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If a chiral molecule reacts with an achiral molecule will the product be chiral or achiral?
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My book says that a pair of enantiomers will react the same way with an achiral substance.
How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?
organic-chemistry stereochemistry isomers chirality
edited Aug 11 at 10:40
Rodrigo de Azevedo
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asked Aug 10 at 12:41
Hema
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My book says that a pair of enantiomers will react the same way with an achiral substance.
How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?
organic-chemistry stereochemistry isomers chirality
edited Aug 11 at 10:40
Rodrigo de Azevedo
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asked Aug 10 at 12:41
Hema
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Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30
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up vote
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up vote
9
down vote
favorite
My book says that a pair of enantiomers will react the same way with an achiral substance.
How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?
organic-chemistry stereochemistry isomers chirality
edited Aug 11 at 10:40
Rodrigo de Azevedo
674717
asked Aug 10 at 12:41
Hema
25210
My book says that a pair of enantiomers will react the same way with an achiral substance.
How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?
organic-chemistry stereochemistry isomers chirality
edited Aug 11 at 10:40
Rodrigo de Azevedo
674717
asked Aug 10 at 12:41
Hema
25210
edited Aug 11 at 10:40
Rodrigo de Azevedo
674717
edited Aug 11 at 10:40
Rodrigo de Azevedo
674717
edited Aug 11 at 10:40
Rodrigo de Azevedo
674717
674717
asked Aug 10 at 12:41
Hema
25210
asked Aug 10 at 12:41
Hema
25210
asked Aug 10 at 12:41
Hema
25210
25210
1
Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30
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1
Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30
1
1
Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30
Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30
add a comment |Â
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The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$ce($S$)-A + B -> [($S$,$R$)-AB]^ddagger + [($S$,$S$)-AB]^ddagger$$
These are diastereomeric and the rates through these two transition states is different.
But $ce($R$)-A$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:
- $ce[($S$,$R$)-AB]^ddagger$
- $ce[($S$,$S$)-AB]^ddagger$
- $ce[($R$,$S$)-AB]^ddagger$
- $ce[($R$,$R$)-AB]^ddagger$
Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.
answered Aug 10 at 13:39
Zhe
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Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).
answered Aug 10 at 13:14
ringo
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It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.
answered Aug 10 at 15:23
Patrick Graham
1412
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
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Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.
answered Aug 10 at 17:02
David Richerby
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1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers
Image source
and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:
- $R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
- $R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$ceCBrClFH+R_2 to ceCBrClF +R_2ceH,$$ where the reactant hydride $R_2ceH$ is achiral and the remaining scaffolding $ceCBrClF$ is now planar and therefore achiral.
(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)
answered Aug 10 at 21:12
E.P.
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$ce($S$)-A + B -> [($S$,$R$)-AB]^ddagger + [($S$,$S$)-AB]^ddagger$$
These are diastereomeric and the rates through these two transition states is different.
But $ce($R$)-A$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:
- $ce[($S$,$R$)-AB]^ddagger$
- $ce[($S$,$S$)-AB]^ddagger$
- $ce[($R$,$S$)-AB]^ddagger$
- $ce[($R$,$R$)-AB]^ddagger$
Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.
answered Aug 10 at 13:39
Zhe
11.3k11847
add a comment |Â
up vote
9
down vote
accepted
The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$ce($S$)-A + B -> [($S$,$R$)-AB]^ddagger + [($S$,$S$)-AB]^ddagger$$
These are diastereomeric and the rates through these two transition states is different.
But $ce($R$)-A$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:
- $ce[($S$,$R$)-AB]^ddagger$
- $ce[($S$,$S$)-AB]^ddagger$
- $ce[($R$,$S$)-AB]^ddagger$
- $ce[($R$,$R$)-AB]^ddagger$
Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.
answered Aug 10 at 13:39
Zhe
11.3k11847
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$ce($S$)-A + B -> [($S$,$R$)-AB]^ddagger + [($S$,$S$)-AB]^ddagger$$
These are diastereomeric and the rates through these two transition states is different.
But $ce($R$)-A$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:
- $ce[($S$,$R$)-AB]^ddagger$
- $ce[($S$,$S$)-AB]^ddagger$
- $ce[($R$,$S$)-AB]^ddagger$
- $ce[($R$,$R$)-AB]^ddagger$
Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.
answered Aug 10 at 13:39
Zhe
11.3k11847
The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$ce($S$)-A + B -> [($S$,$R$)-AB]^ddagger + [($S$,$S$)-AB]^ddagger$$
These are diastereomeric and the rates through these two transition states is different.
