2006-06-15 08:47:58 · 25 answers · asked by polish_forever 1 in Food & Drink ➔ Cooking & Recipes
no cinnamon is a spice brown sugar is a type of unrifined sugar .cinnamon is bit like ground nutmeg
2006-06-15 08:51:34 · answer #1 · answered by Nutty Girl 7 · 3⤊ 1⤋
Nope. Cinnamon is a hard sweet/spicy stick that they grind up to make into powder or you can buy the sticks whole. Brown sugar is just another type of sugar. Just like they made sugar extra fine to make powdered sugar. The consistencies are different between the two too. Hope I was helpful!
2006-06-15 15:53:19 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Cinnamon is a spice. Brown sugar is just sugar with molasses.
2006-06-15 17:17:03 · answer #3 · answered by angieasee64 6 · 0⤊ 0⤋
No, brown sugar is sugar that is brown. Cinnamon is a spice which is not at all sweet but has a nice flavor.
2006-06-15 16:20:20 · answer #4 · answered by Lolita 2 · 0⤊ 0⤋
No. While they may look similar, brown sugar is just that, sugar, while cinnamon is a spice.
2006-06-15 15:52:53 · answer #5 · answered by catfight 1 · 0⤊ 0⤋
They are all wrong, Brown Sugar is made from sugar cane juice and molasses. In the island of Trinidad it is known by its Spanish name "Muscovado" it is very delicious and is generally considered to be more healthy than white sugar which is made from Beet.
Cinnamon is a spice
2006-06-15 16:01:28 · answer #6 · answered by Boyèl 2 · 0⤊ 0⤋
No. It is less refined sugar, with some molasses still in it. Cinnamon is the powdered bark of the tree of the same name.
2006-06-15 15:51:26 · answer #7 · answered by silvercomet 6 · 0⤊ 0⤋
no brown sugar is brown sugar cinnamon is cinnamon
2006-06-15 15:52:14 · answer #8 · answered by tameko_marie 3 · 0⤊ 0⤋
no, you can buy brown sugar, then you can buy cinnamon they are two seperate flavors... thus the names! if you are cooking, I would not substitute cinnamon with brown sugar!
2006-06-15 15:50:57 · answer #9 · answered by sweetgurllexi 3 · 0⤊ 0⤋
No, it's brown sugar.
2006-06-15 15:51:40 · answer #10 · answered by SweetPea 5 · 0⤊ 0⤋