Download Problem Set 7: Stereochemistry-ANSWER KEY and more Summaries Stereochemistry in PDF only on Docsity!
Problem Set 7: Stereochemistry-ANSWER KEY Chemistry 260 Organic Chemistry
The answer is (2). Circled isomers have a stereogenic carbon (*) and hence a stereogenic centre.
Note: a star (*) is used to denote a stereogenic centre (tetrahedral geometry, bonded to 4 different groups) in the structures in this question. (a) chiral (c) chiral (b) not chiral - no stereogenic centre (d) not chiral - no stereogenic centre
CH 3
CH 3 CH 2 CHCH 2 CH 2 CH 3
C C
Cl
H CH 3
H
CH
H
CH 2 CH 3
CH 3 CH 2 CHCH 2 CH 2 CH 3
C C
CH 3
H CH 3
H
CH
H
CH 3
CH 3
CH 3
CH 3
CH 3
CH 3 CH 3 CH 3
CH 3 CH 2 CH 2 CH 2 CHCH 3
CH 3
CH 3
CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3
CH 3
CH 3 CH 2 CH 2 C
H 3 C
CH 3 CH 2 CHCHCH 3 CH 3 CHCH 2 CHCH 3
CH 3
CH 3
CH 3 CH 2 CHCH 2 CH 3
CH 3 CH 2 CCH 2 CH 3
CH 2 C H 3
CH 3 CHC
CH 3
CH 3 CH 2 CH 2 CHCH 2 CH 3
- (1S, 3R)- 1 - chloro- 3 - methylcyclopentane NOTE: (cis)- 1 - chloro- 3 - methylcyclopentane does not include enough information about the absolute configuration of the stereogenic centers)
- The answers are (2), (3) and (4). The pairs in (2), (3) and (4) all have a stereogenic carbon and are enantiomers (nonsuperimposable mirror images). The structures in choice (1) do not have a stereogenic carbon and therefore cannot have optical isomers.
- Structure (4) is chiral. To see this clearly redraw the cyclohexane rings in a planer representation and look for a plane of symmetry.
- The answer is (4). Both of these structures represent cis-1,2-dihydroxycyclohexane and they are identical (again there is no possibility of enantiomers since the molecule is achiral).
- The answer is (4). Both of these structures represent cis-1,2-dihydroxycyclohexane and they are identical (again there is no possibility of enantiomers since the molecule is achiral).
HO OH OH
OH
OH
HO
OH
OH
1,4 cis
achiral
1,2 cis
achiral
1,3 cis
achiral
1,2 trans
no plane of symmetry
chiral
- There are four isomers with formula C 3 H 6 Cl 2 are:
CH 3 CH 2 C
Cl
H
Cl CHCH 2 Cl
Cl
CH 3 C
Cl
Cl
CH 3 ClCH 2 CH 2 CH 2 Cl CH 3
1 , 1 - dichloropropane 1 , 2 - dichloropropane 1 , 3 - dichloropropane 2 , 2 - dichloropropane
These must be A, B, C, and D, but we still have to determine which particular isomer is A, which is B, etc. We do this by considering how many trichloroproducts we get from the chlorination of each one. Note: in this question we consider only the chlorination to produce a trichloro product (i.e. we do not consider tetra-, penta-, etc. chlorinated products that could result from extensive chlorination reactions). Since 2,2-dichloropropane gives one trichloro product, it must be A. Similarly,
Cl
Cl
Cl
CH 3 CH 3
Cl
Cl
H
Cl
Cl
Cl
Cl
Cl
H
1,1-dichloropropane
(D)
Cl
H
CH 3 CH 2 C
Cl
Cl
1,2-dichloropropane
(C)
CH 3 CHCH 2 Cl
1,3-dichloropropane
(B)
ClCH 2 CH 2 CH 2 Cl
Cl
Cl 2
Cl 2
Cl
Cl 2
Cl 2
Cl
2,2-dichloropropane
(A)
C
CH 3 CH 2 C CH 3 CHC
Cl
Cl
+^ CH 2 CH 2 C^ Cl
CH 3 CHCHCl
CH 3 CCH 2 Cl
Cl
Cl
CH 3 CH 2
ClCH 2 CHCH 2 Cl
Cl
C
Cl
ClCH 2 CH 2 CHCl
Cl
ClCH 2 CHCH 2 Cl
1,1,2-trichloropropane
(E)
1,3-dichloropropane gives two trichloro products and so must be B. Both 1,1- dichloropropane and 1,2-dichloropropane give three products, but only 1, 2 - dichloropropane is chiral (it has a chiral carbon, designated by a *), so it must be C. Of the three products of C, the only optically active one is 1,1,2- trichloropropane, so it must be E.
- ( R )-Epinephrine has a specific rotation of - 51˚. A sample that contains a mixture of R- and S-isomers was found to have a rotation of +11.2˚. What is the percentage of R and S isomers in this sample? Rotation of sample = [mole fraction of R in the sample ∙ rotation of sample of pure R isomer] + [mole fraction of S in the sample ∙ rotation of sample pure S isomer] Rotation of sample = +11. Rotation of sample of pure R isomer = - 51 Rotation of sample of pure S isomer = + So, let x = mole fraction of R and 1-x = mole fraction of S 11.2 = [x ∙ (-51)] + [1-x ∙ (+51)] 11.2 = - 51x + 51 - 51x
- 39.8 = - 102x x = 0. Thus, there is 39% of the R-isomer and 61% of the S-isomer in the sample.
- This is a modified version of question #8 from the Stereochemistry- 1 problem set. Given the following five structures, indicate for each pair listed below whether the structures are identical, enantiomers or diastereomers. (a) A and B are diastereomers
- Fluticasone propionate (marketed under the tradename FloventTM) is a steroid based drug that is used in many inhalers as a treatment during an asthma attack. (a) How many asymmetric centres are located in Flovent? 8 (b) Draw the enantiomer of Flovent
- (-) – Paroxetine, also known as Paxil (shown below) is a selective serotonin re- uptake inhibitor that is used as an antidepressant. (-) – Paroxetine has affinity for various protein receptors. (a) Draw the enantiomer of (-) – Paroxetine.
(b) Would you expect (-) – Paroxetine and its enantiomer to have the same biological effect? (In other words, would (-) – Paroxetine and its enantiomer have the same interaction with the same protein target?) Briefly justify your answer. No: enantiomers have different shapes and fit into the enzyme site/ biological receptor (which are chiral) in a different manner. Hence their interaction with the same protein target will probably be different (c) Would you expect (-) – Paroxetine and its enantiomer to have the same boiling point? Briefly justify your answer. Yes, enantiomers have the same chemical and physical properties except optical activity. (d) (-) – Paroxetine has a specific rotation of - 24.3 at 298K. What is the specific rotation of its enantiomer under the same conditions? +24. (e) What would be the specific rotation of a sample that contains 39% of (-) – Paroxetine and 61% of its enantiomer? Observed specific rotation = (0.39) x (-24.3) + (0.61) x (24.3) = +5.
- Draw the enantiomer of each of the following molecules.
- Below are eight pairs of structural formulas. In the box to the right of each pair, place the number (from the six terms listed below) that BEST describes the relationship between the two structures. NOTE: Each term may be used more than once and not all terms need be used. ALSO, CIRCLE ALL THE MESO COMPOUNDS.
- Identical 4. Constitutional Isomers
- Diastereomeers 5. Enantiomers
- Conformers 6. None of the above
