Destroyer Road Map 2 Help

[busupshot83]

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I have a question on the Destroyer Road Map 2 (page 104). On the far left side of the page, it has the following reaction:
R-CH2-CH3 + (K2Cr2O7, H+) --> R-COOH

An alkane can become a carboxylic acid by reacting with a harsh oxidizing agent? I thought alkanes first had to be coverted to alcohols in order to do this?

Thanks.

Bus

 

[Lonely Sol]

I have a question on the Destroyer Road Map 2 (page 104). On the far left side of the page, it has the following reaction:
R-CH2-CH3 + (K2Cr2O7, H+) --> R-COOH

An alkane can become a carboxylic acid by reacting with a harsh oxidizing agent? I thought alkanes first had to be coverted to alcohols in order to do this?

Thanks.

Bus

Well the alkane you are talkin about is attached to the benzene, which makes a big difference, the K2Cr2O7 is a very stong oxidizing agent, it will replace any benzyl carbons into COOH, even if there was 2 chains attached, then both chains would be COOH!!

Hope it helps!

 

[busupshot83]

Well the alkane you are talkin about is attached to the benzene, which makes a big difference, the K2Cr2O7 is a very stong oxidizing agent, it will replace any benzyl carbons into COOH, even if there was 2 chains attached, then both chains would be COOH!!

Hope it helps!

Oh I see... this reaction relies on the aromatic to occur. Therefore, a regular alkane could not undergo this transformation, right?

 

[Lonely Sol]

Oh I see... this reaction relies on the aromatic to occur. Therefore, a regular alkane could not undergo this transformation, right?

Yep!!

 

[busupshot83]

Yep!!

I have so much love for you, in a brotherly sort of way of course. 👍

 

[Lonely Sol]

I have so much love for you, in a brotherly sort of way of course. 👍

loll, Iight i'm glad to help!

 

[busupshot83]

loll, Iight i'm glad to help!

I have another question Sol...

On that same road map, there is the reagent (NBS, ROOR, heat). If an aromatic benzene ring had two methyl groups (the one on that roadmap has only one), does each group bond to the Br? Or in otherwords, does (NBS, ROOR, heat) apply to ALL the substituents of the benzene simutaneously, like the reagent (K2Cr2O7)?

 

[Lonely Sol]

I have another question Sol...

On that same road map, there is the reagent (NBS, ROOR, heat). If an aromatic benzene ring had two methyl groups (the one on that roadmap has only one), does each group bond to the Br? Or in otherwords, does (NBS, ROOR, heat) apply to ALL the substituents of the benzene simutaneously, like the reagent (K2Cr2O7)?

Well, in that case, its going to be a little complicated!! You will have a mixture of two products, one with both of the methyl brominated (major) while another (i think minor) would be only one brominated.

I think it would be pointless to use NBS as a reagent when it comes to real situation for a compound with multiple groups, because you will end up with a mixture and no organic chemist want a mixture, (its more work to purify it). IF I wanted only one methy brominated then, I would start with benzene and reach with ch3cl/alcl3 first then get toluene then use NBS to get benzylic bromine, now I can use Ch3cl/alcl3 again to get the para subsituted methylated!

Someone correct me if I am wrong!

 

[busupshot83]

Well, in that case, its going to be a little complicated!! You will have a mixture of two products, one with both of the methyl brominated (major) while another (i think minor) would be only one brominated.

I think it would be pointless to use NBS as a reagent when it comes to real situation for a compound with multiple groups, because you will end up with a mixture and no organic chemist want a mixture, (its more work to purify it). IF I wanted only one methy brominated then, I would start with benzene and reach with ch3cl/alcl3 first then get toluene then use NBS to get benzylic bromine, now I can use Ch3cl/alcl3 again to get the para subsituted methylated!

Someone correct me if I am wrong!

Thanks buddy, that helps a lot. These roadmaps sure are helpful.

 

[poc91nc]

Well, in that case, its going to be a little complicated!! You will have a mixture of two products, one with both of the methyl brominated (major) while another (i think minor) would be only one brominated.

I think it would be pointless to use NBS as a reagent when it comes to real situation for a compound with multiple groups, because you will end up with a mixture and no organic chemist want a mixture, (its more work to purify it). IF I wanted only one methy brominated then, I would start with benzene and reach with ch3cl/alcl3 first then get toluene then use NBS to get benzylic bromine, now I can use Ch3cl/alcl3 again to get the para subsituted methylated!

Someone correct me if I am wrong!

Careful with that second alkylation! you have Benzylbromine....lot's of lone pairs!!! Very nucleophillic...AlCl3 is a lewis acid!!! I'm thinking no reaction

 

[poc91nc]

I mean AlBr3...

 

[Lonely Sol]

Careful with that second alkylation! you have Benzylbromine....lot's of lone pairs!!! Very nucleophillic...AlCl3 is a lewis acid!!! I'm thinking no reaction

Oh k, I was trying to make an example that you can create a compound alternate ways!!

I guess i am no organic chemist!😀

 

[poc91nc]

Look at the mechanism...why is AlCl3/FeCl3 etc make a good catalyst? Electrophillic aromatic substitution...how do we generate the electrophile?

 

[skyisblue]

Yeah, you would get an attack on both methyl grps. For example, if you have 1,4-dimethylbenzene and add 2NBS/(PhCO2)2, you'd get 1,4-bromomethylbenzene (nomenclature is probably off here.)

I don't know if the attack changes if the positions of the methyl groups changes....,but NBS bromination of an alkylbenzene is at the benzylic position of the side chains...so I'm thinking any benzylic position. there may be minor and major products like lonely has stated, but not sure....we need an orgo master here to clarify the details.....hmmm where is streetwolf?

 

[AK28]

Well the alkane you are talkin about is attached to the benzene, which makes a big difference, the K2Cr2O7 is a very stong oxidizing agent, it will replace any benzyl carbons into COOH, even if there was 2 chains attached, then both chains would be COOH!!

Hope it helps!

I know that KMnO4 needs benzyllic hydrogens in order to do side chain oxidation- does K2Cr2O7 need that as well or will it turn a t-butyl group into COOH?

 

[Likkriue]

I know that KMnO4 needs benzyllic hydrogens in order to do side chain oxidation- does K2Cr2O7 need that as well or will it turn a t-butyl group into COOH?

It will need it as well. Also this thread is almost 9 years old. So the original folks are long gone.