Question

Alkyl halides may undergo elimination reactions with BrØnstead-Lowry bases in which the halide and an adjacent proton are lost to form a new π bond. Because of the loss of a proton and a halide anion the reactions are termed ‘dehydrohalogenation’. The two most common associated mechanisms are designated as either unimolecular (E1) or bimolecular (E2) elimination reactions based on reaction rate studies. E1 reactions proceed via a two-step mechanism that involves the cleavage of the leaving group (here the halide) carbon bond to form a carbonium ion intermediate (slow step) that is followed by the transfer of a β-proton to a base (which can be quite weak and is usually the solvent) and the formation of the new π bond (fast step). Since the first step of the E1 is rate determining, the reaction is dependent only upon the concentration of the alkyl halide and not the concentration of the base. However, dehydrohalogentation reactions most commonly proceed via the E2 mechanism where the base abstracts a β-hydrogen atom at the 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 multiplication product 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 a better description of the reaction. These are the strength of the base, concentration of attacking species, 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 atoms on 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 in the lab period two weeks previous (Experiment 5, Part 1) to synthesize the alkyne 1,2- diphenylacetylene. The final product will be characterized by TLC and UV-Vis spectroscopy. An overview of UV-Vis spectroscopy is provided in the textbook (9.2.4 in Ogilvie, 2018; 15.1-15.3 in Karty, 2014). The characteristic features of each compound in UV-Vis spectroscopy are the wavelength of maximum light absorbance (λmax) and the molar absorptivity or extinction coefficient (ε or sometimes a), as given by the Beer-Lambert law (or Beer’s Law): A = εbc and A = absorbance (unitless), b (sometimes l) = path length (in cm), and c = concentration (in M). The units for ε are therefore M-1 cm-1 . Alkynes are shorter than alkene bonds, are shorter than alkenes. Why?

Answer 1

Alkynes form shorter bonds than alkenes because in alkynes carbon with two pi bonds is sp hybridised where as in alkenes carbon with one pi bond is sp2 hybridised hence alkyne carbon has more s character than alkene carbon . More s character means more electronegativity or the electrons are more frimly held by sp orbitals of alkynes making the bond shorter.

Alkenes form shorter bond than alkanes because alkene are sp2 hybridised where as alkanes are sp3 hybridised hence alkenes has more s character than alkanes . That means alkenes are more electronegative than alkanes or the electrons are more firmly held by sp2 orbitals than sp3 orbitals having more p orbitals and therefore more shielded from the nucleus .