NCERT Solutions for Class 12 Science Chemistry Chapter 11 - Alcohols, Phenols And Ethers

Answer:

Monochlorination of toluene in the presence of sunlight yields benzyl chloride. On hydrolysis of benzyl chloride mucleophilic substitution reaction takes place and benzyl alcohol is obtained as the final product.

Hence, the correct answer is option (iv)

Answer:

The structural isomers of butanol (C₄H₁₀O) are as follows:
 
 
 
So, among the isomers of  butanol, only one isomer is chiral i.e., Butam-2-ol.

Hence, the correct answer is option (i).

Answer:

The given nucleophilic substitution reaction takes through the formation of a intermediate carbocation and among the carbocations tertiary carbocation (3º) is most stable followed by secondary (2º) and primary (1º) respectively. The higher the stability of carbocation higher will be the reactivity of reactant molecule. Therefore, the order of reactivity of alcohol for the given reaction is 3º > 2º> 1º.

Hence, the correct answer is option (iii).

Answer:

Partial oxidation of alcohol to obtain aldehyde is done in the presence pyridinium chlorochromate (PCC).


Hence, the correct answer is option (iii).
 

Answer:

The conversion of alkyl halide to alcohol is an example of substitution reaction. In this reaction, the halide group of the alkyl halide is replaced by – OH

Hence, the correct answer is option (ii).

Answer:

Phenol is also knowns as carbolic acid and is not considered as aromatic alcohol. Also, any alcohol in which the hydroxyl group is attached to a carbon which itself is bonded to an aromatic ring is regarded as aromatic in nature. Therefore, benzyl alcohol and its derivatives are aromatic in nature. Thus, compounds (B) and (C) are aromatic in nature. 

Hence, the correct answer is option (iii).

Answer:

The IUPAC name of the given compound is 5-chlorohexan-2-ol.

Hence, the correct answer is option (iii).

Answer:

The structure of m-cresol is as follows:

IUPAC Name: 3-Methylphenol
Hence, the correct answer is option (i)

Answer:

The IUPAC name of the given compound is 2-Methoxypropane
Hence, the correct answer is option (iii)

Answer:

The conjugate base for a weak acid is quite strong. Amongst the given options, ROH is the weakest acid, therefore its conjugate base RO^– the strongest base.

Hence, the correct answer is option (ii).

Answer:

Phenol due to its acidic nature reacts readily with sodium hydroxide solution in water to give sodium phenoxide ion 

Hence, the correct answer is option (i)

Answer:

The presence of electron withdrawing group at ortho or para position increases the acidic strength of phenol. Therefore, o-nitrophenol is more acidic than phenol, due to the presence of electron withdrawing nitro-group (–NO₂) at ortho position.

Hence, the correct answer is option (ii).

Answer:

Due to the higher electronegativity of sp² hybridised carbon of phenol to which –OH is attached, electron density decreases on oxygen. This increases the polarity of O–H bond and results in an increase in ionisation of phenols than that of alcohols. The acidity of phenol is more than the alcohols due to delocalisation of electrons over the benzene ring.  Also, the presence of an electron withdrawing group such as –Cl contributes in increasing the acidity of phenol. Therefore, m-chlorophenol is most acidic amongst the given options.

Hence, the correct answer is option (iv).

Answer:

The presence of electron withdrawing group like –NO₂ on phenol increases the acidic strength of phenol whereas the presence of electron donating group like –OCH₃ decreases the acidic strength of phenol. The presence of electron withdrawing group has a more pronounced effect, if present at ortho and para position. Therefore, the correct order of decreasing acidic strength is as follows:
(b) > (d) > (a) > (c) > (e)

Hence, the correct answer is option (ii).

Answer:

The substitution reaction of alcohol with HBr/HCl takes place through the formation of carbocation. Higher the stability of carbocation, the more favourable is the reaction. Also, the presence of an electron withdrawing group contributes in decreasing the reactivity of the alcohol for the substitution. Since –NO₂ group is stronger electron withdrawing group than –Cl, therefore, the stability of carbocation increases in the order as follows:
(b) < (c) < (a)

Hence, the correct answer is option (iii).

