Organic Chemistry: Reactions help!! (1 Viewer)

Nash__

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Quick question regarding some reactions, how can you form a primary alcohol from an alkene? eg form 1-propanol from propene. My understanding is that if you were to hydrate propene in the presence of dilute H2SO4 the hydroxyl group will add to the second carbon (by Markovnikov's rule, at least this product will dominate) thus forming 2-propanol which cannot then be transformed into a carboxylic acid for making an ester (as an example). The only other method I can think of is first reacting the alkene with H2 to produce the corresponding alkane and then react that with X2 in the presence of UV light to produce a haloalkane which then can be hydrated to form a specific alcohol. But I am not sure whether to use this process because I dont know if the product of the halogenation of propane (say with bromine) will be 1-bromopropane or 2-bromopropane, in the case of the latter I am then unable to make my primary alcohol that I want for the reaction. Any suggestions would be appreciated!
 
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jazz519

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In the HSC you don't really need to worry about Markovnikov's rule as it's not specified in the syllabus. If the reaction can possibly produce the molecule then even if it's not the major product it's fine

The actual methods for doing these things in real life is not in the scope of the syllabus because in real life we wouldn't use dilute h2so4 to make propan-1-ol as you said it makes the propan-2-ol instead. Similarly, the halogenation process is not selective and you will get a mixture of the two things and 2-bromopropane will once again be the dominant product.

In real life we would use something called a Grignard reagent, on methanal to make a primary alcohol. There are other possible pathways as well but are outside the scope of the syllabus
 

asiansubjects

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Quick question regarding some reactions, how can you form a primary alcohol from an alkene? eg form 1-propanol from propene. My understanding is that if you were to hydrate propene in the presence of dilute H2SO4 the hydroxyl group will add to the second carbon (by Markovnikov's rule, at least this product will dominate) thus forming 2-propanol which cannot then be transformed into a carboxylic acid for making an ester (as an example). The only other method I can think of is first reacting the alkene with H2 to produce the corresponding alkane and then react that with HX in the presence of UV light to produce a haloalkane which then can be hydrated to form a specific alcohol. But I am not sure whether to use this process because I dont know if the product of the halogenation of propane (say with bromine) will be 1-bromopropane or 2-bromopropane, in the case of the latter I am then unable to make my primary alcohol that I want for the reaction. Any suggestions would be appreciated!
I don't think Markovnikov's rule is really tested in the syllabus right now, so it's probably okay to say that 1-propanol can be formed by hydrating propene. If you really wanted a proper way to make 1-propanol from propene, then you'd use a hydroboration reaction that doesn't follow Markovnikov's rule: masterorganicchemistry.com/reaction-guide/hydroboration-of-alkenes/
 

Nash__

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In the HSC you don't really need to worry about Markovnikov's rule as it's not specified in the syllabus. If the reaction can possibly produce the molecule then even if it's not the major product it's fine

The actual methods for doing these things in real life is not in the scope of the syllabus because in real life we wouldn't use dilute h2so4 to make propan-1-ol as you said it makes the propan-2-ol instead. Similarly, the halogenation process is not selective and you will get a mixture of the two things and 2-bromopropane will once again be the dominant product.

In real life we would use something called a Grignard reagent, on methanal to make a primary alcohol. There are other possible pathways as well but are outside the scope of the syllabus
Thank you this is good to hear.
 

Nash__

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I don't think Markovnikov's rule is really tested in the syllabus right now, so it's probably okay to say that 1-propanol can be formed by hydrating propene. If you really wanted a proper way to make 1-propanol from propene, then you'd use a hydroboration reaction that doesn't follow Markovnikov's rule: masterorganicchemistry.com/reaction-guide/hydroboration-of-alkenes/
Cheers. Thanks for the link, I‘ll look into hydroboration!
 

