Steroids are lipids derived from the triterpenoid lanosterol Structures are based on a tetracyclic ring system Four rings designated A, B, C, and D Numbering begins in A ring A, B, and C rings adopt chair conformations but do not undergo cyclohexane ring-flips Steroids

In systematic nomenclature of the R group at position 17 determines the base name of an individual steroid.

CONFORMATIONS: There are four rings in a steroid skeleton and hence there are three fusion points. A/B, B/C and C/D rings share two carbons each (fusion). Every fusion center can either be cis- or trans-fused Two cyclohexane rings can be joined in either a manner Cis fusion gives cis –decalin - Both groups at ring-junction positions (angular positions) are on same side of two rings. Trans fusion gives trans –decalin - Both Groups at ring junction are on opposite sides.

CONFORMATIONS: Steroids can have either a cis or trans fusion of the A and B rings Other ring fusions (B-C and C-D) are usually trans. A-B trans steroid has C19 angular methyl group up, and the C5 hydrogen atom down, on opposite sides of the molecule. A-B cissteroid has both C19 angular methyl group and C5 hydrogen atom on the same side the molecule.

The structures most likely feasible are : trans-trans-transtrans-trans-trans (most natural and synthetic steroids have this skeleton, e.g., 5a-dihydrotestosterone) cis-trans-transcis-trans-trans (some natural steroids have this skeleton, e.g., cholic acids) cis-trans-cis (few natural steroids have this skeleton, e.g. cardiac glycosides)cis-trans-cis

CONFIGURATION: The steroid skeleton (all carbons) with its methyls (18 and 19-CH3) oriented as coming towards the viewer (represented as a bold line) is the reference plane in defining whether a particular hydrogen or a substituent is towards the viewer (b) or going away from the viewer (cross-hatched, a). Thus in a two-dimensional representation, with the 18- and 19-CH3 bold, all bold substituents are b and all cross-hatched substituents are a. When α and β designation are applied to the hydrogen atom at position 5,the ring system in which the A, B ring junction is trans become the 5 α series; and the ring system in which the A, B ring junction is cis becomes the 5 β series.

Substituent groups on steroid ring system can be either axial or equatorial Equatorial substitution is generally more stable for steric reasons Cholesterol has C3 hydroxyl group equatorial

Cholesterol:  It forms about 17% of human brain and nervous tissue.first isolated from human gall stone deposited in bile ducts so known as cholesterol. (chol – from bile acid, stereos – solid, ol - alcohol).  It is white crystalline optically active solid with melting point c and specific rotation (+)39 o.  Source: from human gall stone,fish liver oils,brain and spinal cord of cattle,lanoline.  It is present in the form of palmitate,stearate,oleate esters.

Numbering:

Isolation : Source of cholesterol + ethanol Ethanolic solution of cholesterol + ethanolic solution of digitonin White precipitate of cholesterol digitonide Breakdown of cholesterol and digitonin Digitonin precipitate Ether evaporation cholesterol Ether solution cooling Cholesterol is deposited pyridineDimethyl sulfoxide / heating

Constitution of cholesterol: Structure of nucleus. Position of hydroxyl group and double bond. Nature of position and side chain. Position of angular methyl groups.

Structure of nucleus: ۩ Molecular formulae: C 27 H 46 O. ۩ Forms monoacetate on acetylation indicating one hydroxyl group. ۩ It adds 2 bromine atoms forming cholesterol dibromide indicating presence of one double bond. ۩ A to B proves presence of double bond. ۩ B to C shows presence of secondary alcohol. ۩ The saturated parent hydrocarbon D of cholesterol corresponds to general formulae (CnH2n-6) for tetra cyclic compounds hence it is tetra cyclic alcohol. Total no. of rings is given by: (No. of carbons + 1 ─ no. of hydrogens / 2).

۩ On selenium distillation at 360 o c cholesterol gives diels hydrocarbon and chrysene, the formation of diels hydrocarbon suggests that cholesterol has diels hydrocarbon nucleus in its structure.

Position of hydroxyl group and double bond: ۩ Cholestanone on oxidation with nitric acid gives a dicarboxylic acid which on pyrolysis yields a ketone. From the above reactions the following reactions are concluded: ۩ The oxidation of Cholestanone to dicarboxylic acid having same no. of carbon atoms as original ketone indicates that keto group must be present in the ring. If in the side chain it gives acid with less no. of carbon atoms.

