How far would you place a screen from the slit of the previous problem so that the second minimum is a distance of 2.5 mm from the center of the diffraction pattern?
Pg – 1 Benzene (looks like 1,3,5 – cyclohexatriene) o Uniform C-C bond length (1.39 Angstrom) 2 resonance structures extra stable o M.O. Diagram 6 p orbitals combine and make 6 MO orbitals 3 bonding and 2 anti-bonding Pi molecular Orbitals of Benzene o When comparing energy of electrons in benzene to 6 electrons in other alkenes, benzene is the most stable. o Heats of hydrogenation can be used to compare stability of nonconjugative, conjugated and aromatic systems. Heats of Hydrogenation (~29 kcal/mol for each double bond slightly lower than expected) Resonance stabilizes benzene Resonance energy or delocalization energy lowers ΔH Actual ΔH for benzene = -49.3 kcal/mol Calc valve = - 82.2 kcal/mol 1 In order for the ΔH to show conjugation, that the rotation orbitals must isn’t free line up Heats of Hydrogenation Benzene b/c of its resonance is 33.3 kcal more stable! There’s a BIG benefit to bring aromatic Requirements of Aromaticity Structure must be cyclic w/ conjugated pi bonds Each atom in the ring must have and unhybridized p orbital P orbitals must overlap continuously around the ring Ex) Must obey Hüchel’s Rule o Count the electrons in Heat in the pi system (not all electrons)