Question & Answer: Indicate whether each of the following molecules or ions would or would not be aromatic……

Rules for Aromaticity, Antiaromaticity and Nonaromaticity

A molecule is aromatic if:

– It must be a PLANAR moelcule, i.e. 2-D; should not have a single sp3 hybridized carbon

– Must follow Huckel’s Rule, which state that the conjugated electrons must fit 4n+2 rule… that is

n (0,1,2,3,4…) then, total electrons E = (2, 6, 10, 14, 18…)

– Must be CYCLIC, so the electrons can actually flow in the “ring”

– Most of the species must be Sp2 hybridized (not sp3)

An antiaromatic species is similar to aromatic… it should only fail in:

– It must be a PLANAR moelcule, i.e. 2-D; should not have a single sp3 hybridized carbon

– Must NOT follow Huckel’s Rule, which state that the conjugated electrons must fit 4n+2 rule… that is

n (0,1,2,3,4…) then, total electrons E = (2, 6, 10, 14, 18…); if this is anti aromatic, the lone pairs should be all but those set of numbers…

– Must be CYCLIC, so the electrons can actually flow in the “ring”

– Most of the species must be Sp2 hybridized (not sp3)

Finally… NON–Aromatic are those that will fail at least other than huckels rule:

i.e.

If this molecule is NOT 2-D, i.e. not planar, then this is antiaromatic

If this is not a cycle, then this is NOT aromatic

There is any SP or SP3 hybridized…

Knowing this:

a)

this is cyclic, and 2-D

total electrons in conjugated system = 6 double bonds = 6×2 = 12 electrons

this does not fit hucke’s rule, therefore

antiaromatic

b)

This appears to be 2-Dimensional

but it is not, the central carbon is SP3 hybridized, therefore this is NONAROMATIC

c)

This must be planar, and SP2 hybridized

total electrons in conjguated system = 3xdouble bonds + lone pair = 3*2 + 1*2 = 6+2 = 8

this does NOT fit huckels rule, therefore, ANTIAROMATIC

d)

Planar, 2-D and sp2 hybridization

then

total double dbond s= 8; total electrons = 8*2 = 16

once again this fails huckel rule, so this must be antiaromatic

e)

this is planar, due to the double bonds between both rings

lone pairs = none

doulbe bonds = 6; therefore conjguated system is 6×2 =12

this is antiaroamtic

f)

double bonds = 6×2 = 12

triple bonds ( accounts for only 2 for cnojugated system) = 2

total electrons in system = 12+2 = 4

this applies and staisfies huckel rule

this is planar

therefore AROMATIC

g)

ttal lone pairs = 1

doulbe bonds = 4

total e- = 1×2 + 4×2 = 10

this is then AROMATIC

h)

this cant be aromatic, due to the CH2 present, i.e. SP3 hybridization

this is NONAROMATIC