Chapter 11 Alcohols, Phenols and Ethers

In the presence of sunlight the toluene gets monochlorated.

When toluene reacts with aq NaOH the Cl^- iongoes with Na⁺ and OH^- gets attached to alky group.

Thus, the product is benzyl alcohol.

Total of four alcohols can be made with a similar chemical formula. Which are butan-1-ol , butan-2-ol, 2-Methylpropan-1-ol, 2-Methypropan-2-ol.

Out of this four only butan-2-olhas a chiral (four different groups are attached) carbon atom.

Thus, all other options are wrong.

butan-2-ol structure

butan-1-ol structure

2-Methylpropa-1-ol structure

2-Methylpropan-2-ol structure

In Tertiary alcohols (3^o) the hydroxyl group is easily substituted by the halide group due to greater stability of the carbocation in comparison to primary carbocation and secondary carbocationformed during the reaction.

The greater stability is due to the more inductively donating alkyl groups.The hyperconjugation effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Thus, the correct order of the reaction is 3^o>2^o>1^o in increasing order.

Thus, all other options are wrong.

Alcohols can be converted into aldehydes by making them react with the oxidizing agents like pyridinium chlorochromate (PCC) and CrO₃.

A very strong oxidizing agent like KMnO₄ converts alcohols to carboxylic acids.

LiAlH₄ and catalytic hydrogenation is reducing in nature.

The process of conversion of alkyl halides to alcohols involves a simple mechanism of nucleophilic substitution.

The halide group (-Cl, F, I) is substituted by the hydroxyl group (-OH).

All other reactions do not convert alkyl halide to alcohol.

Aromatic alcohols are those chemical compounds in which a hydroxyl group (-OH) is directly attached to an aromatic hydrocarbon group.

Only figure A follows the above statement.

All other figures are an example of benzyl alcohol.

Thus, the option (iv) is correct.

According to IUPAC nomenclature, we start by selecting the longest Carbon chain in the given organic compound.

Here, the longest carbon chain is of 6 members, so we use hexane. (–OH) Alcohols are given priority over halogens (-Chloro here) according to IUPAC. So, here we start naming the chain from –OH side .

Combining these, we name the given organic compound as:

5-Chlorohexan-2-ol

m-cresol has the following structure according to IUPAC the preference is given to hydroxyl group in respect to the CH₃ group thus the CH₃ is attached to 3 carbon in the benzyl chain and -OH to 1^st carbon atom in the benzyl chain.

According to IUPAC nomenclature, we start by selecting the longest Carbon chain in the given organic compound.

Now, here the longest chain is of 3 members (i.e. Propane) with a methoxy group being attached to the second position.

Hence, the overall nomenclature with prefix and suffix becomes:

2-methoxypropane

Due to the higher electronegativity of sp2 hybridized carbon of phenol to which –OH is attached, electron density decreases on oxygen.

This increases the polarity of the O–H bond and results in an increase in ionization of phenols than that of alcohols and water.

when a group like -NO₂ is on the phenol the electron density of oxygen decreases more due to -R effect. Hence its tendency to attract the H⁺ increases more.

Thus, option (iv) is the strongest base (tendency to attract H⁺ ion).

The above reaction is basically an acid-base reaction more acidic is phenol.

Phenols are more acidic than the alcohols due to to the more stability of phenoxide ion than alkoxide ion due to the resonance of electron pairs of oxygen with the aromatic ring.

Phenols are more acidic than benzyl alcohols due to the pi electrons of oxygen contribution to the aromatic pi systems. But this cannot happen in benzyl alcohols.

Thus, phenol is more acidic.

The presence of electron-withdrawing group(e.g. nitro group), enhances the acidic strength of phenol.

It is due to the effective delocalization of negative charge in phenoxide ion. This makes the OH bond weaker.

If an electron releasing group (example alkoxy group) is present it makes the O-H bond stronger.

Thus H⁺ ion can be easily removed.

Phenols are more acidic than the benzyl alcohols and the cyclohexanol due to the contribution of the pi electrons of the oxygen in the aromatic pi system of phenol making phenoxide ion stable.

When an electron-withdrawing group like Cl is present in phenol structure its acidity increase more due to the due to the effective delocalization of negative charge in phenoxide ion.

These groups also weaken the O-H bond. Thus, H⁺ can be easily donated.

Thus, m-chlorophenol is more acidic than others.

The presence of electron-withdrawing group(e.g. nitro group), enhances the acidic strength of phenol.It isdue to the effective delocalization of negative charge in phenoxide ion.

On the other hand, electron releasing groups (e.g. -OCH₃) in general, do not favour the formation of phenoxide ion resulting in a decrease in the acidic nature.

The effect is more when the substituent group is more at the ortho position or para position in comparison to the meta position.

The reaction is a acid base reaction.

Thus, after the donation of the OH^- ion formation of benzyl carbocation takes place.

When an electron-withdrawing group (NO₂, Cl) is present on the aromatic ring the carbocation stability decreases.Since the electron-withdrawing property of NO₂ is greater than Cl, the carbocation will be highly unstable.

Thus, an order of reactivity is b<c<a in increasing order.

The boiling points of alcohols and phenols increase with an increase in the number of carbon atoms. This is because of the increase in van der Waals forces.

In alcohols, the boiling points decrease with the increase of branching in the carbon chain because of the decrease in van der Waals forces with a decrease in surface area.

Thus b.p increases with a number of carbon atoms in the chain and for the same number of atoms, it depends on branching as in the case of butan-2-ol and butan-1-ol.

H₂/Pd, LiAlH₄ and NaBH₄ are used to convert RCHO into RCH₂OH.

Aldehydes can be converted to their corresponding alcohols by Hydrogen in the presence of a catalyst such as Pd, Pt or Ni.

Aldehydes can be converted to their corresponding primary alcohols by Lithium aluminium hydride or Sodium borohydride.

Hence, options (i), (ii) and (iii) are correct.

Option (iv) is incorrect because reaction of RCHO with RMgX followed by hydrolysis, results in the formation of secondary alcohol.

When Chlorobenzene is fused with NaOH at 300 atm and 623K followed by acidification, it results in the formation of Phenol. Hence, the option (i) is correct.

When Aniline is treated with NaNO₂ + HCl,a Diazonium salt is formed. Then, Diazonium salts are hydrolyzed by warming with water to Phenols. Hence, option (ii) is correct.

