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# theoretical yield for bromination of trans stilbene?

calculate the theoretical yield for the bromination of trans-stilbene, assuming an excess if Br2 and using the amount of trans-stilbene .5g.

2ml of Br2/CH2Cl2 was used.

how do I find theoretical yield??

### 4 Answers

- IggyLv 78 years agoFavorite Answer
Your theoretical yield is the yield you would obtain if the reaction resulted in a complete transformation of all reagents.

Each mole of Stilbene will react with one mole of Bromide [1]

MW of Stilbene is 180.25 g/mol and its density of trans Stilbene is 0.970 g/mL [1].

Therefore 2ml is 0.970 * 2 = 1.94g. This amount is 1.94 / 180.25 ~ 0.01 mole

MW or Bromine is 79.904 g/mol and its density is 3.1028 g / ml [2]

Therefore 2ml is 3.1028 * 2 = 6.2056 g which is 6.2056 / 79.904 ~ 0.07766 mol.

The lower amount (the trans Stilbene) is the limiting reagent.

Therefore we are looking at 0.01 mole of 1,2-dibromo-1,2-diphenylethane product.

The MW of 1,2-dibromo-1,2-diphenylethane = 180.25 - 2 + 79.904 = 258.154 g/mol

So 0.01 mole is 0.01 * 258.154 ~ 2.582 g which is the theoretical yield.

Enjoy :)

Source(s): [1] http://www.bc.edu/schools/cas/chemistry/undergrad/... [2] http://en.wikipedia.org/wiki/Bromine - 5 years ago
RE:

theoretical yield for bromination of trans stilbene?

calculate the theoretical yield for the bromination of trans-stilbene, assuming an excess if Br2 and using the amount of trans-stilbene .5g.

2ml of Br2/CH2Cl2 was used.

how do I find theoretical yield??

Source(s): theoretical yield bromination trans stilbene: https://trimurl.im/c38/theoretical-yield-for-bromi...