a) Draw the reaction mechanism for the reaction.b) Determine the limiting reagent. IntroductionAlkyl halides may undergo elimination reactions with Brønstead-Lowry bases in which the halideand an adjacent proton are lost to form a new π bond. Because of the loss of a proton and ahalide anion the reactions are termed 'dehydrohalogenation'. The two most common associatedmechanisms are designated as either unimolecular (E1) or bimolecular (E2) elimination reactionsbased on reaction rate studies. E1 reactions proceed via a two-step mechanism that involves thecleavage of the leaving group (here the halide) carbon bond to form a carbonium ionintermediate (slow step) that is followed by the transfer of a B-proton to a base (which can bequite weak and is usually the solvent) and the formation of the new π bond (fast step). Since thefirst step of the E1 is rate determining, the reaction is dependent only upon the concentration ofthe alkyl halide and not the concentration of the base. However, dehydrohalogentation reactionsmost commonly proceed via the E2 mechanism where the base abstracts a B-hydrogen atom atthe same time as the halide ion is leaving to generate the new double bond. This one-step(concerted) mechanism proceeds through a single transition state and E2 reactions are termed'bimolecular' because the rate at which they proceed is proportional to the multiplicationproduct of the concentrations of both the alkyl halide and the base.Several factors must be considered to predict which elimination mechanism, E1 or E2, is likely abetter description of the reaction. These are the strength of the base, concentration of attackingspecies, stability of the leaving group (halide anion), nature of the substrate, and solvent effects.This experiment is a case is an E2 elimination - in fact two of them.The action of potassium hydroxide in alcoholic solution on compounds containing halogen atomson adjacent atoms (1,2-dihalides) results in the elimination of two molecules of hydrogen halide(double elimination) and the formation of two new π bonds to give a carbon-carbon triple bond.In this experiment you will perform a double elimination on the stilbene dibromide produced inthe lab period two weeks previous (Experiment 5, Part 1) to synthesize the alkyne 1,2-diphenylacetylene according to the following reaction equation:HBrtriethylene glycol+ 2 KOH+2KBr + 2H₂OBrdiphenylacetylenemeso-1,2-dibromo-1,2-diphenylethaneThe final product will be characterized by TLC and UV-Vis spectroscopy. An overview of UV-Visspectroscopy is provided in the textbook (9.2.4 in Ogilvie, 2018; 15.1-15.3 in Karty, 2014). Thecharacteristic features of each compound in UV-Vis spectroscopy are the wavelength ofmaximum light absorbance (max) and the molar absorptivity or extinction coefficient ( orsometimes a), as given by the Beer-Lambert law (or Beer's Law):A = Ebcand A = absorbance (unitless), b (sometimes I) = path length (in cm), and c = concentration (inM). The units for e are therefore M-¹cm¹.CH

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Answered: a) Draw the reaction mechanism for the…

Science

Chemistry

a) Draw the reaction mechanism for the reaction.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter7: Alkynes

Section: Chapter Questions

Problem 7.35P

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