Answer:(d) Benzyl alcohol

Explanation:

Monochlorination of toluene in sunlight gives benzyl chloride. On hydrolysis with aq. $NaOH$, benzyl chloride shows nucleophilic substitution reaction to give benzyl alcohol.

So, the correct option is (d).

Now, consider the incorrect options:

(a) o-cresol: Monochlorination of toluene in sunlight does not lead to the formation of o-cresol. o-Cresol is an ortho-substituted hydroxytoluene, which would require a hydroxyl group to be directly attached to the benzene ring in the ortho position relative to the methyl group. This is not achieved through monochlorination followed by hydrolysis.

(b) m-cresol: Similar to o-cresol, m-cresol is a meta-substituted hydroxytoluene. The monochlorination of toluene followed by hydrolysis does not introduce a hydroxyl group directly onto the benzene ring in the meta position relative to the methyl group.

(c) 2, 4-dihydroxytoluene: This compound has two hydroxyl groups attached to the benzene ring in the 2 and 4 positions relative to the methyl group. The monochlorination of toluene followed by hydrolysis does not introduce multiple hydroxyl groups onto the benzene ring.

Answer:(a) 1

Explanation:

The three isomers of butanol are possible. Structural formula of these isomers are given below

(i) $\underset{\text { Butan-1-ol }}{CH_3 CH_2 -CH_2-CH_2 OH}$

No carbon is chiral in this compound as none of the four carbon is bonded to four different substituents.

In this compound, asterisk marked carbon is chiral carbon as all four substituents, attached to it are different.

Here, again carbon is not chiral in nature.

So, only one alcohol is chiral in nature and the correct option is (a).

Now, consider the incorrect options:

(b) is incorrect because there is only one chiral alcohol with the molecular formula $C_4 H_{10} O$, not two. The other isomers do not have a carbon atom bonded to four different substituents.

(c) is incorrect because there are not three chiral alcohols with the molecular formula $C_4 H_{10} O$. Only one of the isomers has a chiral center.

(d) is incorrect because there are not four chiral alcohols with the molecular formula $C_4 H_{10} O$. Only one isomer has a chiral center.

Answer:(c) $3^{\circ}>2^{\circ}>1^{\circ}$

Explanation:

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.

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.

So, the order of reactivity of alcohols in the given reaction is $3^{\circ}>2^{\circ}>1^{\circ}$ and correct option is (c).

Now, consider the incorrect options:

(a) $1^{\circ}>2^{\circ}>3^{\circ}$: This option is incorrect because it suggests that primary alcohols are more reactive than secondary and tertiary alcohols. However, in the presence of $ZnCl_2$, tertiary alcohols form more stable tertiary carbocations, which makes them more reactive than secondary and primary alcohols.

(b) $1^{\circ}<2^{\circ}>3^{\circ}$: This option is incorrect because it implies that secondary alcohols are more reactive than tertiary alcohols. In reality, tertiary alcohols form more stable carbocations compared to secondary alcohols, making tertiary alcohols more reactive in this reaction.

(d) $3^{\circ}>1^{\circ}>2^{\circ}$: This option is incorrect because it suggests that primary alcohols are more reactive than secondary alcohols. However, secondary alcohols form more stable carbocations than primary alcohols, making secondary alcohols more reactive than primary alcohols in this reaction.

Answer:(c) treatment with pyridinium chlorochromate

Explanation:

Pyridinium chlorochromate (PCC), a complex of chromium trioxide with pyridine and HCl gives good yield of aldehydes and prevents further oxidation to carboxylic acids.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow{\mathrm{PCC}} \mathrm{CH}_3 \mathrm{CHO} $

Now, consider the incorrect options:

(a) Catalytic hydrogenation: This process involves the addition of hydrogen to a molecule, typically reducing double or triple bonds. It does not oxidize ethanol to ethanal; instead, it would reduce any present double or triple bonds, which is not applicable in this context.

