Question

The molecule 2-butene is able to undergo a process called cis-trans isomerization, where the molecule switches from being a cis-alkene to a trans-alkene. This transformation can be induced by light. What is the hybridization of the two Central carbon atoms in 2-butene? The isomerization requires breaking the π bond. Use the table of bond energies to determine the approximate amount of energy (in joules) required to break the C-C π bond in 2-butene, both per mole and per molecule. What is the wavelength of light that would be required to perform the cis-trans isomerization of one molecule of 2-butene? Which region of the electromagnetic spectrum is this wavelength of light within? 

Answer 1

ANSWERS:

sp2

explanation:

The molecular formula for 2- butene can be written as

CH3−CH=CH−CH3

count the carbon from left to right increasing order

First and fourth carbon formed covalent bond with three hydrogen and a carbon atom. In these positions carbon has undergone by SP3 hybridsation giving 4 hybrid orbital, hence forming four sigma bonds

Second and third (central carbon) carbon has three sigma bonds (two hydrogen and carbon) and one pi bond. Three sigma bonds are formed due to SP2 hybridization. Pz orbital from each central carbon atom under go lateral over lap giving rise a pi bond between second and third carbon atom

The minimum energy required for isomerization is 267 000 J/mol or 4.43×10-19lJ/molecule

EXPLANATION:

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

On a molecular level, the bond energy of a π bond is

267 000 J/1 mol×1 mol/6.022×1023molecules=4.43×10-19lJ/molecule

A quantum of light can excite a π electron to a π* orbital and allow free rotation about the central C-C bond.

λ=448nm

EXPLANATION:

E=h/f

f=E/h=4.43×10^-19J/6.626×10^-34J⋅s=6.69×10^14lHz

WE HAVE,

λ = v/f

λ=3*10^8/6.69*10^14 m

λ=448nm

VISIBLE

EXPLANATION:

Visible (Wavelength ~ 10^-7 meters)

Visible light covers the range of wavelengths from 400 – 750 nm or 0.4 to 0.75 micrometers. This is the only region in spectrum that human eyes are sensitive to. The Sun emits the most radiation in the visible portion of the spectrum.