Answer:

Boiling point increases with increase with increase in the molecular mass of alcohol. It also decreases with branching due to the decrease in surface area on branching. Therefore, the correct order of increasing boiling points for the given alcohols is as follows:
Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol

Hence, the correct answer is option (i).

Answer:

Aldehydes can be converted into alcohol through reduction reaction. The reduction can be carried out by adding hydrogen in the presence of a catalysts like Pd, Pt or Ni or by using a reducing agent like NaBH₄ or LiAlH₄.

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

Answer:

The reactions which would yield phenol as a product are as follows:






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

Answer:

The reagents that would oxidise primary alcohols to aldehydes are as follows:
(i) CrO₃ in anhydrous medium

(ii) Pyridinium chlorochromate

(iii) Heat in the presence of Cu at 573 K

Hence, the correct answers are options (i), (iii) and (iv).

Answer:

Phenol on reaction with neutral FeCl₃ gives violet color due to formation of a complex.
$3\mathrm{Ar}-\mathrm{OH} + {\mathrm{FeCl}}_3\left(\mathrm{neutral}\right) \to {(\mathrm{Ar}-\mathrm{O})}_3-\mathrm{Fe} + 3\mathrm{HCl}$
It also reacts with bromine water form a white precipitate of tribromo derivative of phenol.

Ethanol does not give these tests.

Hence, the correct answers are options (i) and (iii).

Answer:

For an alcohol to be benzylic alcohol, the −OH group is attached to sp³ hybridised carbon atom next to an aromatic ring. Amongst the given compounds, compounds (ii) and (iii) have –OH group attached. to sp³ hybridised carbon atom next to aromatic ring. Therefore, they are benzylic alcohols.

Hence, the correct answers are options (ii) and (iii). 

Answer:

The structure and IUPAC name of glycerol are as follows:

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

Answer:

(A)
 
3-Ethyl-5-methylhexan-2, 4-diol

(B)
 

Answer:

The IUPAC name of the compound is given below.

Answer:

Alcohol forms hydrogen bonds with the molecules which contributes to the solubility of alcohols in water. The solubility of alcohol decreases with the increases in the size of alkyl, aryl group attached to the main chain.

Answer:

Commercially used alcohol is made unfit for drinking by mixing some methanol to it. This is called denaturation of alcohol and the alcohol obtained is called denatured alcohol.

Answer:

PCC (pyridinium chlorochromate) is a complex of chromium trioxide with pyridine and HCl. It is a good reagent used to convert primary alcohols to aldehydes.

Answer:

The acidic character of alcohols is due to the polar nature of O–H bond. Chlorine atom is an electron withdrawing group which increases the polarity of O–H bond. Thus, 2-Chloroethanol is more acidic than ethanol due to −I effect of chlorine atom.

Answer:

PCC (pyridinium chlorochromate) can be used as a reagent to convert to ethanol to ethanal
$\underset{\mathrm{Ethanol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}} \stackrel{\mathrm{PCC}}{\to } \underset{\mathrm{Ethanal}}{{\mathrm{CH}}_3\mathrm{CHO}}$

Answer:

Ethanol can be converted to ethanoic acid using a strong oxidising agent like acidified K₂Cr₂O₇ or KMnO₄.
$\underset{\mathrm{Ethanol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}} \underset{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7/{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{{\mathrm{KMnO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }} \underset{\mathrm{Ethanoic} \mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}}$

Answer:

Ortho-nitrophenol is more volatile due to intra-molecular hydrogen bonding than para-nitrophenol as in p-nitrophenol, inter-molecular hydrogen bonding exists.

Answer:

The acidity of alcohol is due to polar nature of O — H bond. Higher the polarity of O — H bond, more will be the acidity.
o-Nitrophenol is more acidic than o-cresol due to –I and –R of nitro group that contributes in decreasing the electron density on oxygen tending to increase the polarity of O — H bond and leads to the effective delocalisation of negative charge in phenoxide ion.

On the other hand, methyl group in o-cresol shows +I and +R effect increases electron density on oxygen tending to decrease the polarity of O — H bond and do not favour the formation of phenoxide ion. Thus, o-cresol is less acidic than o-nitrophenol.

