Morality of HCI: 12M (mol/L) Number of mole of HCI used: = 15mL x 1L/1000mL x 12mol/1L= 0.18 mole HC1 wwwm Molecular Weight of (CH3)3COH: 74.12 g/Mol Number of mole of (CH3)3COH(tert-butanol) used: = 4.0g x 1mol/ 74.12g = 0.0539 mole (CH3)3COH Based on the above calculations, we know the limiting reagent is tert-butanol Based on our balanced equations for the reaction, we have a ration of 1:1 mole to find our theoretical yield for our product: (CH3)3COH(aq) + HCI (aq) — (CH3)3CC1 (aq) + H20 (1) Molecular weight of (CH3)3CCI: 92.75% g/mol Theoretical Yield of (CH3)3CCI (tert-butyl chloride): = 0.0539 mol (CH3)3COH x 1 mole (CH3)3CC1/ 1mole (CH3)3COH x 92.75g(CH3)3CCI/1mol of (CH3)3CCI=4.99g| Percent Yield: Percent Yield = (experimental value/ theoretical yield) x 100% = (1.89g/4.99g) x 100% = 37.88%
Morality of HCI: 12M (mol/L) Number of mole of HCI used: = 15mL x 1L/1000mL x 12mol/1L= 0.18 mole HC1 wwwm Molecular Weight of (CH3)3COH: 74.12 g/Mol Number of mole of (CH3)3COH(tert-butanol) used: = 4.0g x 1mol/ 74.12g = 0.0539 mole (CH3)3COH Based on the above calculations, we know the limiting reagent is tert-butanol Based on our balanced equations for the reaction, we have a ration of 1:1 mole to find our theoretical yield for our product: (CH3)3COH(aq) + HCI (aq) — (CH3)3CC1 (aq) + H20 (1) Molecular weight of (CH3)3CCI: 92.75% g/mol Theoretical Yield of (CH3)3CCI (tert-butyl chloride): = 0.0539 mol (CH3)3COH x 1 mole (CH3)3CC1/ 1mole (CH3)3COH x 92.75g(CH3)3CCI/1mol of (CH3)3CCI=4.99g| Percent Yield: Percent Yield = (experimental value/ theoretical yield) x 100% = (1.89g/4.99g) x 100% = 37.88%
Morality of HCI: 12M (mol/L) Number of mole of HCI used: = 15mL x 1L/1000mL x 12mol/1L= 0.18 mole HC1 wwwm Molecular Weight of (CH3)3COH: 74.12 g/Mol Number of mole of (CH3)3COH(tert-butanol) used: = 4.0g x 1mol/ 74.12g = 0.0539 mole (CH3)3COH Based on the above calculations, we know the limiting reagent is tert-butanol Based on our balanced equations for the reaction, we have a ration of 1:1 mole to find our theoretical yield for our product: (CH3)3COH(aq) + HCI (aq) — (CH3)3CC1 (aq) + H20 (1) Molecular weight of (CH3)3CCI: 92.75% g/mol Theoretical Yield of (CH3)3CCI (tert-butyl chloride): = 0.0539 mol (CH3)3COH x 1 mole (CH3)3CC1/ 1mole (CH3)3COH x 92.75g(CH3)3CCI/1mol of (CH3)3CCI=4.99g| Percent Yield: Percent Yield = (experimental value/ theoretical yield) x 100% = (1.89g/4.99g) x 100% = 37.88%
In the reaction observed in part III, where is the slow step? do you expect the reaction in part III to be faster for tert-butyl chloride or 2-chloropropane and why?
Part III Silver Nitrate Test for Tertiary Alkyl Halides
Set up the reaction.
Obtain three 13 100 test tubes and label 13.
Pipet 10 drops of distilled water into Test Tube 1.
Pipet 10 drops of tert-butyl chloride into Test Tube 2
Pipet 10 drops of your product into Test Tube 3.
Add approximately 1 mL of the 1% ethanolic silver nitrate solution to each test tube. Swirl each test tube to mix the contents. The appearance of a white precipitate indicates the presence of a 3° halide. Record your observations in the data table. Note: Avoid getting the AgNO3 solution on your skin.
At the end of the experiment, discard the solutions as directed by your instructor.
IMAGES PROVIDED
Transcribed Image Text:Part III Silver Nitrate Test for Alkyl Halides
Test Tube
1
2
3
Compound
Water
Tert-Butyl Chloride
Product
Observations
No change
Cloudy, White
Precipitation
Transcribed Image Text:Theoretical Yield:
Morality of HCI: 12M (mol/L)
Number of mole of HCI used: = 15mL x 1L/1000mL x 12mol/1L= 0.18 mole HC1
Molecular Weight of (CH3)3COH: 74.12 g/Mol
Number of mole of (CH3)3COH(tert-butanol) used: = 4.0g x 1mol/ 74.12g = 0.0539 mole (CH3)3COH
Based on the above calculations, we know the limiting reagent is tert-butanol
Based on our balanced equations for the reaction, we have a ration of 1:1 mole to find our theoretical
yield for our product:
(CH3)3COH(aq) + HCI (aq) — (CH3)3CC1 (aq) + H20 (1)
Molecular weight of (CH3)3CCI: 92.75% g/mol
Theoretical Yield of (CH3)3CCI (tert-butyl chloride):
= 0.0539 mol (CH3)3COH x 1 mole (CH3)3CC1/ 1mole (CH3)3COH x 92.75g(CH3)3CCI/1mol of
(CH3)3CCI= 4.99g|
Percent Yield:
Percent Yield = (experimental value/ theoretical yield) x 100%
= (1.89g/4.99g) x 100%
= 37.88%
Definition Definition Organic compounds in which one or more hydrogen atom in an alkane is replaced by a halogen atom (fluorine, chlorine, bromine, or iodine). These are also known as haloalkanes.
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