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Help with racemates and enantiomer question

2.State whether the following preparations will produce a racemate or a single enantiomer:
a. butan-2-ol from but-2-ene
b.butan-2-ol from 2-bromobutane
c.2-hydroxybutanenitrile from propanal

i just dont get it at all, i dont think ive learned it yet

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Original post by Ace.ms
2.State whether the following preparations will produce a racemate or a single enantiomer:
a. butan-2-ol from but-2-ene
b.butan-2-ol from 2-bromobutane
c.2-hydroxybutanenitrile from propanal

i just dont get it at all, i dont think ive learned it yet

Can you define the terms ‘enantiomer’ and ‘racemate’?

Are you also familiar with the electrophilic addition, sN1/sN2 and nucleophilic addition mechanisms?

You don’t have to draw them out. Briefly describe what happens in each and make it clear what intermediate forms (i.e carbanion, carbocation etc)
(edited 1 year ago)
Reply 2
Original post by TypicalNerd
Can you define the terms ‘enantiomer’ and ‘racemate’?

Are you also familiar with the electrophilic addition, sN1/sN2 and nucleophilic addition mechanisms?

You don’t have to draw them out. Briefly describe what happens in each and make it clear what intermediate forms (i.e carbanion, carbocation etc)

I can define those terms and ywah I'm familiar with the mechanisms. I didn't thibk to try with mechanisms, will do that and see what happens . thanks!!!
Original post by Ace.ms
I can define those terms and ywah I'm familiar with the mechanisms. I didn't thibk to try with mechanisms, will do that and see what happens . thanks!!!

Yeah, it certainly isn’t an obvious thing to think about. Once you have reached a conclusion, please do post it and I’ll give further prompts/assistance if/where needed.

I don’t expect you to draw out the mechanisms (though it may be useful practice for you to do so).
Reply 4
Original post by TypicalNerd
Yeah, it certainly isn’t an obvious thing to think about. Once you have reached a conclusion, please do post it and I’ll give further prompts/assistance if/where needed.

I don’t expect you to draw out the mechanisms (though it may be useful practice for you to do so).

Im kinda stuck as i understand that 2 bromobutane has a chiral carbon, and so does butan 2 ol, so i would have thought its a racemate because it can make either optical isomer? it just feels like the question should mention if 2 bromobutane is R or S 2 bromobutane?? apologies if im completely wrong
Original post by Ace.ms
Im kinda stuck as i understand that 2 bromobutane has a chiral carbon, and so does butan 2 ol, so i would have thought its a racemate because it can make either optical isomer? it just feels like the question should mention if 2 bromobutane is R or S 2 bromobutane?? apologies if im completely wrong

Think about the intermediates in each mechanism.

You also shouldn’t need to worry about which of the R or S enantiomer forms (assuming a single enantiomer only), as the question asks if either a racemate or a single enantiomer forms
(edited 1 year ago)
Reply 6
Original post by TypicalNerd
Think about the intermediates in each mechanism.

You also shouldn’t need to worry about which of the R or S enantiomer forms (assuming a single enantiomer only), as the question asks if either a racemate or a single enantiomer forms

im completely lost, if i had to guess for A id go with single enantiomer but im kinda clueless, would really appreciate if you could explain from the beginning, or link me to some helpful website
Original post by Ace.ms
im completely lost, if i had to guess for A id go with single enantiomer but im kinda clueless, would really appreciate if you could explain from the beginning, or link me to some helpful website

Ok. I’ll go through the first parts of the first example.

When but-2-ene is converted to butan-2-ol, the reaction is electrophilic addition. In this reaction, a carbocation intermediate forms.

On the positive carbon, there are three bonding pairs and zero lone pairs. This implies a trigonal planar shape (i.e it is flat), so it is equally likely to be attacked either side by the electrophile.

Given this information, can you infer whether one enantiomer will be favoured over the other or whether both will form in a 1:1 ratio?
Reply 8
Original post by TypicalNerd
Ok. I’ll go through the first parts of the first example.

When but-2-ene is converted to butan-2-ol, the reaction is electrophilic addition. In this reaction, a carbocation intermediate forms.

On the positive carbon, there are three bonding pairs and zero lone pairs. This implies a trigonal planar shape (i.e it is flat), so it is equally likely to be attacked either side by the electrophile.

Given this information, can you infer whether one enantiomer will be favoured over the other or whether both will form in a 1:1 ratio?

1:1 id assume
Original post by Ace.ms
1:1 id assume

Correct.

Now for the rest of the reactions, can you identify which type of intermediate forms and what effect that has on the possible product(s) that would form?

Note that for the last one, it’s better to consider the starting molecule, rather than the intermediate.
(edited 1 year ago)
Reply 10
Original post by TypicalNerd
Correct.

Now for the rest of the reactions, can you identify which type of intermediate forms and what effect that has on the possible product(s) that would form?

Note that for the last one, it’s better to consider the starting molecule, rather than the intermediate.

for the other 2 id guess single enantiomer, but im not very confident in that answer for C
Original post by Ace.ms
for the other 2 id guess single enantiomer, but im not very confident in that answer for C


For (c), what shape is the carbon atom in the carbonyl group? Will that shape make it susceptible to attack from just one side or both?

