2-butene reacts with hydrogen chloride gas

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When 2-butene reacts with hydrogen chloride (HCl), it produces a single product, 2-chlorobutane, due to the specific positioning of the double bond between the second and third carbons. In contrast, 1-butene reacts with HCl to yield two distinct products because the double bond is located between the first and second carbons, allowing for two possible arrangements of H and Cl attachment. The reaction follows Markovnikov's rule, which typically favors hydrogen attachment to the carbon with fewer alkyl substituents. The discussion highlights the importance of understanding the structure and reactivity of alkenes in addition reactions. Overall, the distinct outcomes of these reactions illustrate the influence of molecular structure on chemical behavior.
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When 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarly, two products are usually found. Explain this.

I know that when 2-butene reacts with hydrogen chloride that the bond breaks and the Cl2 adds on one atom to each carbon in a addition reaction (resulting in 2-chlorobutane). I know the double bond will be between carbons 1 and 2 now in 1-butene, but I don't know how, and why there is two products from this.

Can someone explain this please.
 
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I would draw out the products of both reactions and count the distinct products formed.

Also you say that Cl2 adds one atom to each carbon, but the problem says that HCl is reacting. That's probably just a typo though.
 
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No, what I wrote was correct in that it is reacting, although I meant to put HCl not Cl2. Below is what is shows in my textbook.


2-butene + HCl(g) ---> 2-chlorobutene

I tried making them in the window but looked weird, view the pics and it shows the structure of each, but I don't know what the products are for 1-butene + HCl(g) (suppose to have 2 products).
 

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Well while we wait for the attachments to be...attached let me say this much.

In 2-butene, the double bond is across the 2nd and 3rd carbon, right? (4 carbons total). That means in order to break the double bond, you need to attach something to each of the two carbons since in an organic compound carbon wants 4 bonds. What can you attach? You have H+ and Cl-. That means you can either have an H on the 2nd carbon, and Cl on the 3rd or vice versa.

Now draw out those two compounds. Are they two different compounds or two distinct ones?

Now repeat the process for 1-butene. Since the double bond is across the 1st and 2nd carbon, you need to attach something to each of those two carbons. Once again draw the two products and note if they are different or not.

Note: What I described is kind of a simplied version of what actually happens. Depending on the scope of your course, your instructor might want you to draw out the actual mechanism of the reaction--that is, the step-by-step reaction.

Note 2: In reality, Markovnikov's rule says that the hydrogen will usually, if not exclusively, attach to the carbon with the least alkyl substituents. I'm just saying this as a note for the future. It doesn't really apply to this problem because the problem itself assumed that two products are formed. I don't really know how much of each product is produced in reality for the 1-butene reaction.
 
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I already drew them and they did not line up properly when I posted my reponse, so I attached the pictures; and yes you can still click on them and they will show up, even though it says pending approval.
 
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