Answer:

The reaction that takes place when phenol is treated with bromine water is as follows:

Answer:

The presence of electron donating alkyl group in o-cresol intensifies the charge on phenoxide ion and reduces the stability of o-cresol as compared to phenol. On the other hand, the presence of electron withdrawing nitro group (–NO₂) disperses the negative charge and stabilizes the phenoxide ion. As a result, o-nitrophenol is more acidic than phenol. Hence, increasing order of acidity is a follows:
o-Cresol < Phenol < o-Nitrophenol

Answer:

Alcohol reacts with active metals like Na, K etc. to form corresponding alkoxides. The reaction is as follows:

The order of reactivity of alcohol towards the active metals is as follows:
Primary > Secondary > Tertiary
This can be explained on the basis of cleavage of O – H bond in alcohol. The presence alkyl group contributes in increasing the electron density around the oxygen atom of O – H bond due to +I effects shown by alkyl groups. As a result, the O – H bond is hard to cleave. Therefore, greater the number of alkyl group present, lesser will be reactivity of alcohol.

Answer:

Phenol is obtained as a product on heating benzene diazonium chloride with water.

Answer:

A stronger acid has the ability to displace a weaker acid from its salt.
On treating RO⁻Na⁺ with H₂O, it gives ROH. Hence, water is a stronger acid than ROH.
$$\begin{gathered} \mathrm{HOH} + {\mathrm{RO}}^{-}{\mathrm{Na}}^{+} \underset{}{\to } \mathrm{NaOH} + \mathrm{ROH} \\ \mathrm{Stronger} \mathrm{acid} \mathrm{Weaker} \mathrm{acid} \end{gathered}$$
Also, on treating sodium ethynide (CH≡C⁻Na⁺) with water and alcohol, acetylene (CH≡CH) is obtained as a product.
$$\begin{gathered} \mathrm{HOH} + \mathrm{HC}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+} \underset{}{\to } \mathrm{NaOH} + \mathrm{HC}\equiv \mathrm{CH} \\ \mathrm{Stronger} \mathrm{acid} \mathrm{Weaker} \mathrm{acid} \end{gathered}$$
$$\begin{gathered} \mathrm{ROH} + \mathrm{HC}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+} \underset{}{\to } \mathrm{NaOR} + \mathrm{HC}\equiv \mathrm{CH} \\ \mathrm{Stronger} \mathrm{acid} \mathrm{Weaker} \mathrm{acid} \end{gathered}$$
So, acetylene (CH≡CH) is least acidic.
Therefore, the deceasing order of acidity of the given compounds is .

Answer:

Invertase enzyme is used in the conversion of sucrose to glucose and fructose.
$\underset{\mathrm{Sucrose}}{{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}} + {\mathrm{H}}_2\mathrm{O} \stackrel{\mathrm{Invertase}}{\to } \underset{\mathrm{Glucose}}{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6} + \underset{\mathrm{Fructose}}{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}$
Glucose and fructose undergo fermentation in the presence of zymase enzyme, which is present in yeast.
$\underset{}{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6} \stackrel{\mathrm{Zymase}}{\to } \underset{\mathrm{Ethanol}}{2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} + }2{\mathrm{CO}}_2$

Answer:

The conversion of propan-2-one to tert-butyl is shown below.

Answer:

The isomeric structures of alcohol with molecular formula C₄H₁₀O are as follows
 
 
 
The compound consisting of Chiral carbon are usually optically active compounds. Hence, from the isomers of C₄H₁₀O, only Butan-2-ol is optically active.

Answer:

In the structure of phenol, the lone pairs of electron of oxygen atoms of –OH group are in resonance with the π-electrons of the benzene ring as shown below.

As depicted from the resonating structures, in structure II, III and IV, bond between carbon and oxygen acquires partial double bond character due to resonance. But in alcohols, carbon-oxygen bond is purely single bond.
Therefore, the –OH group in phenol is more strongly held as compared to –OH group in alcohol.

