Optical rotation Exercise \(\PageIndex{1}\) The (+) enantiomer of compound A has an optical rotation of 125 o . If a pure sample of compound A has an optical rotation of 100 o , what is the composition of the sample? Answer Pure = 125 o Optical purity = \(\frac{100}{125} = 0.80\) \[\% Major = 80 + \frac{20}{2} = 90 \% \nonumber\] \[\% Minor = 100-90 = 10 \nonumber\] Exercise \(\PageIndex{2}\) The (+) enantiomer of compound B has an optical rotation of 100 o . If a pure sample of compound B has an optical rotation of 95 o , what is the composition of the sample? Answer Pure = 100 o Optical purity = \(\frac{95}{100} = 0.95\) \[\% Major = 95 + \frac{5}{2} = 97.5\% \nonumber\] \[\% Minor = 100 -97.5 = 2.5 \% \nonumber\] Exercise \(\PageIndex{3}\) The (+) enantiomer of compound C has an optical rotation of 18 o . If a pure sample of C contains 30% of the (+) enantiomer and 60% of the (-) enantiomer, what is the optical rotation value? Answer Pure = 18 o \[\% Major = 60 \% \nonumber\] \[\% Minor = 40 \% \nonumber\] Optical purity \(= \frac{X}{18} = 20 \%\) Solve for X: \[X = 3.6^{o} \nonumber\] Exercise \(\PageIndex{4}\) The (+) enantiomer of compound D has an optical rotation of 25 o . If a pure sample of D contains 80% of the (+) enantiomer and 20% of the (-) enantiomer, what is the optical rotation value? Answer Pure = 25 o \[ \% Major = 80 \% \nonumber\] \[ \% Minor = 20 \% \nonumber\] Optical purity \( = \frac{X}{25} = 80-20 = 60 \%\) Solve for X: \[X = 15^{o} \nonumber\] The polarimetry experiment Exercise \(\PageIndex{5}\) A pure sample of the naturally-occurring, chiral compound A (0.050 g) is dissolved in water (2.0 mL) and the solution is placed in a 0.5 dm cell. Three polarimetry readings are recorded with the sample: 0.625 o , 0.706 o , 0.682 o . What is [a]? Answer \[[a] = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[c = \frac{0.050 g}{2.0mL} = 0.025 \frac{g}{mL} \nonumber\] Average observed rotation \(= \frac{0.625 + 0.706 + 0.682}{3} = 0.671^{o}\) \[[a] = \frac{0.671^{o}}{(0.025 \frac{g}{mL})(0.5 dm)} = 53.68^{o} \nonumber\] Exercise \(\PageIndex{6}\) A pure sample of the chiral compound B (0.540 g) is dissolved in ether (2.0 mL) and the solution is placed in a 1.0 dm cell. Three polarimetry readings are recorded with the sample: 1.225 o , 1.106 o , 1.182 o . What is [a]? Answer \[[a] = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[c = \frac{0.540g}{2.0 mL} = 0.27 \frac{g}{mL} \nonumber\] Average observed rotation \(= \frac{1.225 + 1.106 + 1.182}{3} = 1.171^{o}\) \[[a] = \frac{1.171^{o}}{(0.27 \frac{g}{mL})(1.0 dm)} = 4.34^{o} \nonumber\] Exercise \(\PageIndex{7}\) A pure sample of the (+) enantiomer of compound C shows [a] = 42 o . What would be the observed a if a solution of the sample was made by dissolving 0.250 g in 2.0 mL of acetontirile and was then placed in a 0.5 dm cell? Answer \[[a] = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[c = \frac{0.250g}{2.0 mL} = 0.125 \frac{g}{mL} \nonumber\] \[42^{o} = \frac{observed \: rotation}{(0.125 \frac{g}{mL})(0.5dm)} \nonumber\] observed rotation = 2.625 o Exercise \(\PageIndex{8}\) A student has prepared compound D in lab. She is sure the compound contains no impurities; a number of physical analyses have confirmed the structure and purity of the compound. A sample of compound D (0.10 g) is dissolved in methanol (2.0 mL) and the solution is placed in a 1.0 dm cell. Three polarimetry readings are recorded with the sample: 0.995 o , 0.904 o , 0.936 o . What is [a]? The optical rotation of D has previously been reported as 25 o . What is the optical purity of this sample? What is the enantiomeric excess of this sample? What is the composition of this sample? Why did the previous analyses show that there was only one compound present? Answer a \[[a] = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[c = \frac{0.10g}{2.0 mL} = 0.05 \frac{g}{mL} \nonumber\] Average observed rotation = \(frac{0.995 + 0.904 + 0.936}{3} = 0.945^{o}\) \[[a] = \frac{0.945^{o}}{(0.05 \frac{g}{mL})(1.0 dm)} = 18.9^{o} \nonumber\] Answer b % optical purity \(= \frac{(100)(18.9)}{25} = 75.6 \% \) Answer c enantiomeric excess = 75.6% Answer d \[\frac{100-75.6}{2} = 12.2 \% \nonumber\] 12.2% one enantiomer 87.8% other enantiomer Answer e Enantiomers differ in how they interact with plane polarized light, but not in other physical analyses. Exercise \(\PageIndex{9}\) Suppose a student prepared a sample of a compound with known [a] = 40. The student did not realize that her sample contained 10% (by mass) of a non-chiral impurity. After correcting for c and l, what did she measure for [a]? Answer \[[a] = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[40^{o} = \frac{observed \: rotation}{(l)(c)} \nonumber\] \[40^{o} = \frac{observed \: rotation}{(l)(1.1c)} \nonumber\] Observed rotation = 44 o diastereomers and physical properties Exercise \(\PageIndex{10}\) From the following group of molecules, select: a pair that are the same compound. a pair that would have the same physical properties but opposite optical activities. a pair that have different physical properties but the same connectivity. a pair that have different connectivity. a compound that contains chiral centers but has no optical activity. Exercise \(\PageIndex{11}\) From the following group of molecules, select: a pair that are the same compound. a pair that would have the same physical properties but opposite optical activities. a pair that have different physical properties but the same connectivity. a pair that have different connectivity. Exercise \(\PageIndex{12}\) From the following group of molecules, select: a pair that are the same compound. a pair that would have the same physical properties but opposite optical activities. a pair that have different physical properties but the same connectivity.