Preparation of Alkyl Halides by Nucleophilic Aliphatic Substitution new PDF

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k~$paration of Ethyl Acetate and Butyl Acetate by Simple Distillation and Analysis of Fractions by Gas...

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Preparation of Alkyl Halides by Nucleophilic Aliphatic Substitution. NaI and AgNO3 Tests for Alkyl Halides Riyushi Mahadik March 20, 2014 Methods and Background

The objective of this lab is to prepare tertiary alkyl halide by SN1 reaction, primary alkyl halide by SN2 reaction. The other objectives are to isolate a liquid product by simple distillation, to properly use a serpartory funnel and to distinguish between tertiary, secondary and primary alkyl halides through NaI and AgNO3 tests and finally to calculate the percent yield of a isolated product through a synthesis reaction. As shown in Figure 1, nucleophilic aliphatic substitution reaction involves the conversion between different functional groups in which Nu: represents a nucleophile and L: represents a leaving group. Nucleophiles are either neutral or negatively charged. The nonbonding pairs of electrons on the nucleophile are donated to an electrophilic atom which results in the formation of a new covalent bond during the process of substitution reaction. This reaction is like a Lewis acid-base reaction in which the electrophilic carbon atom acts as a Lewis acid and nucleophile acts as a Lewis base. The leaving group may be neutral or negatively charged and must accept the pair of bonding electrons from the carbon atom when the C-L bond breaks. Thus, the rate of substitution reaction depends on the leaving ability of a particular group L: If the better the leaving group is better at leaving the substance, the faster the reaction will occur. Similarly, conjugate bases of strong acids are also known for good leaving groups whereas those of weak acids are poor leaving groups.

Figure 1: Nucleophilic Aliphatic Substitution Reaction

Figure 2: Summary of Reactions in this lab

As shown in Figure 2, nucleophilic substitution reaction is classified into two types based on their different mechanistic pathways: SN1 and SN2 reaction. SN1 reaction is an endothermic process in which the first step of SN1 reaction is much slower than the second step and involves heterolytic cleavage or ionization of the C-L bond to generate an unstable carbocation. The breakage of the C-L bond requires higher energy. The intermediate carbocation formed during the first step can undergo rearrangement to form a more stable carbocation, form an alkene through elimination reaction or react with the nucleophile to form a stable substitution product. The second step of SN1 reaction is relatively fast because it is an exothermic process which involves bond formation between the carbocation and the nucleophile. The first step of the SN1 reaction is the rate-determining step (rds) and the rate of the overall reaction depends only on the concentration of the substrate R-L which is completely independent of the concentration of the nucleophile. Thus, SN1 reaction is called a unimolecular reaction because its rate determining step involves only one species which is the substrate R-L. This concept is further explained by a rate equation Rate = k1[R-L] where k1 is the first order rate constant. Since the transition state in SN1 reaction involves the formation of a stable carbocation, these types of reactions favors compounds with a leaving group from tertiary-bonded carbons due to its stability, which is tertiary > secondary >>primary. According to Le Chatelier’s principle, if equilibrium is disturbed, then conditions in the reaction will change to keep it at equilibrium. Therefore, if the concentration of the reactants increases, the reaction will shift to the right resulting in the formation of the product.

Figure 3: SN1 versus E1 As shown in Figure 3, unimolecular elimination reaction, E1, competes with SN1 substitution reactions. The reaction depends on the nature of the nucleophile. Substitution can occur in the weakly basic conditions and also highly polarizable nucleophiles such as I-, Br-, Cl-, H2O, and Ch3CO2-. For elimination, strongly basic and only slightly polarizable nucleophiles condition should be met when RO-, H2N-, H-, and HO- are used. Moreover, larger nucleophiles tend to favor elimination because the hydrogen atom is more sterically available than is the carbon atom bearing the leaving group. In this experiment, tertiary alkyl halide, 2-chloro-2-methylbutane will be prepared through SN1 reaction by reacting 2-methyl-2-butanol with hydrochloric acid. In this reaction,–OH is a poor leaving group which is protonated by hydrochloric acid and oxonium ion is formed. Since H2O is a good leaving group, it results in the formation of a carbocation which at the end is attacked by the nucleophilic chloride ion. The below Sn1 reaction mechanism is shown in Figure 4.