(b) Treatment with $LiAlH_4$: Lithium aluminium hydride is a strong reducing agent, which is used to reduce aldehydes, ketones, and carboxylic acids to alcohols. It does not oxidize ethanol to ethanal; rather, it would reduce any oxidized form back to ethanol or even further to ethane.

(d) Treatment with $KMnO_4$: Potassium permanganate is a strong oxidizing agent that typically oxidizes primary alcohols all the way to carboxylic acids. It would not stop at the aldehyde stage, thus converting ethanol to acetic acid rather than ethanal.

Answer:(b) substitution reaction

Explanation:

The process of converting alkyl halides into alcohols can be represented as:

$ R-X \rightarrow R-O H $

since X - is replaced by $\mathrm{OH}^{-}$ group it is a nucleophilic substitution reaction.

Therefore the process of converting alkyl halides into alcohol involves substitution reaction.

Now, consider the incorrect options:

(a) Addition reaction: An addition reaction involves the combination of two or more molecules to form a larger molecule, typically converting unsaturated compounds to saturated compounds. This is not applicable to the conversion of alkyl halides to alcohols, which involves replacing a halogen atom with a hydroxyl group.

(c) Dehydrohalogenation reaction: Dehydrohalogenation is a reaction where an alkyl halide loses a hydrogen atom and a halogen atom, resulting in the formation of an alkene. This process does not produce alcohols but rather alkenes.

(d) Rearrangement reaction: A rearrangement reaction involves the reorganization of the molecular structure to form isomers of the original compound. This does not apply to the conversion of alkyl halides to alcohols, which involves the substitution of a halogen atom with a hydroxyl group rather than a structural rearrangement.

Answer:(c) $B, C$

Explanation:

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.

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).

Now, consider the incorrect options:

(a) is incorrect because it includes compounds (A) and (D), which are phenols and not aromatic alcohols. Phenols have the -OH group directly attached to the benzene ring, making them a separate class of compounds.

(b) is incorrect because it includes compound (A), which is phenol and not an aromatic alcohol. Phenol has the -OH group directly attached to the benzene ring, making it a phenol rather than an aromatic alcohol.

(d) is incorrect because it includes only compound (A), which is phenol and not an aromatic alcohol. Phenol has the -OH group directly attached to the benzene ring, making it a phenol rather than an aromatic alcohol.

Answer:(c) 5-chlorohexan-2-ol

Explanation:

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

Now, consider the incorrect options:

(a) 2-chloro-5-hydroxyhexane: This name is incorrect because the hydroxyl group (-OH) should be given priority over the chloro group (Cl) in the numbering of the carbon chain. The hydroxyl group should be assigned the lowest possible number, which is not the case here.

(b) 2-hydroxy-5-chlorohexane: This name is incorrect for the same reason as option (a). The hydroxyl group should be given priority and assigned the lowest possible number, which is not reflected in this name.

(d) 2-chlorohexan-5-ol: This name is incorrect because the numbering of the carbon chain should prioritize the hydroxyl group, resulting in the hydroxyl group being at position 2 and the chloro group at position 5, not the other way around.

Answer:(a) 3-methylphenol

Explanation:

The structure of $m$-cresol is

IUPAC name is 3-methylphenol because - $OH$ is the functional group and the methyl is substituent.

Now, consider the incorrect options:

(b) 3-chlorophenol: This option is incorrect because it suggests the presence of a chlorine substituent at the 3-position on the benzene ring, whereas $m$-cresol has a methyl group at the 3-position, not a chlorine atom.

(c) 3-methoxyphenol: This option is incorrect because it suggests the presence of a methoxy group $ (-OCH_3) $ at the 3-position on the benzene ring, whereas $m$-cresol has a methyl group at the 3-position, not a methoxy group.

(d) benzene-1, 3-diol: This option is incorrect because it suggests the presence of two hydroxyl groups (-OH) at the 1 and 3 positions on the benzene ring, whereas $m$-cresol has only one hydroxyl group at the 1-position and a methyl group at the 3-position.

