There was a review for thier new cd in our local newspaper that made that comparrison...
2006-12-10 08:17:26 · 11 answers · asked by Anonymous in Entertainment & Music ➔ Music
NO. Whoever wrote that article was on crack.
2006-12-10 08:45:53 · answer #1 · answered by ethersflame 4 · 0⤊ 0⤋
Is this even a serious question??? MCR is nowhere near as talented, legendary or inspirational as the Rolling Stones. That is one of the most disgusting comparisons anyone can make. I'm not even a fan of the Rolling Stones and I can tell you right now that in 5-7 years My Chemical Romance will no longer be a part of the music business, yet the Rolling Stones will still be one of the most influential bands.
2006-12-10 16:24:34 · answer #2 · answered by mal_crane 2 · 0⤊ 1⤋
Love MCR but that is a bit much to say yet. They have only been around approx 6 years. They need to do a lot more work to even attempt that high of a distinction. (I don't mean to be as great as the Rolling Stones, I mean to reach that high of a level of fame and longevity). Do I think they can accomplish this? Dunno. They are doing great so far but can they keep it up?
2006-12-12 00:12:29 · answer #3 · answered by ♫ Insane_Princess ♪ 5 · 0⤊ 0⤋
I understand exactly what your asking, and the answer is No. I hate to say it, because mcr is a great band....
2006-12-10 16:28:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋
No. Those two bands don't even have a similar sound
2006-12-10 16:20:13 · answer #5 · answered by december_chick_89 4 · 0⤊ 0⤋
hell no!!!!! i like them but they still have more to prove to me as a band
2006-12-10 16:20:14 · answer #6 · answered by namso141 3 · 0⤊ 0⤋
if the rolling stones are whiny pussy emo bitchs then yes they are.
2006-12-10 16:19:47 · answer #7 · answered by anarchy4ever91 2 · 0⤊ 2⤋
no but i get what ur saying
2006-12-10 16:45:22 · answer #8 · answered by banana10678 2 · 0⤊ 0⤋
God no. Not even close.
2006-12-10 16:18:51 · answer #9 · answered by Hillaryforpresident 5 · 0⤊ 0⤋
No.
2006-12-10 16:23:28 · answer #10 · answered by ms_mystic_d 1 · 0⤊ 0⤋