos.plus.x86_64 ro
root=UUID=92dcc576-4e61-41f3-ae42-d06b3aec01ae rd_NO_LUKS
KEYBOARDTYPE=pc KEYTABLE=cf rd_NO_MD LANG=fr_FR.UTF-8
SYSFONT=latarcyrheb-sun16 rd_NO_LVM rd_NO_DM crashkernel=auto
        initrd /boot/initramfs-2.6.32-754.3.5.el6.centos.plus.x86_64.img
}

When I try to boot, I get a message that "linux" and "initrd" aren't
valid commands. If I change them to "linuxefi" and initrdefi", I get
this error:

error: kernel too old
error: you need to load the kernel first

The information in the entry seems correct (partition, UUID, etc.).

I've searched on Google quite a bit, but didn't find anything useful.
Are there any suggestions?

-- 
Yves Bellefeuille
<yan at storm.ca>