۩ pyrolysis of dicarboxylic acid to ketone with less no. of carbon atoms reveals that dicarboxylic acid is 1,6 or 1,7 dicarboxylic acid (blancs rule). ۩ But opening of ring to yield dicarboxylic acid occurs due to hydroxyl group so that hydroxyl group not in the ring D if it is in the ring D then it gives 1,5 dicarboxylic acid, so hydroxyl group is in ring A. ۩ When cholestanone is oxidized actually 2 isomeric dicarboxylic acids are obtained which indicates that keto group is flanked on either side by methylene group so that CH2-CO-CH2.

 If proposed structure for cholesterol is examined such an arrangement is only possible if hydroxyl group is present in ring A.

۩ The position of hydroxyl group at position 3 is further confirmed by following reactions The above reaction is formulated as follows if hydroxyl group is present in position 3 which corresponds with the position 7 in 3,7-dimethyl cyclopentenophenanthrene.

All the above reactions are given as:

Position of double bond: From the above reactions the following reactions are concluded: ۩ The conversion of A to B represents the hydroxylation of double bond. ۩ B on oxidation gives a diketone indiating that in B two of OH groups are secondary in nature and third is tertieary one (which resists oxidation). ۩ The oxidation of D to E with out loss of any cartbon atom suggests that two ketonic groups in D are present in different rings i,e the double bond and OH group are present in two different rings and as we have already seen that OH group is present in ring A double bond must be present in ring B,C or D.

۩ Since D can form a pyridazine derivative with hydrazine the two ketonic groups of D are in γ position with respect to each other which is only possible only if double bond is present in between C5 and C6 and all above reactions are written as

۩ The position of double bond is further proved by following set of reactions ۩ From above reaction oxidation of B to C with loss of one carbon atom indicates that keto group and double bond in B are in same ring. ۩ uv absorption spectrum of B λmax 240nm shows that keto group and double bond are conjugated.  These results can be explained on assumption that there is a migration of double bond of cholesterol during formation of B which is possible only if the double bond is present in between C5 and C6.

Nature of position and side chain: ۩ Cholesterol acetate on oxidation with chromic acid forms an acetate of hydroxyl ketone and a steam volatile ketone,(isohexyl methyl ketone). ۩ The above reaction shows that isohexyl methyl ketone forms the side chain of cholesterol and is attached to nucleus of cholesterol through a carbon atom oxidized to ketone group.

The nature of side chain can further be elucidated by application of barbier-wieland degradation. Cholesterol is converted to 5β- cholestane.

These degradations lead to following conclusions: ۩ The formation of acetone from coprostane indicates that side chain terminates in a isopropyl group. ۩ The conversion of D to ketone E indicates that α carbon of acid is secondary in nature so that there is alkyl group on α carbon in D.(if no alkyl group on α carbon then acid has been formed instead of ketone ). ۩ Oxidation of E to F with loss of one carbon atom suggests that the alkyl group in E and D is methyl. ۩ Oxidation of G to H without any loss of carbon atom suggests that in the former the ketonic group is present in the ring.

From the above we came to know that coprostane contains a side chain of 8 carbon atoms arranged in the following order. Point of attachment of side chain to nucleus:  The dicarboxylic acid formed on heating with aceticanhydride forms an anhydride suggesting that it is dicarboxylic acid (blanc rule).  1,5- dicarboxylic acid can only be obtained from 5 membered ring hence the side chain is attached to 5 membered ring (but actual attachment point is not revealed).  On selenium distillation at 360c cholesterol gives diels hydrocarbon indicating that side chain is attached to C17 in cholesterol because selenium dehydrogenation may degrade a side chain to a methyl group.  X-ray crystallographs also support 17 th position of side chain.

The following reaction as follows:

Position of angular methyl groups :  Ring with 17 carbon atoms and side chain with 8 carbon atoms but another 2 carbon atoms.  The keto acid on clemmenson reduction followed by twice B W degradation gives a tertiary acid so one of the angular methyl groups must be present on C10.  On selenium distillation at 360c cholesterol gives diels hydrocarbon and chrysene, the formation of chrysene is explained that there is an angular methyl groups at either C13 or C14 which enters 5 membered ring to form a 6 membered (this evidence not complete because here C 14 also explained chrysene.)  Acotobilianic anhydride on distillation with selenium gives 1,2-dimethyl phenanthrene indicating that methyl group is present in position 13, had it been on c14,then only monomethyl phenanthrene wiuld have been formed.

Angular methyl at C13: Angular methyl at C14:

A = (Ac o ) 2 o /zinc-(Ac o ) 2 o B = claisen condensation C = michael condensation D = oso4/ KOH E = CH 3 ONa/C 2 H 5 COOH/CH 3 COCH 3

Reference: Chemistry of natural products vol Chatwal Chemistry of natural products vol Agarwal Text book of organic chemistry vol I.L.Finar

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