Alkyl Halide: A Nucleophilic Substitution Reaction
Synthesis of an Alkyl Halide: A Nucleophilic Substitution Reaction
Unknown Letter: B
23 April 2013
Academic Integrity Statement:
“Experimental data may be collected with other students in organic chemistry labs. However I understand that sharing information required for a lab report (including but not limited to word processing or spreadsheet files, calculations, graphs, conclusions and additional problems at the end of the lab report) with other students is a violation of the University Academic Integrity
Code.” __________________________________
(Signature)
Purpose:
The purpose of the experiment was to perform a synthesis of an alkyl halide by a nucleophilic substitution reaction. The alkyl halide, along with the starting …show more content…
alcohol, was determined through NMR spectroscopy, gas chromatography, and IR spectroscopy.
Chemical Equation:
Experimental Procedure: In a 100 mL round bottom flask, containing a Teflon stir bar, 17.123 grams of sodium bromide, 17 mL of distilled water, and 8.6 mL of unknown alcohol were combined.
Through the top of the condenser, 14 mL of concentrated sulfuric acid was added drop wise over a period of ten minutes. The addition of the acid created a change in color of the solution to a rustic orange color. The mixture was refluxed for 30 minutes and the solution changed to a faded yellow-orange color. Once reflux was complete, a simple distillation was set up, with the 25 mL collection graduated cylinder placed in an ice bath. The mixture was distilled and combined with the hood partners’ distilled mixture. The distillate was placed in a separatory funnel with 10 mL of distilled water. Once separated, the distillate was added to the funnel again with 9 M sulfuric acid. The organic layer was still in the funnel after separation, as distilled water was added to the solution and then 10 mL of a saturated solution of sodium bicarbonate. The organic solution was dried with anhydrous sodium sulfate for a week. The liquid product was transferred to a 25 mL round bottom flask equipped with a stir bar and a simple distillation was set up. The solution was distilled and the density of the bromide was calculated at 1.222 grams/mL with a boiling range of 87-92◦C. The theoretical yield was calculated at 12.847 grams and the percent yield was 72.3 percent. Next was the characterization and assessment of the purity of the …show more content…
alkyl bromide using IR spectroscopy, NMR spectroscopy, and gas chromatography.
References: Mohrig, J. R.; Morrill, T. C.; Hammond, C. N.; and Neckers, D. C.; Experimental
Organic Chemistry; W. H. Freeman and Company: New York, NY, 1997; pp. 76-81.
Gilbert, J. C.; and Martin S. F. Experimental Organic Chemistry, 2nd Edition; Saunders College
Publishing, Harcourt Brace College Publishers, 1998; pp. 383-385.
Results:
Calculation of Theoretical Yield: 2-Butanol alcohol: 8.6 mL * 0.808 g/mL = 6.9488 grams 6.9488 g/74.12 grams of 2-Butanol = 0.0938 moles 2-Butanol 17.123 grams of NaBr / 102.9 g/mol of NaBr = 0.166 moles of NaBr Alcohol is the limiting reagent. 1 moles of alcohol / 1 mole of 2-bromobutane = 0.0938 moles 2-bromobutane 0.0938 moles 2-bromobutane * 137.03 grams/moles = 12.847 grams of 2-bromobutane and theoretical yield.
Calculation of Percent Yield: 9.287 grams actually obtained / 12.847 grams theoretical yield = 0.7229 0.7229 * 100 = 72.3 % yield
Density of Alkyl Halide: Grams of distillate in graduated cylinder/ volume of distillate in mL = 9.287 grams / 7.60 mL = 1.222 g/mL of alkyl halide
Conclusion:
The experiment performed was able to successfully identify a starting alcohol and alkyl halide using reflux, distillation, washing, and separatory funnel. Due to tests, such as gas chromatography, IR spectroscopy, boiling point, and NMR spectroscopy, being conducted, results for the experiment showed 2-bromobutane as the identified alkyl halide.