Benzene is first sulphonated with Oleum to form benzene sulphuric acid and then, it is converted to sodium phenoxide by heating benzene sulphuric acid with molten NaOH. Acidification of sodium phenoxide results in the formation of phenol. Hence, the option (iii) is correct.

Option (iv) is incorrect because of Chlorobenzene when fused with NaOH at 623K and 300 atmospheric pressure followed by acidification results in the formation of Phenol.

CrO₃ in anhydrous medium can be used to oxidize primary alcohols to aldehydes. Here, CrO₃ acts as an oxidizing agent. Hence, option (i) is correct.

Pyridinium chlorochromate (PCC) can also be used to oxidize primary alcohols to their corresponding aldehydes. Hence, option (iii) is correct.

Primary alcohols can also be oxidized to their corresponding aldehydes by dehydrogenation or heating them in the presence of Cu at 573K. Hence, option (iv) is correct.

We know,KMnO₄ is a strong oxidizing agent. Thus, when Primary alcohols are treated with KMnO₄ in acidic medium,they are oxidized to a mixture of carboxylic acids. Hence, Option (ii) is incorrect.

• Phenol can be distinguished from ethanol by treating them with Bromine water. Phenol reacts with Br₂/water and is precipitated as 2,4,6-tribromophenol. The White precipitate is formed in case of Phenol.

No white precipitate is formed in case of Ethanol. Hence, the option (i) is correct.

• Phenol reacts with neutral FeCl₃ to form a violet coloured complex whereas Ethanol does notgive any reaction with neutral FeCl₃. Hence, the option (iii) is correct.

• Phenol reacts with Sodium metal to yield Sodium phenoxide and Hydrogen gas. Ethanol reacts with Na to form Sodium ethoxide and Hydrogen gas. Both of them show the same kind of products. Thus, one cannot distinguish Phenol and Ethanol using Na. Hence, options (ii) and (iv) are incorrect.

We know, Benzylic alcohols are compounds having alcohol as a functional group on a carbon atom that is directly attached to the Benzene ring.

Thus, options (ii) and (iii) are Benzylic alcohols because the –OH groups are attached on a carbon atom that is directly attached to the Benzene ring.

Options (i) and (iv) are incorrect because the –OH group in both the options is not attached to a carbon atom that is directly attached to the Benzene ring. Hence, options (ii) and (iii) are incorrect.

Structure of Glycerol:

IUPAC name: Propane-1,2,3-triol

Here, the –OH groups are the functional groups. The number of carbon atoms in the above structure is 3 and all the bonds between the Carbon atoms are single bonds. So, the IUPAC name starts with Propane. Three –OH groups are attached to the Carbon atoms at position 1, 2 and 3. Therefore, the IUPAC name of glycerol is Propane-1,2,3-triol.

(A) IUPAC name of the compound is 3-Ethyl-5-methylhexane-2,4-diol.

Here, the –OH groups are the functional groups. The number of carbon atoms in the longest chain is 6 and all the bonds between the carbon atoms are single bonds. So, the IUPAC name of the longest carbon chain is Hexane. The counting of the position starts from the right-hand side so that all the substitute groups get the lowest position numbers. Two–OH groups are attached to the carbon atoms at2^nd and 4^thpositions, one methyl group at 5^th position and one ethyl group are attached to the 3^rd position. Therefore, the IUPAC name of the above compound is 3-Ethyl-5-methylhexane-2,4-diol.

(B)

IUPAC name of the compound is1-Methoxy-3-nitrocyclohexane.

The number of longest carbon atoms in a chain in the above structure is 6 and they in present in a cyclic manner. All the bonds between the carbon atoms are single bonds. So, the IUPAC name of the longest carbon chain is cyclohexane. Among nitro and methoxy groups,methoxy is written first, then nitro. Therefore, the IUPAC name of the above compound is 1-Methoxy-3-nitrocyclohexane.

IUPAC name of the compound is 3-Methylpent-2-ene-1,2-diol.

The number of longest carbon atoms chain in the above structure is 5. All the bonds between the carbon atoms are single bonds except one, a double bond between C₂ and C₃ atoms. So, the IUPAC name of the longest carbon chain is pent-2-ene. Two –OH groups are attached to the Carbon atoms at position 1 and 2 and one methyl group at 3^rd position. Therefore, the IUPAC name of the above compound is 3-Methylpent-2-ene-1,2-diol.

The factors responsible for the solubility of alcohols in water are:

I) Hydrogen bonds

II) Size of the alkyl or aryl groups

III) Molecular mass of the Alcohols.

Alcohols are soluble in water due to their ability to form hydrogen bonds with the water molecules or the intermolecular hydrogen bonding.

With increase in the size of the alkyl or aryl groups, the solubility of alcohols in water decreases. The low molecular mass alcohols are soluble in water due to the presence of intermolecular hydrogen bonding between alcohol molecules.

With an increase in the alkyl group of alcohol or in case of high molecular mass alcohols, it suppresses the effect of polar nature of –OH group of alcohol. Thus, the solubility of alcohol decreases with increases in molecular size and therefore, alcohols with lower molecular masses are more soluble than the alcohols with higher molecular masses.

Denatured alcohol is commercial alcohol. Alcohols that are used for drinking are made unfit for human consumption by mixing alcohol with some copper sulfate and pyridine, which gives the colour and a foul smell to the liquid respectively. This is called denatured alcohol.

The above chemical reaction basically shows the oxidation of secondary alcohol into a ketone. This can be easily achieved by using oxidizing agents like chromic anhydride (CrO₃), Pyridinium chlorochromate (PCC), etc.

The suggested reagents for the above conversion are PCC or CrO₃.

2-chloroethanol, Alcohols are acidic in nature due to the presence of one or more polar groups, i.e., -OH group.

Out of 2-chloroethanol and ethanol, 2-chloroethanol is more acidic because of the presence of an electron withdrawing group, Chlorine atom. The presence of the electron withdrawing group results in negative inductive effect and thus, the electron density in the –O-H bond decreases. It stabilizes the alkoxide ion and therefore, 2-chloroethanol can easily release a proton.

PCC or Pyridinium chlorochromate, The above conversion basically shows the oxidation of primary alcohol to an aldehyde. This can be easily achieved by using mild oxidizing agents like chromic anhydride (CrO₃) and Pyridinium chlorochromate (PCC).