Answer:

Due to resonance and dispersal of negative charge over the benzene ring in phenols, the nucleophile is repelled from attacking the electron rich ortho and para positions. As a result, phenols do not give nucleophilic substitution reactions.

Answer:

The mechanism of preparation of alcohols from alkenes is as follows:

Step 1 − Protonation of alkene by electrophilic attack of H­₃O⁺ to form carbocation

Step 2 − Nucleophilic attack of water on the carbocation

Step 3 − Deprotonation to form alcohol

Answer:

The dipole moment is CO₂ is zero due to its linear geometry, the polarities of two C=O bonds get cancelled and the molecular is non-polar.

The structure of ether is similar to water and has bent structure. Thus, the polarity of two R – O groups does not get cancelled. Hence, the molecule is polar.

Answer:

The reaction of alcohol with Lucas reagent (Conc. HCl and ZnCl₂) proceeds through the formation of carbocation intermediate.
The stability of carbocation intermediate determines the rate of the reaction, i.e., more stable the intermediate carbocation, more reactive is the alcohol.
Tertiary carbocation being most stable among three class of carbocations are found to be most reactive towards the Lucas reagent.
Thus, the order of reactivity of three classes of alcohols is 3° > 2° > 1°.

Answer:

The reactions for conversion of phenol to aspirin are as follows:

Answer:

Nitration of phenol is easy as compared to benzene because the presence of –OH group in phenol increases the electron density at ortho and para position of the benzene ring due to the +R effect. As a result, nitronium electrophile can readily attack the benzene ring of phenols.

Answer:

In phenoxide ion, the negative charge of oxygen is transferred to benzene ring through resonance. This helps the weak electrophile CO₂ to easily get attached to the benzene ring, finally giving salicylic acid. Therefore, phenoxide ion is more reactive than phenol towards electrophilic substitution reaction and used in Kolbe’s reaction.

Answer:

In case of phenol, the  –OH group is attached to the carbon of benzene ring, which is sp² hybridised and thus, benzene ring produces an electron withdrawing effect. Whereas, in case of methanol, the –OH group is attached to the carbon atom which is sp³ hybridised and thus, it produces an electron releasing effect. Hence, the C–O bond in phenol is less polar than the C–O bond in methanol and the dipole moment of phenol is smaller than that of methanol.

Answer:

Alkoxides are not only nucleophiles but also strong bases. When they react with tertiary alkyl halides, they leads to elimination reaction.
When tert-butyl-bromide react with sodium-tert-butoxide, elimination takes place instead of substitution. Thus, isobutylene is formed instead of di-tert-butyl ether.

Answer:

The tetrahedral bond angle is 109.5°. The C — O — H bond angle in alcohols is slightly less than tetrahedral angle due lone pair -lone pair repulsions between the unshared electron pairs of oxygen. For example, the C — O — H bond angle in methanol is 108.9°.
But in case of ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement and the C — O — C bond angle is slightly greater than tetrahedral angle because of greater repulsion between the two bulky alkyl groups. The C — O — C bond angle is 111.7° in ethers.

Answer:

The solubility of alcohols in water depends upon the extent of hydrogen bonding. The alcohols with lower molecular mass will have smaller hydrocarbon part (hydrophobic part) and thus, the tendency to form hydrogen bonds is much higher in such alcohols and therefore, they are more soluble in water.

Answer:

The nitro group in p-Nitrophenol is an electron withdrawing group. So, it produces –I and –R effect. Thus, reducing the electron density on the oxygen of the O — H bond of phenol and due to the effective delocalisation of negative charge in phenoxide ion, making it more acidic.

Answer:

Ethers have low polarity in comparison to alcohols. As a result, they do not show association by inter-molecular hydrogen bonding. Therefore, ethers have low boiling points than alcohols of comparable molecular masses.

Answer:

In phenol the oxygen is attached to sp² hybridised and is more electro-negative as compared to sp³ hybridised carbon atom in methanol. As a result, the C — O bond in phenols is more polar as compared to C — O bond in methanol. Also, in phenol, the unshared electron pairs over oxygen conjugates with carbon of benzene ring, which develops a partial double bond character of  C — O bond. 

