Optical rotation question

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arc5005

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How can a compound with an optical rotation of +233.0° be discerned from a compound with an optical rotation of -127.0°?

A.
The intensity of the light is greater with the positive optical rotation.
B. The sample with +233.0° optical rotation when diluted to half of its original concentration would show an optical rotation of +116.5°.
C. The larger the absolute value of the optical rotation, the greater the density of the compound.
D. It is not possible to distinguish the two compounds from one another.




Answer:
B) The sample with +233.0° optical rotation when diluted to half of its original concentration would show an optical rotation of +116.5°.

When using a polarimeter, an observed optical rotation of +233.0° and -127.0° would result in the same reading (given that a full circle is 360°). To discern one optical rotation from the other, the sample should be diluted to reduce the magnitude of the observed rotation. If the actual optical rotation were in fact +233.0°, then the lower concentration would show a rotation less than +233.0° (less clockwise). If the actual optical rotation were in fact -127.0°, then the lower concentration would show a rotation of lesser magnitude than -127.0° (less counterclockwise). If the solution concentration were cut in half for instance, the observed rotation would be either +116.5° or -63.5°. The change in rotation can therefore determine the original rotation value. The only answer that indicates changing the concentration is choice B. The intensity of the light depends on absorption, not rotation of plane-polarized light, so it's not choice A. The magnitude of rotation does not depend on the solution density or compound density, so eliminate choice C.


I'm having a hard time understanding their reasoning

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Hi @arc5005 ,

The essence of this question is that +233.0° and -127.0° would appear to be the same readings, so you have to figure out some way of altering the setup to distinguish between the two compounds. For a substance in solution, the equation for the specific rotation of polarized light at a given temperature and wavelength (which you can think of as a 'standardized' value) is α / (C * l), where α is the observed angle, l is the path length, and C is the concentration of that substance in solution. We're interested in changing α, and a little bit of algebra will tell us that α is proportional to (C * l). This means that diluting the solution will reduce the angle of optical rotation. Reducing the angle of optical rotation will make the (+) angle less + and the (-) angle less minus. In other words, both angles will get closer to the 0° degree point, but in opposite directions, making it possible to tell them apart. I'm including a basic sketch of this (angles not perfectly to scale), which should help you visualize this more clearly.

rsz_sdn_illustration.png


Great question & best of luck studying!
 
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Hi @arc5005 ,

The essence of this question is that +233.0° and -127.0° would appear to be the same readings, so you have to figure out some way of altering the setup to distinguish between the two compounds. For a substance in solution, the equation for the specific rotation of polarized light at a given temperature and wavelength (which you can think of as a 'standardized' value) is a / (C * l), where α is the observed angle, l is the path length, and C is the concentration of that substance in solution. We're interested in changing a, and a little bit of algebra will tell us that αa is proportional to (C * l). This means that diluting the solution will reduce the angle of optical rotation. Reducing the angle of optical rotation will make the (+) angle less + and the (-) angle less minus. In other words, both angles will get closer to the 0° degree point, but in opposite directions, making it possible to tell them apart. I'm including a basic sketch of this (angles not perfectly to scale), which should help you visualize this more clearly.

View attachment 224151

Great question & best of luck studying!


Okay so... I just want to make sure I understand this correctly. When compounds are in a polarimeter, and you have one molecule that is (+) and one molecule that is a (-), then the (+) molecule goes clockwise around the 360° circle (polarimeter) and can have a measurement of 0° to +360°? On the other hand, the (-) molecule can then go counterclockwise around the circle, thus showing a measurement of 0° to -360°? When diluted, both molecules will then show a decreased angle in opposite directions (both towards their respective 0°)?

[a] = a / (C * l) therefore: a = (C * l * [α]) ?

1) And if α is directly proportional to C * l, then diluting the solution would reduce the angle of optical rotation.
2) Reducing the angle through dilution will result in the (+) angle less + and the (-) angle less minus.
3) Both will get closer to the 0° degree point, but in opposite directions
4) The +233.0° when diluted to half, will go from +233° to +116.5°
5) The -127.0° when diluted to half, will go from 127.0° to -63.5°.
6) This results in the image you showed, which would mean that the two compounds could be discerned via polarimeter/optical rotation.


Follow-up questions:
1) If these molecules are enantiomers, then the result would be a 0° of optical rotation if there is no enantiomeric excess? Even though each enantiomer would have an identical optical rotation with opposite signs?

2) So then, if you dilute a racemic mixture, would both compounds decrease their optical rotation? (with the overall optical activity being 0° because they'd both decrease in the same amounts?
 
Hi @arc5005,

Yes, sounds like you got it (steps 1-6). In terms of the follow-up questions, it sounds like you're on track, but there are two major things you have to keep in mind: (1) racemic mixtures have an overall optical activity of 0° by definition, no matter what; (2) enantiomers (under the same experimental conditions, including concentration) show optical rotation that is equal in magnitude but opposite in direction, no matter what. Don't let any complicated problem set-up confuse you about those two points! That said, you're right in terms of your prediction of what would happen if you dilute a racemic mixture, with the small caveat that it'd be very hard to measure, because you'd keep getting an overall result of 0°. If you happen to have some spare time, it's interesting to read about Pasteur's experiments in which this phenomenon was first clarified -- very neat stuff!

Hope this is helpful!
 
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Hi @arc5005,

Yes, sounds like you got it (steps 1-6). In terms of the follow-up questions, it sounds like you're on track, but there are two major things you have to keep in mind: (1) racemic mixtures have an overall optical activity of 0° by definition, no matter what; (2) enantiomers (under the same experimental conditions, including concentration) show optical rotation that is equal in magnitude but opposite in direction, no matter what. Don't let any complicated problem set-up confuse you about those two points! That said, you're right in terms of your prediction of what would happen if you dilute a racemic mixture, with the small caveat that it'd be very hard to measure, because you'd keep getting an overall result of 0°. If you happen to have some spare time, it's interesting to read about Pasteur's experiments in which this phenomenon was first clarified -- very neat stuff!

Hope this is helpful!

thank you!
 
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