Hi, a few days ago i took an exam and there was an exercise that required to do a scale of stability of the alkenes. 

I had to do an E2 with NH3 on the 2-metyl-3R-Cl-4S-F pentane. I obtaneid these two. Which is more stable and why?

Sorry i forgot the F on the second one.

The attachment  is broken. Can not open.

5 hours ago, chenbeier said:

The attachment  is broken. Can not open.

What attachment?

9 hours ago, vv3rtigo said:

Hi, a few days ago i took an exam and there was an exercise that required to do a scale of stability of the alkenes. 

I had to do an E2 with NH3 on the 2-metyl-3R-Cl-4S-F pentane. I obtaneid these two. Which is more stable and why?

Sorry i forgot the F on the second one.

What does "do an E2 with NH3 on the 2-metyl-3R-Cl-4S-F pentane"mean? E2 I assume means an elimination reaction, somehow involving ammonia, but what are the substituents on pentane and what were the 2 products? There seems to be a .jpg file attached which we can't see. 

If you use a few more words, it may help to clarify what this is about.  

1 hour ago, exchemist said:

What does "do an E2 with NH3 on the 2-metyl-3R-Cl-4S-F pentane"mean? E2 I assume means an elimination reaction, somehow involving ammonia, but what are the substituents on pentane and what were the 2 products? There seems to be a .jpg file attached which we can't see. 

If you use a few more words, it may help to clarify what this is about.  

Yes it's an elimination reaction (E2) with NH3 as a solvent. The substrate is: 2-methyl-3-Chloro-4-Fluoropentane.

The products are two alkenes (i put them in the photo but I don't why you can't see it). One of the alkenes has the the double bond on the C2-C3. The other one on the C3-C4. 

My professor wanted to know which product is more stable.

2 hours ago, vv3rtigo said:

What attachment?

Your photo, I cannot open. 404 file not found on unbuntuu

Edited September 17, 20223 yr by chenbeier

9 minutes ago, chenbeier said:

Your photo, I cannot open. 404 file not found on unbuntuu

 it's an elimination reaction (E2) with NH3 as a solvent. The substrate is: 2-methyl-3-Chloro-4-Fluoropentane.

The products are two alkenes (i put them in the photo but I don't why you can't see it). One of the alkenes has the the double bond on the C2-C3. The other one on the C3-C4. 

My professor wanted to know which product is more stable.

30 minutes ago, vv3rtigo said:

Yes it's an elimination reaction (E2) with NH3 as a solvent. The substrate is: 2-methyl-3-Chloro-4-Fluoropentane.

The products are two alkenes (i put them in the photo but I don't why you can't see it). One of the alkenes has the the double bond on the C2-C3. The other one on the C3-C4. 

My professor wanted to know which product is more stable.

Ah, is this about the Zaitsev Rule?  

15 minutes ago, exchemist said:

Ah, is this about the Zaitsev Rule?  

I don't think so, we didn't study this rule.

34 minutes ago, vv3rtigo said:

I don't think so, we didn't study this rule.

OK but won't the Me group influence the proportions of 2-3 vs. 3-4 double bonds? What effect do you expect it to have? 

1 hour ago, vv3rtigo said:

 it's an elimination reaction (E2) with NH3 as a solvent. The substrate is: 2-methyl-3-Chloro-4-Fluoropentane.

The products are two alkenes (i put them in the photo but I don't why you can't see it). One of the alkenes has the the double bond on the C2-C3. The other one on the C3-C4. 

My professor wanted to know which product is more stable.

 

 

I don't see the chloro unit in your picture  ?

38 minutes ago, studiot said:

 

 

I don't see the chloro unit in your picture  ?

That's because it's been eliminated in the formation of the double bond. 

The first molecule will be built because after Elimination the Cl. The carbanion is stabilized through the methyl group.

39 minutes ago, chenbeier said:

The first molecule will be built because after Elimination the Cl. The carbanion is stabilized through the methyl group.

If it's E2, surely there is no charged intermediate, is there?

If it's E1, I think I'd expect Cl- to leave and the resulting carbocation to be stabilised by Me. Or am I getting mixed up? 

Edited September 17, 20223 yr by exchemist

You are right,if it is E2. I was thinking it is E1.

1 hour ago, chenbeier said:

You are right,if it is E2. I was thinking it is E1.

But with E1 it will be a carbocation, surely, which Me can stabilise as alkyl groups are slightly electron-donating? (I admit it's many years since I did this stuff.) 

17 hours ago, exchemist said:

But with E1 it will be a carbocation, surely, which Me can stabilise as alkyl groups are slightly electron-donating? (I admit it's many years since I did this stuff.) 

It's an E2. I wanted to know which alkene is more stable. Only this

1 minute ago, vv3rtigo said:

It's an E2. I wanted to know which alkene is more stable. Only this

And what do you think of the answers you have been given here? Do you now understand?

Edited September 18, 20223 yr by exchemist

On 9/18/2022 at 10:31 AM, exchemist said:

And what do you think of the answers you have been given here? Do you now understand?

So the one with the Fluoro isn't stable?

1 hour ago, vv3rtigo said:

So the one with the Fluoro isn't stable?

No, you need to take a more nuanced approach to this. Nobody is saying the 3-4 alkene is "not stable". It's more subtle than that.

What we've been saying, and what the Zaitsev (or Saytzeff) Rule is saying [suggest you look it up], is that these elimination reactions, where there is more than one possible product, tend to favour the option with the more highly alkyl-substituted double bond.  You will in general get a mixture of products containing both options, but there will be a higher yield of the one with the methyl substituted double bond.

The reason is that alkyl groups are slightly electron-donating. You may be aware that alkyl groups are ortho/para directing in aromatic substitutions for instance. (F, by contrast, will be somewhat electron withdrawing.) More highly alkyl-substituted alkenes have greater stability (stronger bonds) than unsubstituted ones, as a result. How this occurs is a bit complicated, involving something called hyperconjugation, which gets into MO theory: you may or may not be covering this kind of thing in your course.  

But the question of "stability" is also worth thinking through a bit further, from the point of view of kinetics vs. thermodynamics. Both products are, so far as I can see,  thermodynamically stable, albeit the Me-substituted one is of somewhat lower energy - the more stable of the two. So that answers the direct question you asked. But which product is favoured in the course of a synthesis reaction can be a question of kinetics as well as thermodynamics. The explanation of the Zaitsev or Saytzeff rule given in my old synthesis textbook is that the electron-donating (hyperconjugation) character of the alkyl group stabilises the transition state, thus lowering the activation energy and causing that product to form faster than the other one. This may be something to argue out with your prof. 

  

Edited September 19, 20223 yr by exchemist