Conversion of Alcohol to Alkyl Halides Essay - 2563 Words

Conversion of Alcohols to Alkyl Halides Ankita Patel August 6, 2013 Introduction This lab consisted of the conversion of alcohols into alkyl halides through common substitution methods. These methods include SN1 and SN2 mechanism, both of which can occur for this type of reaction. For both reactions, the first step of protonation will be to add hydrogen to the –OH group and then the rest of the reaction will proceed according to the type of mechanism. SN1 reactions form a cation intermediate once the H2O group leaves, then allowing a halide (such as Br) to attack the positively charged reagent1. On the other hand, SN2 reactions are one-step mechanism in which no intermediate is formed and the halide attaches as the leaving†¦show more content†¦Thus, one could safely assume the product from 1-propanol was 1-bromopropane. This is mainly due the C-H wag around 1260cm-1 indication it was a terminal alkyl halide. This reaction went through SN2 mechanism not only because the alcohol was primary but also because there were no rearrangeme nts. If a rearrangement would have occurred, it would have indicated that it was a SN1 mechanism. Further analysis was then done to determine the exact identity of the product and the chemical makeup. B B C C A A B B C C A A Figure 2: NMR Spectrum for 1-propanol The results from the NMR of 1-propanol showed 3 different prominent peaks with the peak at 2.2 cm-1 being the acetone. Because 1-bromopropane has three non-equivalent hydrogens it was found to represent this set of NMR data. The other product, 2-bromopropane only had 2 different types of hydrogens and would have only had 2 peaks. Further analysis of the structure of 1-bromopropane showed that the hydrogens closest the bromine group were an indication of peak A in the graph. Because of the electronegativity of the bromine, this peak was located further downfield. There were 2 neighboring hydrogens so using the n+1 rule gave the 3 peaks. Going down peak B showed the next carbon which had 5 neighboring hydrogens thus giving 6 peaks. Finally, the carbon furthest away from the bromine was found at peak C. It had 2 neighboring hydrogens and provided 3 peaks.Show MoreRelatedConversion of Alcohol to Alkyl Halides716 Words   |  3 PagesIntroduction In this experiment, Conversion of Alcohol to Alkyl Halides and alcohol is converted to an alkyl halide through SN1 or an SN2 mechanism. This is done by using 1-propanol and 2-pentantol with HBr, Hydrobromic acid. Only half of the groups will use 1-propanol, and 2-pentantol. All results are analyzed using NMR and IR. An SN1 reaction, requires two steps. The first step, using an alcohol as an reactant, is the pronation of the –OH group from the R group. 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Unlike the alkyl halidesRead MoreDiscussion and Conclusion: Preparation of 1-Bromobutane1035 Words   |  5 Pagesï » ¿ Discussion and Conclusion: Preparation of 1-Bromobutane The purpose of this experiment was to demonstrate the conversion of a primary alcohol, 1-butanol, to a primary bromoalkane, 1-bromobutane, a SN2 reaction. The conversion of 1-butanol to 1-bromobutane relies on sulfuric acid which plays two important roles. First, it protonates the alcohol of 1-butanol to form an oxonium ion which is a good leaving group. Secondly, it produces the hydrobromic acid, the nucleophile, which attacks 1-butanolRead MorePreparation of 1-Bromobutane from 1-Butanol by Sn2 Reaction2275 Words   |  10 Pagesfrom 1-butanol by SN2 reaction Objective: 1. To study the preparation of 1-bromobutane from 1-butanol by an SN2 reaction 2. To study the method of purification of an organic compound by simple extraction 3. To study the test of identification of alkyl halide Chemicals and Apparatus: ~18.0 g 1-butanol, ~20.0 g sodium bromide, 15 ml of concentrated sulfuric acid, anhydrous magnesium sulfate, ~10 ml of 5% aqueous sodium bicarbonate, ~1 ml of sodium iodide – acetone reagent, 1 ml of bromine in chloroformRead MoreRing Contraction Of A Ring1556 Words   |  7 Pageshydroxide, to yield a carboxylic acid but most of the time either an alkoxide base or an amine to yield an ester or an amide.2. and 5. This rearrangement proceeds through a carbanion which attacks an endocyclic carbon and expels a leaving group (a halide) forming a bicyclic molecule with rings smaller than the original substrate.4 The bicycle is then opened by nucleophilic attack on the ketone to give the contracted product.4 Scheme 2: A Typical Favorskii Rearrangement20 Scheme 3: A typical Favorskii