Benzene word notes PDF

Preview

Summary

notes for first year chem...

Description

F324, Rings, acids and amines

S.C

(a) compare the Kekulé and delocalised models for benzene in terms of p-orbital overlap forming bonds; Kekule’s structure of benzene: -

A 6-membered carbon ring; the carbon atoms are arranged in a hexagonal shape. Alternating single and double bonds (3 double bonds and 3 single bonds).

The delocalised model of benzene: -

-

-

A cyclic hydrocarbon with 6 carbon atoms and 6 hydrogen atoms. The 6 carbon atoms are arranged in a planar hexagonal ring. Each carbon atom is bonded to 2 other carbon atoms and 1 hydrogen atom. The shape around a carbon atom is trigonal planar and the bond angle is 120 o. Each carbon atom has 4 outer shell electrons; 2 of which are used to bond with a carbon atom on each side and 1 of which is used to bond to a hydrogen atom. The bonds in this plane are called sigma bonds. This leaves 1 outer shell electron in the 2p orbital above and below the plane of carbon atoms. The electron in a p-orbital of a carbon atom overlaps with the electrons in the p-orbitals of the carbon atoms on either side. This results in a ring of electron density above and below the plane of carbon atoms. This overlap produces a system of pi-bonds which spread over all 6 carbon atoms. The p electrons are no longer shared between just 2 carbon atoms, but are spread all over the whole ring and are said to be delocalised.

(b) review the evidence for a delocalised model of benzene in terms of bond lengths, enthalpy change of hydrogenation and resistance to reaction [see also (e) below]; Experimental evidence

X-ray studies showed that the length of carbon-carbon bonds were all the same, being somewhere in between the length of a C---C bond and a

How does it support the How does it show that Kekule’s delocalised model of benzene? structure of benzene is incorrect? The delocalised system means Kekule’s structure contained that the bonds are all the same both single and double bonds, type (in between single and which are different lengths. double) and therefore have the same length.

F324, Rings, acids and amines C===C bond. The enthalpy change of hydrogenation of benzene is -208 kJ mol-1 and not -360 kJ mol-1, which means that benzene is more stable than expected.

Benzene does not decolourise bromine water under normal conditions, and needs a halogen carrier in order to do so.

S.C

The delocalisation of the pielectrons would stabilise the structure and lower the energy released when it is hydrogenated.

The delocalised pi-electron system has insufficient electron density to polarise a bromine molecule. A catalyst is needed to generate a bromide ion, which can then react with the delocalised electrons.

The enthalpy change of hydrogenation of cyclohexene (with 1 double bond) is -120 kJ mol-1, so the enthalpy change of hydrogenation of Kekule’s structure of benzene (with 3 double bonds) is expected to be -360 kJ mol-1. If benzene did have 3 double bonds, it would have enough electron density to polarise a bromine molecule and then react with it. No catalyst would be required.

(c) describe the electrophilic substitution of arenes with (i) concentrated nitric acid in the presence of concentrated sulfuric acid, (ii) a halogen in the presence of a halogen carrier; Nitration of benzene: -

Benzene is nitrated using a nitrating mixture of concentrated sulphuric acid and concentrated nitric acid at a temperature below 50oC. The sulphuric acid acts as a catalyst. A hydrogen atom on the benzene ring is replaced by a nitro (-NO 2) group.

C6H6 + HNO3  C6H5NO2 + H2O

+ H2O -

The temperature must be kept constant, as if it is allowed to increase; more than 1 nitro group may be substituted onto the benzene ring.

Halogenation of benzene: -

Benzene can be halogenated using a halogen, under room temperature and pressure in the presence of a halogen carrier. A halogen carrier is a special type of catalyst. Common examples include FeCl 3, AlCl3, FeBr3, AlBr3 and iron metal. A hydrogen atom is replaced by a halogen atom.

Eg. Chlorination of benzene Chlorine reacts with benzene at room temperature and pressure in the presence of FeCl 3, AlCl3 or iron (III) metal to produce chlorobenzene. C6H6 + Cl2  C6H5Cl +HCl

F324, Rings, acids and amines

S.C

(d) outline the mechanism of electrophilic substitution in arenes, using the mononitration and monohalogenation of benzene as examples (see also unit F322: 2.1.1.h–j); Electrophilic substitution in benzene: -

-

-

-

Benzene does not take part in addition reactions, the typical reactions of alkenes with localised pi-bonds. This is because in an addition reaction, electrons from the delocalised structure would need to bond to the atom or group of atoms being added. This would result in the product being less stable than benzene, which is not an energetically favourable reaction. Addition reactions disrupt the delocalisation of the ring structure. Instead, benzene takes part in electrophilic substitution reactions where one of its hydrogen atoms is replaced by another atom or group of atoms. The organic product formed retains the delocalisation and therefore the stability of the benzene ring. The electron-dense ring in benzene attracts an electrophile – the electrophile accepts a pair of pi-electrons from the delocalised ring and forms a covalent bond. An intermediate forms, containing both the hydrogen atom and the electrophile – the delocalised pi-electron cloud has been disrupted and the intermediate is less stable than benzene. The unstable intermediate rapidly loses the hydrogen atom as a hydrogen ion. The delocalised ring of electrons is reforms and the stability is restored.