But no such conjugation is observed in methanol. Hence, the carbon-oxygen bond in phenol is slightly stronger than that in methanol.

Answer:

Phenoxide ion obtained after the removal of H⁺ from phenol and is resonance stabilized.

Ethoxide ion will be obtained by the removal of H⁺ and is destabilised by +I effect of ethyl group which increases electron density on oxygen tending to decrease the polarity of O-H bond. 
 
So, ethanol is less acidic than phenol.
Also, water is a better proton donor (i.e., stronger acid) than ethanol.
$\underset{\mathrm{Base}}{{\mathrm{C}}_2{\mathrm{H}}_5\stackrel{-}{\mathrm{O}} }+ \underset{\mathrm{Acid}}{\mathrm{H}-\underset{··}{\stackrel{··}{\mathrm{O}}}-\mathrm{H}} \to \underset{\mathrm{Conjugate} \mathrm{acid}}{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} } + \underset{\mathrm{Conjugate} \mathrm{base}}{\stackrel{-}{\mathrm{O}}\mathrm{H}}$
In the above reaction, it is observed that an ethoxide ion is a better proton acceptor than hydroxide ion, which suggests that ethoxide is a stronger base than hydroxide.
Hence, the increasing of acidity is ethanol < water < phenol.

Answer:

Column I	Column II
(i)	(d) o-cresol
(ii)	(c) Catechol
(iii)	(f) Resorcinol
(iv)	(a) Hydroquinone
(v)	(g) Anisole
(vi)	(b) Phenetole

Answer:

	Column I		Column II
(i)		(d)
(ii)		(e)
(iii)		(b)
(iv)		(a)

Answer:

(i) A small percentage of ethylene glycol is used as antifreeze in car engines.
(ii) Ethanol is used as a solvent in perfumes as it can dissolve fatty and waxy substances very easily. These fatty and waxy substances provide odour to perfumes. Apart from being a good solvent, it cause less irritation to the skin.
(iii) On nitration with conc. HNO₃, phenol is converted into picric acid.

(iv) Methanol is known as wood spirit as it is obtained by destructive distillation of wood.
(v) Neutral ferric chloride changes its color to purple when treated with phenol. This test is used as a detection test for phenols.
(vi) Soaps are prepared by the reaction of fatty acids with NaOH. 

Answer:

(iii) 

(v) 

(vi) 

Answer:

The product of addition reaction of water to but-1-ene is butan-2-ol because the reaction proceeds in acidic medium through the formation secondary carbocation in accordance with Markovnikov’s rule.
$\underset{\mathrm{But}-1-\mathrm{ene}}{{\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_3} +{\mathrm{H}}_2\mathrm{O} \stackrel{{\mathrm{H}}^{+}}{\to } \underset{\mathrm{Butan}-2-\mathrm{ol}}{{\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_2-{\mathrm{CH}}_3}$

Hence, the correct answer is option (ii).

Answer:

The nitro group in p-nitrophenol is an electron withdrawing group. So, it produces –I and –R effect. Thus, reducing the electron density on the oxygen of the O — H bond of phenol and due to the effective delocalisation of negative charge in phenoxide ion, making it more acidic than phenol.


Hence, the correct answer is option (i).

Answer:

According to IUPAC system of nomenclature, ethers are regarded as hydrocarbon derivatives in which a hydrogen atom is replaced by an –OR or –OAr group, where R and Ar represent alkyl and aryl groups, respectively. The larger (R) group is chosen as the parent hydrocarbon.

The IUPAC name of the compound is 2-propoxypropane.

Hence, the correct answer is option (iv).

Answer:

In ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle (109^o-28′) due to the repulsive interaction between the two bulky (–R) groups. The C — O — C bond angle is 111.7° in methoxymethane.


Hence, the correct answer is option (iv).

Answer:

Alcohols of comparable molecular mass have higher boiling points than the corresponding ethers due to the formation of intermolecular hydrogen bonding in alcohols but not in ethers.


Hence, the correct answer is option (ii).

Answer:

The bromination  of benzene takes place in the presence of a Lewis acid, such as FeBr₃, which polarises the bromine molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring.

