Draw the structure of metenolone from the information given below: 1) A –CH 3 group at C1. 2) A double bond between C1 and C2. 3) A ketone at C3. 4) An acetate group at C17 in β configuration.

In this practice question, it says, draw the structure of metal no loan, from the given information, that's below. Alright. So, first we need to do is draw the tetracyclic structure for our steroid. So we have 3 Cyclohexanes. Next, we should number them. So we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And this is 11, 12, 13, 14, 15, 16, and 17. Alright. So it says we have a methyl group on carbon 1. So on carbon 1, there's a methyl group. We have a double bond between carbons 12, which is good because that means we don't need to provide stereochemistry to the methyl group we just added on carbon 1. A ketone on carbon 3, so there's a ketone here, carbonyl group. And then in a acetate group at carbon 17 in the beta configuration. Right. So here to do that, beta would mean that it's wedged, and your acetate, we have oxygen, and then there goes your acetate group. So, this wouldn't be the complete thing because we technically need to draw the stereochemistries of other groups attached. We would have a methyl group on carbon 1013 which remember those are going to be wedged. Then we have a hydrogen here that is dashed. And then we're at carbon 10, and we're going to the right of it. We're gonna have alpha hydrogen, which is down, Beta Hydrogen which is up, Alpha Hydrogen which is down. So here this would represent our complete structure based on the four criteria and also based on what we know as the basically the baseline skeletal structure of a given steroid. Alright. So this will be our final answer.