IR Spectroscopy- the IR spectroscopy test was run in order to compare the functional groups of the starting alcohol along with the alkyl halide synthesized.
The starting alcohol had the below functional groups appear on the IR spectrometer.
O-H – Alcohol | 3400, broad | C-H- Alkenes | 2900, strong and sharp | C-H – t butyl group | 1450, sharp, medium | C-O- Acids | 1000, sharp, medium |
In comparison to the functional groups of the starting alcohol, which showed evidence of a t-butyl group, the functional groups of the alkyl halide were examined. C-H – Alkenes | 2900, strong and sharp | C-H – Alkyl Halides | 1200, sharp and medium | C=O- Carbonyl | 1700, sharp and low |
Both the alcohol and the alkyl halide contained evidence of alkenes, in almost the exact same form. The alcohol had hydrogen bonds that were not represented in the alkyl halide product. The alcohol and the alkyl halide contained various carbon-hydrogen bonds but differed in the functional group represented. The IR spectrums of the two products were similar in bonds but differed in placement, concluding that one was an alcohol and the other was a bromide that formed.
1 H NMR Spectroscopy- the NMR spectroscopy was used to determine the chemical shifts, splitting patterns, and the integration (number of hydrogen atoms) for the starting alcohol as well as the alkyl halide.
The NMR table for the starting alcohol was analyzed first.
Chemical Shift | 3.75 shift
1.5 shift
1.3 shift
0.8 shift | Splitting Pattern | 3.75 proton- quartet split
1.5 proton- triplet to singlet
1.3 proton- doublet
0.8 proton- triplet | Integration | 10 Hydrogen |
The NMR for the alkyl halide was analyzed below. Chemical Shifts | 1.0 shift of second Carbon
4.1 shift on the proton carbon to bromine
1.6 and 1.8 shifts protons close to bromine-shifted down | Splitting Pattern | 4.0 protons – split into a quartet
1.8 protons-split into quartet and doublet
1.6 protons- split into doublet
1.0 protons – split into triplet | Integration | 9 Hydrogen | There were similar shifts in some areas, such as the shifts at 0.8 ppm and 1.0 ppm and both the alcohol and the alkyl halide had shifts far down field near the 4.0 ppm range representing methylene hydrogens. Both represented methyl hydrogens upfield near the 1.21 ppm range as well. 2-Butanol contains methyl and is often referred to as 2-methyl-2-propanol.
Gas Chromatography- gas chromatography was used to determine the purity of the sample. The standards and the sample from the experiment were placed in the table below with the retention times to determine the identity and purity of the alkyl halide. 1-bromopropane | 1.541 | 2-bromopropane | 1.498 | 1-bromobutane | 1.713 | 2-bromobutane | 1.635 | 1-bromopentane | 2.071 | Experiment alkyl halide | 1.636 |
The gas chromatography for the alkyl halide in the experiment was almost exact with the 2-bromobutane.
Since the closeness of two, the sample could be considered fairly pure. The area percentage of 1.636 retention time was listed as 95 percent, with only a little less than 5 percent of the area being impure.
Physical Properties- the physical properties were not the main source for determining the identity of the alkyl halide and starting alcohol because of the usual inaccuracy, however the results for boiling point and density proved close to 2-bromobutane and 2-Butanol. The boiling range from the week one experiment was 76-118 ◦C, which was too large of a range and the alkyl halide had not been completely purified. However, the boiling point for 2-butanol alcohol was about 83 ◦C, which lies within the range of the starting boiling range.
The boiling range for the product was 87-92 ◦C which closely related to boiling point of 2-bromobutane at 91 ◦C. The density was calculated to 1.222 g/mL with 2-bromobutane being 1.255 g/mL. The density of the product was not significant in identifying the halide because it was too closely related with 1-bromopentane as well. Physical properties are helpful in determining the identity of products but are never significant enough to confirm the identity of the
product.
The results from the NMR spectroscopy, IR spectroscopy, and gas chromatography lead to confirmation of 2-bromobutane as the alkyl halide produced during the substitution reaction.
