Chemistry-class-12-html_C-12-ALCOHOLS-PHENOLS-AND-ETHERS

<p><Success style="float:center; font-size: 25px; text-align:center;"><b> <strong>ALCOHOLS PHENOLS AND ETHERS</strong></b></Success></p>
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<p><strong>Answer</strong> (d)</p>
<p>Monochlorination of toluene in sunlight gives benzyl chloride. On hydrolysis with aq. $NaOH$, benzyl chloride shows nucleophilic substitution reaction to give benzyl alcohol.</p>
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<p>So, the correct option is (d).</p>
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<p><strong>Answer</strong> (a)</p>
<p>The three isomers of butanol are possible. Structural formula of these isomers are given below</p>
<p>(i) $\underset{\text{ Butan-1-ol }}{CH_3 CH_2 -CH_2-CH_2 OH}$</p>
<p>No carbon is chiral in this compound as none of the four carbon is bonded to four different substituents.
(ii)</p>
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<p><strong>Answer</strong> (c)</p>
<p>The given reaction is nucleophilic substitution reaction in which $-OH$ group is replaced by $-Cl$. Tertiary alcohols, when react with $HCl$ in presence of $ZnCl_2$, form tertiary carbocation.</p>
<p>This intermediate $3^{\circ}$ carbocation is more stable than $2^{\circ}$ carbocation as well as $1^{\circ}$ carbocation. Higher the stability of intermediate, higher will be the reactivity of reactant molecule.</p>
<p>So, the order of reactivity of alcohols in the given reaction is $3^{\circ}>2^{\circ}>1^{\circ}$ and correct option is (c).</p>
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<p><strong>Answer</strong> (c)</p>
<p>Ethanal $(CH_3 CHO)$ is an oxidised product of ethanol. Pyridinium chlorochromate $(C _5 H _5 \stackrel{+}{N} HCl \overline CrO _3)$ oxidises primary alcohols to aldehydes. Strong oxidising agents such as $KMnO_4$ are used for getting carboxylic acid from alcohols.</p>
<p>The oxidation process can be stopped at the aldehyde stage if $Cr(VI)$ reagents such as Collin’s reagent $(CrO_3 \cdot 2 C_5 H_5 N)$, Corey’s reagent or pyridinium chlorochromate and pyridinium dichromate $[(C_5 H_5 NH_2)^{2+} Cr_2 O_7^{2-}]$ in anhydrous medium are used as the oxidising agent. So, the correct option is (c).</p>
<p>$\underset{Ethanol}{CH_3CH_2OH} \xrightarrow{PCC} \underset{Ethanal}{CH_3CHO}$</p>
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<p><strong>Answer</strong> (b)</p>
<p>An addition reaction is a reaction where two or more molecules combine to form a larger one. These reactions occur to change the unsaturated compound to saturated compound. In dehydrohalogenation reaction, alkyl halides give alkenes.</p>
<p>Rearrangement gives the structural isomers of the reactant while in substitution reaction one of the group or an atom is replaced by other group. Therefore, the process of converting alkyl halides into alcohols involves substitution reaction.</p>
<p>Reaction of alkyl halides</p>
<p>$
\underset{\text{ Alkyl halide }}{R-X} \xrightarrow{OH^{-}}  \underset{\text{ Alcohol }}{R-OH}
$</p>
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<p><strong>Answer</strong> (c)</p>
<p>Phenol is also known as, carbolic acid’ cannot be considered as aromatic alcohol. It is quite separate branch of compound called phenols. So, compound (A) i.e., phenol and compound $(D)$ i.e., a derivative of phenol cannot be considered as aromatic alcohol.</p>
<p>On the other hand, compound $(B)$ and $(C),-OH$ group is bonded to $s p^{3}$ hybridised carbon which inturn is bonded to benzene ring. Hence, the correct option is (c).</p>
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<p><strong>Answer</strong> (a)</p>
<p>Phenol is more acidic in nature because by the loss of one proton, it gives phenoxide ion. This phenoxide ion is resonance stabilised. As phenoxide ion is a stable intermediate so, the tendency to give proton is more in phenol than the others. Phenols being more acidic than alcohols, dissolves in $NaOH$.</p>