For (b), it’s harder to tell, as it’s a haloalkane to an alcohol, but I’d probably agree that it favours one enantiomer over the other. Any ideas why?
(edited 1 year ago)
I would just look at the products, butan-2-ol and 2-hydroxybutanenitrile both have chiral carbon atoms so can exist as a pair of enantiomers. However you always get racemic mixtures if you start with achiral materials, since all these reactions start with achiral materials and do not involve enzymes/specific catalysts you will I think always get racemic mixtures and NOT single enantiomers.

Can you think of one example in your a-level course where you made a one single enantiomer from any reaction? I doubt there is any!
Original post by Scanjo63
I would just look at the products, butan-2-ol and 2-hydroxybutanenitrile both have chiral carbon atoms so can exist as a pair of enantiomers. However you always get racemic mixtures if you start with achiral materials, since all these reactions start with achiral materials and do not involve enzymes/specific catalysts you will I think always get racemic mixtures and NOT single enantiomers.

Can you think of one example in your a-level course where you made a one single enantiomer from any reaction? I doubt there is any!


Whilst it’s a good point and (if correct) perhaps better for future reference, in the event that the OP is asked to explain why either a single enantiomer forms or whether a racemic mixture forms, it’s perhaps better to think about it in terms of the mechanism.
(edited 1 year ago)
Reply 14
Original post by TypicalNerd
For (c), what shape is the carbon atom in the carbonyl group? Will that shape make it susceptible to attack from just one side or both?

For (b), it’s harder to tell, as it’s a haloalkane to an alcohol, but I’d probably agree that it favours one enantiomer over the other. Any ideas why?

my only reasoning for B was that the bromine there means only 1 bond is freed up from the carbon so it has to go in one place, flawed logic ik haha 😭
for c id guess it is susceptible from both sides after looking at the structure because it has a chiral carbon and thus both enantiomers have an equal chance of forming because there are two possible reactant isomers?!?!?
Original post by Ace.ms
my only reasoning for B was that the bromine there means only 1 bond is freed up from the carbon so it has to go in one place, flawed logic ik haha 😭
for c id guess it is susceptible from both sides after looking at the structure because it has a chiral carbon and thus both enantiomers have an equal chance of forming because there are two possible reactant isomers?!?!?


For (c), you’ve got it. It will form both enantiomers in a 1:1 ratio.

For (b), even I’m uncertain, as haloalkanes can undergo both sN1 and sN2 nucleophilic substitution. In the sN1 route, you get a carbocation, so a racemate should theoretically form. In the sN2 route, a transition state forms and this means that there is only one place in which the haloalkane is attacked (hence only one enantiomer).
Reply 16
Original post by TypicalNerd
For (c), you’ve got it. It will form both enantiomers in a 1:1 ratio.

For (b), even I’m uncertain, as haloalkanes can undergo both sN1 and sN2 nucleophilic substitution. In the sN1 route, you get a carbocation, so a racemate should theoretically form. In the sN2 route, a transition state forms and this means that there is only one place in which the haloalkane is attacked (hence only one enantiomer).

this is only a random homework so its not a biggie if lose that mark on b, ill put single enantiomer and hope for the best.
Thanks so much for the help, i was looking at ur profile and saw you got accepted to Oxford, congrats, you clearly deserve it based on how you managed to basically teach me the whole topic just now :biggrin:
Original post by Ace.ms
this is only a random homework so its not a biggie if lose that mark on b, ill put single enantiomer and hope for the best.
Thanks so much for the help, i was looking at ur profile and saw you got accepted to Oxford, congrats, you clearly deserve it based on how you managed to basically teach me the whole topic just now :biggrin:


Thanks for the congratulations.

By any chance, are you doing A level or uni chemistry?

If you are doing A level, then the exam board may actually be what gives away the answer (OCR A doesn’t require you to know about sN1, so your answer will involve sN2). If it’s uni level, maybe think about steric hindrance from the groups and whether that makes it easier to go through sN1 or sN2
Reply 18
Original post by TypicalNerd
Thanks for the congratulations.

By any chance, are you doing A level or uni chemistry?

If you are doing A level, then the exam board may actually be what gives away the answer (OCR A doesn’t require you to know about sN1, so your answer will involve sN2). If it’s uni level, maybe think about steric hindrance from the groups and whether that makes it easier to go through sN1 or sN2

A level AQA, teacher hasnt properly gone over the topic (only done nucleophilic, not really explained sN1 and sN2) in class so im not sure what which of those we are meant to know
Original post by Ace.ms
A level AQA, teacher hasnt properly gone over the topic (only done nucleophilic, not really explained sN1 and sN2) in class so im not sure what which of those we are meant to know

Looking at the specification at a glance and the notes by PMT, I see no indication that you need to know about sN1. Assume it’s sN2 and that you are right

Also, I just found this when looking up which mechanisms you are expected to be able to draw and it’s amazing https://chemrevise.files.wordpress.com/2017/02/aqa-mechanisms-a-level-summary.pdf
(edited 1 year ago)

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