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21 Cards in this Set

  • Front
  • Back
bond strength
triple>double>single
bond length
single>double>triple
hybridization
valence orbitals of carbon
2s, 2px,2py, 2pz need to recombinded in different was to generate new orbitals that point in the right directions

SP SP2 SP3
sigma bond
end to end overlap of 2 hybridized orbital

S
Pi bond
side to side overlap of two p orbitlas

one Pi bond has 2 pi electron
Single bond
are sigma bond
double bond
one sigma and one pi
triple
one sigma and two pi
electrons in ()bonds can be delocalized
pi
conjugation or delocalization of electrons usually makes a molecule more??
stable
Huckel's rule
determines whether ring is aromatic or not


4n+2 pi e-
tolune
benzene and CH3
phenol
benzene and OH
Aniline
benzene and NH2
Anisole
benzene and OCH3
antiaromatic
4n electron

destabilized
4n+2
A number of π delocalized electrons that is even, but not a multiple of 4. This is known as Hückel's rule.

Permissible numbers of π electrons include 2, 6, 10, 14, and so on

A number of π delocalized electrons that is even, but not a multiple of 4. This is known as Hückel's rule. Permissible numbers of π electrons include 2, 6, 10, 14, and so on
physical property

as carbon chain length inc (MW inc) and what happens to b.p, m.p, and density
b.p inc
m.p inc
density inc
physical property

as branching of C inc
what happens to the bp. mp and density
b.p dec
m.p dec
density dec
aromatic must satisfy following
closed ring
stable
delocalized(means conjugated, which means there is pi bond)
follow hukel's rule
(4n+2)

C must be sp2

radicals (free e-) only added as Pi bond in order to satisfy huckel's rule
solubility
molecule that can H bond (have OH or polar) dissolve better thus more soluble

carbon chain inc=dec in solubility