The chemical reaction involved in this conversion is:

Acidified KMnO₄, The above conversion basically shows the oxidation of primary alcohol to a carboxylic acid. This can be easily achieved by using strong oxidizing agents like acidified KMnO₄, K₂Cr₂O₇ etc.

The chemical reaction involved in this conversion is:

o-nitrophenol , o-nitrophenol is more volatile in nature due to the presence of an intramolecular Hydrogen bonding between -NO₂ and –OH groups. They also form intermolecular Hydrogen bonding.

In p-nitrophenol, there is no intramolecular Hydrogen bonding due to large separation between the –NO₂ and –OH groups, but they can form intermolecular Hydrogen bonding. Therefore, p-nitrophenol is less volatile.

o-nitrophenol , Ortho-nitrophenol is more acidic because of the presence of an electron-withdrawing group (–NO₂) at an ortho position with respect to the –OH group, which enhances the acidic strength of the compound by stabilizing the phenoxide ion and therefore, o-nitrophenol can easily release a proton.

o-cresol is less acidic due to the presence of an electron releasing group (alkyl group). They do not favour the formation of phenoxide ion. Thus, decreases the acidic strength of the compound.

The name of the compound formed in this reaction is 2,4,6-tribromophenol. The structure of the compound formed is:

Phenol reacts with Br₂/water (Bromine water) and yields 2,4,6-tribromophenol. 2,4,6-tribromophenol is observed as a white coloured precipitate. The chemical reaction involved in this conversion is:

increasing order of the acidity of the given compounds: o-cresol < Phenol < o-nitrophenol.

The presence of an electron-withdrawing group (–NO₂) at an ortho position with respect to the –OH group, enhances the acidic strength of the compound by stabilizing the phenoxide ion, whereas, the presence of electron releasing group (alkyl group) do not favour the formation of phenoxide ion. Thus, an electron releasing group decreases the acidic strength of the compound. Thus, o-nitrophenol is most acidic and o-cresol is the least acidic in nature among the three given compounds.

The decreasing order of reactivity of sodium metal towards alcohols: Primary alcohols > secondary alcohols > tertiary alcohols

Alcohol reacts with Sodium metal to yield corresponding Sodium phenoxide/alkoxide and Hydrogen gas.

The reactivity of sodium metal towards tertiary alcohols is lowest mainly due to two reasons. First,steric hindrance of the alkyl groups. Second, an increase in the electron density on Oxygen atom in the hydroxyl bond.

With an increase in the number of alkyl groups, the electron density in the O-H bond increases and therefore, the release of a proton is not that easy. Thus, the reactivity of sodium metal towards tertiary alcohols is least among primary, secondary and tertiary alcohols.

Phenol is formed.

• When benzene diazonium chloride is heated with water, Phenol is formed along with the by-products, Nitrogen gas and Hydrochloric acid.

• The chemical reaction involved is:

• This reaction is commonly used for the synthesis of phenol from Aniline. When Aniline is treated with NaNO₂/ HCl, a Diazonium salt is formed.

The decreasing order of the acidity of the given compounds:

H₂O > ROH > HC ≡ CH

HC ≡ CH is less acidic because the carbon atoms here is sphybridized, so the electron density is higher on the carbon atom and therefore, the release of a proton is not easy. Thus, they are the least acidic compound among the given three compounds.

ROH is less acidic than water because of the alkyl group (electron releasing group).They do not favour the formation of alkoxide ion. Thus, decreases the acidic strength of the compound.

Water among the three given compounds is most acidic.

Names of the enzymes involved in the preparation of ethanol from sucrose by fermentation are invertase and zymase.

Invertase converts sucrose into glucose and fructose. Then, glucose and fructose undergo fermentation in the presence of zymase and ethanol is produced.

Reactions involved in the preparation of ethanol from sucrose by fermentation are:

Propan-2-one can be converted into tert- butyl alcohol by using Grignard reagent.

Propan-2-one is treated with CH₃MgBr in the presence of dry ether (Grignard reagent), followed by the hydrolysis to yield tert- butyl alcohol.

• The chemical reaction involved is:

C₄H₁₀O means the isomers have 4 carbon atoms and one –OH group with all the sigma bonds between the adjacent carbon atoms.

Structures of the isomers of alcohols with molecular formula C₄H₁₀O:

a)

Butan-1-ol

b)

Butan-2-ol

c)

2-Methylpropan-1-ol

d)

` 2-Methylpropan-2-ol

We know, optical activity is shown by the compounds having a chiral center or a carbon atom attached to the four different groups. They are capable of rotating the plane of polarized light.

Here, only Butan-2-ol exhibits optical activity because it has a chiral carbon atom.

The –OH group in phenol is directly attached to the sp²hybridized carbon atom of the benzene ring. The carbon-oxygen bond length in phenol is smaller as compared to the carbon-oxygen bond length in alkyl alcohol and this is due to the partial double bond character or due to the resonance and charge distribution in phenol. Therefore, the OH group in phenols is more strongly held as compared to OH group in alcohols.

The resonance structures in phenol:

The –OH group in phenol is directly attached to the sp² hybridized carbon atom of the benzene ring. Due to the resonance, the carbon-oxygen bond length in phenol is smaller and therefore, is difficult to break this bond.

Due to the resonance, the ortho- and para-positions in the benzene ring becomes electron-rich and therefore, activates it towards electrophilic substitution reaction. Thus, nucleophilic substitution reactions are not very common in phenols.

The resonance structures in phenol:

Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. The mechanism of this reaction involves the following three steps:

Step 1) Protonation of alkene and formation of carbocation

Here, carbocation formation takes place by the electrophilic attack of H₃0⁺ on alkene carbon atom and water molecule is released.

STEP 2) Nucleophilic attack of water

Step 3) Deprotonation occurs and an alcohol is formed. H₃0⁺ is released.

O=C=O nonpolar because the dipole moment of the two C=O bonds are exactly equal and opposite of each other. Therefore, they cancel each other and so, the net dipole moment of O=C=O is zero. Whereas, the net dipole moment of R—O—R is not equal to zero because the shape of this compound is non-linear and thus, R—O—R is polar in nature.

The reactivity of all the three classes of alcohols with concentrated HCl and ZnCl₂ (Lucas reagent) is different because of the steric hindrance of the alkyl groups and the stability of the carbocation.

We know, the stability of carbocation is in the following order:

Tertiary > secondary > primary

Primary alcohol does not show any reaction at room temperature because the 1° carbocation is least stable. Thus, they cannot form halides easily.

