Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active Solution Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active.

1. Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active
Solution
Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active