Alkenes from Alcohols: Analysis of a Mixture by Gas Chromatography

Topics: Alkene, Organic reaction, Chromatography Pages: 3 (774 words) Published: July 16, 2008
Alkenes from Alcohols: Analysis of a Mixture by Gas Chromatography

Aim: To analyze a mixture of alkenes by gas chromatography.

Introduction: In this lab, we specifically used elimination reaction; however we only used the E1 reaction. In the presence of strong acids, alcohols protonate to form a good leaving group, namely water. Upon loss of a proton to a good leaving group, an introduction of unsaturation (a double bond) can be preformed. According to Wikipedia, an E2 reaction is typically of secondary and tertiary substituted alkyl halides. An E2 reaction results in formation of a Pi bond. The reason we only used an E1 reaction is because the alcohol functional group was attached to a tertiary carbon, which makes it very compatible to have the compound go through an E1 reaction. In addition, we were using sulphuric acid with heat, it is considered as a weak base making the reactivity for an E1 reaction strong. The reaction also would favor a protic solvent in order for it to be an E1 reaction. When the compound goes through the E1 reaction, it forms a carbocation, and in some cases it could be formed on a secondary carbon. This is when you will see rearrangement of hydrogen to put the compound in more stable alkenes. The stability of the carbocation plays an important role in the amount of alkenes formed. In an E2 reaction, we use a compound that is attached to a secondary or primary carbon. The reactivity is better in those conditions and also when heat is used.

When the product is formed, you want it to follow Zaitsev’s Rule. When the more stable product of an elimination reaction predominates the reaction is said to obey Zaitsev’s Rule. An E1 m reaction will obey Zaisev’s Rule. In this experiment two forms of alkenes are produced, disubstituted and trisubstituted alkene is more stable than the disubstituted. The di- and tri- are very important to understand. They refer to double bond and how it is attached to whatever number...
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