Secondary alcohol does not show any turbidly at room temperature but shows turbidity upon heating.

Tertiary alcohol immediately shows turbidity after addition of Lucas reagent as they can form halides easily due to the higher stabilityof the carbocation.

Aspirin is also known as acetylsalicylic acid. Phenol is treated with NaOH to produce phenoxide ion. Phenoxide ion then undergoes electrophilic substitution with CO₂ to yield salicylic acid as the major product. This is called kolbe’s reaction.

Then, acetylation of Salicylic acid produces Aspirin as a major product.

Out of benzene and phenol, phenol is more easily nitrated. Phenol is more easily nitrated than benzene because of the presence of the hydroxyl group in phenol.Due to the resonance effect caused by –OH group, the ortho- and para-positions in the benzene ring becomes electron rich and therefore, activates it towards electrophilic substitution reaction. Thus, Nitration, an aromatic electrophilic substitution occurs at a position where the electron density is high.

In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide (a weak electrophile) because phenoxide ion is more reactive towards electrophilic aromatic substitution.Due to the resonance effect caused by –OH group, the ortho- and para-positions in the benzene ring becomes electron rich and therefore, activates it towards electrophilic substitution reaction.

Dipole moment of phenol is smaller than that of methanol due to the electron withdrawing effect of the phenylring. Due to the resonance, the polarity of the C-O bond in phenol decreases.

In methanol, due to the presence of electron releasing methyl group, the electron density on the oxygen atom increases and thus, the C-O bond in methanol is more polar.

Ethers can be prepared by Williamson synthesis in which an alkyl halide is reacted with sodium alkoxide. Di-tert-butyl ether can’t be prepared by this method because in this case, elimination is more favoured over substitution. In other words, elimination competes over the substitution unlike in case of primary alkyl halides.So, alkene is formed as the major product and ether is not formed.

The major product formed in this reaction is 2-Methylprop-1-ene, an alkene and not ether.

The C—O—H bond angle in alcohols is slightly less than the tetrahedral angle due the repulsion between the unshared pair of electrons or the lone pair of electrons on the oxygen atom, whereas, the C—O—C bond angle in ether is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky alkyl groups.

The low molecular mass alcohols are soluble in water due to presence of intermolecular hydrogen bonding between alcohol molecules and this is possible due to the presence of polar –OH group in alcohol.

With increase in the alkyl group of alcohol or in case of high molecular mass alcohols, it suppresses the effect of polar nature of –OH group of alcohol. Thus, solubility of alcohol decreases with increases in molecular size.

Para-nitrophenol is more acidic than phenol due to the presence of an electron withdrawing group, -NO₂ group, which enhances the acidic strength of the compound by stabilizing the phenoxide ion and the ortho- and para-positions in the benzene ring becomes electron rich. Therefore, p-nitrophenol can easily release a proton as compared to phenol.

Alcohols and ethers of comparable molecular mass have different boiling points because of the presence of intermolecular hydrogen bonding in alcohols.

There is no intermolecular hydrogen bonding in ethers and thus, ethers have low boiling point as compared to the alcohols of the same molecular mass.

The carbon-oxygen bond in phenol is slightly stronger than that in methanol because of two reasons:

i) Due to the resonance, it develops apartial double bond character in carbon-oxygen bond. Thus, decrease in the size of carbon-oxygen bond in phenol.

ii) In phenol, Oxygen is directly attached to a sp²hybridized carbon atom, whereas in methanol, Oxygen is directly attached to a sp³ hybridized carbon atom. Thus, the bond formed between oxygen and sp² hybridized carbon atom is slightly stronger than that in methanol.

Increasing order of acidity is: Ethanol < water < phenol

After deprotonation, in case of phenol, phenoxide ion is formed and gets stabilized by the resonance. Thus, phenol is most acidic among the three given compounds.

The ethoxide ion produced after the removal of proton from ethanol gets destabilized by the positive inductive effect of the ethyl group. Therefore, ethanol is less acidic than phenol. The electron releasing group of ethanol increases the electron density on oxygen and hence, the removal of proton from ethanol is difficult. Water is better proton donor than ethanol. Thus, ethanol is weaker acid than water.

(i) ) - (d) o-Cresol

So what are cresols? Cresols are nothing but methyl phenols. It means that a methyl group has substituted a Hydrogen atom in a phenol molecule.

Now, depending upon the position of the methyl group with respect to the –OH group of phenol, the compound name is given accordingly. For eg. if the methyl group is situated just adjacent to the –OH group of phenol, it is ortho cresol(o-cresol). IUPAC name is 2-methylphenol.

If the methyl group occupies the position next to the ortho, the compound is meta cresol(m-cresol). IUPAC name is 3-methylphenol.

If the methyl group is situated vertically opposite to the –OH group of phenol, the compound is called para cresol(p-cresol). IUPAC name is 4-methylphenol.

So, we can conclude that the structure of the compound given to us is o-cresol.

(ii)

- (c) Catechol

So what are catechols? Catechols are nothing but a part of benzene-diols. So they are formed when a hydroxyl group(-OH) substitutes a Hydrogen atom in a phenol molecule at the ortho position . Thus they are named diols due to the presence of two –OH groups in the molecule – one present phenol and the other, the substituted one at o-position. IUPAC name for catechol is 1,2-dihydroxybenzene. It is used in the production of pesticides, perfumes and pharmaceuticals.

Catechol.

(iii)

- (f) Resorcinol.

Resorcinols are somewhat similar to catechols. They also belong to the family of benzene-diols.

As we discussed earlier that catechols are ‘ortho substituted hydroxy phenols’. Similarlyresorcinols are also hydroxyl phenols, but they are meta substituted. So they are formed when a hydroxyl group(-OH) substitutes a Hydrogen atom in a phenol molecule at the metaposition . Thus they are named diols due to the presence of two –OH groups in the molecule – one present phenol and the other, the substituted one at m-position. IUPAC name for catechol is 1,3-dihydroxybenzene. It is used to treat acne and various other skin disorders.

Resorcinol.

(iv)

- (a)Hydroquinone

Hydroquinones are similar to that of resorcinols and catechols. As we know that resorcinols are meta substituted hydroxyl phenol and catechols are ortho substituted hydroxyl phenols, similarly hydroquinones belong to the family of benzene-diols.

