Dehydration Of Cyclohexanol A. Dehydration Reaction: Write A Balanced Chemical Equation For Dehydration Of Cyclohexanol: .OH H+ +H2O Percent Yield Calculations: 0.685 Amount Of Cyclohexanol Used: 1.5 ML = 8 Moles Of Cyclohexanol Used – 0.006% MoL Moles Of Cyclohexene Expected (Theoretical Yield): Amount Of Cyclohexene Isolated (Isolated Yield):
The Correct Answer and Explanation is:
A. Balanced Chemical Equation for the Dehydration of Cyclohexanol
The acid-catalyzed dehydration of cyclohexanol to form cyclohexene is an elimination (E1) reaction that involves the removal of a water molecule: C6H11OH→H+, ΔC6H10+H2O\text{C}_6\text{H}_{11}\text{OH} \xrightarrow{\text{H}^+, \ \Delta} \text{C}_6\text{H}_{10} + \text{H}_2\text{O}
Balanced Chemical Equation: C6H11OH→H+, heatC6H10+H2O\text{C}_6\text{H}_{11}\text{OH} \xrightarrow{\text{H}^+,\ \text{heat}} \text{C}_6\text{H}_{10} + \text{H}_2\text{O}
Percent Yield Calculations
Given:
- Amount of cyclohexanol used = 1.5 mL
- Density of cyclohexanol ≈ 0.962 g/mL
- Molar mass of cyclohexanol = 100.16 g/mol
Step 1: Convert volume to mass 1.5 mL×0.962 g/mL=1.443 g1.5\ \text{mL} \times 0.962\ \text{g/mL} = 1.443\ \text{g}
Step 2: Convert mass to moles 1.443 g100.16 g/mol=0.0144 mol cyclohexanol\frac{1.443\ \text{g}}{100.16\ \text{g/mol}} = 0.0144\ \text{mol cyclohexanol}
Since the dehydration is 1:1:
- Theoretical yield of cyclohexene = 0.0144 mol
Step 3: Use the provided percent yield (0.685% or 0.00685 as a decimal) to calculate actual yield: Moles of cyclohexene (actual)=0.0144 mol×0.00685=9.864×10−5 mol\text{Moles of cyclohexene (actual)} = 0.0144\ \text{mol} \times 0.00685 = 9.864 \times 10^{-5}\ \text{mol}
Step 4: Convert moles of cyclohexene to mass (molar mass = 82.15 g/mol): 9.864×10−5 mol×82.15 g/mol=0.0081 g (≈ 8.1 mg)9.864 \times 10^{-5}\ \text{mol} \times 82.15\ \text{g/mol} = 0.0081\ \text{g (≈ 8.1 mg)}
Explanation
The dehydration of cyclohexanol to cyclohexene is a classic organic reaction used to demonstrate elimination mechanisms and alkene synthesis. This reaction typically occurs in the presence of a strong acid catalyst (such as sulfuric or phosphoric acid) and requires heat to proceed. Under acidic conditions, the hydroxyl group (–OH) of cyclohexanol is protonated to become a better leaving group (water). This leads to the formation of a carbocation intermediate, which then loses a proton to form the double bond of cyclohexene.
In this lab scenario, 1.5 mL of cyclohexanol was used. After converting to mass and then to moles, we find approximately 0.0144 mol of cyclohexanol were used. Since the reaction is 1:1, the theoretical yield of cyclohexene is also 0.0144 mol.
The percent yield is calculated based on the actual amount of cyclohexene isolated versus the theoretical maximum. In this case, the percent yield is extremely low (0.685%), which might be due to several factors: incomplete reaction, side reactions, poor separation or distillation, or loss during transfer and purification steps.
The actual amount of cyclohexene isolated was approximately 8.1 mg (0.0081 g), which aligns with a 0.685% yield. Such a low yield suggests that experimental conditions may need improvement or that significant product loss occurred.
Understanding and optimizing this reaction is important for organic synthesis, especially when scaling up or improving efficiency in industrial or academic settings.A. Balanced Chemical Equation for the Dehydration of Cyclohexanol
The acid-catalyzed dehydration of cyclohexanol to form cyclohexene is an elimination (E1) reaction that involves the removal of a water molecule: C6H11OH→H+, ΔC6H10+H2O\text{C}_6\text{H}_{11}\text{OH} \xrightarrow{\text{H}^+, \ \Delta} \text{C}_6\text{H}_{10} + \text{H}_2\text{O}
Balanced Chemical Equation: C6H11OH→H+, heatC6H10+H2O\text{C}_6\text{H}_{11}\text{OH} \xrightarrow{\text{H}^+,\ \text{heat}} \text{C}_6\text{H}_{10} + \text{H}_2\text{O}
Percent Yield Calculations
Given:
- Amount of cyclohexanol used = 1.5 mL
- Density of cyclohexanol ≈ 0.962 g/mL
- Molar mass of cyclohexanol = 100.16 g/mol
Step 1: Convert volume to mass 1.5 mL×0.962 g/mL=1.443 g1.5\ \text{mL} \times 0.962\ \text{g/mL} = 1.443\ \text{g}
Step 2: Convert mass to moles 1.443 g100.16 g/mol=0.0144 mol cyclohexanol\frac{1.443\ \text{g}}{100.16\ \text{g/mol}} = 0.0144\ \text{mol cyclohexanol}
Since the dehydration is 1:1:
- Theoretical yield of cyclohexene = 0.0144 mol
Step 3: Use the provided percent yield (0.685% or 0.00685 as a decimal) to calculate actual yield: Moles of cyclohexene (actual)=0.0144 mol×0.00685=9.864×10−5 mol\text{Moles of cyclohexene (actual)} = 0.0144\ \text{mol} \times 0.00685 = 9.864 \times 10^{-5}\ \text{mol}
Step 4: Convert moles of cyclohexene to mass (molar mass = 82.15 g/mol): 9.864×10−5 mol×82.15 g/mol=0.0081 g (≈ 8.1 mg)9.864 \times 10^{-5}\ \text{mol} \times 82.15\ \text{g/mol} = 0.0081\ \text{g (≈ 8.1 mg)}