Answer:(c) 2-methoxypropane

Explanation:

IUPAC name of the above compound is 2-methoxypropane.

Now, consider the incorrect options:

(a) 1-methoxy-1-methylethane: This name suggests that the methoxy group $ (-OCH_3) $ is attached to the first carbon of a two-carbon chain (ethane) with a methyl group also attached to the same carbon. However, the compound in question has a three-carbon chain (propane) with the methoxy group attached to the second carbon, not the first.

(b) 2-methoxy-2-methylethane: This name implies that the methoxy group and a methyl group are both attached to the second carbon of a two-carbon chain (ethane). This is incorrect because the compound has a three-carbon chain (propane) with the methoxy group attached to the second carbon, and there is no additional methyl group attached to the second carbon.

(d) isopropylmethyl ether: This name describes an ether with an isopropyl group $ (CH_3-CH-CH_3) $ and a methyl group $ (CH_3) $ attached to the oxygen atom. However, the compound in question is an ether with a methoxy group $ (-OCH_3) $ attached to the second carbon of a propane chain, not an isopropyl group.

Answer:(b) ${ }^{\ominus} OR$

Explanation:

Weakest acid has the strongest conjugate base. $ROH$ is the acid of ${ }^{\ominus} OR$ conjugate base, HOH is the acid of $^-OH,C_6H_5OH$ is the acid of $C_6H_5O^-$ and

is the acid of .

Among all these acids, ROH is the weakest acid.

Therefore, the strongest base is ${ }^{\ominus} OR$ .

Now, consider the incorrect options:

(a) ${ }^{\ominus} OH$: The conjugate acid of ${ }^{\ominus} OH$ is water (HOH), which is a stronger acid compared to ROH. Therefore, ${ }^{\ominus} OH$ is a weaker base compared to ${ }^{\ominus} OR$.

(c) ${ }^{\ominus} OC_6 H_5$: The conjugate acid of ${ }^{\ominus} OC_6 H_5$ is phenol ($C_6H_5OH$), which is a stronger acid compared to ROH. Therefore, ${ }^{\ominus} OC_6 H_5$ is a weaker base compared to ${ }^{\ominus} OR$.

Answer:(a) $C_6 H_5 OH$

Explanation:

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 $NaOH$.

Now, consider the incorrect options:

(b) $C_6 H_5 CH_2 OH$: Benzyl alcohol ($C_6 H_5 CH_2 OH$) is not acidic enough to react with sodium hydroxide. It does not form a stable ion like phenoxide ion, and thus does not dissolve in $NaOH$.

(c) $(CH_3)_3 COH$: Tertiary butyl alcohol ($(CH_3)_3 COH$) is also not acidic enough to react with sodium hydroxide. It does not have the ability to form a stable ion upon losing a proton, and therefore does not dissolve in $NaOH$.

(d) $C_2 H_5 OH$: Ethanol ($C_2 H_5 OH$) is a typical alcohol and is not acidic enough to react with sodium hydroxide. It does not form a stable ion upon losing a proton, and thus does not dissolve in $NaOH$.

Answer:(b) $o$ -nitrophenol

Explanation:

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.

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.

Now, consider the incorrect options:

(a) Ethanol: Ethanol is less acidic than phenol because the hydroxyl group in ethanol is attached to an alkyl group, which is electron-donating. This decreases the ability of the oxygen to stabilize the negative charge on the conjugate base, making ethanol less acidic than phenol.

(c) o-Methylphenol: In o-methylphenol, the methyl group is an electron-donating group. The presence of this group at the ortho position increases the electron density on the phenol ring, making it less able to stabilize the negative charge on the conjugate base, thus decreasing its acidity compared to phenol.

(d) o-Methoxyphenol: In o-methoxyphenol, the methoxy group is also an electron-donating group. The presence of this group at the ortho position increases the electron density on the phenol ring, making it less able to stabilize the negative charge on the conjugate base, thus decreasing its acidity compared to phenol.