They are para substituted hydroxyl phenols. So they are formed when a hydroxyl group(-OH) substitutes a Hydrogen atom in a phenol molecule at the paraposition . Thus they are named diols due to the presence of two –OH groups in the molecule – one present phenol and the other, the substituted one at p-position. IUPAC name for catechol is 1,4-dihydroxybenzene.

Its other names are benzene-1,4-diol and quinol. It is a white granular solid.

Hydroquinone is a depigmenting agent used to lighten areas of darkened skin such as freckles, age spots, chloasma, and melisma caused by pregnancy, birth control pills, hormone medicine, or injury to the skin. However, they are banned in some parts of the world as it is found to be a carcinogen or cancer-causing chemical.

Hydroquinone.

(v)

- (g) Anisole

Anisole is nothing but methoxybenzene. So this compound is an ether.

It is formed by the substitution of a Hydrogen atom from a benzene molecule by a methoxy group(-OCH₃). It is a colorless liquid with a smell reminiscent of anise seed, and in fact many of its derivatives are found in natural and artificial fragrances.

Anisole.

(vi)

- (b) Phenetole

Similar to anisole, phenetole is also an ether. More precisely, phenetole is ethoxybenzene. ether. It is formed by the substitution of a Hydrogen atom from a benzene molecule by a ethoxy group(-OC₂H₅).

It is also known as ethylphenyl ether. It is volatile in nature and is explosive. It will dissolve in many non-polar compounds e.g. ethanol or ether but not in polar substances like water. Anisole is a colorless to yellowish oily liquid.

Anisole.

(i) CH₃—O—CH₃ - (d) CH₃-OH + CH₃-I

When methoxymethane(CH₃OCH₃) reacts with HI, methanol(CH₃OH) and methyl iodide(CH₃I) are produced.

If we want to see how it happened then here it goes. CH₃OCH₃ , the reactant breaks into CH₃O^- and CH₃⁺ (due to more electronegative Oxygen atom, negative charge is acquired) . Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen). Now, CH₃O^- attacks H⁺ to form CH₃OH and CH₃⁺ attacks I^- to form CH₃I as unlike charges attract. This is the explanation to how this reaction proceeds.

(ii)

- (e)

When 2-methoxypropane() reacts with HI, 2-hydroxypropane() and methyl iodide(CH₃I) is produced.

Now lets see what happens in the reaction. 2-methoxypropane() breaks into a stable 2° carbanion() and CH₃⁺ . This is because due to the –I inductive effect of the –CH₃ group the negative charge on the electronegative Oxygen atom increases. Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen). Now, the carbanion attracts the H⁺ to form 2-hydroxypropane() and CH₃⁺ attacks I^- to form CH₃I as unlike charges attract. This is the explanation to how this reaction proceeds.

(iii)

- (b)

When tert-butyl methyl ether() reacts with HI, tert-butyl iodide ()is produced along with methanol (CH₃OH).

Now, lets see what happens in this reaction.

tert-butyl methyl ether() breaks into a very stable tertiary(3°) carbocation, which is the tert-butyl carbocation() and methoxy ion(CH₃O^-). Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen).

As we know –CH₃ groups show –I inductive effect, due to the presence of 3 such methyl groups the delta positive charge on the carbocation starts diminishing due to which the carbocation becomes highly stable. Thus it would now prefer the attack of the I^- ion and it would result in the formation of tert-butyl iodide (). Along with that the methoxy ion(CH₃O^-) reacts with H⁺ to form methanol (CH₃OH) as unlike charges attract. This is the explanation to how this reaction proceeds.

Note:

tert stands for tertiary.

(iv)

- (a)

When anisole () reacts with HI, phenol () and methyl iodide (CH₃I) are produced.

Lets see how the reaction goes. Anisole breaks down into phenoxide ion (C₆H₅O^-). As we know, phenoxide ion is resonance stabilized , the formation of the phenoxide ion is highly preferred. So we get to know that any substance always want to acquire a state where it gains maximum stability.

Along with that CH₃⁺ carbocation is also formed. Similarly HI breaks into H⁺ and I^- (as Iodine is more electronegative than Hydrogen).

Now, the phenoxide ion() reacts with H⁺ to form a well known compound called phenol (). Also CH₃⁺ carbocation andI^- interact amongst themselves to form methyl iodide(CH₃I) as unlike charges attract. This is the explanation to how this reaction proceeds.

(i) Antifreeze used in car engine - (e) Ethleneglycol

Antifreeze are substances which when added to water lowers the freezing point. Thus they are mostly used in the radiators of a motor vehicle to prevent them from freezing in winters.

Ethylene glycol or ethane-1,2-diol is used as an antifreeze in car engine. It is used as an coolant in winter for car radiators due to its lower freezing point than water.

In summers also, ethylene glycol acts as a coolant to the car radiator and absorbs the heat.

Ethylene glycol

(ii) Solvent used in perfumes – (f) Ethanol

Ethanol (C₂H₅OH) is a good solvent for fatty and waxy substances. They are the main carriers of fragrance oils. When ethanol comes in contact with skin, it gets warmed by the temperature and gets evaporated quickly thereby releasing the fragrance evenly. Moreover, it is not so irritating to the skin. Thus they are used as solvent used in perfumes and so they are known as perfumers alcohols.

(iii) Starting material for picric acid – (d) Phenol

Phenols give 2,4,6-trinitrophenol(Picric acid) when it reacts with a mixture of conc. HNO₃ and H₂SO₄. This is because of the production of negative charge on the three locations of the benzene ring- ortho and para (as shown in the diagram) due to the strong activating effect(+R effect) of the –OH group (due to the presence of lone pairs on Oxygen atom) on the benzene ring towards electrophilic reaction where the electrophile(-NO₂ ) can easily attack.

Such is the effect of –OH group on benzene that instead of producing mono nitro derivative on the ring, the reaction readily proceeds to form tri nitro derivative that is 2,4,6-trinitrophenol.

(iv) Wood spirit – (c) Methanol

Methanol (CH₃OH) is toxic in nature. It is called wood alcohol or wood spirit because earlier it was produced chiefly by the destructive distillation of wood. But today, it is produced industrially by the hydrogenation of carbon monoxide gas.

(v) Reagent used for detection of phenolic group – (a)Neutral ferric chloride

The test for the detection of phenolic group with neutral ferric chloride(FeCl₃) is known as Ferric chloride test.