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<p>Therefore, option (a) is the correct.</p>
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<p><strong>Answer</strong> (b)</p>
<p>In o-nitrophenol, nitro group is present at ortho position. Presence of electron withdrawing group at ortho position increases the acidic strength. On the other hand, in o-methylphenol and in o-methoxyphenol, electron releasing group $(-CH_3,-OCH_3)$ are present.</p>
<p>Presence of these groups at ortho or para positions of phenol decreases the acidic strength of phenols. So, phenol is less acidic than o - nitrophenol.</p>
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<p><strong>Answer</strong> (d)</p>
<p>Alpha carbon of benzyl alcohol and cyclohexanol is $s p^{3}$ hybridised while in phenol and $m$-chlorophenol, it is $s p^{2}$ hybridised. In $m$-chlorophenol electron withdrawing group $(-Cl)$ is present at meta position.</p>
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<p>Presence of electron withdrawing group increases the acidic strength. So, $m$-chlorophenol is most acidic among all the given compounds. The correct option is (d).</p>
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<p><strong>Answer</strong> (b)</p>
<p>Presence of electron withdrawing group on phenols, increases its acidic strength. So, both compounds i.e., $p$-nitrophenol (II) and $m$-nitrophenol (IV) are stronger acid than (I). If this $-NO_2$ group is present at $p$-position, then it exerts both $-I$ and $-R$ effect but if it is present at meta position, then it exerts only $-I$ effect. Therefore, $p$-nitrophenol is much stronger acid then $m$-nitrophenol.</p>
<p>On the other hand, presence of electron releasing group on phenol, decreases its acidic strength. If $-OCH_3$ group is present at meta position, it will not exert $+R$ effect but exert $-I$ effect.</p>
<p>But, if it is present at para position, then it will exert $+R$ effect. Therefore, $m$-methoxy phenol is more acidic than $p$-methoxy phenol.</p>
<p>Hence, the correct option is (b).</p>
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<p><strong>Answer</strong> (c)</p>
<p>Reaction of the given compounds with $HBr / HCl$ is a nucleophilic substitution reaction. It follows $S_N{ }^{1}$ mechanism. $S_N{ }^{1}$ mechanism depends upon the stability of carbocation. Presence of electron withdrawing group decreases the stability of carbocation. In compound (II) and (III) EWG is present at para position.</p>
<p>Since, $-NO_2$ group is a stronger EWG than $-Cl$. So,</p>
<p>$NO_2-C_6 H_5-\stackrel{+}{C} H_2$ carbocation will be less stable than</p>
<p>$Cl-C_6 H_5-\stackrel{+}{C} H_2$ carbocation.</p>
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<p><strong>Ques.</strong> Arrange the following compounds in increasing order of boiling point. Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol</p>
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<p>(a) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol</p>
<p>(b) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol</p>
<p>(c) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol</p>
<p>(d) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol</p>
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<p><strong>Answer</strong> (a)</p>
<p>Boiling point increases with increase in the number of carbon atoms because of increase in molecular mass. So, the boiling point of pentan-1-ol is more than that of all other given compounds. Further, among isomeric alcohols, $1^{\circ}$ alcohols have higher boiling points than $2^{\circ}$ alcohols because of higher surface area in $1^{\circ}$ alcohols. Therefore, boiling point increase in the order</p>
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**Answer** $(a, b, c)$

