3BL-Lecture05 - Welcome to Chemistry 3BL Lecture 5 David...

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Welcome to Chemistry 3BL Lecture 5 David Rabuka
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Friedel-Craft’s Lab Reflections • Questions for the Friedel-Craft’s lab are posted on the Chem 3BL Blackboard site. •Let’s examine the work-up procedure to recover the product.
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A Standard Recrystallization impure solid add solvent at low temperature at lower temperature not all of the material will dissolve heat at higher temperature all of the material will dissolve...releases impurities to solvent cool as solution cools slowly crystal lattice can carefully assemble... selective for desired compound
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Recrystallization - Solvent Characteristics 0 20 40 60 80 100 120 0 10 20 30 40 50 60 70 80 90 100 temperature (deg. Celcius) solubility (mass/volume) low ideal high
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Recrystallization - Mixed Solvent 0 20 40 60 80 100 120 0 10 20 30 40 50 60 70 80 90 100 temperature (deg. Celcius) solubility (mass/volume) EtOH mixture CH2Cl2
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One Product - Why? Cl (2 equiv) AlCl 3 (catalytic) • The H-NMR data is consistent with the product of the above reaction. • Why is para addition to the phenyl rings favored?
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One Product - Why? Consider ortho addition: H Steric repulsion disfavors "ortho" addition. The activation energy (E a ) for ortho addition is higher than that for other modes of addition by steric repulsion.
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One Product - Why? Consider meta addition: Meta addition gives only three resonance structures. All secondary, allylic carbocations.
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One Product - Why? Consider para addition: Para addition gives six resonance structures, spreading the positive charge into the adjacent phenyl group. The E a of this process is much lower than the activation energies for meta or ortho addition.
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One Product - Why? Consider the second addition: Steric repulsion disfavors "ortho" addition with respect to the alkyl group (on the same ring).
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