Phenol gives a violet-coloured water soluble complex with ferric chloride. The complex formed is a coordination compound in which iron is hexavalent.

In general all compounds having the enolic group (having ene and -ol groups) respond to this test.

(vi) By products of soap industry used in cosmetics – (b) Glycerol

Glycerol(propan-1,2,3-triol) is a thick, transparent liquid that is commonly found in soaps. It is a bye product formed during the saponification(process where soaps are produced) process. It is also used in cosmetics.

(i) Methanol – (d) Wood spirit

Methanol (CH₃OH) is toxic in nature. It is called wood alcohol or wood spirit because earlier it was produced chiefly by the destructive distillation of wood. But today, it is produced industrially by the hydrogenation of carbon monoxide gas.

(ii) Kolbe’s reaction - (a) Conversion of phenol to o-hydroxysalicylic acid

In Kolbe’s reaction, first phenol is converted to sodium phenoxide. Then sodium phenoxide is heated with CO₂ at 400K and at a pressure of 4-7atm, due to which sodium salicylate is formed as the major product. This on acidification gives salicylic acid(2-hydroxybenzoic acid) or o-hydroxybenzoic acid.

First phenol reacts with sodium hydroxide (NaOH). Like a normal acid-base reaction, salt and water are produced where salt is sodium phenoxide. Sodium phenoxide on heating with CO₂ at 400K and at a pressure of 4-7atm, sodium salicylate is formed.

This is because CO₂ attacks the ortho position of sodium phenoxide initially and then rearrangement occurs. As CO₂ attacks, the ortho-Hydrogen atom is substituted by the COO^-group(from CO₂) and the H⁺ and Na⁺(from sodium phenoxide) interchange their position to form sodium salicylate.

The acidification occurs (attack of H⁺) and salicylic acid is formed.

(iii) Williamson’s synthesis – (f) Reaction of alkyl halide with sodium alkoxide

Williamson’s synthesis is based on the nucleophilic substitution of alkyl halides(R-X) in which halogen(X) are replaced with alkoxy group (-OR’) where R and R’ are alkyl groups. Thus we get R-OR’ which is an ether. So this method is used in the preparation of ethers.

This method involves the treatment of an alkyl halide with sodium or potassium salt of an alcohol or phenol.

Note:

The halide used in this reaction should be primary otherwise secondary or tertiary halides readily undergo elimination reaction in the presence of a strong base to form alkenes instead of ethers which is not desired.

(iv) Conversion of 2° alcohols to ketones – (e) Heated copper at 573K

When 2° alcohols are subjected to heated Cu turnings at 573K through a tube then dehydrogenation occurs (oxidation reaction).

2° alcohols are oxidized to form ketones having the same number of Carbon atom as the initial alcohol.

Note:

• When primary alcohol is dehydrogenated with Cu at 573K under same conditions then an aldehyde is produced having same number of Carbon atoms as the initial alcohol.

• Tertiary alcohol however does not get hydrogenated due to the absence of α-hydrogen (Hydrogen atom present at the carbon atom adjacent to the carbon atom having the functional group – OH). Instead due to dehydration , an alkene is produced having same number of Carbon atoms as the initial alcohol.

(v) Reimer-Tiemann reaction – (c) Conversion of phenol to salicylaldehyde or 2-hydroxybenzaldehyde.

Treatment of phenol with chloroform(CHCl₃) in the presence of an aqueous alkali at 340K results in the formation of salicylaldehyde as major product and the para isomer 4-hydroxybenzaldehyde as minor product.

This is the step by step reaction mechanism for the reaction.

First an electrophile dichlorocarbene (:CCl₂) is formed from CHCl₃.

After the generation of electrophile, the electrophile attacks the ortho position of the phenoxide ion(which is resonance stabilized). Now is formed and NaOH is made to react with it. Due to which, dehydration takes place and ultimately by acidification, salicylaldehyde is produced.

Reimer Tiemann Reaction.

Note:

Whenever, 2 –OH groups are attached to same Carbon atom, the molecule itself is unstable. To regain its stability, the molecule loses 1 water molecule.

(vi) Fermentation – (b) Ethyl alcohol

Fermentation of carbohydrates (like sugar) yields ethanol containing water. This solution on fractional distillation forms an azeotropic mixture (mixture of liquids that have a constant boiling point) containing 95% ethanol and 5% water. This 95% ethanol is called rectified spirit.

Sucrose is first converted to glucose and fructose with enzyme invertase.

Sucrose Glucose Fructose

Enzyme zymase(present in yeast)then converts glucose and fructose to ethanol.

Fermentation is carried out in anaerobic conditions (in absence of air).

Assertion and reason both are wrong statements.

Addition of water in acidic medium to but-1-ene yields butan-2-ol and not butan-1-ol. This is due to the formation of more stable intermediate 2° carbocation instead of the less stable 1° carbocation as the reaction proceeds in normal acidic condition. This is because it follows Markovnikoff’s Rule of addition.

According to it, it can be deduced that the hydroxyl group (from water) adds to the carbon of alkene having fewer number of hydrogen substituents and the hydrogen ion (from water) adds to the carbon of the alkene with more hydrogen substituents. Thus the formation of butan-2-ol is preferred.

This is the step by step explanation about the formation of butan-2-ol.

When a +R (Positive resonance effect) showing functional group joins phenol or benzene ring, the acidic nature decreases. These groups have their lone pairs with them and they participate in resonance or in delocalization.They show electron releasing tendency. The electrons are released towards benzene ring through delocalization. As a result of which the acidic strength of the compound is reduced due to the destabilization of the phenoxide ion due to concentration of negative charge. So, the acidic strength is much lesser than that of phenol. Eg. –NH_2, -OR, R, OH, where R is an alkyl group. These are called ortho, para directing groups.

Similarly for –R (Negative resonance effect) showing functional groups, when they join phenol, acidic nature of the resulted compound is more. This is because the phenoxide ion is stabilized by the dispersal of negative charge as these groups show electron withdrawing tendency. Eg. –NO_2, -CHO, -CN, -COOH, etc. These are called meta directing groups.

From the equivalent resonance structures above, in the second, third and fourth figure, we observe negative charge is developed at certain position. The position of the charge at the second and fourth position is same and it is termed as ‘ortho’(o) position and the position at the third figure is termed as ‘para’(p) position.