Answer:(d) $m$ - chlorophenol

Explanation:

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.

Presence of electron withdrawing group increases the acidic strength. So, $m$-chlorophenol is most acidic among all the given compounds.

Now, consider the incorrect options:

(a) Benzyl alcohol: The alpha carbon of benzyl alcohol is $sp^3$ hybridized, which makes it less capable of stabilizing the negative charge on the conjugate base after deprotonation. Additionally, benzyl alcohol lacks any electron-withdrawing groups that could enhance its acidity.

(b) Cyclohexanol: Similar to benzyl alcohol, the alpha carbon of cyclohexanol is $sp^3$ hybridized, which does not favor the stabilization of the conjugate base. Cyclohexanol also lacks any electron-withdrawing groups that could increase its acidic strength.

(c) Phenol: Although phenol has an $sp^2$ hybridized carbon, which is better at stabilizing the negative charge on the conjugate base compared to $sp^3$ hybridized carbons, it does not have an additional electron-withdrawing group like $m$-chlorophenol. This makes phenol less acidic than $m$-chlorophenol.

Answer:(b) II $>$ IV $>$ I $>$ III $>$ V

Explanation:

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.

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.

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.

Hence, the correct order is II $>$ IV $>$ I $>$ III $>$ V .

Now, consider the incorrect options:

(a) V $>$ IV $>$ II $>$ I $>$ III:

This option incorrectly places $m$-methoxy phenol (V) as the strongest acid. The $-OCH_3$ group is an electron-donating group, which decreases the acidity of phenol. Therefore, $m$-methoxy phenol cannot be stronger than $p$-nitrophenol (II) and $m$-nitrophenol (IV), which have electron-withdrawing $-NO_2$ groups.

(c) IV $>$ V $>$ III $>$ II $>$ I:

This option incorrectly places $m$-methoxy phenol (V) as stronger than $p$-nitrophenol (II). The $-NO_2$ group in $p$-nitrophenol exerts both $-I$ and $-R$ effects, making it a stronger acid than $m$-methoxy phenol, which only has a $-I$ effect and no $+R$ effect.

(d) V $>$ IV $>$ III $>$ II $>$ I:

This option incorrectly places $m$-methoxy phenol (V) as the strongest acid. As mentioned, the $-OCH_3$ group is an electron-donating group, which decreases the acidity of phenol. Therefore, $m$-methoxy phenol cannot be stronger than $p$-nitrophenol (II) and $m$-nitrophenol (IV), which have electron-withdrawing $-NO_2$ groups.

Answer:(c) II $<$ III $<$ I

Explanation:

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.

Since, $-NO_2$ group is a stronger EWG than $-Cl$. So, $NO_2-C_6 H_5-\stackrel{+}{C} H_2$ carbocation will be less stable than $Cl-C_6 H_5-\stackrel{+}{C} H_2$ carbocation.

Thus, the order of stability of carbocation is

Therefore, compound (II) is least reactive and correct order is II $<$ III $<$ I .

Now, consider the incorrect options:

(a) I < II < III:

This option suggests that compound I is the least reactive and compound III is the most reactive. However, compound I has no electron-withdrawing group, making its carbocation the most stable and thus the most reactive. Therefore, this order is incorrect.

(b) II < I < III:

This option suggests that compound II is the least reactive and compound III is the most reactive, with compound I in between. However, compound I should be the most reactive due to the absence of electron-withdrawing groups, making its carbocation the most stable. Therefore, this order is incorrect.

(d) III < II < I:

This option suggests that compound III is the least reactive and compound I is the most reactive, with compound II in between. While it correctly places compound I as the most reactive, it incorrectly places compound III as the least reactive. Compound II should be the least reactive due to the stronger electron-withdrawing effect of the $-NO_2$ group compared to the $-Cl$ group in compound III. Therefore, this order is incorrect.