Conversion of aldehyde into alcohol is a reduction reaction. This reduction can be carried out by adding hydrogen in presence of finely divided metal catalyst such as platinum, palladium or nickel.

$
RCHO+\xrightarrow{H_2 / Pd} R CH_2 OH
$

It can also be prepared by using $NaBH_4$ and $LiAlH_4$ as a reducing agent.

$ RCHO \xrightarrow{NaBH_4} RCH_2 OH $

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Haloarenes are less reactive towards nucleophilic substitution reaction. Therefore, it requires very high temperature and pressure. Chlorobenzene does not undergo hydrolysis on treatment with aq. $NaOH$ at $298 K$ and $1 atm$.

Therefore, correct options are (a), (b) and (c).

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**Answer** $(b, c)$

In benzylic alcohols, the $-OH$ group is attached to a $s p^{3}$ hybridised carbon atom next to an aromatic ring. In compound (a) and (d), the $-OH$ group is attached to a $s p^{3}$ hybridised carbon atom but this carbon is not attached to the benzene ring.

On the other hand, in compound (a) and (c), the $-OH$ group is attached to a $s p^{3}$ hybridised carbon atom next to an aromatic ring.

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**Answer**

Solubility of alcohols in water depends upon the two factors.

(i) Hydrogen bonding Higher the extent of hydrogen bonding, higher is the solubility. The hydrogen group of alcohol form hydrogen bonding. Solubility increases with increase in the number of $-OH$ groups in alcohols of comparable molecular masses.

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/467d8c5f-800b-466e-b843-145c3469b300/public)

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**Answer**

Alcohol is used in large quantities in the manufacture of alcoholic liquors. Its continuous use damages the various vital organs. Therefore, to refrain the people from drinking alcohol, heavy excise duty is levied on the sale of alcoholic beverages. But, it is used in various industries as it is a very good solvent.

Therefore, industrial alcohol must be cheap. Thus, to provide cheaper alcohol to industries and to refrain people from drinking alcohol, it is mixed with some copper sulphate, pyridine, methyl alcohol or acetone.

Alcohol is made unfit for drinking by mixing some quantity of any of these substances in it. This is called denatured alcohol.

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**Answer**

2-chloroethanol is more acidic than ethanol. Due to $-I$ effect (electron withdrawing group) of the $Cl$-atom electron density in $O-H$ bond decreases. So, $O-H$ bond of 2-chloroethanol becomes weaker than $O-H$ bond of ethanol. Thus, 2-chloroethanol is more acidic than ethanol.

$
\mathrm{Cl} \leftarrow \mathrm{CH}_2 \leftarrow \mathrm{CH}_2 \leftarrow \mathrm{O} \leftarrow \mathrm{H}>\mathrm{CH}_3 \rightarrow \mathrm{CH}_2 \rightarrow \mathrm{O} \rightarrow \mathrm{H}
$

Stronger acid due to $-I$ effect of $Cl$.

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**Answer**

o-nitrophenol is more volatile than $p$-nitrophenol due to presence of intramolecular hydrogen bonding. In para nitrophenol intermolecular hydrogen bonding is present. This intermolecular hydrogen bonding causes the association of molecules.

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**Answer**

The acidic character of alcohols is due to the polar nature of $O-H$ bond. Higher the polarity of $O-H$ bond, more will be the acidic strength. Due to $-I$ and $-R$ effect of nitro group, electron density decreases in the $O-H$ bond of $o$-nitrophenol and thus polarity increases.

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**Answer**

Nitro group shows $-I$ and $-R$ effect as follows

Due to this $-I$ and $-R$ effect of o-nitrophenol, it is a stronger acid than phenol. On the other hand, $-CH_3$ group produces $+I$ effect $-I$ and $-R$ effect increases the acidic strength by increasing the polarity of $-OH$ bond while $+I$ effect decreases the polarity due to increase in electron density on $-OH$ bond. So, o-cresol is a weaker acid than phenol. Thus, the correct order is o-cresol $<$ phenol $<0$-nitrophenol.

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**Answer**

An electron releasing group produces $+I$ effect so it increases the electron density on oxygen and decreases the polarity of $O-H$ bond. As the number of alkyl group increases from $1^{\circ}$ to $3^{\circ}$ alcohols, the electron density on the $O-H$ bond increases.

It will finally decrease the polarity and increase the strength of $O-H$ bond in going from $1^{\circ}$ to $3^{\circ}$ alcohols. Thus, acidic strength decreases in the following order

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As we know that, sodium metal is basic in nature and alcohols are acidic in nature. Thus, reactivity of alcohol with sodium metal decreases with decrease in acidic strength. Therefore, the correct order is $1^{\circ}>2^{\circ}>3^{\circ}$.

**Explanation**

This question is based on the concept of the acidity and reactivity. The acidic character of alcohols is due to the polar nature of the $O-H$ bond.

