Week 9 - Dehydration of 4-Methyl-2-Pentanol by Acid...

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Unformatted text preview: Dehydration of 4-Methyl-2-Pentanol by Acid Catalyzed Elimination Reaction. Bromine and Baeyer Tests for Unsaturation Mark Gumapas November 15, 2007 Methods and Background The purpose of this lab is to dehydrate 4-methyl-2-pentanol by an acid catalyzed elimination reaction to form a mixture of alkenes. The product was then verified and identified through the bromine and Baeyer tests for unsaturation. The favored product in this lab is the formation of an alkene. Alkenes are organic compounds that possess a polarizable carbon-carbon -bond (double bond) as the functional group. Elimination reactions are the most common way to produce an alkene. Processes such as dehydrohalogenation and dehydration of alcohols both follow the elimination pathway. In dehydrohalogenation, a transformation of an alkyl halide to alkene is observed. In this reaction the electronegative halogen atom causes the carbon-halogen bond to become polarized, which can be inductively transmitted through the -bond network to increase the acidity of hydrogen atoms on the carbon atom (-carbon atom). A concerted reaction is then observed where a proton from the -carbon atom is abstracted by a strong base while the halide ion, the leaving group, departs from the -carbon atom, and a carbon-carbon double bond is formed. The transition state of the E2 reaction should preferably have an 180 dihedral angle between the -hydrogen being removed and the leaving halide. Due to the fact that the reaction is concerted, the rate of reaction is dependent upon the concentrations of both reactants involved in the transition state of the rate-determining step. This rate law can be illustrated as: Rate = k 2 [alkyl halide][B:]. If the halogen is unsymmetrically located on the carbon skeleton, dehydrohalogenation may produce a mixture of products, and since this reaction is normally irreversible under these conditions, the produced alkenes do not equilibriate, so the products are determined by the relative rates of formation, which are determined by the relative free energies of the transition states, instead of the alkenes themselves. According to Zaitsevs rule, the major product in an E2 elimination reaction is the more highly substituted alkene. This is due to an increase in the number of alkyl substituents on the double bond, which usually increases the stability of an alkene (lowers the free energy). Steric factors may also influence the relative free energies of the transition states by raising the free energy in them. This may happen when the substituents on the -carbon are sterically demanding, therefore hindering the approaching base. If substituents on the -carbon atom sterically provide hindrance to the approaching base, removal from a less substituted carbon atom to yield a less substituted alkene will become favorable, as the energy of the transition state leading to the more highly substituted alkene becomes higher than that of the less highly substituted alkene. Similarly, if bulky bases are used, more of the less substituted alkene will form because abstraction of the more accessible proton will be favored. Dehydrohalogenation of Alcohols...
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Week 9 - Dehydration of 4-Methyl-2-Pentanol by Acid...

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