2 of the delocalised electrons are donated to the electrophile, forming a covalent bond.

The C-H bond breaks and 2 electrons are returned to the delocalised ring.

Nitration of benzene:

F324, Rings, acids and amines

S.C

Formation of catalyst: H2SO4 + HNO3  NO2+ + HSO4- + H2O

Regeneration of the catalyst: HSO4- + H+  H2SO4

Halogenation of benzene: Formation of catalyst: AlCl3 + Cl2  AlCl4- + Cl+

Regeneration of the catalyst: AlCl4- + H+  AlCl3 + HCl (e) explain the relative resistance to bromination of benzene, compared with alkenes, in terms of the delocalised electron density of the bonds in benzene compared with the localised electron density of the C=C bond in alkenes; In alkenes… -

-

The pi-bond contains localised electrons that are situated above and below the 2 carbon atoms in the double bond. This produces a region of high electron density above the 2 carbon atoms. When a non-polar molecule, such as bromine, approaches an alkene the electrons in the pibond repel the electrons in the bromine molecule. This induces a dipole in the bromine molecule, causing one end of the bromine molecule to be slightly more positive than the other end. The bromine molecule is now polarised. The pi-electron pair from the double bond is attracted to the slightly positive bromine atom, causing the double bond to break. A covalent bond forms between the carbon atom and the bromine atom, forming a positively-charged carbocation. The Br---Br bond breaks by heterolytic fission, resulting in a bromide ion. The bromide ion is attracted to the positively charged carbocation, forming a covalent bond.

F324, Rings, acids and amines

S.C

Bromination of benzene and alkenes -

-

Benzene contains delocalised pi-electrons spread over all 6 carbon atoms in the ring structure. Alkenes have localised pi-electrons above and below just 2 carbon atoms. This means that alkenes have a higher pi-electron density above and below any 2 carbon atoms, whereas benzene has a lower pi-electron density. When a non-polar molecule approaches benzene, it doesn’t have sufficient pi-electron density above and below any 2 carbon atoms, to polarise the molecule. This makes benzene resistant to reactions with non-polar halogens. A halogen carrier is needed to generate the more powerful electrophile, which can then attract the pi-electrons from benzene so that a reaction can take place. Molecules in which the electrons are evenly distributed over their whole structure tend to be more stable than molecules with localised electrons. They need more energy to react. In the case of benzene, energy is needed to disrupt the delocalised pi-electron cloud.

(f) describe the reactions of phenol: (i) with aqueous alkalis and with sodium to form salts, (ii) with bromine to form 2,4,6-tribromophenol; Reaction with sodium hydroxide: - When phenol is dissolved in water, it forms a weak acidic solution by losing H + from it’s –OHgroup: C6H5OH + aq ⇌ C6H5O- + H+ - Phenol is neutralised by aqueous sodium hydroxide to form the salt sodium phenoxide and water. C6H5OH + NaOH  C6H5O-Na+ + H2O

Reaction with sodium:

F324, Rings, acids and amines

S.C

-

When phenol reacts with a reactive metal, such as sodium, the metal effervescences and produces hydrogen gas. The organic product, sodium phenoxide, is a salt. 2C6H5OH + 2Na  2C6H5O-Na+ + H2

Reaction with bromine: - Phenol takes part in electrophilic substitution when reacted with bromine. NO catalyst is required. - When phenol is reacted with bromine, a thick white precipitate of 2,4,6tribromophenol forms.