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**Answer**

A stronger acid displaces a weaker acid from its salt. When $RONa$ is treated with $H_2 O$, it forms $R O H$. So, water is a stronger acid than $R O H$.

$
\underset{\text{ Stronger acid }}{HOH}+RONa \longrightarrow NaOH+\underset{\text{ Weaker acid }}{ROH}
$

Similarly, when sodium ethynide is treated with water and alcohol, then acetylene is obtained.

$\underset{\substack{\text { Stronger } \\ \text { acid }}}{\mathrm{HOH}}+\mathrm{CH} \equiv \mathrm{CNa} \longrightarrow \underset{\text { Weak acid }}{\mathrm{HC}} \equiv \mathrm{CH}+\mathrm{NaOH}$
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**Answer**

Sucrose is converted to glucose and fructose in the presence of an enzyme, invertase or sucrase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase. Both these enzymes are present in yeast.

$
\begin{aligned}
& \underset{\text{ Sucrose }}{C    _{12} H   _{22} O   _{11}}+H _2 O \xrightarrow[\text{ Invertase/sucrose }]{C _6 H   _{12} O _6}+\underset{\text{ Glucose }}{C _6 H   _{12} O _6} \\\\\\
& \underset{\text{ Glucose/Fructose }}{C _6 H    _{12} O _6} \xrightarrow[\substack{\text{ Ethanol }}]{\text{ Zymase }} \underset{2}{2 CH _3 CH _2 OH}+2 CO _2
\end{aligned}
$

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The asymmetry of the molecule is responsible for the optical activity in a molecule. If all the four substituents attached to the carbon are different then the carbon is called asymmetric or chiral carbon and such a molecule is called asymmetric molecule.

In the above explained structure, it is only butan-2-ol which contains a chiral carbon and hence it is optically active.

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Out of these five resonating structures, II, III and IV structures contain a carbon-oxygen double bond character. In other words, carbon-oxygen bond in phenol acquires a partial double bond character due to resonance.

But in alcohols carbon-oxygen bond in alcohols is purely single bond. Therefore, $-OH$ group in phenols is more strongly held as compared to $-OH$ group in alcohols.

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**Answer**

Resonance is an important factor in phenols. During resonance $-OH$ group in phenol gives its electrons to the benzene ring. As a result of this, the electron density on benzene ring is very high. This increased electron density repels nucleophiles.

Therefore, nucleophiles cannot attack the benzene ring and phenols usually do not give nucleophilic substitution reaction.

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**Answer**

$CO _2$ is a linear molecule. The dipole moment of two $C=O$ bonds are equal and opposite and they cancel each other and hence the dipole moment of $CO _2$ is zero and it is a non-polar molecule.

$\mathrm{O} \xlongequal[\mu=0]{{\leftarrow+}{}} \mathrm{C} \xlongequal{+\rightarrow} \mathrm{O}$

While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipoles and hence $R-O-R$ is a polar molecule.

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**Answer**

The reaction of alcohols with Lucas reagent (conc. $HCl$ and $ZnCl _2$ ) follow $S   _{N}{ }^{1}$ mechanism. $S  _{N}{ }^{1}$ mechanism depends upon the stability of carbocations (intermediate). More stable the intermediate carbocation, more reactive is the alcohol.

Tertiary carbocations are most stable among the three classes of carbocations and the order of the stability of carbocation is $3^{\circ}>2^{\circ}>1^{\circ}$. This order, inturn, reflects the order of reactivity of three classes of alcohols i.e., $3^{\circ}>2^{\circ}>1^{\circ}$.

Thus, as the stability of carbocations are different so the reactivity of all the three classes of alcohols with Lucas reagent is different.

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**Answer**

Aspirin can be prepared by the reaction of salicylic acid with acetic anhydride. Salicylic acid is prepared by the reaction of phenol with $CO _2$ and $NaOH$.

This process is known as Kolbe's reaction. The product salicylic acid is used in the preparation of aspirin. After wards, when salicylic acid is treated with acetic anhydride then acetyl group replaces the hydrogen of $-OH$ group i.e., acetylation occurs at $-OH$ group of salicylic acid. Reaction is as follows

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**Answer**

In phenoxide ion, the ability to give lone pair of electrons to the benzene ring is more in comparison to phenols. Therefore, the reactivity of phenoxide ion towards electrophilic substitution reaction is more in comparison to phenols.

