Reac 714 Studying Sn1 and Sn2 Reactions: Nucleophilic Substitution at Saturated Carbon
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REAC 714 Studying SN1 and SN2 Reactions: Nucleophilic Substitution at Saturated Carbon
Date of Experiment: February 6, 2008
Objective:
The objective of this laboratory experiment is to study both SN1 and SN2 reactions. The first part of the lab focuses on synthesizing 1-bromobutane from 1-butanol by using an SN2 mechanism. The obtained product will then be analyzed using infrared spectroscopy and refractive index. The second part of the lab concentrates on how different factors influence the rate of SN1 reactions. The factors that will be examined are the leaving group, Br – versus Cl-; the structure of the alkyl group, 3◦ versus 2◦; and the polarity of the solvent, 40 percent 2-propanol versus 60 percent 2-propanol.
Discussion/Theory/Reaction Mechanisms
SN2 Reaction Mechanism:
This reaction takes place by an SN2 mechanism. The reaction must be carried out in a very acidic solution because the OH group of the alcohol is not a weak base, and therefore, not a good leaving group. By conducting the reaction in a very acidic environment, the alcohol group is converted into water, which is a weak base, and thus a good leaving group. Due to lack of steric interference, the primary carbon atom is now open to back-side attack. The bromide ion, which acts as the nucleophile attacks from the back-side and bonds to the carbon atom. As a result, 1-bromobutane is formed and water is given off as the leaving group.
SN1 Reaction Mechanism:
The reactions that take place in the second part of the experiment all take place by way of an SN1 reaction. All of the reactions proceed with the same mechanism. SN1 reactions occur under solvolysis conditions, where the solvent serves as the nucleophile.
The slow, rate-determining step occurs as the bromine leaves the 2-bromo-2-methylpropane, forming a 3◦ carbocation. The water, which acts as the nucleophile and the solvent, attaches to the carbocation at the central carbon. Another water molecule then removes a hydrogen from the positively charged oxygen atom, thus leaving 2-methyl-2-propanol. In this reaction, HBr is formed as a byproduct. It is a strong acid and completely ionizes in solution. Due to this acidity, as the reaction occurs, the acidity of the solution increases.
(Post Laboratory Questions)
3) The 2-bromo-2-methylpropane reacted much faster than the 2-choloro-2-methylpropane in the SN1 experiment. The reason for this is due to the fact that bromine is a better leaving group than chlorine. The bromine reaction was completed in only 2 minutes and 5 seconds, while the chlorine reaction took 41 minutes and 20 seconds to complete.
4) Bromine is the better leaving group. This is because it is a weaker base than the chlorine ion. Since the best leaving groups are weak bases, bromine is a weaker base than chlorine and it functions as a better leaving group.
5) The 2-bromo-2-methylpropane reacted faster in the SN1 reaction than the 2-bromopropane. This is due to the fact that after the first, slow, rate-determining step of the reaction occurs, and the bromine ion leaves, the resulting carbocations differ in stability. A 3◦ carbocation is formed with the 2-bromo-2-methylpropane. On the other hand, a secondary carbocation is formed with the 2-bromopropane. Since tertiary carbocations are more stable than secondary carbocations, the 2-bromo-2-methylpropane reacts faster.
6) The 40 % mixture 2-propanol with water is more polar. Water is more polar than 2-propanol, and thus the mixture with the most water will be the most polar. The 40% mixture contains 60% water as opposed to the 60% mixture which only contains 40% water.
7) The 2-bromo-2-methylpropane reacted faster in the 40% 2-propanol mixture. SN1 reactions occur best in polar protic solvents. Since the 40% mixture was more polar than the 60% one, the reaction occurred faster in the 40% mixture.
8) Solvent polarity has the greatest impact on the rate of an SN1 reaction because it caused the fastest results.
Data (SN2)
Mass of 1-butanol used: 1.011 grams
Mass of product (1-bromobutane): 0.275 grams
Theoretical Yield: 1.870 g 1-bromobutane
Percent Yield: 14.7 %
Corrected Refractive Index: 1.6055
Calculations
Theoretical Yield of 1-bromobutane:
(1.011 g 1-butanol) (1 mol 1-butanol) (1 mol 1-bromotubane) (137 g 1-bromobutane) = 1.870 g 1-bromobutane
( 74.1 g 1-butanol) (1-mol 1-butanol) (1 mol 1-bromotubane)
Percent Yield:
(Mass of product crystals) (0.275 g 1-bromobutane) X 100% = 14.7%
(Theoretical Yield)
(1.870 g 1-bromobutane)
Refractive Index
n20 = nT + 0.00045 (T – 20○ C)
n20 = 1.6052 + 0.00045 (20.6 – 20 ○ C)
= 1.6055
Data (SN1)
Time required for solution to become colorless
Rxn with 2-bromo-2-methylpropane
2min 5 sec
Rxn with 2-chloro-2-methylpropane
41min 20 sec
Rxn with 2-bromo-2-methylpropane
2min 8 sec
Rxn with 2-bromopropane
3min 50 sec
Rxn using 40% mixture 2-propanol in water
55 sec