Question

Laboratory MOUTU ICEF MCPHS University - Professor Z. I. Po CHE 231L.1-7 - Organic Chemistry I - Fall 2019 E or Zand the aliene as Eor Zand () shown below. Identify Examine the cholesterol-lowering drug Rosuvastatin (Crestor absolute configuration of the 2 alcohol as Rors. Eor 2 Rosuvastatin Crestor d to treat high holes e and r and related conditions, and to prevent cardiovascular disease JER Ja ZS bZR [CES ſe] E, achiral 7. Examine the molecule piperine, an alkaloid responsible for the punency of black pepper and long pepper. Assign the carbon-carbon double bonds numbered C-2 and C-4 as either for Z spicy component of black peper Id C-22:04Z a C-23:0-42 d) C-2 E:C-4E [b] C-22:04 lej C-2 neither EnerZ: C-4E 8. Isotretinoin (Accutane) a prescription medication primarily used to treat severe acne. Assign the carbon-carbon double bonds numbered C-9 and C-13 as either or Z Isotretinoin (Acevane diterpene medication primarily used to treat severe cystic acne о он (Note: ring is historie, not IUPAC) (a) C-9 E: C-13 Z fbl C-920-13 E (c) C-9 E:C-13 Id C-92: C-13Z lej C-9 neither Enor 2: C-13Z 9. An eleven-carbon hydrocarbon with three degrees of unsaturation ( 3) will have how many hydrogen atoms fa) 16 [b] 18 (c) 20 [d] 22 [el 24 10. Afteen-carbon hydrocarbon with two degrees of unsaturation (9 2) and two oxygen atoms will have how many hydrogen atoms! (a) 24 [b] 26 [c] 28 [d] 30 fel 32 11. The acid-catalyzed addition of water to (6) 2,2-dimethyl-3-hexene gives which of the following as the major organic product? (a)2-methyl-2-hexanol (CZ -dimethyl-3-hexanol 2,3-dimethyl-3-hexanol d]2,3-dimethyl-3-hexanol [e] Both A and D 12. Which of the following products of an electrophilic addition of water/1H50 to an alkene will be formed following Markovnikow's rule aj [b Ic a secondary alcohol from a trisubstituted alkene tertiary alcohol from a transdisubstituted alkene a tertiary alcohol from a cis-disubstituted alkene tertiary alcohol from a gem-disubstituted alkene a primary alcohol from a tetra substituted alkene d a

Answer 1

Q.6.Ans :- Option (d) i.e. E,R is the answer

Explanation :-

because Highest priority groups EL ties on obbosite side. i OH 04 0 NY because o R I clockwise rotation

Cahn-Ingold-Prelog Priority Rules :-

1.Select the highest to lowest priority group or atom on the basis of atomic mass i.e. atom having more atomic mass should get first number.

2. If atomic mass or number of very first element is same then notice the atomic mass of the next atom and so on.

Also,

R-S configuration :-

(1). If lowest priority group lies on dash position then rotate the molecule from I to II and II to III.

(2). If lowest priority group do not lies on dash position then rotate the molecule from III to II and II to I.

(3). For clockwise direction of the arrow use the R-isomer and for anti-clockwise direction of the arrow use the S-isomer

E-Z configuration :-

Heavier groups or lowered groups are assigned on the basis of priority rule i.e. atom or first atom of the group having more atomic mass than other should get the highest priority.

E-Configuration :- Is used when the heavier groups or lowered groups attached to the adjacent carbon atoms lies on the opposite side.

Z-Configuration :- Is used when the heavier groups or lowered groups attached to the adjacent carbon atoms lies on the same side.

Hence, Option (d) is the answer.