Figure 4: SN1 Reaction Mechanism In SN2 reaction, primary alcohols reacting with a hydrogen halide H-X (X= Cl, Br, or I) to obtain primary alkyl halides. In compare to SN1 reaction, SN2 reaction consists of only one step which is also the rate-determining step and does not go through carbocation formation. In SN2, reaction, the nucleophile attacks the carbon atom on the opposite side of the leaving group. When the nucleophile attaches the carbon atom, the leaving group cleaves the carbon atom. The carbon atom takes bonding pair of electrons. The rate determining step of this reaction is dependent on the concentrations of both the substrate, R-L, and the nucleophile, Nu:, since this reaction is bimolecular reaction. This concept is explained by a rate equation Rate = k2[R-L][Nu:]where k2 is the second order rate constant. In these reactions, the nucleophile attacks the substrate which is directly behind the leaving group. Therefore, the more sterically hindered a carbon atom is, the less chances to undergo an SN2 reaction. Therefore, SN2 reactions favor compounds with a leaving group from primary-bonded carbons due to its stability, which is primary > secondary >> tertiary. Based on the stability of the transition state, leaving groups from secondary-bonded carbon, the reaction can proceed either pathway, SN1 or SN2. According to Le Chatelier’s principle, the reaction will be favored by increasing the concentration of the reactants. By increasing the amount of nucleophile, the concentration of the compound in the transition state increases which increases the rate of product formation. The SN2 reaction prefers leaving groups that are weaker bases, since they have greater leaving ability.

Figure 5: SN2 versus E2

As shown in Figure 5, SN2 reaction competes with bimolecular elimination reaction to produce alkenes. In this experiment, primary alkyl halide, 1-bromobutane will be prepared through SN2 reaction by reacting 1-butanol with sodium bromide and sulfuric acid. –OH is a poor leaving group which is protonated by sulfuric acid and thus oxonium ion is formed. Since H2O is a good leaving group, the nucleophilic chloride ion does a backside attack on the molecule. The reaction mechanism for S N2 is shown in Figure 6.

Figure 6: SN2 Reaction Mechanism In order, to verify whether the substitution product is primary, secondary, or tertiary alkyl halide, the sodium iodide (NaI) and the silver nitrate (AgNO3) test will be done. The reaction of silver nitrate and alky halide is an SN1 reaction because its rate-determining step involves only one molecule. Silver chloride is formed as one of the products, which precipitates in ethanol as shown in Figure 7. Formation of the precipitate determines whether or not a halide is present. The rates of precipitation of silver halide in this test increases with increasing carbocation stability and substitution (tertiary > secondary >> primary). Another method that distinguishes halides into primary, secondary, and tertiary halides is the sodium iodide test dissolved in acetone (Figure 8). In this test sodium bromide and chloride are not very soluble in acetone in compare to sodium iodide. The reaction of sodium iodide with alkyl halides is SN2 reaction in which iodide ion is the nucleophile. The order of reactivity of this reaction is primary > secondary >> tertiary. In this test, sodium bromide quickly forms a precipitate at room temperature, whereas primary and secondary chlorides must be heated to form precipitates. Secondary and tertiary alkyl halides also react at higher temperatures but tertiary chlorides does not react in relative time.

