Draw the structures of trans-stilbene (trans-2-diphenylethene) and meso-stilbene dibromide (1,2-dibromo-1,2-diphenylethane), the product formed when trans-stilbene adds Br2.

Draw the structures of trans-stilbene (trans-2-diphenylethene) and meso-stilbene dibromide (1,2-dibromo-1,2-diphenylethane), the product formed when trans-stilbene adds Br2. Take care to show the detail of the stereochemistry of the dibromide (no stereochemistry, no marks). Calculate the Theoretical Yield of the reaction by completing the Reagent Table below: Write the molecular formula in the standardized Hill System Order – the elements and their counts should be in the following form: carbon first, then hydrogen, then other elements (if present) in alphabetical order – e.g. CxHyBr2 Name Reactant Mass used (g) Molecular formula Molecular weight (g/mol) Moles used Molar ratio/Stoichiometry Limiting reagent (yes or no) Theoretical yield (moles) Theoretical yield (g) Reactant trans-stilbene 2.70g C14H12 180.25 0.015 1:1 No 0.015 2.70 Reactant Br2 2 Br2 159.81 0.015 1:1 No 0.015 4.80 Product meso-stilbene dibromide C14H10Br2 334.14 0.015 1:1 Yes 0.015 5.01 Calculate approximately the volume of the 10% w/v solution of bromine in glacial acetic acid that would have been required to consume all the 2.7g of the trans-stilbene? For each of the following, draw a circle around the correct answer: A pure sample of a compound is expected to have a higher melting point range than an impure sample of the same compound. The Rf of a compound is the distance traveled by the solvent front divided by the distance traveled by the solute spot. On a silica plate with a non-polar mobile phase, the Rr of 1,2-dibromo-1,2-diphenylethane is expected to be lower than that of trans-stilbene because the: true false true false true false

The Correct Answer and Explanation is:

1. Structures of trans-Stilbene and Meso-Stilbene Dibromide:

• Trans-Stilbene (Trans-2-diphenylethene): The structure of trans-stilbene involves two phenyl groups (C6H5) attached to a double-bonded ethene (C=C) group. In the trans form, the two phenyl groups are opposite to each other across the double bond. Structure: iniCopyEdit Ph-CH=CH-Ph Where Ph represents a phenyl group (C6H5). The two phenyl groups are oriented in opposite directions.
• Meso-Stilbene Dibromide (1,2-dibromo-1,2-diphenylethane): The addition of Br2 to trans-stilbene leads to a dibromide product where the bromines add across the double bond, creating a cyclic intermediate. The stereochemistry of the dibromide is meso, meaning it has two identical halves and is optically inactive due to internal symmetry. Structure: scssCopyEdit Ph-CH(Br)-CH(Br)-Ph The bromines add to opposite sides of the original double bond in a trans fashion, producing the meso stereoisomer with two chiral centers, but with overall symmetry, which makes it achiral.

2. Reagent Table Calculation:

Given Data:

• Trans-Stilbene: 2.70g
• Br2: 2.70g
• Molecular weight of trans-stilbene (C14H12): 180.25 g/mol
• Molecular weight of Br2: 159.81 g/mol
• Molecular weight of meso-stilbene dibromide (C14H10Br2): 334.14 g/mol

Step-by-step calculation:

1. Moles of Trans-Stilbene: Moles of trans-stilbene=MassMolar Mass=2.70g180.25 g/mol=0.015 mol\text{Moles of trans-stilbene} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.70g}{180.25 \, \text{g/mol}} = 0.015 \, \text{mol}Moles of trans-stilbene=Molar MassMass​=180.25g/mol2.70g​=0.015mol
2. Moles of Br2: Moles of Br2=MassMolar Mass=2.70g159.81 g/mol=0.0169 mol\text{Moles of Br2} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.70g}{159.81 \, \text{g/mol}} = 0.0169 \, \text{mol}Moles of Br2=Molar MassMass​=159.81g/mol2.70g​=0.0169mol Since the molar ratio between trans-stilbene and Br2 is 1:1 in this reaction, trans-stilbene is the limiting reagent.
3. Theoretical Yield (moles) of meso-stilbene dibromide:
   Since the reaction is 1:1, the theoretical yield in moles is the same as the moles of trans-stilbene. Theoretical yield (moles)=0.015 mol\text{Theoretical yield (moles)} = 0.015 \, \text{mol}Theoretical yield (moles)=0.015mol
4. Theoretical Yield (grams) of meso-stilbene dibromide: Theoretical yield (grams)=Moles×Molar Mass=0.015 mol×334.14 g/mol=5.01 g\text{Theoretical yield (grams)} = \text{Moles} \times \text{Molar Mass} = 0.015 \, \text{mol} \times 334.14 \, \text{g/mol} = 5.01 \, \text{g}Theoretical yield (grams)=Moles×Molar Mass=0.015mol×334.14g/mol=5.01g

3. Volume of 10% w/v Bromine Solution:

To calculate the volume of a 10% w/v (weight/volume) solution of bromine required to react with 2.7g of trans-stilbene, we need to use the following:

Given:

• 10% w/v solution of bromine means 10g of Br2 in 100 mL solution.
• Moles of trans-stilbene = 0.015 mol.
• Molar ratio of trans-stilbene to Br2 = 1:1.

Since moles of Br2 needed = moles of trans-stilbene = 0.015 mol, we can calculate the grams of Br2 required:Grams of Br2 needed=0.015 mol×159.81 g/mol=2.40 g\text{Grams of Br2 needed} = 0.015 \, \text{mol} \times 159.81 \, \text{g/mol} = 2.40 \, \text{g}Grams of Br2 needed=0.015mol×159.81g/mol=2.40g

Now, to find the volume of the 10% w/v solution required to provide 2.40g of Br2:Volume (mL)=Mass of Br2Concentration (g/mL)=2.40 g(10 g/100 mL)=24 mL\text{Volume (mL)} = \frac{\text{Mass of Br2}}{\text{Concentration (g/mL)}} = \frac{2.40 \, \text{g}}{(10 \, \text{g}/100 \, \text{mL})} = 24 \, \text{mL}Volume (mL)=Concentration (g/mL)Mass of Br2​=(10g/100mL)2.40g​=24mL

So, 24 mL of the 10% w/v solution is required.

4. True/False Questions:

• A pure sample of a compound is expected to have a higher melting point range than an impure sample:
  True. Impurities disrupt the orderly arrangement of molecules in a solid, lowering the melting point and broadening the melting range.
• The Rf of a compound is the distance traveled by the solvent front divided by the distance traveled by the solute spot:
  False. The Rf value is the ratio of the distance traveled by the solute spot to the distance traveled by the solvent front, not the other way around.
• On a silica plate with a non-polar mobile phase, the Rf of 1,2-dibromo-1,2-diphenylethane is expected to be lower than that of trans-stilbene because the:
  True. In chromatography, with a non-polar mobile phase, more polar compounds (like 1,2-dibromo-1,2-diphenylethane) tend to interact more strongly with the polar silica stationary phase and move slower, resulting in a lower Rf.