Dehydration and Gc Lab Report

Introduction In an E1 reaction, where E stands for emptying and 1 stands for unibulwarkecular. The breaking of the C-LV truss is completed before any reaction occurs surrounded by the base to lose a hydrogen and form the carbon-carbon double bond 1. When the more substituted alkene is the dominant intersection, the reaction follows Zaitsevs rule. Zaitsevs rule states that the major product of a ? -elimination reaction is the most stable alkene 1. Acid-Catalyzed Dehydration is the elimination of a molecule of weewee from adjacent carbon atoms.An alcohol smoke be converted to an alkene by dehydration, which is often brought on by heating the alcohol with either 85% phosphoric acid or concent treadd sulfuric acid 1. The objective of this look into is to dehydrate 3-methyl-3-pentanol to get hold the product mixture of i several(prenominal)ric alkenes 3-methyl-2pentene and 2-ethyl-1-butene. Then use the ordnance chromato interprety to damp the product mixture and analyze the comp osition 2. pic pic pic Figure1 Table of Reagents Name molecular Weight Density Melting Point Boiling Point 2-methylcyclohexnol 114. 19 g/mol 0. 921 g/cm3 -9. 5oC 165oC Phosphoric Acid 98 g/mol 1. 88 g/cm3 42. 35oC 158oC Calcium Chloride 110. 98 g/mol 2. 15 g/cm3 772oC 1,935oC Fig The chart shows the reagents use in the research laboratory experiment and information regarding the solvents. Experimental The experiment started by gathering the supplies for distillment. In the vial, 2mL of 2-methylcyclohexanol and 1mL of phosphoric acid is added. The solvents be thoroughly mixed and a fewer boiling chips is added to help with the boil. The vertebral column bath is set up with the thermometer and the sand will be heated up to 100oC and the vial containing the solvents will be placed on the sand and let to boil. The process should take almost 30 minutes to start boiling.A beaker with a flask will be covered in ice and the water would be sucked out with the Pasteur pipette. When t he distillation process is completed, the distillated liquid would be saved. CaCl2 would be added to the liquid to prevent drying up and allows distillate to dry over drying agent. Then the little container will be weighed empty, and then the container with the liquid will be weight. Results ****** I cant figure out how to insert our graph. When I copy it my estimator says it is as well big to paste in a word document. So Im going to pretend the graph is in this spot, and put the results of the graph under here.Hopefully some peerless else in the group can use their computer to insert the graph or we can print it off and the graph will just have its own page. A gas chromatogram is a plot of a response against the safekeeping time. Chemical substances as gases ar retained on the liquid column (stationary phase) with a flow gas being the carrier (mobile phase) through the column. At the end of the column is some means of detection. The peak height or the peak sector is used to qu antitate the amount of substance. The number of peaks can also help determine wh ethyl ether or not one is dealing with a pure compound. poster one has a retention time of . 29 seconds and a peak area of 999. 00. Peak two was found to have a retention time of . 37 seconds and a peak area of 5067. 00. The percent composition for peak one is 16. 46884% and the percent composition for peak two is 83. 53116%. The percent yield is given up by the ratio of the experimental yield to the theoretical yield, which was found to be 52%. Discussion All possible products were observed. Under kinetic control, a significant yield of 3-methlycyclohexane is expected when 2-methylcyclohexane is dehydrated.The kinetic product is known as Hofman product. The mechanism of this dehydration involves the formation of a tertiary carbocation intermediate. The observed products do support E1 mechanism. E1 mechanism indicates an elimination, unimolecularreaction, where rate = k R-LG. This implies that the rate determining measuring of the mechanism depends on the decomposition of a single molecular species. The nerve tract involves two critical steps, which are the loss of the leaving group to draw acarbocation intermediate, thenthe loss of a proton from the carbocation to form a pi-bond.In this experiment the slow step in this elimination is the loss of a water molecule (the leaving group) from the oxonium ion to form the carbocation intermediate. This unimolecular rate-determining step makes this an E1 mechanism. The percent composition of a compound is a relative measure of the mass of each distinct element present in the compound. It gives the composition of the sample that was injected into the gas chromatograph, which in the long run helps determine alkene stability. In an error analysis of this experiment, peak outcome is critical.In a good chromatographic judicial separation, the components of the sample are completely disjunct from each other in the chromatogram. Unfortu nately, an incomplete separation of the components in a sample gives poor peak resultant role and there is an overlap among adjacent peaks in the chromatogram. When the peaks in a chromatogram are poorly resolved, it is needed to adjust one or more of the separation parameters until baseline resolution is obtained. Reference 1 Brown, William Henry. Organic Chemistry. Belmont, CA Brooks/Cole Cengage Learning, 2009. Print. Page 149-157 2 Hill, Richard K. , and rump Barbaro. Experiments in Organic Chemistry. Raleigh, NC Contemporary Pub. of Raleigh, 2005. Print. Page E8-13 to E8-15 Questions (T8-9) 1)When peaks in the gas chromatograms are poorly separated, it is best to change one or more separation parameters so the baseline resolution is obtained. 2)Benzene o-xylene p-xylene toluene 3)Cyclohexyl methyl ether would have a shorter retention time because of the increase in temperature. 4)A. guardianship will increase when the temperature column is decreasedB. When you increase the length of a column, the retention time would increase because of the longer distance traveled. C. Increasing flow rate of carrier gas will decrease the retention time 5)105oC 6) secondary to no partitioning of components in the sample will occur and hence giving poor to no separation. 7)29mm2, 210mm2, 136mm2 Mole %=(Area under individual Peaks)/(Total area under all the peaks) x 100% %=29mm/375mm x100%=7. 73% %=210mm/375mm x100%=56% %=136mm/375mm x100%=36. 27% Total area under peak= 29+210+136= 375mm