Question 1

Which of the following is the highest priority in the Cahn-Ingold-Prelog system of prioritizing groups for R/S nomenclature?  &#10;A) Propyl group&#10;B) Methyl group&#10;C) Bromine&#10;D) Chlorine

Accepted Answer

Priority is determined by atomic number, Br = 35, Cl = 17, C = 6, C = 6. Therefore, Br has the highest priority.

Question 2

Using the priorities assigned to carbon 2, what is the stereochemical designation of carbon 2?  &#10;A) R&#10;B) S&#10;C) neither

Accepted Answer

Orient the molecule with the methyl facing away from you, the bromine is 1, then you go counterclockwise to the chlorine and then continue counterclockwise to the propyl. Therefore, the designation is S.

Question 3

What is the stereochemical designation of carbon 2?  &#10;A) R&#10;B) S&#10;C) neither

Accepted Answer

Bromine is priority 1, propyl is 2, and methyl is 3. Hydrogen, the 4 priority group, is already facing away from you. Starting at bromine, you go counterclockwise to get to the propyl. Therefore, the designation is S.

Question 4

What is the stereochemical designation of carbon 2?  &#10;A) R&#10;B) S&#10;C) neither

Accepted Answer

Bromine is priority 1, propyl is 2, and methyl is 3. Hydrogen, the 4 priority group, is facing towards you. When the 4 group is facing towards you, start at 1 priority and determine if you go counterclockwise or clockwise just like you would normally. Starting at bromine, you go counterclockwise to get to the propyl. Therefore, the designation would normally be S. Since the 4 group is facing you, whatever orientation you get (which is S in this case), change it to the other one (which would be R). Therefore, this is R. Bottom line: when 4 group faces you, figure out R/S as usual and then change it to the opposite orientation.

Question 5

What is the stereochemical designation of the following molecule? (The roman numerals, I-IV are just used to indicate the different attachments NOT to designate priority).  &#10;A) R&#10;B) S&#10;C) neither

Accepted Answer

All four attached groups are carbons, which have the same priority. When you have groups with the same atomic , especially carbon, you proceed to the other attached groups to the carbon. In this case we have a CH₃, CH₂, CH₂, CH at I, II, III and IV respectively. This means IV is highest priority and I is lowest priority. II is CH₂CH₃, III is CH₂CH₂CH₃. If you look at each one's second carbon you are comparing a CH₃ to a CH₂, CH₂ has less hydrogen so is higher priority. This means II is third in priority and III is second in priority. Going from the highest priority, IV, to the second highest priority, III, you move clockwise. But since the lowest priority is facing you, the correct designation is S.

Question 6

What is the stereochemical designation of the following molecule?  &#10;A) R&#10;B) S&#10;C) neither

Accepted Answer

In assigning priority, the OH is 1 and the hydrogen is 4. But the carbon clockwise from the "stereocenter" is CH double bonded to a CH to a C with two methyls. The carbon counterclockwise from the "stereocenter" is CH double bonded to CH to a C with two methyls. In other words, you cannot decide which side has priority. In addition, this molecule has a plane of symmetry, which destroys stereochemistry. Therefore, the designation is neither R nor S.

Question 7

If a sample of the pure R enantiomer of a molecule has a specific rotation of - 40&#176; and your mixture of that molecule has an observed optical rotation of + 22&#176;, what is the % ee of your sample?&#10;A) 55 % ee R&#10;B) 55 % ee S&#10;C) 18 % ee R&#10;D) .55 % ee R&#10;E) none of the above

Accepted Answer

% ee = (observed/specific) × 100.% ee = (22/40) × 100; 0.55 × 100 = 55 %. Since the pure R rotates the polarized light counterclockwise or levorotatory (which is determined by the rotation being negative) and your observed rotation is positive, that means your sample has more S than R since S is positive in this case.

Question 8

If a sample is 50 % ee of R stereoisomer, what is the % R in the mixture?&#10;A) 50&#10;B) 100&#10;C) 25&#10;D) 75

Accepted Answer

If a sample is 50 % ee R, that means 50 % is pure R. The other 50 % is 1/2 R and 1/2 S.1/2 of 50 is 25 %. Therefore, the mixture is 75 % R(50 + 1/2(50)) and 25 %S (1/2 (50)).

Question 9

If a chemical reaction produces a mixture of 80% R and 20 % S, what is the % ee?&#10;A) 80&#10;B) 60&#10;C) 20&#10;D) 4

Accepted Answer

% ee = % R - % S or 80 - 20 = 60 % ee.

Question 10

If a sample of the pure R enantiomer of a molecule has a specific rotation of - 40&#176; what is the specific rotation of pure S?&#10;A) -60&#176;&#10;B) +60&#176;&#10;C) -40&#176;&#10;D) +40&#176;&#10;E) cannot be determined

Accepted Answer

Whatever the specific rotation is for one isomer, the other isomer will be the same value with the opposite sign. Therefore, -40 is R so +40 is S.

Question 11

If the specific rotation of (R)-2-methylhexan-2-ol is -35&#176;, what is the specific rotation of (S)-hexan-2-ol?&#10;A) -35&#176;&#10;B) +35&#176;&#10;C) it is negative but value cannot be determined&#10;D) it is positive but value cannot be determined&#10;E) it cannot be determined

Accepted Answer

The answer of If the specific rotation of (R)-2-methylhexan-2-ol is...

Question 12

If a sample of the pure R enantiomer of a molecule has a specific rotation of - 40&#176; and your mixture of that molecule has an 82% ee S, what is the observed specific rotation of your sample?&#10;A) +40&#176;&#10;B) +8.2&#176;&#10;C) +32.8&#176;&#10;D) -48.7&#176;&#10;E) cannot be determined

Accepted Answer

The answer of If a sample of the pure R...

Question 13

Indicate the relationship of the pair of molecules shown.  &#10;A) same molecules&#10;B) enantiomers&#10;C) diastereomers&#10;D) different molecules&#10;E) constitutional isomers

Accepted Answer