Figure 7: Silver Nitrate Test

Figure 8: Sodium Iodide Test Procedure: Synthesis of 1-Bromobutane: 11.11g of sodium bromide was transferred to a 100mL round bottom flask with 10 mL of water, 10 mL of 1-butanol, and 1-2 boiling stones. The contents of the flask were mixed by gently swirling it, and the flask was then placed in an ice-water bath. Then, 10 mL of sulfuric acid was added to the chilled reaction mixture slowly while swirling it. The reflux apparatus shown in Figure was set up. The flask was warmed gradually until heating under for 45 minutes. After 45 minutes, the water hoses from the Hempel column were removed and the water was drained from the outside of the column. Then, the apparatus for simple distillation technique was set up as show in figure to distill the product mixture until the distillate in the receiving flask was clear. The flask was kept in ice-water bath at all times. Finally, the distillate was then transferred to separatory funnel and washed with 10 mL of water. 1-bromobutane (organic layer) is the denser bottom layer. The two layers were separated and the organic layer was returned to the separatory funnel which was then sequentially washed with 8 mL of concentrated sulfuric acid, two x 5mL of 2 M NaOH, 10 mL of water, and 10 mL of saturated sodium chloride (brine). Since concentrated sulfuric acid is denser than 1-bromobutane, the organic layer was the top layer for this washing step. NaOH, water, and brine are less dense than 1-bromobutane, and so the organic layer was the bottom layer for these washing steps. 1-Bromobutane was then transferred to a clean flask and water was absorbed by using 1.5g of sodium sulfate. The dried 1-bromobutane was weighed and percent yield was calculated.

Figure 9: Heating under reflux apparatus

Figure 10: Simple Distillation Apparatus Synthesis of 2-Chloro-2-Methylbutane: In a separatory funnel, 10 mL of 2-chloro-2-methylbutanol and 25 mL of concentrated (12M) hydrochloric acid were placed and swirled gently without the stopper. After 1 min, stopper was placed on the separatory funnel and then inverted at times venting out the excess pressure. The stopcock was again closed, the funnel was shaken, and the mixture was allowed to separate into two layers. The bottom organic layer was sequentially washed with 10 mL portions of saturated aqueous sodium chloride and cold saturated aqueous sodium bicarbonate. When sodium bicarbonate was added, vigorous gas evolution which was controlled by gently swirling the funnel. The funnel was then stoppered, shaken vigorously, and vented frequently to release any gas pressure. The organic layer was separated and washed sequentially with 10 mL portions of

water and aqueous sodium chloride. The aqueous layer was carefully removed and 2-chloro-2methylbutanol was dried using 1.5 g of sodium sulfate. The final product was transferred to round bottom flask for simple distillation. The fraction was collected when the head temperature was greater than 75ºC (760 Torr) to a receiver place in an ice-water bath. For the tests, silver nitrate and sodium iodide tests were performed on the final products. A silver nitrate test was performed to verify the obtained alkyl halide. Three small test tubes were prepared with 5 drops of each of the primary, secondary, and tertiary alkyl halides were added to 2 mL of a 0.1 M solution of silver nitrate in 95% ethanol. Different colored precipitates should be formed that is silver chloride is white, silver bromide is pale yellow, and silver iodide is dark yellow. For sodium iodide test, the presence of alkyl chlorides and bromides was checked. Three small test tubes were prepared with 5 drops of each of the primary, secondary, and tertiary alkyl halides were added to 1 mL of the sodium iodide-acetone. Precipitate should form within 3 min at room temperature. Data Acquisition/Calculation: Observations: Products

AgNO3 Test

2-chloro-2methylbutane (3°)

Cloudy, white liquid at Yellow cloudy but became clear the top and white precipitate at the bottom of the test tube. No reaction (clear) No reaction(clear)

2-chlorobutane (2°) 1-bromobutane (1°)

Cloudy liquid, no precipitation

NaI Test

Partially cloudy liquid, white precipitate forms

Percent Yield The percent yield will help in determining the success of this experiment. The formula for calculating the percent yield as follows, Percent Yield = [(Actual Yield) / (Theoretical Yield)] x 100

Chemical Density Molar Mass Used

Table 1: Calculating % Yield for Sn2 Reaction 1-butanol 1-bromobutane 3 0.81 g/cm 1.27 g/cm3 74.12 g/mol 137.02 g/mol 10 mL from ntp.niehs.nih.gov

Theoretical Yield of 1-bromobutane: Sn2 (10 mL 1-butanol)(0.81g/mL 1-butanol)(mol/74.12g 1-butanol) = 0.109 mol 1-butanol 1 mole 1-butanol makes 1 mole 1-bromobutane (based on stoichiometry--this is a 1:1 reaction)

