Question #4312

Suppose that you wanted to synthesize a polyene that has its lowest excitation energy at a wavelength of 600 nm. What would the structure of molecule be, according to free electron molecular orbital theory?

Expert's answer

Free electron molecular orbital therby approximates polyene absorption as a particle in a box. The best results are obtained using not the literal additive length of the bonds but rather a = 1.4(k+2) Å, where k is the number of double bonds.

hc/l = DE = (h2/8ma2)[(k+1)2-k2] = (h2/8ma2)(2k+1)

Substituting for a,

DE = [h2/8m(1.4Å)2][(2k+1)/(k+2)2].

Solving for l,

l = (8m(1.4Å)2/h2)(hc)[(k+2)2/(2k+1)] = [(8m*(1.4x10-10 m)*c)/h][(k+2)2/(2k+1)].

The pertinent mass is the mass of an electron, 9.11x10-31 kg; c=3.00x108 m s-1; h = 6.63x10-34 J/s.

Substituting values,

l = [(8*9.11x10-31 kg*(1.4x10-10 m)2*3.00x108 m s-1)/6.63x10-34 Js][(k+2)2/(2k+1)] = (6.46x10-8 m)[(k+2)2/(2k+1)] = (64.6 nm)[(k+2)2/(2k+1)].

Using guess and check methodology,

k l л/nm

2 206

14 570

15 602

16 634

The structure must have 15 double bonds: CH_{2}=CH-(CH=CH)_{13}-CH=CH_{2}

hc/l = DE = (h2/8ma2)[(k+1)2-k2] = (h2/8ma2)(2k+1)

Substituting for a,

DE = [h2/8m(1.4Å)2][(2k+1)/(k+2)2].

Solving for l,

l = (8m(1.4Å)2/h2)(hc)[(k+2)2/(2k+1)] = [(8m*(1.4x10-10 m)*c)/h][(k+2)2/(2k+1)].

The pertinent mass is the mass of an electron, 9.11x10-31 kg; c=3.00x108 m s-1; h = 6.63x10-34 J/s.

Substituting values,

l = [(8*9.11x10-31 kg*(1.4x10-10 m)2*3.00x108 m s-1)/6.63x10-34 Js][(k+2)2/(2k+1)] = (6.46x10-8 m)[(k+2)2/(2k+1)] = (64.6 nm)[(k+2)2/(2k+1)].

Using guess and check methodology,

k l л/nm

2 206

14 570

15 602

16 634

The structure must have 15 double bonds: CH

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