Unknown Letter: B
23 April 2013
Academic Integrity Statement:
“Experimental data may be collected with other students in organic chemistry labs. However I understand that sharing information required for a lab report (including but not limited to word processing or spreadsheet files, calculations, graphs, conclusions and additional problems at the end of the lab report) with other students is a violation of the University Academic Integrity
Code.” __________________________________
(Signature)
Purpose:
The purpose of the experiment was to perform a synthesis of an alkyl halide by a nucleophilic substitution reaction. The alkyl halide, along with the starting …show more content…
alcohol, was determined through NMR spectroscopy, gas chromatography, and IR spectroscopy.
Chemical Equation:
Experimental Procedure: In a 100 mL round bottom flask, containing a Teflon stir bar, 17.123 grams of sodium bromide, 17 mL of distilled water, and 8.6 mL of unknown alcohol were combined.
Through the top of the condenser, 14 mL of concentrated sulfuric acid was added drop wise over a period of ten minutes. The addition of the acid created a change in color of the solution to a rustic orange color. The mixture was refluxed for 30 minutes and the solution changed to a faded yellow-orange color. Once reflux was complete, a simple distillation was set up, with the 25 mL collection graduated cylinder placed in an ice bath. The mixture was distilled and combined with the hood partners’ distilled mixture. The distillate was placed in a separatory funnel with 10 mL of distilled water. Once separated, the distillate was added to the funnel again with 9 M sulfuric acid. The organic layer was still in the funnel after separation, as distilled water was added to the solution and then 10 mL of a saturated solution of sodium bicarbonate. The organic solution was dried with anhydrous sodium sulfate for a week. The liquid product was transferred to a 25 mL round bottom flask equipped with a stir bar and a simple distillation was set up. The solution was distilled and the density of the bromide was calculated at 1.222 grams/mL with a boiling range of 87-92◦C. The theoretical yield was calculated at 12.847 grams and the percent yield was 72.3 percent. Next was the characterization and assessment of the purity of the …show more content…
alkyl bromide using IR spectroscopy, NMR spectroscopy, and gas chromatography.
References: Mohrig, J. R.; Morrill, T. C.; Hammond, C. N.; and Neckers, D. C.; Experimental
Organic Chemistry; W. H. Freeman and Company: New York, NY, 1997; pp. 76-81.
Gilbert, J. C.; and Martin S. F. Experimental Organic Chemistry, 2nd Edition; Saunders College
Publishing, Harcourt Brace College Publishers, 1998; pp. 383-385.
Results:
Calculation of Theoretical Yield: 2-Butanol alcohol: 8.6 mL * 0.808 g/mL = 6.9488 grams 6.9488 g/74.12 grams of 2-Butanol = 0.0938 moles 2-Butanol 17.123 grams of NaBr / 102.9 g/mol of NaBr = 0.166 moles of NaBr Alcohol is the limiting reagent. 1 moles of alcohol / 1 mole of 2-bromobutane = 0.0938 moles 2-bromobutane 0.0938 moles 2-bromobutane * 137.03 grams/moles = 12.847 grams of 2-bromobutane and theoretical yield.
Calculation of Percent Yield: 9.287 grams actually obtained / 12.847 grams theoretical yield = 0.7229 0.7229 * 100 = 72.3 % yield
Density of Alkyl Halide: Grams of distillate in graduated cylinder/ volume of distillate in mL = 9.287 grams / 7.60 mL = 1.222 g/mL of alkyl halide
Conclusion:
The experiment performed was able to successfully identify a starting alcohol and alkyl halide using reflux, distillation, washing, and separatory funnel. Due to tests, such as gas chromatography, IR spectroscopy, boiling point, and NMR spectroscopy, being conducted, results for the experiment showed 2-bromobutane as the identified alkyl halide.
IR Spectroscopy- the IR spectroscopy test was run in order to compare the functional groups of the starting alcohol along with the alkyl halide synthesized.