Thus, phenoxide ion being a stronger nucleophile reacts easily with $CO _2$ (weak electrophile) than phenols in Kolbe's reaction.

**Note** Kolbe's process is also known as Kolbe - Schmitt reaction. This reaction is precursor to aspirin.

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<div  style='float: right; width:45%; text-align: left;font-size:21px'>

![alt text](http://13.201.94.86:82//subject-images/chemistry/p8.png)

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**Answer**

Dipole moment depends upon the polarity of bonds. Higher the polarity of bonds in molecule, higher will be its dipole moment. In phenol carbon is $s p^{2}$ hybridised and due to this reason benzene ring is producing electron withdrawing effect.

On the other hand, carbon of methanol is $s p^{3}$ hybridised and produces electron releasing effect (+ I effect). Thus, $C-O$ bond in phenol is less polar than $C-O$ bond in methanol and therefore, the dipole moment of phenol is smaller than that of methanol.

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**Answer**

In order to prepare di-tert-butyl ether, sodium tert-butoxide must be reacted with tert-butyl bromide. Alkoxides are not only nucleophiles but they are strong base as well. They react with $3^{\circ}$ alkyl halides leading to the elimination reaction.

When tert-butyl-bromide reacts with sodium tert-butoxide instead of substitution, elimination takes place. As a result of this elimination reaction, Iso butylene is formed instead of di-tert butyl ether.

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**Answer**

The bond angle in $C-\underset{..}{\ddot{O}}-H$ in alcohols is slightly less than tetrahedral angle $(109^{\circ} 28^{\prime})$. It is due to the repulsion between the unshared electron pairs of oxygen. In alcohols, two lone pair of electrons are present. Therefore, there is comparatively more repulsion and less bond angle.

The $C-O-C$ bond angle in ether is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky $(-R)$ groups.

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**Answer**

Solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecule. The hydrocarbon part methoxy methane (i.e., $R$ group) tends to prevent the formation of hydrogen bonds.

Alcohols with lower molar mass will have smaller hydrocarbon part and therefore tendency to form hydrogen bonding is more and they are more soluble in water.

$\underset{\text{(More soluble)}}{CH_3 CH_2 Oh} > \underset{\text{(Less soluble)}}{CH_3 CH_2 CH_2 Oh} $

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**Answer**

Nitro group of phenol produces $-I$ and $-R$ effect. Because of these two effects $-NO _2$ group is electron withdrawing in nature. So, the electron density in the $O-H$ bond of $p$-nitrophenol decreases relative to the $O-H$ bond of phenol.

The decrease in electron density of the $O-H$ bond of $p$-nitrophenol, the polarity of $O-H$ bond is decrease and in turn make it more acidic than phenol.

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**Answer**

Boiling point depends upon the strength of intermolecular forces of attraction. Higher these forces of attraction, more will be the boiling point. Alcohols undergo intermolecular hydrogen bonding. So, the molecules of alcohols are held together by strong intermolecular forces of attraction.

But in ethers no hydrogen atom is bonded to oxygen. Therefore, ethers are held together by weak dipole-dipole forces, not by strong hydrogen bond.

Since, lesser amount of energy is required than to break weak dipole-dipole forces in ethers than to break strong hydrogen bonds in alcohol.

![alt text](http://13.201.94.86:82//subject-images/chemistry/p10.png)

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**Answer**

Positive resonance effect is observed in phenols. Due to this $+R$ effect, lone pair of electrons of $-OH$ group are in conjugation with $\pi$ electrons of the ring and the following resonance hybrid are obtained.

From the above resonating structure, it is very clear that $C-O$ bond of phenol acquires some partial double bond character while the $C-O$ bond of methanol is purely single bond. Therefore, the carbon-oxygen bond in phenol is slightly stronger than that in methanol.