Here, at these positions, the electron density is more and the electrophile is most likely to attack at these positions. So if we substitute any –R effect showing groups(like nitro), at these positions on a phenol, the compound thus formed would be more acidic than phenol as discussed earlier.

So, p-nitrophenol is more acidic than phenol.

Note:

1. There is another position called ‘meta’(m) which lies in between ortho and para.

2. para substituted isomer with electron withdrawing group is more acidic than ortho substituted one. This is due to the more stable intermolecular Hydrogen bonding(bonding with other molecules of the same or different species) for p isomer and intramolecular hydrogen bonding(bonding formed in the same molecule) for o- isomer.

We number the primary chain. We observe a 3 carbon ether group is joined with the second carbon of main chain.

Since ethers have naming conventions like alkoxyalkane where alkoxy represents the carbon number present in the ether part and alkane represents the carbon number joined with it.

So according to it, we get the IUPAC name to be 2-propoxypropane.

And yes, indeed in IUPAC nomenclature, ether is a hydrocarbon derivative where an hydrogen atom is replaced by –OR or –OAr group. As mentioned earlier, these are alkoxy groups as alkane and oxygen both are present simultaneously.

We know ethers are represented by R-O-R (where R is an alkyl group or even Ph group).

We know that the tetrahedral bond angle around oxygen atom is equal to 109°28’.

If we look at the structure, we can observe that the electron pairs(which includes the bond pairs as well as lone pairs) surrounding the Oxygen atom assume tetrahedral arrangement that is sp³ hybrid state. As we know, generally the bond angle observed for sp³ hybrid atom is 109°28’, but in ethers it is not so. In ethers, bond angle is larger. Why is that so?

This deviation in angle caused by the repulsive interactions due to the bulky alkyl groups due to their ‘electron pushing nature’ as we shown in the diagram. This can also be termed ‘steric crowding’ of the alkyl groups.

Bulkier the alkyl group, more would be the repulsion or repulsive interactions and therefore larger would be the deviation in C-O-C angle.

So bond angle comparison is :

ethoxyethane> methoxymethane> methane.

Note:

One would say that we would observe lone pair-lone pair repulsion so if we consider that case the bond angle should definitely be lower than the standard tetrahedral angle.

The explanation is that, yes, we do observe lone pair-lone pair repulsion (lp-lp) but the deviation caused by the steric, bulky alkyl groups are so massive that the lp-lp repulsion cannot overpower it. That is why bond angle is larger.

But if we compare the bond angle of water with that of the standard tetrahedral angle, it is lower due to the same lp-lp repulsion. The reason is same as discussed, that Hydrogen being lighter, experience the repulsions as a result of which the bond angle decreases to around 106°

Boiling points of alcohols are than those of alkanes, haloalkanes or ethers of comparable molecular masses. This is due to the formation of very stable intermolecular hydrogen bonding(bonding with other molecules of the same or different species). because of the presence of electronegative Oxygen atom and –OH group. So due to the intermolecular hydrogen bonding, many similar molecules join due to the formation of these strong bonds. As a result of which, a large amount of energy is required to break these bonds.

Intermolecular Hydrogen bonding in ROH, where R is an alkyl group.

If we see ethers(ROR’), we will find that ethers have much low boiling points as compared to the isomeric alcohols. Unlike alcohols, as we saw earlier, the ethers are not associated by any sort of intermolecular Hydrogen bonding due to the absence of Hydrogen atom with the electronegative Oxygen atom which was present in case of alcohols(-OH).

No H atom is associated with Oxygen atom, so no intermolecular hydrogen bonding observed.

The inter-particle forces existing there are weak dipole-dipole forces.

Due to all these reasons, ethers have low boiling point as compared to their isomeric alcohols.

So what is a Lewis acid?

Lewis acids are electron deficient species. They contain an empty orbital which are capable of accepting an electron pair, from Lewis base(electron enriched species) thereby forming Lewis adduct. Eg. FeCl₃, FeBr_3, BF₃, anhydrous AlCl₃, etc.

Now, we know that benzene reacts by electrophilic substitution reaction. In the bromination reaction of benzene, we need to produce an electrophile which can attach itself on the benzene ring due to which reaction would proceed.

So, how can we generate this electrophile?

For the generation of electrophile, we need Lewis acid(FeBr₃).Since the Lewis acid need electrons, it gets them when the Br-Br bond breaks and FeBr₃ accepts the Br^- and we get FeBr₄^- along with the generated electrophile Br⁺.

Now, this electrophile attacks the benzene ring and through resonance and removal of a H⁺ from the ring, we get bromobenzene.

Now, we are left with a H⁺ and FeBr₄^- . They combine among themselves thereby forming AlCl₃ and HBr.

Since FeBr₃ is formed again, it is said to be a catalyst.

This was the mechanism of bromination of benzene.

The step by step explanation of the formation of bromobenzene.

We got to see that the Lewis acid actually polarizes bromine molecule into Br⁺ and Br^-.

Now, in case of bromination of phenol, we do not need any Lewis acid. This is because of the strong activating effect(+R effect) of the –OH group (due to the presence of lone pairs on Oxygen atom) on the benzene ring towards electrophilic reaction. Thus Lewis acids are not required here to produce electrophiles.

Such is the effect of –OH group on benzene that instead of producing mono bromo derivative on the ring, the reaction readily proceeds to form tri bromo derivative that is 2,4,6-tribromophenol which is observed as a white precipitatewhen the reaction happens with bromine water.

As we discussed earlier the three positions where negative charges are largely located, we can relate here that the electrophiles would love to attack at these three positions thereby giving the tribromo derivative.

Here is the figure once again.

It is not because of the fact that m-nitrophenol and p-nitrophenol exist as associated molecules is the reason why their solubility is higher than o-nitrophenol, although the fact is correct but it is not the proper explanation which suits here.

As we discussed earlier that o-nitrophenol forms intramolecular Hydrogen bonding(bonding formed in the same molecule) due to the closeness of the –NO₂ group with the –OH group. This inhibits its hydrogen bonding with water and reduces its solubility in water.

Thus we get to know that more the stable bond formed between a compound and water, more is its solubility in water.

Now, if we consider p- or m-nitrophenol we observe the formation of intermolecular hydrogen bonding(bonding with many molecules of same or different species). The intermolecular hydrogen bondings thus formed with are more stable and due to which their solubility gets increased.

Now if we look we might think what is an acid?

