Solvent and Chloride Percent Yield
Title: Prep of t-Butyl Chloride via SN1 Reaction
Purpose:
The purpose of this experiment is to synthesize tert-butyl chloride via an SN1 reaction. t-Butyl Chloride was synthesized from t-Butyl Alcohol using hydrochloric acid in separatory funnel; isolation of t-Butyl Chloride was done under distillation conditions. The experiment resulted in 8.29grams of purified compound, which is a 66.27 percent yield.
Procedure:
As per handout with changes
Equation:
Mechanism:
Results: (Scan IR spectrum)
Compounds t-Butyl Chloride(Experimental) t-Butyl Chloride(Book Value)
Molecular Weight (g/mol) 92.5681 92.5681
Density (g/cm3) 0.842 0.842
Boiling Point (°C) 48 50
Refractive Index 1.3855 1.3857
Theoretical Yield:
Let A = t-Butyl Alcohol
Let C = t-Butyl Chloride
10.02gA x 1 mol A x 1 mol C x 92.57 g C = 12.51g C 74.12g A 1 mol A 1 mol C
Theoretical Yield = 12.51 grams t-Butyl Chloride
Actual Yield = 8.29 grams t-Butyl Chloride
Percent Yield = 8.29/12.51 x 100 = 66.27%
Discussion:
To produce t-Butyl Chloride from t-Butyl alcohol using hydrochloric acid, an SN1 type reaction is necessary. Optimal SN1 dehydration conditions require the reaction to take place in cool temperatures and in acidic conditions. SN1 type reactions have the fastest rate and the highest yield with tertiary alcohols. The first step of the SN1 type dehydration reaction is the protonation of the alcohol to form an alkyloxonium ion and free the halide ion to become the nucleophile. The alkyloxonium ion then loses the oxonium group and becomes the carbocation intermediate. Following its formation, the carbocation is captured by the halide and the resultant alkyl halide is formed. The limiting reagent in this reaction is the t-Butyl Alcohol that dictates the maximum possible product formation. As we can see the percent yield for my results was not so sufficient. A percent yield of 66.27% indicates that a lot of the product was lost. However I think when I was
Purpose:
The purpose of this experiment is to synthesize tert-butyl chloride via an SN1 reaction. t-Butyl Chloride was synthesized from t-Butyl Alcohol using hydrochloric acid in separatory funnel; isolation of t-Butyl Chloride was done under distillation conditions. The experiment resulted in 8.29grams of purified compound, which is a 66.27 percent yield.
Procedure:
As per handout with changes
Equation:
Mechanism:
Results: (Scan IR spectrum)
Compounds t-Butyl Chloride(Experimental) t-Butyl Chloride(Book Value)
Molecular Weight (g/mol) 92.5681 92.5681
Density (g/cm3) 0.842 0.842
Boiling Point (°C) 48 50
Refractive Index 1.3855 1.3857
Theoretical Yield:
Let A = t-Butyl Alcohol
Let C = t-Butyl Chloride
10.02gA x 1 mol A x 1 mol C x 92.57 g C = 12.51g C 74.12g A 1 mol A 1 mol C
Theoretical Yield = 12.51 grams t-Butyl Chloride
Actual Yield = 8.29 grams t-Butyl Chloride
Percent Yield = 8.29/12.51 x 100 = 66.27%
Discussion:
To produce t-Butyl Chloride from t-Butyl alcohol using hydrochloric acid, an SN1 type reaction is necessary. Optimal SN1 dehydration conditions require the reaction to take place in cool temperatures and in acidic conditions. SN1 type reactions have the fastest rate and the highest yield with tertiary alcohols. The first step of the SN1 type dehydration reaction is the protonation of the alcohol to form an alkyloxonium ion and free the halide ion to become the nucleophile. The alkyloxonium ion then loses the oxonium group and becomes the carbocation intermediate. Following its formation, the carbocation is captured by the halide and the resultant alkyl halide is formed. The limiting reagent in this reaction is the t-Butyl Alcohol that dictates the maximum possible product formation. As we can see the percent yield for my results was not so sufficient. A percent yield of 66.27% indicates that a lot of the product was lost. However I think when I was