1. What is the maximum volume (at STP) of the butene mixture that could be obtained by the dehydration of 81 mg of 2-butanol?
2. When 2-methylpropene is bubbled into dilute sulfuric acid at room temperature, it appears to dissolve. What new substance has been formed?
Responses
* O Chem - DrBob222, Monday, January 5, 2009 at 8:52pm
" 1. Write the equation for the dehydration of 2-butanol and balance it. You need not worry about where the double bond goes, just the empirical formula is all you need.
2. mols = 0.081g/molar mass 2-butanol.
3. Using the coefficients in the balanced equation, (I think it will be 1:1), convert mols 2-butanol to the product.
4. Convert mols to liters remembering that 1 mol occupies 22.4 L at STP. I note the problem talks about a "mixture of products" but you can only work this problem for ONE product unless you know the percentages of the various products.
For 2. I suspect that you will hydrate the double bond. Look in your book to find if this follows the Markovnikov's rule or not. "
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For 1 is it just 81 mg/ (74.12 g/mol x 100) = 1.09 mol x 22.4 L= -->0.024L ???
For 2 is it just gunna be 2-methylpropane instead of propene because the double bond is taken off?
For 1 is it just 81 mg/ (74.12 g/mol x 100) = 1.09 mol x 22.4 L= -->0.024L ???
You're answer is correct. Your math is lousy. 81 mg x (1 g/1000 mg) x (1 mol/74.12 g) = 0.00109 mol butene (or butenes if it is a mixture) and that times 22.4 L/mol = 0.02448 which rounds to 0.024 L).
For 2 is it just gunna be 2-methylpropane instead of propene because the double bond is taken off?
I'm just not sure of this so take the following with a grain a salt since I'm not an organic chemist and my organic chemistry is about 60 years old. At any rate, the best reference I can find states that concentrated cold H2SO4 with an alkene produces the alkyl hydrogen sulfate which means that H and OSO3H add across the double bond and it does follow M's rule which means the H adds to the terminal C and the OSO3H adds to the middle C. I don't know if room T is cold H2SO4 or not. However, I also found in the same reference that if the alkyl hydrogen sulfate is heated, with water, (which dilute H2SO4 should have in it) then it forms the alcohol (which is what I proposed first). The product would be 2 propanol. But two things bother me.
1. The reference shows that 98%H2SO4 works on ethene, 80% H2SO4 works on propene and 63% H2SO4 works on isobutylene (isn't that 2-methylpropene?) to form the alkyl hydrogen sulfate and it must be heated to form the alcohol.
2. Dr Russ answered your first post, also, and suggested the formation of alkyl hydrogen sulfate although he left it as a question and not a response. Dr. Russ has answered other organic chemistry questions; I'm sure he knows more about org chem that I.
At any rate, the decision is yours, based on all of this information, but I would go with the formation of the alkyl hydrogen sulfate as a final product. The reference I looked in goes to great length talking about the solubility of the sulfate in H2SO4 and how easily it dissolves. The only question in my mind is if the alkyl hydrogen sulfate will hydrolyze to 2-propanol. Based on Dr Russ post plus the fact that my reference shows 63% H2SO4 doing the trick with isobutylene then I would go with the alkyl hydrogen sulfate as the final product. I hope Dr Russ reads this and will comment.
Dr Russ apparently did not read this. About 12 hours after the last post I made, I called a friend of mine who IS an organic chemist and he thinks I was right the first time. He thinks the alkyl hydrogen sulfate is formed first BUT that the reaction will generate enough heat to hydrolyze (in dilute H2SO4) the intermediate sulfate to 2-methyl-2-propanol. He thinks it very unlikely that the reaction will (or can) stop at the alkyl hydrogen sulfate intermediate stage. I hope this helps. Also note that in my response above I inadvertently wrote 2-propanol instead of 2-methyl-2-propanol (four lines from the end) AS WELL AS the line just before I start the reasons I was confused (1, 2, 3).
alright thank you so much for your help!
1. Well, you've got the right idea, but let's calculate it together in a slightly different way. First, we need to find the number of moles of 2-butanol. So it's 81 mg / (74.12 g/mol x 1000 mg/g) = 0.00109 mol. Now, according to the balanced equation, the molar ratio between 2-butanol and butene is 1:1. Therefore, we have 0.00109 mol of butene. And finally, we can convert this to liters at STP: 0.00109 mol x 22.4 L/mol = 0.0244 L. So, the maximum volume of the butene mixture is approximately 0.0244 liters.
2. Ah, the mysterious disappearance of 2-methylpropene. Well, when it dissolves in dilute sulfuric acid at room temperature, a new substance is formed called 2-methylpropane. It's like 2-methylpropene, but without the double bond. So, it went from being a propene to a propane. It's like a magician turned the double bond into thin air. Poof!
For question 1:
To find the maximum volume of the butene mixture obtained by the dehydration of 81 mg of 2-butanol, you need to follow these steps:
1. Write the balanced equation for the dehydration of 2-butanol: C4H10O -> C4H8 + H2O
2. Determine the number of moles of 2-butanol:
mols = mass / molar mass = 0.081 g / 74.12 g/mol = 0.00109 mol
3. Since the balanced equation shows a 1:1 ratio between 2-butanol and the product (butene), the number of moles of butene formed is also 0.00109 mol.
4. Convert the number of moles of butene to volume at STP:
1 mol of gas occupies 22.4 L at STP, so the volume of butene produced is: volume = 0.00109 mol x 22.4 L/mol = 0.0244 L
Therefore, the maximum volume (at STP) of the butene mixture obtained by the dehydration of 81 mg of 2-butanol is approximately 0.0244 L.
For question 2:
When 2-methylpropene is bubbled into dilute sulfuric acid at room temperature, it forms a new substance called 2-methylpropan-2-ol (also known as tert-butyl alcohol or t-butanol).
The reaction involves the hydration of the double bond in 2-methylpropene, which follows the Markovnikov's rule. The H+ ion from the sulfuric acid adds to the carbon atom with fewer hydrogen atoms attached, resulting in the formation of the alcohol compound 2-methylpropan-2-ol.
So, the new substance formed when 2-methylpropene reacts with dilute sulfuric acid is 2-methylpropan-2-ol.