Question

1. Write a molecule (with its open formula), which is aromatic for each of the following items, with a total of 14 carbons, together with the side groups. Name each structure you write according to the IUPAC system. a) Write and name an aromatic molecule with a fused ring. b) Write and name a dialkylbenzene molecule. c) Write and name the molecule of an alkylanil. d) Write and name an aromatic dinitrile molecule. e) Write and name a heteroaromatic molecule. f) Name and name the aromatic molecule containing an alkenyl group in a side group. g) Write and name an molecule of alkynylphenol. h) Write and name a derivative of benzaldehyde. i) Write and name a biphenyl molecule. j) Write and name the thiol molecule containing an aromatic group. 2. Write by proving and explaining whether the following molecules are aromatic, anti- aromatic or non-aromatic. b) c) .6.0

Answer 1

1 All the below compound contain 14 carbons and their structures and IUPAC names are

a) b) C) H anthracene 1,2-Dibutylbenzene 1-([1,1'-biphenyl]-3-yl)-N-methylmethanimine CN e) f) d) NC [1,1'-biphenyl]-4,4-dicarbonitrile 2-methylacridine 4-vinyl-1,1'-biphenyl g OH i) h) H 4,4'-dimethyl-1,1'-biphenyl 4-ethynyl-[1,1'-biphenyl]-3-ol 4'-methyl-[1,1'-biphenyl)-3-carbaldehyde SH j) 2-(4-hexylphenylethane-1-thiol

Conditions for aromatic, Non aromatic aand antiaromatic

(1) Aromatic Compound (a) Huckel rule (i) Monocyclic system (ü) Polycyclic benzenoid (b) Craig's rule Polycyclic non-benzenoid (2) Anti-aromatic compound 4n te delocalized n=1, 2, 3, 4, 5, ...... n = 1 = 4e (3) Non-aromatic compound restriction of delocalization Tes n=2 = 8e non-planar structure Presence of Sp3 carbon n= 3 = 12 te system planar (sp and spa-hybridized) (system conjugated)

This is non benzoid molecule, crag's rule is applicable Craig's Rule: This rule is applicable for polyclic non-benzenoid compounds. If molecule contain C -axis then count total number of double bonds (N) and calculate value of N-1 which decide aromaticity in compound. (1) If N-1 = odd, compound is non-aromatic (2) IfN-1=even, compound is aromatic. N=5 resonating structure C2 (A) (B) N-1 = 41t bond (aromatic)

H H Belongs ( 4n+2) Te system. electron does not delocalise then compound will be non-aromatic (Repulsion between H-hydrogen atoms, causes the compound to be non planar. So, compound is non-aromatic)

6 pi electrons 'S Belongs to (4n+2) Te system. Aromatic

+ 6 pi electrons Belongs to (4n+2) Te system. Electrons are delocalised, therefore Aromatic

Sp3 carbon Restriction on the delocalization of electrons due to presence of sp3 carbon in the ring,, therefore NON Aromatic