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**Answer**

The phenoxide ion obtained after the removal of a proton is resonance stabilised as follows

Whereas, the ethoxide ion obtained after the removal of a proton is not stabilised but destabilised due to $+I$ effect of $-C _2 H _5$ group. Therefore, phenol is a stronger acid than ethanol. $+I$ effect of $CH _3-CH _2$ group increase the electron density on $O-H$ bond in ethanol. As the electron density on the $O-H$ bond of ethanol is more than that of water. So, ethanol is a weaker acid than water. Thus, the increasing order of acidity is

ethanol $<$ water $<$ phenol

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<div style='float: left; width:100%; text-align: left;font-size:22px'>
 
 
A. Cresols are organic compounds which are methyl phenols. There are three forms of cresol-o-cresol, $p$ - cresol and $m$ - cresol.

B. Catechol is also known as pyrocatechol. Its IUPAC name is 1,2-dihydrobenzene. It is used in the production of pesticides, perfumes and pharmaceuticals.

C. Its IUPAC name is 1, 3-dihydroxybenzene. Resorcinol is used to treat acne, seborrheic dermatitis and other skin disorder.

D. Hydroquinone is also known as quinol. Its IUPAC name is 1, 4- dihydroxybenzene. It is a white granular solid. It is a good reducing agent.
 
E. Anisole or methoxy benzene, is a colourless liquid with a smell reminiscent of anise seed.

F. Phenetole is an organic compound. It is also known as ethylphenyl ether. It is volatile in nature and its vapours are explosive in nature.

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B. In $(CH _3) _2 CH-O-CH _3$ unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows $S   _{N}{ }^{2}$ mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are

$
\begin{aligned}
& H _3 C \\\\\\
& H _3 C
\end{aligned}>CHOH+CH _3 I
$

C. In this case, one of the alkyl group is tertiary and the other is primary. It follows $S   _{N}{ }^{1}$ mechanism and halide ion attacks the tertiary alkyl group and the products are $(CH _3) _3 C-I$ and $CH _3 OH$.

D. Here, the unsymmetrical ether is alkyl aryl ether. In this ether $O-CH _3$ bond is weaker than $O-C _6 H _5$ bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are $C _6 H _5-OH$ and $CH _3 I$.

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**Ques.** Match the items of Column I with items of Column II.

| |Column I |  | Column II |  
| :--- | :--- | ---: | :--- |
| A. | Antifreeze used in car engine | 1. | Neutral ferric chloride |
| B. | Solvent used in perfumes | 2. | Glycerol |
| C. | Starting material for picric acid | 3. | Methanol |
| D. | Wood spirit | 4. | Phenol |
| E. | Reagent used for detection of phenolic group | 5. | Ethylene glycol |
| F. | By product of soap industry used in cosmetics | 6. | Ethanol |

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A. IUPAC name of ethylene glycol is ethane $-1,2$ - diol. It is primarily used as raw material in the manufacturing of polyester fibers and fabric industry. A small percentage of it is used in antifreeze formulations.

B. Ethanol is a good solvent for fatty and waxy substances. Fats and waxes provide odour to the perfumes. Apart from being a good solvent, it is less irritating to the skin. So, it is used in perfumes.

C. Phenol is converted into picric acid (2, 4, 6-trinitro-phenol) by the reaction of phenol with conc. $HNO _3$.

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D. Methanol, $CH _3 OH$ is also known as 'wood spirit' as it was produced by the destructive distillation of wood.

E. Neutral ferric chloride give purple/red colour when treated with phenols. It is the reagent used for detection of phenolic group.

F. Soaps are prepared by the reactions of fatty acid with $NaOH$.

This glycerol (propan -1, 2, 3 - triol) is the by product of soap industry and used in cosmetics.

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**Ques.** Match the items of Column I with items of Column II.

|  | Column I | |Column II |
| :---: | :---: | :---: | :---: |
| A. | Methanol | 1.| Conversion of phenol to <br> o-hydroxysalicylic acid |
| B. | Kolbe's reaction | 2.| Ethyl alcohol |
| C. | Williamson's synthesis | 3. |Conversion of phenol to <br> salicylaldehyde |
| D. | Conversion of $2^{\circ}$ alcohol to ketone | 4.| Wood spirit |
| E. | Reimer-Tiemann reaction | 5. |Heated copper at $573 K$ |
| F. | Fermentation | 6.| Reaction of alkyl halide <br> with sodium alkoxide |

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C. Williamson synthesis is an important method for the preparation of ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

$
R-X+R-ONa \longrightarrow ROR+NaX
$

D. When a $2^{\circ}$ alcohol is allowed to pass over heated copper at $573 K$, dehydrogenation takes place and an ketone is formed.

