Organic Chemistry Qns
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Hi to all experts:
I have 2 A-Level organic chem qns to ask here on behalf of my friend 'cos I have difficulty solving it too:
Q1) What relative amounts of 1-bromobutane and 2-bromobutane are actually formed when butane is brominated?
Q2) If all C-H bonds had the same homolytic dissociation energy, what would be the relative amounts of 1-bromobutane and 2-bromobutane produced in the bromination of butane?
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Originally posted by lianamaster:
Hi to all experts:
I have 2 A-Level organic chem qns to ask here on behalf of my friend 'cos I have difficulty solving it too:
Q1) What relative amounts of 1-bromobutane and 2-bromobutane are actually formed when butane is brominated?
Q2) If all C-H bonds had the same homolytic dissociation energy, what would be the relative amounts of 1-bromobutane and 2-bromobutane produced in the bromination of butane?
Do u need a proper estimation? I doubt whether that is possible but 1 bromo should always be more than 2-bromo if i never remember wrong since each new bromination requires more energy to occur but being a substituition reaction, you cannot really control and stop it anywhere so i doubt you can really get a clear amount of how much is formed but can only compare whether its more or less than another.
I didn't do A's chem so i dunno whether they got teach any wierd ways to determine it but for a reaction you cannot control, asking for amount seems illogical to me.
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I just need the ratio will do.
Apparently, we need to think about the stability of the radicals such that:
Stability of radicals: Tertiary > Secondary > Primary
It all depends on the site of attack of the Br radical on butane and how exothermic and endothermic the reaction based on the difference in energy released/absorbed when breaking the C-H bond to form the C-Br bond.
But I got stuck in the computation of the relative abundance of the isomers.
For example, propane reacts with chlorine in UV to give 45% 1-chloropropane and 55% 2-chloropropane but I dunno how to get to that percentage.
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Originally posted by lianamaster:
I just need the ratio will do.
Apparently, we need to think about the stability of the radicals such that:
Stability of radicals: Tertiary > Secondary > Primary
It all depends on the site of attack of the Br radical on butane and how exothermic and endothermic the reaction based on the difference in energy released/absorbed when breaking the C-H bond to form the C-Br bond.
But I got stuck in the computation of the relative abundance of the isomers.
For example, propane reacts with chlorine in UV to give 45% 1-chloropropane and 55% 2-chloropropane but I dunno how to get to that percentage.
Ok, i misread earlier it seems.. Somehow my brain was thinking it as mono sub, di sub etc instead of the position of the Br...
Sounds like something familiar but i 4got how to do already since that is not the focus of uni chemistry.
Wait for UltimaOnline to solve this then :p. A's chem is always a problem for me since i dunno wat they teach since a decent amount of stuff are not used or bothered about in uni.
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Originally posted by dkcx:
Ok, i misread earlier it seems.. Somehow my brain was thinking it as mono sub, di sub etc instead of the position of the Br...
Sounds like something familiar but i 4got how to do already since that is not the focus of uni chemistry.
Wait for UltimaOnline to solve this then :p. A's chem is always a problem for me since i dunno wat they teach since a decent amount of stuff are not used or bothered about in uni.
No problem. I had difficulty in this qn too... I wasn't really taught on the quantitative aspects of free-radical substitution. We were just taught on how to write mechanisms, comment on the feasibility of the reaction and prediction of products.
Besides, there are countless of other more important reactions to learn other than this free-radical substitution.
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Originally posted by lianamaster:
No problem. I had difficulty in this qn too... I wasn't really taught on the quantitative aspects of free-radical substitution. We were just taught on how to write mechanisms, comment on the feasibility of the reaction and prediction of products.
Besides, there are countless of other more important reactions to learn other than this free-radical substitution.
