The iodide ion is an excellent nucleophile, and the nonpolar solvent, acetone, favors the Sn2 reactions; it does not favor ionization of the alkyl halide. Perpendicular to the plane formed by the three sp 2 hybrid orbitals is an empty, unhybridized p orbital. You will use your data to calculate the activation energy of this reaction. To each tube rapidly add 1mL of an 18% solution of sodium iodide in acetone. In the S N1 reaction, the carbocation species is a reaction intermediate.
In 1-chlorobutane and 1-bromobutane, the leaving group was attached to a primary carbon, or primary electrophile. A potential energy diagram for an S N1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states. Because tert-butyl chloride is insoluble in water, but soluble in acetone, so you need some acetone for the reaction to work. The precipitate would be AgCl. Silver nitrate is a weak nucleophile. The displaced halogen atom becomes a halide ion. S N2: Compound 7 did react, whereas compound 1 did not.
The principal product in this case is R-Nuc. I am stuck writing a conclusion for this lab because I really didn't understand how mechanisms Sn1 and Sn2 relate to the lab. Compound 8 reacted as well, but took a longer time. Same as original but with 800 μL of t-butyl chloride. Part 1: Effect of Structure of the Alkyl Halide on the Relative Rates of S N2 Reactions Measure 2 mL of 15% sodium iodide in acetone into each of three clean, dry 10-cm test tubes.
In-Lab Questions Please print the worksheet for this lab. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction. If a large number of groups are bonded to the same carbon that bears the leaving group, the. The others are much simpler. So I- is better than Br-, which is better than Cl- etc. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
Rinse the test tubes with acetone and dispose of the acetone rinse in the same container. This is because they are both involved in the action step. Acetone, with a dielectric constant of 21, is a relatively nonpolar solvent that will readily dissolve sodium iodide. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. Pre-Lab Answer all assigned WebAssign questions. The stepwise unimolecular mechanism proceeds through a carbocation intermediate.
Procedure: Following the lab manual. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. These reactions are promoted by a polar protic solvent and are favored by tertiary electrophiles. Compound 1 the corresponding saturated alkylhalide reacted in 2 minutes. Let me rephrase what I just said: Most of the differences between S N1 and S N2 reactions relate to carbocation formation.
Around room temperature, every 10 °C increase approximately doubles the rate of reaction. If water molecules were present, the NaBr precipitate would dissolve in the water molecules, therefore making the precipitate unobservable and yielding erroneous results. Do not put them down the drain. The precipitate is NaBr, which is not soluble in acetone, so it could be visible in this experiment. These syntheses are often carried out by nucleophilic substitution reactions in which the halide is replaced by some nucleophile. On the other hand, bromine makes for a much better leaving group in 1-bromobutane, than chlorine does in 1-chlorobutane. Below are the reaction equations.
All the reactions save 7 display second order kinetics, reaction 7 is first order. There are two different types of substitution reactions. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile. Percent yield, which you need to calculate, is simply a an expression we use to describe what percentage of that you actually get after doing the experiment. The molecularity of a reaction is defined as the number of molecules or ions that participate in the rate determining step.