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?
3 answers
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.