0.109 mol 1-butanol (137.02g/mol 1-bromobutane) = 14.97 g 1-bromobutane Actual Yield of 1-bromobutane: 6.99g Percent Yield of 1-bromobutane: [(6.99g) / (14.97)] x 100 = 46.6% Table 2: Calculating % Yield for Sn1 Reaction Chemical 2-methyl-2-butanol Density 0.805 g/mL Molar Mass 88.15 g/mol Used 10 mL

2-chloro-2-methylbutane 0.866 g/mL 106.59 g/mol from www.chemicalbook.com

Theoretical Yield of 2-methyl-2-butanol: SN1 (10 mL 2-methyl-2-butanol)(0.805 g/mol 2-methyl-2-butanol)(mol/88.15g 2-methyl-2-butanol) = 0.0913 mol 2-methyl-2-butanol Same thing as Sn1. It is a 1:1 reaction 0.0913 mol 2-methyl-2-butanol (106.59g/mol 2-chloro-2-methylbutane) = 9.73 g 2-chloro-2methylbutane Actual Yield of 2-chloro-2-methylbutane: 33.62 g – 16.43 g =17.19 g Percent Yield of 2-chloro-2-methylbutane:

[(17.19g) / (9.73g)] x 100 = 177%

Conclusion: The objective of this lab is to prepare tertiary alkyl halide by SN1 reaction, primary alkyl halide by SN2 reaction. The other objectives are to isolate a liquid product by simple distillation, to properly use a serpartory funnel and to distinguish between tertiary, secondary and primary alkyl halides through NaI and AgNO3 tests and finally to calculate the percent yield of a isolated product through a synthesis reaction. Simple distillation was also performed and 2-chloro-2methylbutane was collected at a temperature greater than 75ºC. 1-Brmobutane was prepared by adding 1-butanol and sulfuric acid with sodium bromide through the SN2 mechanism of the substitution reaction. In this lab, silver nitrate reacted with tertiary alkyl halide, then primary and secondary alky halide because silver nitrate is a good Lewis acid that can easily cleave the bond between carbon and halogen atoms resulting in the formation of a carbocation. On the other hand, sodium iodide is not a good Lewis acid, so therefore it attacks the carbocation at the same time the chloride ion is cleaved. 2-chloro-2-methylbutane is tertiary alkyl halide that when tested with AgNO3, white liquid and white precipitated very quickly. When 2-chloro-2-methylbutane was tested with sodium iodide, yellow cloudy precipitation was seen. When 2-chlorobutane was tested with sodium iodide and silver nitrate, precipitation didn’t occur. It needs relative time to show white

percipitates. When 1-bromobutane was tested with AgNO3, it didn’t form a precipitate but cloudy liquid was observed. On the contrast when 1-bromobutane was tested with silver nitrate, a cloudy white precipitate quickly formed. The percent yield of the primary and tertiary alkyl halides were calculated to determine the success of the experiment. The percent yield of 2-chloro-2-methylbutane is 177% and 1bromobutane was 46.6%. There was an error made in weighing the round bottom flask without the product and with the product. Thus, this happened to produce significantly large percent yield for 2-chloro-2-methylbutane. The low percent yield for 1-bromobutane could be due to the extraction step. Some of the 1-bromobutane was not extracted with aqueous solution; due to volatility of the products. Even though an ice bath was used to prevent the product from evaporating, there might be possibility some of it evaporated. Since alcohol is soluble in water, some of the reactants might be washed away with water which resulted in less amount of product formation. Finally, some of the solution of 1-bromobutane was also spilled on the table and was taken of care of. Thus, there might be several reasons that why the product was very low in amount giving the low percent yield. This process can be explained that when alcohol is protonated and water is produced as a leaving group, mixing alcohol with the water. Thus, it lowers the concentration of carbocation in the solution. Despite the significantly wrong percent yields of the products formed, this experiment was performed systematically preventing any other errors. Reference Gilbert, John C., and Stephen F. Martin. Experimental Organic Chemistry. Cengage Learning, Massachusetts, 2011, 5th Ed, pp. 75-81, 153-161, 93-100....