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 question 1, your calculation is almost correct. Let's go through the steps again:
1. Write the balanced equation for the dehydration of 2-butanol.
C4H10O -> C4H8 + H2O
2. Calculate the number of moles of 2-butanol.
mols = mass / molar mass = 0.081g / 74.12 g/mol = 0.001094 mol
3. The balanced equation tells us that the molar ratio between 2-butanol and butene is 1:1. So, the number of moles of butene produced is also 0.001094 mol.
4. Convert the moles of butene to liters at STP.
1 mol of any gas occupies 22.4 L at STP.
volume = moles x 22.4 L/mol = 0.001094 mol x 22.4 L/mol = 0.0245 L or 24.5 mL
So, the maximum volume (at STP) of the butene mixture obtained by the dehydration of 81 mg of 2-butanol is approximately 0.0245 L or 24.5 mL.
For question 2, the reaction of 2-methylpropene with dilute sulfuric acid at room temperature results in the formation of a new substance. This reaction is an example of hydration, where the double bond in 2-methylpropene is converted to an alcohol functional group (-OH) through the addition of a water molecule (H2O). The specific product formed is 2-methylpropan-2-ol (also known as tert-butyl alcohol or t-butanol).
So, the new substance formed when 2-methylpropene is bubbled into dilute sulfuric acid at room temperature is 2-methylpropan-2-ol (t-butanol).