Why does 2-butene react differently than 1-butene with hydrogen chloride gas?

[walker]

Yeah I'm pretty much stumped on this one...

Basically I'm supposed to explain the following:

When 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarly two products are usually found.

My best guess is the difference in position of the C=C bond between the 2-butene and 1-butene alkenes. But I have no idea how to explain it. If anyone wants to give me a hint or something that would be great.

Thanks

 

[so-crates]

Would that be cis- or or trans- 2-butene ?

 

[walker]

I thought it might be, but I don't think the number of isomers has anything to do with the number of products 2-butene creates when it is combined with hydrogen chlroide. Plus it is already stated that 2-butene only forms one product (2-chlorobutane) when it is combined with hydrogen chloride. So I really am stumped on this one. I believe that 1-butene might also form 2-chlorobutane but I'm not 100% sure and I have no idea what other product it would form... frustrating!

 

[so-crates]

OK, a little hint : what intermediate is formed after the first step in the reaction for each reaction ?

 

[walker]

2-butene

CH3CH=CHCH3 + HCl -> CH3CH2-CHCH3+Cl

1-butene

CH2=CHCH2CH3 +HCl -> CH3-CHCH2CH3+Cl

is that correct?

 

[chem_tr]

Well, 2-butene has two same carbanion intermediate, either $H_3C-CH^--CH^+-CH_3$ or $H_3C-CH^+-CH^--CH_3$. In each case, the result does not change; the product will contain $Cl^-$ where the $CH^+$ lies, and $H^+$ at the position of $CH^-$.

The case for 1-butene is different. Here, two kinds of carbanion may be produced; the one with higher yield is $H_2C^--CH^+-CH_2-CH_3$ and the other, low-yield-one is $H_2C^+-CH^--CH_2-CH_3$. The products after reacting with HCl may be written by referring to the paragraph above. Do you have an idea why I wrote "high-yield" and "low-yield" for these two compounds? If so, you have understood the phenomenon, aka "stable carbocation".

 

[ShawnD]

I wish I understood what chem_tr was talking about

Basically I'm supposed to explain the following:

When 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarly two products are usually found.

Each addition reaction has 2 possibilities

2-butene CH3-CH=CH-CH3
It's important to notice that if you draw it like a gimp it looks symmetrical. In one case, the chloride goes to carbon 2 and the hydrogen goes to carbon 3. In the other case, chloride goes to carbon 3 and hydrogen goes to carbon 2. Just rotate those products and you'll see they are the same thing, so that's why there's only 1 product. Both are 2-chlorobutane.

1-butene CH2=CH-CH2-CH3
In one case, chloride goes to carbon 1 and hydrogen goes to carbon 2; 1-chlorobutane. In the other case, chloride goes to carbon 2 and hydrogen goes to carbon 1; 2-chlorobutane.
Following markovnikov's rule, the 2-chlorobutane will be the major product and 1-chlorobutane will be the minor product.

 

[walker]

Alright that makes sense. So can it be said that the location of the C=C bond results in 1-butene creating two products and 2-butene only creating one?

What you said makes perfect sense. I'm just trying to figure out a way of applying it within the scope of the lesson material.

 

[Gokul43201]

Alright that makes sense. So can it be said that the location of the C=C bond results in 1-butene creating two products and 2-butene only creating one?

Yes, it can.

 

[so-crates]

The mechanism is bimolecular in its rate limiting step. It occurs when the pi electrons of the C=C bond are attracted to the polarized hydrogen atom. The intermediate I was referring to was the carbocation formed when the R-C=CH2 bond is protnated to become $$CH_3-CH_2-CH^+-CH_3$$. As ShawnD noted, this follows Markonikov's rule because it will be the more stable carbocation (secondary as opposed to primary)

Now the Cl- can attack the trigonal planar(flat) carbocation intermediate. It can attack from either direction, which yields a pair of enantiomers 2R chlorobutane and 2S chlorobutane. 1-chlorobutane should be a very minor product.

chem_tr: this reaction does not involve carbanions

Edit: now, actually that I think about it, you get a pair of enantiomers with the 2-butene as well, so the question has not been very well though out! It must be the 1-chlorobutane minor product as others have mentioned.