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clarification: racemic mixtures, optical activity
Started by Unlearner
loveoforganic will give you a better answer but here's my take:
1) Look for the definition of racemic to answer your question. I think racemic mixtures are mixtures of two entantiomers. Each one is optically active but they cancel each other out as they rotate light in opposite directions. Entantiomers are stereoisomers and thus have chiral centers.
If a product does not have chiral centers, i don't think it is a stereoisomer and thus can't be racemic
2)SN1 will product an optically inactive product(s) as long as you have chiral carbons (I am unsure what happens if the molecule doesn't have any with respect to rotating light). Just make sure the carbons have 4 different substituents on them
I am getting the feeling you are memorizing and not understanding enough. I could be wrong but thats the feeling I'm getting. As far as orgo goes, this is important stuff so make sure understand it very well. Just a heads up
To add further to what sv3 said, SN1 reactions create a racemic mixture of product with respect to any chiral sites in the reactant. This means there will be products that retain the same orientation at each chiral site from the reactant AND products that have an inverted orientation at that chiral site. So one half of product will be one enantiomer and the other half the opposite enantiomer.
So in direct answer to your 2nd question, the mixture of products of an SN1 reaction involving a reactant with a chiral site as a whole will be optically inactive. However, individual products can be optically active, if they can be separated or made into unequal proportions. This is assuming the product isn't a meso compound, in which case it isn't optically active.
I'm making these clarification because a question regarding this on the MCAT may involve different situations and you have to be very clear on what they're looking for specifically. Being able to apply the facts to different situations is the most important thing on the MCAT from what I've seen so far.
So in direct answer to your 2nd question, the mixture of products of an SN1 reaction involving a reactant with a chiral site as a whole will be optically inactive. However, individual products can be optically active, if they can be separated or made into unequal proportions. This is assuming the product isn't a meso compound, in which case it isn't optically active.
I'm making these clarification because a question regarding this on the MCAT may involve different situations and you have to be very clear on what they're looking for specifically. Being able to apply the facts to different situations is the most important thing on the MCAT from what I've seen so far.
SN1 --> 50% R and 50% S products. Optical activity (i.e. bending plane polarized light) thus cancels out and no net plane shift is observed. Pretty sure it only counts for chiral pairs. There are probably exceptions that ochem nerds can address, but generally SN1 (supposedly) creates an sp2 hybridized intermediate that fluctuates/vibrates, thus I think most (if not all) SN1 rxns create optically inactive solutions.
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You have to remember that if the electrophile contains 2 chiral centers, the nucleophile could attack at a major position and a minor position due to steric hindrance. The compound would be a diastereomer ===> not a racemic mixture. Just so you know that Sn1 does not always result in racemic mixtures...
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i think i might be interchangeably using products with carbon centers as being either chiral or achiral. so to clarify, a racemic mixture makes optically inactive products which are achiral as a whole, but its enantiomers (S/R) has carbon centers that are chiral. is that correct?
Solid clarification and good point on what the MCAT will try to do.
phEight.......is this common enough an occurence to take into consideration on an MCAT question or will they give you direction?
I look for direction in terms of answer choices when needed. I just haven't had this come up in all the practice i've done so thought i'd ask about the occurence of this. There seem to be exceptions to everything and I just want to make sure I'm not always saying to myself "well you never know, maybe this can happen and so can this"........
thanks
You're really misuing terms here. A molecule is (a)chiral. A solution is racemic, optically active, or optically inactive. Chirality refers to a molecule's lack of a plane of symmetry due to the molecule's chiral centers (points of 4 different substituents). A chiral molecule rotates plane polarized light in a certain direction (+/-) by a certain amount (specific rotation). The enantiomer of a chiral molecule rotates plane polarized light by the same specific rotation but in the opposite direction. A racemic mixture is a 50/50 mixture of OPTICALLY ACTIVE molecules, enantiomers specifically. Because enantiomers rotate plane polarized light by the same amount, but in the opposite direction, if they are combined in equal portions, their effects cancel out and there is no overall rotation.
sv3 - Even if there were two chiral centers in a molecule, both of which were capable of undergoing an SN1, that's not enough to assume you'd end up with an optically active solution. There's actually a good chance you'll end up with an optically inactive solution again, or very near optically inactive depending on steric factors. You WOULD end up with multiple stereoisomers, but assuming there are no huge steric factors in play, you'd also end up with a near equal amount of the enantiomer of each of those stereoisomers and end up with an opitcally inactive solution again. I'm not sure if that's too much info for the MCAT or not, but if I had to guess I'd think so.
Well, I doubt something specific will be asked about the MCAT, but an understanding of this sort of concept certainly seems to drill in concepts of stereochemistry and reactivity that probably will be tested on the MCAT. The solution most probably will not be optically active as loc stated, but it most definitely will not be a racemic mixture. I think it's helpful knowledge to be aware of. I feel it's important to know that not all sn1 reactions lead to racemic mixtures, not because of the fact in itself, but why it is this way, if that makes any sense.
thanks very much, that clears it up nicely. Always good to know these little things now, rather than on the MCAT!
phEight - I know what you mean, know because of the why, not just because. The one thing I want to make sure I get is if a solution has enantiomers and is optically inactive, it is racemic right? In your response you say its likely optically inactive but not racemic? Is this because of distereoisomer presence that your assuming?
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It's just that a racemate is by definiton an equal mixture of 2 enantiomers. If you have multiple sets of enantiomers, it doesn't fall into the definition anymore.
Not only that, if there are multiple chiral centers with an Sn1 reaction, it will result in diastereomers too, not just enantiomers. Sv3, yea that's what I mean, the presence of the diastereomer should mean it's not a racemic solution.
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But if the diastereomers were not present in equal proportion with their enantiomer, the solution would be optically active, so I hold to what I said! 😛
Well, there's no reason not to hold what you said.. I'm not sure if we're talking or thinking about the same things... maybe we are and you could clarify it for me. Basically what I was trying to say was that in cases with multiple stereocenter electrophiles, an Sn1 reaction will yield diastereomers, NOT enantiomers. A diastereomer by definition is an isomer in which at least one, but not all chiral centers are different. An enantiomer has all chiral centers different. What I'm saying is that in some Sn1 reactions, only diastereomers result, not enantiomers and therefore you can't call that a racemic mixture. I'm kind of confused why you mentioned diastereomers being in solution with enantiomers... I feel like I'm not catching something here, and if that's the case itd be nice if you could clarify. 🙂
Ok, say you're running a reaction of cyanide on (R,R)-2,4-diiodoheptane. Assume the reaction proceeds by SN1 (it obviously wouldn't, just assume!). Now also assume that steric factors do not come into play. You'd end up with an equal amount of cyanide adding to form the R and S in the 2 position, and equal amount of cyanide adding to form the R and S in the 4 position, and and equal amount of cyanide adding to both positions, all in perfect mixtures. Therefore, you end up with diastereomers, yes, but equal portion of the enantiomer of each diastereomer.
At any rate, I was just sort of being silly and I think this is a silly argument because I'm pretty sure you're right that we're saying the same thing!
At any rate, I was just sort of being silly and I think this is a silly argument because I'm pretty sure you're right that we're saying the same thing!
Yeah, probably so. Thanks for explaining your reasoning though. I was led to believe in most cases steric issues will play a bit of a role, but I guess one just has to assess the possibilities. All of this is much more simple just looking at a problem rather than trying to create hypothetical situations.Sv3, I hear ya..