But $ce($R$)-A$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:
- $ce[($S$,$R$)-AB]^ddagger$
- $ce[($S$,$S$)-AB]^ddagger$
- $ce[($R$,$S$)-AB]^ddagger$
- $ce[($R$,$R$)-AB]^ddagger$
Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.
answered Aug 10 at 13:39
Zhe
11.3k11847
answered Aug 10 at 13:39
Zhe
11.3k11847
answered Aug 10 at 13:39
Zhe
11.3k11847
answered Aug 10 at 13:39
Zhe
11.3k11847
11.3k11847
add a comment |Â
add a comment |Â
up vote
4
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Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).
answered Aug 10 at 13:14
ringo
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up vote
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Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).
answered Aug 10 at 13:14
ringo
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up vote
4
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up vote
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Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).
answered Aug 10 at 13:14
ringo
19.2k554101
Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).
answered Aug 10 at 13:14
ringo
19.2k554101
answered Aug 10 at 13:14
ringo
19.2k554101
answered Aug 10 at 13:14
ringo
19.2k554101
answered Aug 10 at 13:14
ringo
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19.2k554101
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add a comment |Â
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It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.
answered Aug 10 at 15:23
Patrick Graham
1412
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
add a comment |Â
up vote
4
down vote
It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.
answered Aug 10 at 15:23
Patrick Graham
1412
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.
answered Aug 10 at 15:23
Patrick Graham
1412
It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.
answered Aug 10 at 15:23
Patrick Graham
1412
answered Aug 10 at 15:23
Patrick Graham
1412
answered Aug 10 at 15:23
Patrick Graham
1412
answered Aug 10 at 15:23
Patrick Graham
1412
1412
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
add a comment |Â
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
strike out "it's possible" and replace by "will always"
– Karl
Aug 11 at 9:32
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.
– Patrick Graham
Aug 13 at 0:49
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.
– Karl
Aug 13 at 7:38
add a comment |Â
up vote
3
down vote
Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.
answered Aug 10 at 17:02
David Richerby
335412
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up vote
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Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.
answered Aug 10 at 17:02
David Richerby
335412
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up vote
3
down vote
up vote
3
down vote
Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.
answered Aug 10 at 17:02
David Richerby
335412
Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.
answered Aug 10 at 17:02
David Richerby
335412
answered Aug 10 at 17:02
David Richerby
335412
answered Aug 10 at 17:02
David Richerby
335412
answered Aug 10 at 17:02
David Richerby
335412
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1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers
Image source
and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:
- $R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
- $R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$ceCBrClFH+R_2 to ceCBrClF +R_2ceH,$$ where the reactant hydride $R_2ceH$ is achiral and the remaining scaffolding $ceCBrClF$ is now planar and therefore achiral.
(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)
answered Aug 10 at 21:12
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218110
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1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers
Image source
and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:
- $R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
- $R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$ceCBrClFH+R_2 to ceCBrClF +R_2ceH,$$ where the reactant hydride $R_2ceH$ is achiral and the remaining scaffolding $ceCBrClF$ is now planar and therefore achiral.
(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)
answered Aug 10 at 21:12
E.P.
218110
add a comment |Â
up vote
2
down vote
up vote
2
down vote
1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers
Image source
and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:
- $R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
- $R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$ceCBrClFH+R_2 to ceCBrClF +R_2ceH,$$ where the reactant hydride $R_2ceH$ is achiral and the remaining scaffolding $ceCBrClF$ is now planar and therefore achiral.
(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)
answered Aug 10 at 21:12
E.P.
218110
1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers
Image source
and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:
- $R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
- $R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$ceCBrClFH+R_2 to ceCBrClF +R_2ceH,$$ where the reactant hydride $R_2ceH$ is achiral and the remaining scaffolding $ceCBrClF$ is now planar and therefore achiral.
(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)
answered Aug 10 at 21:12
E.P.
218110
answered Aug 10 at 21:12
E.P.
218110
answered Aug 10 at 21:12
E.P.
218110
answered Aug 10 at 21:12
E.P.
218110
218110
add a comment |Â
add a comment |Â
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1
Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.
– Karl
Aug 11 at 9:30