Hence, the correct answer is option (iv).

Answer:

 o-Nitrophenol undergoes intramolecular hydrogen bonding. As a result, it does not from hydrogen bonds with water and is less soluble than the m- and p-isomers.


But m- and p-Nitrophenol forms hydrogen bonds with water. Due to the presence of intermolecular H-bonding, m-Nitrophenol and p-Nitrophenol exists as associated molecules.


Hence, the correct answer is option (v).

Answer:

Phenol is more acidic than ethanol as phenoxide ion is stabilised by resonance while ethoxide ion is destabilised by electron releasing effect of ethyl group and the negative charge is localised on oxygen. This decreases the acidic strength of ethanol.


Alcohols and phenols react with active metals such as sodium, potassium and aluminium to yield corresponding alkoxides/phenoxides and hydrogen. Sodium ethoxide can be prepared by the reaction of ethanol with sodium as follows:

Hence, the correct answer is option (iii).

Answer:

In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring.
Phenol forms 2, 4, 6–tribromophenol on treatment with Br₂ in water. 


But phenol on treatment with Br₂ in the presence of CS₂ form o-Bromophenol and p-Bromophenol.



Hence, the correct answer is option (ii).

Answer:

Phenol gives o-, p- nitrophenol on nitration with dil. HNO₃ and H₂SO₄ at 298 K.


The –OH group attached to the benzene ring in phenol activates it towards electrophilic substitution. Also, it directs the incoming group to ortho and para positions in the ring as these positions become electron rich due to the resonance effect caused by –OH group. Thus, — OH group in phenol is ortho-pare directing.

Hence, the correct answer is option (iv).

Answer:

The mechanism of the reaction of HI with methoxybenzene is as given below:
Protonation of anisole gives methylphenyl oxonium ion.



In this ion, the bond between O—C₆H₅ is stronger than the bond between O—CH₃. This is due to partial double bond character in O — C₆H₅ due to resonance between the lone pair of electrons on the O atom and the sp² hybridised carbon atom of the phenyl group. Therefore, I⁻ prefers to attack by breaking the weaker O—CH₃ bond. Thus, forming methyl iodide and phenol.



 

Answer:

(a) Cumene (isopropyl benzene) is the starting material used in the industrial preparation of phenol.



(b) Reactions for bromination of phenol are as follows:
(i) In aqeous medium



(ii) In non-aqeous medium



(c) In case of phenol, the highly activating effect of –OH group attached to the benzene ring polarises the bromine molecule to form Br⁺ ion, which acts as an electrophile. Therefore, Lewis acid is not required in the bromination of phenol.

Answer:

Phenol is first converted into salicylic acid using sodium hydroxide and carbon dioxide reagents.


The salicylic acid is treated with acetic anhydride. On acetylation of salicylic acid, aspirin is formed.
 

Answer:

Biocatalysts are the substances obtained from some natural sources that include enzymes from biological source or a whole cell. Enzymes work in enhancing the rate of a chemical reaction.
The process of fermentation of molasses yield ethanol. It occurs by breaking down of large molecules into simple ones in the presence of enzymes. The source of enzymes is yeast in fermentation. Yeast consists of invertase and zymase enzyme.
The enzyme invertase hydrolyses sucrose to glucose and fructose.
$\underset{\mathrm{Sucrose}}{{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{12}}+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{Invertase}}{\to }\underset{\mathrm{Glucose}}{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}+\underset{\mathrm{Fructose}}{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}$
Whereas, enzyme zymase converts glucose into ethanol.
${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\stackrel{\mathrm{Zymase}}{\to }2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+2{\mathrm{CO}}_2$
For wine making, the grapes are the main source of sugar and yeast. As the grapes ripen, the content of sugar increases and the yeast grows on the outer skin. When grapes are crushed, sugar and the enzyme come in contact and fermentation starts. Fermentation takes place in anaerobic conditions i.e. in absence of air. Carbon dioxide is released during fermentation
Further fermentation is inhibited by zymase once the percentage of alcohol exceeds 14 percent in wine.