The point was to generate diphenylacetylene from trans-stilbene with pyridinium

Question

The point was to generate diphenylacetylene from trans-stilbene with pyridinium hydrobromide.

I added 2 g of trans-stilbene and 4 g of pyridinium hydrobromide (plus some other steps) to create stilbene dibrmomide.

I dissolved 1.5 g of potassium hydroxide in 20 ml of ethylene glycol, and the generated stilbene dibromide. (some other steps), then I collected the solid through suction filtration.

I purified the solid, and yielded .97 g.

I need to determine the actual yield and

overall percentage yield. Can you guys help me, and include balanced equations, and clear reaction equation math?

Explanation / Answer

trans stilbene - C14H12 , diphenyl acetlyne - C14 H10

C14H12 + 2 C5H6BrN --------> C14H12Br2   ,

moles of trans stilbene = 2/ 180.25 = 0.0111,

moles of pyridinium hydrobromide = ( 4/ 160) = 0.025,

moles of stilbene dibromide produced = 0.0111, ( from stochiometry in eq 1:1)

moles of KOH = 1.5/56.1 = 0.02674,

C14H12Br2 + 2KOH ---------> C14H10 (diphenyl acetylne) + other produicts

moles of C14H14Br2 reacted = 0.0111,

moles of diphenyl acetylene formed = 0.0111,

mass of product = 0.0111 x 178 = 1.9758 gm = theoretical yiled ,

actual yiled = 0.97 gm

percent yield = ( 0.97/1.9758) x100 = 49.1 %

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