Chemistry 212 Fall Experiment 3: S N 1 Evaluation Summary

Transcription

1 Experiment 3: S N 1 Evaluation Summary Last Name: Atique Date: 11/05/2013 First Name: Anika Lab Day/TA/Group: Wednesday; TA: Patrick Julien; Group C Lab reports must be typed and chemical structures must be drawn with ChemDraw. Report must not exceed five pages (including this page). Page limit does not include any attached spectra or references. ** Deductions for hand written report/structures and exceeding page limit ** Report Breakdown Data/Results: /10 Discussion: /10 Report Total: /20 Other Lab Marks Performance: /6 Prelab Quiz: /3 TA Comments on Report/Performance Report assessment TA (name): TA (signature): Performance assessment TA (name): TA (signature): 1 of 5

2 Experiment 3: S N 1 Report Sheet Last Name: Atique Date:11/05/2013 First Name: Anika Lab Day/TA/Group: Wednesday; Patrick Julien, Group C Objective: (1 mark) (What is the purpose of this experiment?) To investigate the reaction mechanism via SN1 during the conversion of an alcohol (2,5-dimethyl- 2,5-hexanediol) to an alkyl chloride (2,5-dichloro-2,5-dimethylhexane) Introduction: (2 marks) (Brief description of the concept/reaction studied S N 1 reactions and why it is important) SN1 is a nucleophilic substitution reaction where the rate determining step is the formation of the carbocation via the cleavage of the carbon-leaving group. The reaction involves multiple steps, and is unimolecular, with only the concentration of electrophile determining the rate and not the nucleophile. SN1 is important because it provides a pathway for tertiary alcohol which have high steric hindrance to undergo substitution reaction. The high steric hindrance prevents it from going though an SN2 mechanism. In this experiment the OH group is protonated by HCl to make it a good leaving group H2O+. The rate determining step is the cleavage of this leaving group bond with carbon which results in the formation of a carbocation that is rapidly attacked by the nucleophile present in the vessel, Cl-, to form the product. Reaction Equation: (2 marks) (Equation and data [molar mass, concentration, density, volume, mass, moles, etc as appropriate including theoretical yield] for reactants and products) Materials Molar mass (g/mol) Quantity Density (g/ml) Concentrat ion Moles/ mmol 2,5-dimethyl g ,5-hexandiol HCl ml Molar ratio Theoretic al yield 2

3 2,5-dichloro- 2,5- methylhexane g Table 1. Results Yield and Physical Properties of S N 1 Product (2 marks) 2,5-Dichloro-2,5-Dimethylhexane Yield (g).24g Yield (%) 77.4% Appearance Melting Point White solid C Table 2. Results Solubility Tests (1 mark) 2,5-Dimethyl-2,5-hexandiol 2,5-Dichloro-2,5-Dimethylhexane Ethanol Dissolves Dissolves Hexanes Insoluble Dissolves Ethyl Acetate Dissolves Dissolves Figure 1. IR Spectrum (1 mark) Attach IR Spectrum of Your Starting Material (2,5-dimethyl-2,5-hexanediol - provided in lab manual) and Product (2,5-Dichloro-2,5-Dimethylhexane ) label important stretching frequencies Figure 2. TLC Plate (1 mark) Draw a diagram of your TLC plate below. Label spots/solvent front, distance travelled, and solvent used for mobile phase. Calculate R f values. 4.8cm 4.1cm Solvent used: hexanes:ethyl acetate (95:5) Rf value for SM is zero Rf value for the product is 4.1/4.8 =.84 Adobe Acrobat Document Adobe Acrobat Document 3 of 5 PCOO SM

4 Discussion Questions: 10 marks 1. S N 1 and E1 reactions both involved the initial formation of a carbocation intermediate. In this experiment, why is the S N 1 product favoured over the E1 product? In this experiment we use HCl which gives Cl- as the nucleophile. Cl- is a strong nucleophile and favors SN1 reaction. E1 requires a weak nucleophile. 2. Explain the observed difference in solubility between 2,5-dimethyl-2,5-hexandiol and 2,5- dichloro-2,5-dimethylhexane in ethanol. 2,5-dimethyl-2,5-hexandiol has hydrogen bonds between its molecules. Hexane is non polar, so for the alcohol to dissolve in hexane strong hydrogen bonds have to be broken and only weak vander waals forces and permanent dipole-induced-dipole intermolecular bonds are formed. Bond breaking cost energy and energy is released when bonds are formed, the energy needed to break hydrogen bonds is far greater than the energy released by van der waals and permanent dipole-induced-dipole interactions. This is not favorable, so hexandiol is insoluble in hexane. For the chloroalkane the intermolecular bonds being broken and the intermolecular bond formed with hexane are almost of the same strength wan require similar energies during formation and breakage, so solubility is favored. 3. Analyze your TLC plate: Did your reaction go to completion? Why does your product have a larger Rf value than the starting material? The reaction did go to completion because the co-spot shows only the spot for the final products, with no intermediate spots. The product is far less polar the starting material so it interaction more with the non- polar mobile phase than with the polar phase of silica on the TLC plate. This causes the product to travel further in the solvent and thus it has a larger Rf. 4. Analyze the MP and IR spectrum of your product: What evidence is present that supports the formation of 2,5-dichloro-2,5-dimethylhexane? Mp of the product is: C MP of 2,5-dichloro-2,5-dimethylhexane: C 4

5 The melting points are very close to each other and this shows the presence of 2,5-dichloro-2,5- dimethylhexane In the IR spectrum the OH peak is missing from the spectrum of the product this shows the absence of 2,5-dimethyl-2,5-hexandiol, as the alcohol was converted to 2,5-dichloro- 2,5- dimethylhexane. 5. Draw a detailed mechanism for the formation of 2,5-dichloro-2,5-dimethylhexane from 2,5-dimethyl-2,5-hexanediol. 5 of 5

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