Question

please answer all part

7. The following is the structure of natural Menthol, one of the stereoisomers of 2-isopropyl-5 methylcyclohexanol. It is isolated from various mint oils and is used in the treatment of minor throat irritation. a) Draw both chair conformations of menthol and indicate which conformation is more stable. OH becaue in e te stable ars es aunt S b) Identify all the chirality centers in this compound and assign the absolute configuration mast Auial pt (R or S) to each chirality center. c) Draw the enantiomer of the above compound and assign R or S configuration to all of the chirality centers in therenantiomer. 0.5

Answer 1

a) The given Kekul e structure of menthol is сна (UP) Methyl and hydroxy groups are above the plane Isopropyl group is pointed below the plane он (UP) CH3 Нс (DOWN Now, while drawing the conformation in chair form, the UP and DOWN orientati on will stay the same. Hence, the chair forms are (UP) (UP) (UP) CHа Он он CHз ring flip (UP) Нэс DOWN) CHз CHз Нас More stable (DOWN The conformation in the left has all the substituents in axial position. Since the axial position faces 1, 3- diaxial interactions, it is unstable. The conformation on the right on the other hand has all the substituents in the more stable equitorial position which does not face any di axial interacti on. Hence, it is the more stable conformation

b) The given Kekule structure of menthol is CH3 3 Он H "CHз НаС There are three chiral centers in menthol as marked by asterisks On each chiral center, we assign the highest to lowest priority(using CIP rules) of each group attached and assign R configuration to clockwise orientati on of group and S configuration to anticlockwise ori entati on of groups Clockwise from 1 --2-- 3. Hence, R configurati on 3 сна Clockwise from 1 --2-3. Hence, R configuration он CH3 Нас CH3 он 2-3 with H group on front. Clockwise from 1 Hence, S configuration CH3 Нас Hence, this isomer is the (1R, 2S, 5R) isomer

The Kekule structure of the enantiomer of above structure of menthol can be drawn by flipping the c) oreintation of substituents from up to down and down to up CH3 н 3 нон CH3 Н.С On each chiral center, we assign the highest to lowest priority(using CIP rules) of each group attached and assign R configurati on to clockwise orientati on of group and S configuration to anticl ockwise ori entati on of groups Clockwise from 1-2- 3 with H on front Hence, S configuration 3 CH3 H Clockwise from 1 --2--3 with H on front Hence, S configuration iOH CH3 Нас CH3 Clockwise from 1 --2--3 with H on back Hence, R configurati on CH 3 Нс Hence, the enantiomer is (1S, 2R, 5S) isomer 5ш 14

To summarize option b and c, we can assign the configurations as follows:

(S) (R) (S) (R) "он он (R) (S) ПИ