Alkyl halides undergo nucleophilic substitution and elimination reactions. When the kinetics of the reaction are measured, if the rate of the reaction is found to be dependent upon the concentration of the nucleophile as well as the alkyl halide the reaction is second order. The substitution reaction is thus termed SN2, and the elimination reaction is termed E2. These reactions are bimolecular and take place in a single step. In the SN2 reaction, the nucleophile attacks the α-carbon from the backside and displaces the leaving group with an inversion of configuration occurring at the carbon. In the E2 elimination reaction, strong base removes an acidic hydrogen from the B-carbon and an alkene is formed as the leaving group is expelled from the α-carbon. This elimination follows Zaitsev’s rule whereby the more substituted alkene is generally formed. In E2 elimination, both the B-hydrogen and the leaving group must be oriented anti to each other. The same reaction conditions apply for both SN2 and E2, therefore these reactions often compete. Use of a strong, hindered base will tend to favor elimination, while use of a weak base/strong nucleophile will favor substitution. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions

Perimeter? wheres the answers

Both models are effective because they show the valence electrons, and these valence electrons indicate the reactivity of an element.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.