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(i) How many equivalents of base did you add to the reaction mixture, relative to meso-stilbene dibromide? Use your actual experimental amount and show calculations including units, or course.
(ii) What product outcome would you expect if you added exactly 1.5 equivalents of base, relative to the dibromide compound?
(iii) After the reaction mixture was cooled to room temperature, what was the purpose of adding water? Describe how and why this process works in 3 sentences or less.

Respuesta :

Answer:

The question is not whole. It requires relevant date with it.

Explanation:

As I said, the question is incomplete. I will however try to explain as much as I could with experimental supposed amount.

i) Molecular mass of meso-stilbene dibromide = 340g

  Molecular mass of Potassium hydroxide = 56g

As per the equation, one mole of Stilbene dibromide requires two mole of potassium hydroxide to have one mole of diphenylacetylene and two mole of Potassium bromide along with two moles of water.

Now, for experimental purpose, 0.05 moles of stilbene dibromide will be enough for 0.1 moles of Potassium hydroxide (KOH). Therefore, we will require

Amount of meso-stilbene dibromide = 340 x 0.05 = 17g

Amount of KOH = 0.1 x 112 = 11.2g

Experimental amount recovered after experiment = 4g

Experimental molar amount recovered = 4 / 178 = 0.022 moles.

ii) There would have been less amount of product recovered if more base had been added.

iii) Adding cold water dropwise while stirring solution with the glass rod will solublize KBr (by-product) which is water soluble. Solid diphenylacetylene will separate as it does not dissolve.

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