Bromobutane free essay sample

Majority of the time, to prepare alkyl halides is via the nucleophilic substitution reactions of alcohols. SN2 reaction is the type of reaction used in this lab experiment. Our objective was to see how a primary alkyl halide reacted with an alcohol. We did a conversion of n-butanol to 1-bromobutane. Br- ions is the nucleophile for this reaction which is generated from an aqueous solution of NaBr. The catalyst that converted the OH functional group of butanol to a better leaving group (H2O) was the Sulfuric Acid. [pic] Results. Theoretical Yield C4H9OH +  NaBr  -gt; C4H9Br + NaOH From 1-Butanol 20mL (butanol) x 0. 810g (butanol) x 1mole (butanol) x 1mole(bromobutane) x 137. 03g(bromobutane) = 29. 95 g of 1bromobutane 1. 00 mL(butanol) 74. 12g(butanol) 1mole(butanol)1mole (bromobutane) From NaBr 27g (NaBr) x 1mole (NaBr) x 1mole (bromobutane) x 137. 03g(bromobutane) = 35. 95 g of 1bromobutane 102. 91g (NaBr) 1 mole (NaBr) 1mole (bromobutane) % Yield = 4. We will write a custom essay sample on Bromobutane or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page 74g / 29. 95 g = 15. 8% Volume ( 3. 8mL Mass ( 4. 74g 4. 74 / 3. 8 = 1. 246 g/mL After proceed with our four major operations that were refluxing, simple distillation, separation and drying, we could do our corresponding calculations. Refluxing was used to keep a reaction at a constant temperature. Simple distillation was used to help remove and purify a particular substance in that case was 1-bromobutane, from other components in the reaction flask. Separation was carried out because there was a high probability of other components in the distillate. And finally we used CaCl2 as a very good drying agent for a variety of solvents. So for instance, -butanol is the limiting reagent, theoretically 29. 95g of 1-bromobutane. Our percent yield was 15. 8% as showed in the calculations. Our volume was 1. 247 g/mL. And as finally our boliling point was 99C. Questions. †¢ Show the mechanism for the reaction perfomed in the laboratory [pic] †¢ The last step of the mechanism is an example of SN2 reaction. Why is it an SN2 reaction? Because the acid in the reaction protonated the alcoholic oxygen and the positive charge draws electrons away fro the C allowing the negative Bromide Ion to displace (SN2) the protonated oxygen as H2O. So for instance the reaction rate does depend on both the bromide ion and alcoholonium ion the reaction will be second order. †¢ Show by a mechanism how some 2-bromobutane could form as a by-product form this reaction The protonated alcohol can undergo elimination to form 1-butene which can then react with HBR to add by Markownikoff’s rule to the double bond to form 2-bromobutane [pic] †¢ Answer the question in footnote 3 below. The primary purpose of the sulfuric acid wash is to remove any unreacted 1-butanol. The acid protonates the OH group of the alcohol, converting it into its conjugate acid and increasing its solubility in the aqueous acid wash solution significantly. Why ? Butanol is protonated so then for sure it is more soluble in H2O and also is ionic. †¢ T-butyl bromide (2-bromo-2-methylpropane) can be prepared by simply taking t-butyl alcohol and shaking it with an aq solution HBr at room temperature. The reaction is much faster than with n-butyl alcohol and is essentially 100% complete within a few minutes. Give a mechanism for this reaction. What is this type of reaction called. (CH3)C-OH + HBr ===gt; (CH3)3C-OH2(+) + Br- CH3)3C-OH2(+) ===gt; (CH3)3C(+) + H2O (CH3)3C(+) + Br- ===gt; (CH3)3C-Br It is called the SN1 reaction. The great stability of the (CH3)C(+) ion accelerates the reaction. †¢ Remembering that SN2 reactions go with 100% inversion of configuration, while SN1 reactions lead to racemization, explain why the reaction of  ®-2butanol as in this experiment gives a mixture of about 75% (S)-2-bromobutane and about 25%  ®-2-bromobutane Ionizaton of optically pure alkyl halide/alcohols molecule leads to the planar, achiral or symmetrical carbocation with an empty p-orbital perpendicular to the plane. Addition of the nucleophile can take place at both sides of the carbocation with equal ease owing to the symmetry of carbocation, resulting in mixture, equal amounts of R and S. But in actual practice, depending on the nature of substrate, solvent and leaving group, there may be a preferred side for its attack by nucleophile, in resulting of a product that automatically will contain different amounts of the two enantiomers, for then to yield optically active product.