There are also countless stuff i'm learning that i wish never existed like quantum... :p
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Originally posted by dkcx:
There are also countless stuff i'm learning that i wish never existed like quantum... :p
Some ppl just love it but not me, sadly. =D
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Originally posted by lianamaster:
Some ppl just love it but not me, sadly. =D
In our society where the majority normally means the norm, you are normal ;)
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Exactly.
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For Q1. The reactivity (radical intermediate stability) factor depends on temperature, but bromination is a lot more selective than chlorination. At 125 deg C, a tertiary radical is formed 1600x faster, and a secondary radical 82x faster, compared to primary radical. You needn't memorize these values; the exam question will provide the relevant values if they require you to factor this in. Relative amount of 1-bromobutane : Number of H atoms substitutable x reactivity (radical stability) = 6 x 1 = 6 Percentage yield = 6/(328+6) = 1.80% Relative amount of 2-bromobutane : Number of H atoms substitutable x reactivity (radical stability) = 4 x 82 = 328 Percentage yield = 6/(328+6) = 98.2% For Q2. Assuming all C-H bonds had the same homolytic dissociation energy, in other words, ignoring the reactivity (radical stability) factor, simply count the number of hydrogens substitutable to obtain the two isomeric monobrominated products : Relative amount of 1-bromobutane : Number of H atoms substitutable = 6 Percentage yield = 6/(6+4) = 60% Relative amount of 2-bromobutane : Number of H atoms substitutable = 4 Percentage yield = 4/(6+4) = 40%
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is q1 an a level qn, i have never seen it before. particularly regarding "reactivity (radical stability)" factor. only qn 2 seen before...
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It's actually University level Organic Chemistry. At 'A' levels, the exam question will usually allow you to explain and apply only the simpler factor of number of hydrogens substitutable. If the exam question at 'A' levels requires you to factor in primary, secondary, tertiary radical stability (or equivalently, dissociation energies of primary, secondary, tertiary C-H bonds), the question will clue you in with relevant data (a brief explanation, together with relative rates of free radical substitution involving primary, secondary, tertiary alkyl radical intermediates, eg. 5.0 > 3.8 > 1.0 for chlorination at 298K). For the 2007 'A' level H2 Chemistry exam, for instance, the question required the candidate to "describe, explain and calculate based on any ONE factor that determines isomeric product distribution of monochlorination of butane" (where the two factors are of course, number of hydrogens substitutable, and stability of alkyl radical intermediates). So don't worry too much about this.
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Originally posted by UltimaOnline:
For Q1.
The reactivity (radical intermediate stability) factor depends on temperature, but bromination is a lot more selective than chlorination.
At 125 deg C, a tertiary radical is formed 1600x faster, and a secondary radical 82x faster, compared to primary radical.
You needn’t memorize these values; the exam question will provide the relevant values if they require you to factor this in.
Relative amount of 1-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 6×1 = 6Percentage yield = 6/(328+6) = 1.80%
Relative amount of 2-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 4×82 = 328Percentage yield = 6/(328+6) = 98.2%
For Q2.
Assuming all C-H bonds had the same homolytic dissociation energy, in other words, ignoring the reactivity (radical stability) factor, simply count the number of hydrogens substitutable to obtain the two isomeric monobrominated products :
Relative amount of 1-bromobutane :
Number of H atoms substitutable
= 6Percentage yield = 6/(6+4) = 60%
Relative amount of 2-bromobutane :
Number of H atoms substitutable
= 4Percentage yield = 4/(6+4) = 40%
Yup! That's the answer I am looking for.
But how to I know that there are 6 H in 1-bromobutane and 4 H in 2-bromobutane that are sustitutable?
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The 3 H atoms on the 1st C + the 3 H atoms on the 4th C = 6 H atoms. Substituting any one of these H atoms gives you 1-bromobutane. The 2 H atoms on the 2nd C + the 2 H atoms on the 3rd C = 4 H atoms. Substituting any one of these H atoms gives you 2-bromobutane.
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Got it.
Thanks so much.