$ R-\underset{OH}{\underset{|}{CH}} -R^{\prime} \xrightarrow[{ 573 K }]{Cu} R-\underset{O}{\underset{||}{C}}-R
$

E. On treating phenol with chloroform in the presence of $NaOH$, an aldehydic group is introduced at ortho position of benzene ring

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F. Ethanol is prepared by the fermentation of sugars.

$
\begin{matrix} 
C    _{12} H   _{22} O   _{11}+H _2 O \xrightarrow{\text{ Invertase }} C _6 H   _{12} O _6+C _6 H   _{12} O _6 \\\\\\
C _6 H    _{12} O _6 \xrightarrow{\text{ Zymase }} 2 C _2 H _5 OH+2 CO _2
\end{matrix} 
$

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### Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason $(R)$ is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct and reason is correct explanation of assertion.

(b) Assertion and reason both are wrong statements.

(d) Assertion is wrong statement but reason is correct statement.

(e) Both assertion and reason are correct statements but reason is not correct explanation of assertion.

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**Answer**

In case of alkyl aryl ethers, the products are always phenol and an alkyl halide because due to resonance $C _6 H _5-O$ bond has partial double bond character. The mechanism is given below

**Mechanism** Protonation of anisole gives methylphenyl oxonium ion.

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In this ion, the bond between $O-CH _3$ is weaker than the bond between $O-C _6 H _5$ which has partial double bond character. This partial double bond character is due to the resonance between the lone pair of electrons on the $O$-atom and the $s p^{2}$ hybridised carbon atom of the phenyl group. Therefore, attack by $I^{-}$ion exclusively breaks the weaker $O-CH _3$ bond forming methyl iodide and phenol.

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In aqueous solution, phenol ionises to form phenoxide ion. This ion activates the benzene ring to a very large extent and hence the substitution of halogen takes place at all three positions.

On the other hand, in non-aqueous solution ionisation of phenol is greatly suppressed. Therefore, ring is activated slightly and hence monosubstitution occur.

(c) Lewis acid is an electron deficient molecule. In bromination of benzene, Lewis acid is used to polarise $Br _2$ to form $Br^{+}$electrophile.

In case of phenol, oxygen atom of phenol itself polarises the bromine molecule to form $Br^{+}$ion (electrophile). So, Lewis acid is not required in the bromination of phenol.

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**Answer**

Enzymes are biocatalyst. These biocatalysts (enzymes) are used in the industrial preparation of ethanol. Ethanol is prepared by the fermentation of molasses - a dark brown coloured syrup left after crystallisation of sugar which still contains about $40 \%$ of sugar.

The process of fermentation actually involves breaking down of large molecules into simple ones in the presence of enzymes. The source of these enzymes is yeast. The various reactions taking place during fermentation of carbohydrates are

$
\underset{\text{ Sucrose }}{C    _{12} H   _{22} O   _{11}}+H _2 O \xrightarrow{\text{ Invertase }} \underset{\text{ Glucose }}{C _6 H   _{12} O _6}+\underset{\text{ Fructose }}{C _6 H   _{12} O _6}
$

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$
\underset{\substack{\text{ Glucose }}}{C _6 H    _{12} O _6} \xrightarrow{\text{ Zymase }} \underset{\text{ Ethyl alcohol }}{2 C _2 H _5 OH}+2 CO _2
$

In wine making, grapes are the source of sugars and yeast. As grapes ripen, the quantity of sugar increases and 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. $CO _2$ gas is released during fermentation.

The action of zymase is inhibited once the percentage of alcohol formed exceeds 14 per cent. If air gets into fermentation mixture, the oxygen of air oxidises ethanol to ethanoic acid which in turn destroys the taste of alcoholic drinks.

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