The starting alcohol had the below functional groups appear on the IR spectrometer.
O-H – Alcohol | 3400, broad | C-H- Alkenes | 2900, strong and sharp | C-H – t butyl group | 1450, sharp, medium | C-O- Acids | 1000, sharp, medium |
In comparison to the functional groups of the starting alcohol, which showed evidence of a t-butyl group, the functional groups of the alkyl halide were examined. C-H – Alkenes | 2900, strong and sharp | C-H – Alkyl Halides | 1200, sharp and medium | C=O- Carbonyl | 1700, sharp and low |
Both the alcohol and the alkyl halide contained evidence of alkenes, in almost the exact same form. The alcohol had hydrogen bonds that were not represented in the alkyl halide product. The alcohol and the alkyl halide contained various carbon-hydrogen bonds but differed in the functional group represented. The IR spectrums of the two products were similar in bonds but differed in placement, concluding that one was an alcohol and the other was a bromide that formed.
1 H NMR Spectroscopy- the NMR spectroscopy was used to determine the chemical shifts, splitting patterns, and the integration (number of hydrogen atoms) for the starting alcohol as well as the alkyl halide.
The NMR table for the starting alcohol was analyzed first.
Chemical Shift | 3.75 shift
1.5 shift
1.3 shift
0.8 shift | Splitting Pattern | 3.75 proton- quartet split
1.5 proton- triplet to singlet
1.3 proton- doublet
0.8 proton- triplet | Integration | 10 Hydrogen |
The NMR for the alkyl halide was analyzed below. Chemical Shifts | 1.0 shift of second Carbon
4.1 shift on the proton carbon to bromine
1.6 and 1.8 shifts protons close to bromine-shifted down | Splitting Pattern | 4.0 protons – split into a quartet
1.8 protons-split into quartet and doublet
1.6 protons- split into doublet
1.0 protons – split into triplet | Integration | 9 Hydrogen | There were similar shifts in some areas, such as the shifts at 0.8 ppm and 1.0 ppm and both the alcohol and the alkyl halide had shifts far down field near the 4.0 ppm range representing methylene hydrogens. Both represented methyl hydrogens upfield near the 1.21 ppm range as well. 2-Butanol contains methyl and is often referred to as 2-methyl-2-propanol.
Gas Chromatography- gas chromatography was used to determine the purity of the sample. The standards and the sample from the experiment were placed in the table below with the retention times to determine the identity and purity of the alkyl halide. 1-bromopropane | 1.541 | 2-bromopropane | 1.498 | 1-bromobutane | 1.713 | 2-bromobutane | 1.635 | 1-bromopentane | 2.071 | Experiment alkyl halide | 1.636 |
The gas chromatography for the alkyl halide in the experiment was almost exact with the 2-bromobutane.
Since the closeness of two, the sample could be considered fairly pure. The area percentage of 1.636 retention time was listed as 95 percent, with only a little less than 5 percent of the area being impure.
Physical Properties- the physical properties were not the main source for determining the identity of the alkyl halide and starting alcohol because of the usual inaccuracy, however the results for boiling point and density proved close to 2-bromobutane and 2-Butanol. The boiling range from the week one experiment was 76-118 ◦C, which was too large of a range and the alkyl halide had not been completely purified. However, the boiling point for 2-butanol alcohol was about 83 ◦C, which lies within the range of the starting boiling range.
The boiling range for the product was 87-92 ◦C which closely related to boiling point of 2-bromobutane at 91 ◦C. The density was calculated to 1.222 g/mL with 2-bromobutane being 1.255 g/mL. The density of the product was not significant in identifying the halide because it was too closely related with 1-bromopentane as well. Physical properties are helpful in determining the identity of products but are never significant enough to confirm the identity of the
product.
The results from the NMR spectroscopy, IR spectroscopy, and gas chromatography lead to confirmation of 2-bromobutane as the alkyl halide produced during the substitution reaction.