Organic Chemistry: Benzene and Its Derivates
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Organic Chemistry: Benzene and Its Derivates



The organic reactions involving benzene emphasising on the electrophilic substitution on benzene ring, phenol and aniline.

The organic reactions involving benzene emphasising on the electrophilic substitution on benzene ring, phenol and aniline.



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Organic Chemistry: Benzene and Its Derivates Organic Chemistry: Benzene and Its Derivates Presentation Transcript

  • Organic Chemistry IV Benzene and Its Derivates Indra Yudhipratama
  • Outline  Aromaticity  Huckel’s rule  The Reactions (Electrophilic Substitution)  Halogenation  Friedel-Craft’s Reaction  Alkylation and acylation  Nitration and sulphonation  Oxidation and reduction of benzene derivates  Disubstitution (Ortho, meta, para directing groups)  Phenol and aniline  The relative acidity of phenol  The relative basicity of aniline  Diazoniums compounds
  • The Main Features  The bond length is between C – C and C=C (1.38 A)  Due to delocalised electron (resonance structure)
  • The Main Features  The structure is planar  Each carbon has p orbital that forms π bonding  Maximum bonding benzene should planar
  • p Cloud Formation in Benzene
  • Aromaticity (Hückel’s Rules)  Huckel’s rules define the classification of aromatic and non-aromatic molecule.  The criteria of aromatic molecule:  All the atoms are sp2 hybridised and in planar cyclic arrangement. All atoms are sp2 but not a cyclic. Hence, non-aromatic There is non-sp2 atom. Hence, non-aromatic All atoms are sp2 and a cyclic. Hence, could be aromatic
  • Huckel’s rules  Huckel’s rule  Number of π-electrons is (4n+2),  How to calculate π-electrons?   based on the structure, p-orbitals in sp2 arrangement has 1 electron Has 6 π-electrons (4n+2, n=1) Hence, aromatic Has 4 π-electrons (4n, n=1) Hence, anti-aromatic
  • Huckel’s Rule (summary) Is the molecule has no non-sp2 atoms? YES NO How many π- electrons in the molecule? 4n+2 Not 4n+2 aromatic Anti- aromatic non-aromatic
  • Huckel’s rules Porphyrin ring in the haem group
  • Huckel’s rule  Which molecules are aromatic?  Is this molecule aromatic? 6 π-electrons 2 π-electrons
  • The reactions  Benzene undergoes electrophilic substitution  Doesn’t undergo electrophilic addition  The consequence of aromatic properties
  • The reactions - Halogenation  Halogenation. E.g. chlorination  Via:  The presence of Lewis acid (e.g. AlCl3) helps benzene to react with Cl2
  • The Reactions – Friedel-Crafts Reaction  Friedel-Crafts Reaction (Alkylation)  To substitute with hydrocarbon chain  Via: Electrophilic generation
  • The reactions Friedel-Crafts Reaction  There is a problem for this reaction when longer alkyl halide is used  Rearrangement of the electrophile (carbocation)  Trying to find the most stable carbocation
  • The Reactions  Friedel-Crafts Reaction (Acylation)  To substitute with R-CO –  Via:  Electrophilic generation  acylium ion  stabilised by resonance. Both structures are valid.
  • The reactions  Acylation can be used to get around the ‘messy’ long chain alkylation.
  • The Reactions  The nitration (concentrated sulphuric acid as catalyst)  Via:
  • The Reactions  Sulphonation  Via:
  • The reactions  Sulphonation  Producing strong sulphonic acid
  • The Reactions  The Oxidation of toluene  Where R is alkyl group  The Reduction of Aniline R OH O 1) KMnO4, OH- , Heat 2) H3O+ NO2 NH2 Fe HCl aniline
  • The Reactions  Formation of Diazonium salts  Diazonium salts is a good precursor compound for:  Halogenation  formation of phenol  deamination  coupling reaction of arenediazonium salts NH2 N + N Cl - NaNO2, HCl H2O 0 - 15 o C
  • The Reactions
  • The Reactions  Coupling reaction of arenediazonium salts  Where Q is activating group ( –OH, –NR3). E.g.: N + N Cl - + Q N N Q N + N Cl - + OH O + N N H H Cl - OH N N N + N Cl - + OH OH N N
  • Disubstitution of Benzene  The benzene ring can be substituted with another FG more than once.  The second position is determined by the first FG  Three possible positions: ortho (1,2) meta (1,3) para (1,4) CH3 NH2 OH Cl CH3O OHONH2O NH CH3 O R R1 R R1 R R1
  • Disubstitution of Benzene  The determining factor  The nature of FG  electron withdrawing (EW) or electron donating (ED) group  EW: the FG generally has partial positive charge  It deactivate the benzene ring, so it is less reactive  ED: the FG generally has partial negative charge  It activate the benzene ring, so it is more reactive
  • Disubstitution of Benzene
  • Disubstitution of Benzene  E.g. Application for synthesis route
  • Phenol  The structure  The relative acidity  Acidity  The easiness to release H+ (proton)  The stability of the acid conjugate determine the relative acidity.  The comparison with water and alcohol (e.g. ethanol) OH
  • Phenol  Let’s put water as the standard and the conjugate.  More stable the conjugate, more acid the substance.  In ethoxide ions the alkyl group push the electrons  increasing the charge  In phenoxide ions, it forms a bigger resonance structure due to unbonding p-orbital OH O H H CH3 OH O - O - H CH3 O -
  • Phenol  The effect of substituent  The principle: The reduction of the charge  The deactivating benzene substituent will make phenol more acidic  The activating benzene substituent will make phenol less acidic. Phenol 3-methylphenol 3-nitrophenol 3-chlorophenol pKa = 9.89 pKa = 10.01 pKa = 8.28 pKa = 8.80 OH OH CH3 OH Cl OH NO2
  • Phenol  Predict the pKa of 2,4 dinitrophenol. (a) 10.17 (c) 8.11 (d) 3.96 (b) 9.31
  • Phenol  Esterification of Phenol  No reaction with carboxylic acid  Only react with acyl chloride or acetic anhydride OH + CH3 OH O H3O+ No reaction
  • Phenol  Suggest the products from the reactions below
  • Phenol  How to distinguish with alcohol?  Since the phenol is more acidic than alcohol, so it can reacts with weaker base (e.g. NaHCO3)  Both of them can react with Na
  • Aniline  The Basicity of amines  Basicity >< Acidity  Basicity  How easy a compound can accept H+  The case: The relative Basicity of ethylamine, amine, and aniline  The easiness of compound to accept H+  The availability of lone pair electrons on N atom CH3 NH2 NH3 NH2 aniline ammoniamethanamine
  • Aniline  The reactions  Phenylamine cannot react in the similar way like amine.  Phenylamine is not a better nucleophile than amine  the availability of the electrons on N atom to do the reaction HNO3 concd H2SO4 N + O - O Fe/Sn HCl NH2 NaNO2, HCl 0 - 15 o C N + N Cl - This reaction can produce the other amines. Could you draw the other products?
  • Aniline  Phenylamine could form an amide with acyl chloride.  Important synthetic pathway for aniline-based compound