CM_Tutor

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Hydroboration-oxidation is definitely an easy way to selectively form 1-propanol from propene - see https://en.wikipedia.org/wiki/Hydroboration–oxidation_reaction

The reaction occurs in two stages. The first reacts propene with borane as a tetrahydrofuran (THF) etherate, BH3.THF:

3 CH2=CH-CH3 + BH3.THF ----> B(CH2CH2CH3)3 + THF

The resulting tripropylborane is the oxidised by hydrogen peroxide (in acidic conditions) to yield 1-propanol and boric acid:

B(CH2CH2CH3)3 + 3 H2O2 ----> 3 CH3CH2CH2OH + H3BO3
 

CM_Tutor

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The only other method I can think of is first reacting the alkene with H2 to produce the corresponding alkane and then react that with X2 in the presence of UV light to produce a haloalkane which then can be hydrated to form a specific alcohol. But I am not sure whether to use this process because I dont know if the product of the halogenation of propane (say with bromine) will be 1-bromopropane or 2-bromopropane, in the case of the latter I am then unable to make my primary alcohol that I want for the reaction. Any suggestions would be appreciated!
Free radical halogenation of an alkane with X2 and UV light does favour the formation of secondary and tertiary products over primary ones. However, as there are three times as many H atoms in propane that will lead to a primary product as there are H's leading to a secondary product, the yield of 1-bromopropane as opposed to 2-bromopropane will likely be higher from brominating propane than from adding HBr to propene.

Do note, though, that the substitution of the 1-bromopropane to yield 1-propanol is not really a "hydration" as it is a substitution with an alkali (NaOH) rather than a reaction with water:

C3H7Br + NaOH ----> C3H7OH + NaBr
 

Isomorph

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A route to producing 1-propanol from propene can be completed using only syllabus reactions as the major product.

Brominate propene under UV light to form exclusively 3-bromopropene (dilute Br2 conditions to avoid further bromination). This occurs not only because the only saturated site is brominated, but also because the radical formed in the process is stabilised by resonance. React with dilute NaOH or H2SO4 (I believe the latter works, but haven't seen it much in textbooks), then hydrogenate using Pd/C/H2 to form desired 1-propanol.

Edit: I just realised you wouldn't want to use dilute H2SO4 here as it might hydrate the double bond. But perhaps in other scenarios it might work?
 

Isomorph

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Free radical halogenation of an alkane with X2 and UV light does favour the formation of secondary and tertiary products over primary ones. However, as there are three times as many H atoms in propane that will lead to a primary product as there are H's leading to a secondary product, the yield of 1-bromopropane as opposed to 2-bromopropane will likely be higher from brominating propane than from adding HBr to propene.

Do note, though, that the substitution of the 1-bromopropane to yield 1-propanol is not really a "hydration" as it is a substitution with an alkali (NaOH) rather than a reaction with water:

C3H7Br + NaOH ----> C3H7OH + NaBr
Bromination actually produces 97% 2-bromopropane, although the proportion is close for chlorination. The explanation is interesting and can be found here

 

mathsbrain

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hi guys,is there any good past papers website for 2018 onwards new syllabus?
 

CM_Tutor

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Bromination actually produces 97% 2-bromopropane, although the proportion is close for chlorination. The explanation is interesting and can be found here

Isomorph, you are quite right that bromination does favour secondary over primary haloalkane products much more strongly than does chlorination, thanks for the reminder. That explanation, though, goes into areas beyond the HSC syllabus and also skips some technical issues.
 

CM_Tutor

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A route to producing 1-propanol from propene can be completed using only syllabus reactions as the major product.

Brominate propene under UV light to form exclusively 3-bromopropene (dilute Br2 conditions to avoid further bromination). This occurs not only because the only saturated site is brominated, but also because the radical formed in the process is stabilised by resonance. React with dilute NaOH or H2SO4 (I believe the latter works, but haven't seen it much in textbooks), then hydrogenate using Pd/C/H2 to form desired 1-propanol.

Edit: I just realised you wouldn't want to use dilute H2SO4 here as it might hydrate the double bond. But perhaps in other scenarios it might work?
Even with dilute Br2 and UV light, you will still get some addition across the double bond. There may be conditions whereby this is avoided (the processes for preparing vinyl chloride offer some ideas) but I don't think a straight HSC syllabus method for propylene to 1-propanol exists.
 

Isomorph

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Even with dilute Br2 and UV light, you will still get some addition across the double bond. There may be conditions whereby this is avoided (the processes for preparing vinyl chloride offer some ideas) but I don't think a straight HSC syllabus method for propylene to 1-propanol exists.
Yeah I think you are right that there is no straight HSC syllabus method.

I have made a great number of attempts but the core of the problem is that the syllabus offers no way to reduce any species except using Pd/C/H2, which is not helpful for diols and triols, as any attempt to dehydrate them forms a complex mixture of non-desired products.
 

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