(g) explain the relative ease of bromination of phenol compared with benzene, in terms of electron-pair donation to the benzene ring from an oxygen p-orbital in phenol; - A lone pair of electrons occupying the p-orbital of the oxygen atom in the –OH group is drawn into the benzene ring. - This increases the electron density of the ring structure and activates the ring. - This increased electron density polarises bromine molecules, which are then attracted more strongly to the ring structure than in benzene. (h) state the uses of phenols in production of plastics, antiseptics, disinfectants and resins for paints. Phenols can be used in the production of plastics, antiseptics, disinfectants and resins for paints. (a) describe the oxidation of alcohols (see also unit F322: 2.2.1.f) using Cr2O7 2–/H+ (ie K2Cr2O7/H2SO4), including: (i) the oxidation of primary alcohols to form aldehydes and carboxylic acids; the control of the oxidation product using different reaction conditions, (ii) the oxidation of secondary alcohols to form ketones; Oxidation of alcohols: - A suitable oxidising agent is a solution containing acidified potassium dichromate ions, H+/Cr2O72-. - The oxidising mixture can be made from potassium dichromate (K 2Cr2O7) mixed with sulphuric acid (H2SO4). - During the reaction the colour of the acidified potassium dichromate changes from orange to green.

Primary alcohols:

F324, Rings, acids and amines -

S.C

Primary alcohols are oxidised to aldehydes. If they are oxidised any further, they can form carboxylic acids. When preparing an aldehyde, you will need to distil the aldehyde as it is being formed to prevent further oxidation to a carboxylic acid. When making a carboxylic acid, the reaction mixture is heated under reflux before distilling the product off.

CH3CH2OH + [O]  CH3CHO + H2O CH3CHO + [O]  CH3COOH +H2O Secondary alcohols: - Secondary alcohols are oxidised to ketones. - Ketones are not oxidised any further – they are resistant to oxidation.

CH3CHOHCH3 + [O]  CH3COCH3 + H2O (b) describe the oxidation of aldehydes using Cr2O7 2–/H+ to form carboxylic acids; - Aldehydes can be oxidised using acidified potassium dichromate to form carboxylic acids. - The reaction is carried out under reflux. - The colour of the solution changes from orange to green.

(c) describe the reduction of carbonyl compounds using NaBH4 to form alcohols;

F324, Rings, acids and amines -

S.C

Carbonyl compounds are reduced to alcohols. A suitable reducing agent is NaBH4. The reaction is carried out by warming the carbonyl compound with the reducing agent. Water is used as a solvent.

(d) outline the mechanism for nucleophilic addition reactions of aldehydes and ketones with hydrides, such as NaBH4 (see also unit F322:2.1.1.h–j); Aldehydes and ketones react with NaBH 4 in a nucleophilic addition reaction. BH4- is a source of hydride ions, :H-. The hydride ion acts as a nucleophile. - The electron-deficient carbon atom in the polar C=O bond is attacked by the hydride ion, :H -. - The lone pair of electrons from the :H- ion forms a bond with the carbon atom. - At the same time, pi-bond between the C=O to breaks, forming a negatively-charged intermediate. - The intermediate donates an electron pair to a hydrogen atom of an H 2O molecule, forming a dative covalent bond with it, and a hydroxide ion, OH-. - The organic addition product is an alcohol. (e) describe the use of 2,4-dinitrophenylhydrazine to: (i) detect the presence of a carbonyl group in an organic compound, (ii) identify a carbonyl compound from the melting point of the derivative; Detecting the presence of a carbonyl group in an organic compound: - 2,4-DNP (Brady’s reagent) can be used to detect the presence of a carbonyl group in an organic compound. Brady’s reagent is a mixture of methanol and sulphuric acid. - Brady’s reagent is added to the organic compound; the formation of an orange precipitate confirms the presence of the C=O bond found in a carbonyl compound. Identifying a carbonyl compound from the melting point of the 2,4-DNP derivative: - The 2,4-DNP derivative is slightly impure. - The impure product is filtered and recrystallized to produce a purified sample of orange crystals. - The crystals are filtered and allowed to dry. - The melting point of this purified derivative is then measured and recorded. It can be compared to a database or data table to identify the original aldehyde or ketone. (f) describe the use of Tollens’ reagent (ammoniacal silver nitrate) to: (i) detect the presence of an aldehyde group, (ii) distinguish between aldehydes and ketones, explained in terms of the oxidation of aldehydes to carboxylic acids with reduction of silver ions to silver.

F324, Rings, acids and amines

S.C

Tollens’ reagent is a weak oxidising agent which is used to distinguish between aldehydes and ketones. Aldehydes are oxidised by Tollens’ reagent to form a carboxylic acid whereas ketones do not react with Tollens’ reagent. Tollens’ reagent can be made by… - Adding aqueous sodium hydroxide to aqueous silver nitrate until a brown precipitate forms. - Then adding dilute aqueous ammonia until the precipitate just dissolves. During the reaction, aldehydes are oxidised to carboxylic acids, whilst silver ions are reduced to silver metal. The formation of a silver mirror confirms the presence of an aldehyde, whereas there is no reaction with a ketone. RCHO + [O]  RCOOH Ag+(aq) + e-  Ag (s)...