Need help with Hofmann Reactions please

[Swenis]

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Hi fine ladies and gents!

I've had a really busy semester so far so I apologize for being dormant 😳 I'm about to finish up Organic Chem 1 and I just completed the chapter on eliminations, but one thing I really don't understand is Hofmann reactions. My book only has like two paragraphs on it, and all that I know is that it involves a quaternary nitrogen and that the major product is the least substituted alkene. However, that's all that I know and I'm having trouble doing Hofmann reactions on my homework. Could anyone help me with an overview? I haven't had much luck elsewhere.

 

[fenil2005]

hi,
Actually our professor didn't finish the chapter yet, but i will tell you whatever he taught us till now. In hoffmann reaction, you have bulky base which abstract a less hindered hydrogen. In, Zaitsev's product the base is small in size or not bulky, therefore it can abstract the more hindered hydrogen as a result. The best way is write down a reaction which gives both type of products and draw the structure showing all the bonds(c-H) you will see in the Hoffmann product the carbon from where the H is abstracted is less hindered compare to the other one. Its hard to explain this way but i hope this helps you. Its all about the size of the base and the steric hinderence.

 

[beccala33]

Hi fine ladies and gents!

I've had a really busy semester so far so I apologize for being dormant 😳 I'm about to finish up Organic Chem 1 and I just completed the chapter on eliminations, but one thing I really don't understand is Hofmann reactions. My book only has like two paragraphs on it, and all that I know is that it involves a quaternary nitrogen and that the major product is the least substituted alkene. However, that's all that I know and I'm having trouble doing Hofmann reactions on my homework. Could anyone help me with an overview? I haven't had much luck elsewhere.

I can try and help you...
Hoffmann product is the least highly substituted product. The bulkier the base that you are using (ie. tertiary or quartinary C), the smaller the saytzeff product and the more hoffmann product. Here's how it works... let's say that you have a molecule like 2-bromo-2-methyl-butane. The carbon at the 1 position has 3 hydrogens on it, the carbon at the 2 position has a methyl group and a bromine on it, and the carbon at the 3 position has 2 hydrogens on it and is a secondary carbon. The saytzeff product would be if the bromine and one of the hydrogens on the 3rd carbon (more highly substituted carbon, secondary) "came off" (elimination, they don't really come off, the proton is abstracted but you catch my drift) and the 2nd and 3rd carbon became double bonded **sometimes it helps if you draw these** Now the hoffman product is when the bromine and the hydrogen on the 1st carbon "come off" (the 1st carbon is the carbon adjacent to the bromine carbon that is the least highly substituted, primary) and the 1st and 2nd carbons become double bonded.
an example of bases that can be used for more of each product are:
for a greater hoffman yield - OC(CH3)3 BULKY
for a greater saytzeff yield - OCH2CH3

 

[Swenis]

Ok, thanks so far everyone! If anyone would please expand on that, what would be the major Hofmann product(s) for this reaction? (see attachment 🙂 )

 

[beccala33]

Ok, thanks so far everyone! If anyone would please expand on that, what would be the major Hofmann product(s) for this reaction? (see attachment 🙂 )

Ok, since NH2 is a strong base (poor leaving group) it doesn't readily undergo elimination. What happens is that first you need to treat it with something like 3CH3-I to get the 2 Hydrogens that are on the N off and instead the N bonds with (CH3)3. Now N(CH3)3 is a good leaving group because N now has 4 bonds (which it doesn't like to have) so it becomes a better leaving group and the carbon adjacent to the nitrogen double bonds with the carbon adjacent to IT. Your product would be 2-methyl-2pentENE. (a 5 carbon chain with a double bond at the second carbon and a methyl group also attached to the second carbon)

 

[Swenis]

Ok, since NH2 is a strong base (poor leaving group) it doesn't readily undergo elimination. What happens is that first you need to treat it with something like 3CH3-I to get the 2 Hydrogens that are on the N off and instead the N bonds with (CH3)3. Now N(CH3)3 is a good leaving group because N now has 4 bonds (which it doesn't like to have) so it becomes a better leaving group and the carbon adjacent to the nitrogen double bonds with the carbon adjacent to IT. Your product would be 2-methyl-2pentENE. (a 5 carbon chain with a double bond at the second carbon and a methyl group also attached to the second carbon)

Thanks becca! So to see if I understand it, from the next one, I would more or less take off the NH2 like it was a leaving group and bring a hydrogen from the carbon of the phenyl group to make a double bond like this? Ph-=?

Sorry about the lack of nomenclature. I've forgotten some of it since the last exam 🙂

 

[beccala33]

Thanks becca! So to see if I understand it, from the next one, I would more or less take off the NH2 like it was a leaving group and bring a hydrogen from the carbon of the phenyl group to make a double bond like this? Ph-=?

Sorry about the lack of nomenclature. I've forgotten some of it since the last exam 🙂

That's right, the Ph bonded to a C double bonded to a C...
Make sure you remember that the first step is that that the NH2 becomes N(CH3)3 and that's why it leaves with the hydrogen, but your answer is exactly right!