A simple comparative statement would be a substance is said to more acidic if it de-protonates more easily, or in other words it can release H⁺ easily. After the loss of a H⁺ if one molecule is more stable than the other then it is said to be more acidic.

Now taking phenol(C₆H₅OH).

If it loses a H⁺ it turns into phenoxide ion(C₆H₅O^-).Now, this phenoxide ion is stabilized due to resonance due to the delocalization of the negative charge on the entire ring. A resonance stabilized molecule is said to be highly stable.

Now in ethanol(C₂H₅OH), after de-protonation, ethoxide ion is produced(C₂H₅O^-).

Instead of stabilizing , due to the inductive effect of the bulky ethyl groups the ethoxide ion gets destabilized. If an electron withdrawing group(-NO_2,CN) was present then it would have been stabilized to some extent but as we know resonance stabilization provides much more stability than inductive effect.

As a result of which ethanol is less acidic.

As we discussed earlier that in the presence of bromine water how phenols form a tribromo derivative of 2,4,6-tribromophenol.

What we told was in case of bromination of phenol , we do not need any Lewis acid. This is because of the strong activating effect(+R effect) of the –OH group (due to the presence of lone pairs on Oxygen atom) on the benzene ring towards electrophilic reaction. Thus Lewis acids are not required here to produce electrophiles.

Such is the effect of –OH group on benzene that instead of producing mono bromo derivative on the ring, the reaction readily proceeds to form tri bromo derivative that is 2,4,6-tribromophenol which is observed as a white precipitate when the reaction happens with bromine water .

Now, if we treat phenol with carbon disulphide(CS₂) at 273 K, the reaction is somewhat suppressed and instead of a tribromo derivative, a monobromo derivative is produced because CS₂ is a less polar solvent as compared to bromine water. p-bromophenol is produced as a major product and o-bromophenol is also produced but as a minor product.

In phenol, polarization of bromine molecules take place even in the absence of Lewis acids.

We discussed earlier what are ortho and para directing groups.When a +R (Positive resonance effect) showing functional group joins phenol or benzene ring, the acidic nature decreases. These groups have their lone pairs with them and they participate in resonance or in delocalization. They show electron releasing tendency. The electrons are released towards benzene ring through delocalization. As a result of which the acidic strength of the compound is reduced due to the destabilization of the phenoxide ion due to concentration of negative charge. So, the acidic strength is much lesser than that of phenol. Eg. –NH_2, -OR, R, OH, where R is an alkyl group. These are called ortho, para directing groups.

Phenols give 2,4,6-trinitrophenol(Picric acid) when it reacts with a mixture of conc. HNO₃ and H₂SO₄. This is because of the production of negative charge on the three locations of the benzene ring due to the strong activating effect(+R effect) of the –OH group (due to the presence of lone pairs on Oxygen atom) on the benzene ring towards electrophilic reaction where the electrophile(-NO₂ ) can easily attack.

Such is the effect of –OH group on benzene that instead of producing mono nitro derivative on the ring, the reaction readily proceeds to form tri nitro derivative that is 2,4,6-trinitrophenol.

Same thing we observed in case of the reaction of phenol with bromine water.

Methoxybenzene reacts with hydroiodic acid HI to form phenol and iodomethane.

Mechanism of the above reaction is as follows:

Methoxybenzene is an alkyl aryl ether. There are two bonds linked with an oxygen atom in methoxybenzene. One is an oxygen-methyl group( O-alkyl bond) and the other one is Oxygen-Aryl group( O-Aryl bond). Out of these, Oxygen-Aryl bond is more stable due to resonance having partial double bond character. So in this reaction, the oxygen-methyl bond gets cleaved to form methyl iodide and phenol.

The reaction of HI with methoxybenzene takes place in two steps.

Step1:

The oxygen of ether gets protonated (addition of proton) by capturing H⁺ of HI to form protonated ether molecule, known as oxonium ion.

Step 2:

In this step, SN² reaction occurs since Iodide ion (I^-) is a good nucleophile. I^- ion preferably attacks the least substituted carbon of protonated ether molecule ( in this case methyl group) to form methyl iodide and phenol.

(a) The starting material is Cumene (isopropylbenzene) for the industrial preparation of phenol.

(b) The reaction of bromine with phenol is known as bromination of phenol.

Bromination can occur in two ways.

i) In an aqueous medium – In aqueous solution, bromine water( Br₂ + H₂O)is formed which reacts with phenol to form trisubstituted phenol called 2,4,6-tribromo phenol.

ii) In a non-aqueous medium – In absence of water, phenol reacts with low polar solvents like CS₂ or CHCl₃ to form two monosubstituted (ortho and para) bromophenols.

(c) In an Electrophilic substitution reaction, lewis acid is used to generate the required electrophile (E⁺)by means of polarization.During bromination of phenol, the oxygen atom of phenol can polarize Br-Br bond to generate Br⁺ ion. So, the presence of lewis acid is not required.

This process continues which leads to the formation of trisubstituted product 2,4,6-tribromophenol.In aqueous solution, phenoxide ion donates electrons to the benzene ring to a large extent, thus the ring gets activated. But in a non-polar solvent, only -OH group donates electrons to ring to a small extent. Thus monosubstituted product o-bromophenol and p- bromophenol are formed. p- bromophenol is formed as a major product.

The structure of Aspirin is:

Conversion of phenol to Aspirin is done in two steps.

Step1

First, phenol is converted to Salicylic acid by the following reaction:

Step2

Salicylic acid on acetylation( introducing acetyl group –COCH₃) using acetic anhydride produces Aspirin.

Biocatalysts are complex organic compounds. In living organisms, they can act as catalysts for various biochemical reactions. Enzymes are one of the most important biocatalysts which are used for the manufacture of Ethanol.The following method is used to produce ethanol from sugar solution( molasses).

Molasses is a non-crystalline form of sugar which is obtained from mother liquor of sugar solution. Mother liquor can be prepared by crystallization of sugar from the sugar solution. This mother liquor contains about 50% sugar. It is then diluted to almost 10%.In this diluted solution, yeast is added and it is kept for 2 to 3 days. Yeast is a microorganism which produces the enzymes zymase and invertase. These two enzymes perform the following conversions given below:

If the concentration of ethanol exceeds 15 %, the enzyme gets inhibited. This whole process is carried out in the absence of air under anaerobic condition. In aerobic condition(in presence of oxygen) ethanol is oxidized to produce ethanoic acid.