Question

Problem 6. (30 points) The molecule shown below is a common steroid. Redraw this steroid as a careful perspective drawing, using chair conformations for the cyclohexane rings and a reasonable conformation for the cyclopentyl group. The Bring should have the same conformation as the chair cyclohexane you drew in problem 1. You do not need to show all of the hydrogens on the rings; it is only necessary to show the substituents and hydrogens that I have shown explicitly. You may abbreviate the R group on the D ring as "R." M "R" HOMECOH

Answer 1

Solution:

Problem 6:

A chair conformation of given compound is drawn below.

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Problem 6 WI COH me me AD HORA chair conformation: mWme me H •Co₂H iD 9 HOH ON - To start with we need to first draw number it chain form of ring B and - Then place all substituents on it Me H my goji me u me Now connect ring i and place substituent on ringe me t me, 12 Now Coq connect sing a through and Co M OM Me - me k me Now connect sing a through C4 and ₂ R 78113 9 и 14 H H on H A- 16 ON

Problem 1: Only ring B is considered and drawn all the substituents position.

As Ring B is attached with ring A through C3 and C4 comonc carbons thus, Me group at C3 is axial and H at C4 is equatorial.

As Ring B is attached with the ring C through common carbons such as C1and C2, H at C1and C2 are axial.

Problem 1 - me мен - 1), 12 CH 4 17 3 1210 Noh 0 14 H OH 16 Now we will consider ring B only if me, - connecting A H toringe w connecting A to ring a

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