Answer:(a) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol

Explanation:

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:

Propan-1-ol $<$ butan-2-ol $<$ butan-1-ol $<$ pentan-1-ol

Now, consider the incorrect options:

(b) is incorrect because it places butan-1-ol before butan-2-ol. However, butan-1-ol has a higher boiling point than butan-2-ol due to its primary alcohol structure, which has a higher surface area and stronger intermolecular hydrogen bonding compared to the secondary alcohol structure of butan-2-ol.

(c) is incorrect because it repeats the same order as option (a), which is actually the correct order. Therefore, it does not provide a different incorrect order to analyze.

(d) is incorrect because it places butan-2-ol after butan-1-ol. However, butan-2-ol should come before butan-1-ol in the order of increasing boiling points due to the reasons mentioned above regarding primary and secondary alcohols.

Answer:(a, b, c)

Explanation:

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 $

$ RCHO \xrightarrow{LiAlH_4} RCH_2 OH $

Now, consider the incorrect options:

(d) Reaction with $R M g X$ followed by hydrolysis:

This option is incorrect because the reaction of $R MgX$ (Grignard reagent) with an aldehyde other than methanal (formaldehyde) results in the formation of secondary alcohols, not primary alcohols. The question specifically asks for the conversion of $RCHO$ (an aldehyde) into $RCH_2OH$ (a primary alcohol).

Reaction of $R MgX$ with any aldehyde other than methanal gives secondary alcohols not the primary alcohols.

(Where, $R=-C_2 H_5, C_3 H_7$ etc.)

Answer:(a, b, c)

Explanation:

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).

Now, consider the incorrect options:

(d): The reaction conditions are not specified, and without high temperature and pressure, chlorobenzene will not undergo hydrolysis to yield phenol. Haloarenes are less reactive towards nucleophilic substitution reactions under standard conditions.

Answer:(a, c, d)

Explanation:

(a) $CrO_3$ in anhydrous medium is used to oxidise primary alcohols to aldehydes.

$ RCH_2 OH \xrightarrow{CrO_3 \text { (anhydrous) }} RCHO $

(c) Pyridinium chlorochromate $(PCC)$ is a very good reagent for the oxidation of primary alcohols to aldehydes.

$ RCH_2 OH \xrightarrow{PCC} RCHO $

(d) When the vapours of primary alcohols are passed over heated copper at $573 K$, dehydrogenation takes place and an aldehyde is formed.

$ RCH_2 OH \xrightarrow[573 K]{Cu} RCHO $

Now, consider the incorrect options:

(b) Acidic $KMnO_4$ is a very strong oxidising agent. It oxidises primary alcohols into carboxylic acid, not aldehydes.

$ RCH_2 OH \xrightarrow{KMnO_4} RCOOH $

Answer:(a, c)

Explanation:

Ethanol does not react with $Br_2$ /water while phenol gives 2, 4, 6-tribromophenol as white precipitate.

Ethanol and phenol both react with sodium, so phenol cannot be distinguished from ethanol by the reaction with sodium.

Ethanol does not give any reaction with neutral $FeCl_3$ solution while phenol gives purple, blue or red colour when treated with neutral $FeCl_3$.

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

Now, consider the incorrect options:

(b) $Na$: Ethanol and phenol both react with sodium, producing hydrogen gas and forming their respective sodium alkoxides. Therefore, phenol cannot be distinguished from ethanol by the reaction with sodium.

(d) All of these: This option is incorrect because ethanol reacts with sodium, so it cannot be used to distinguish phenol from ethanol. Only reactions with $Br_2$/water and neutral $FeCl_3$ can be used to distinguish phenol from ethanol.

Answer:(b, c)

Explanation:

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

Now, consider the incorrect options:

(a) the $-OH$ group is attached to a $s p^{3}$ hybridised carbon atom, but this carbon is not attached to the benzene ring.

(d) the $-OH$ group is attached to a $s p^{3}$ hybridised carbon atom, but this carbon is not attached to the benzene ring.

Answer

Structure of glycerol is

So, IUPAC name is: propan- 1, 2, 3-triol

Answer

The correct IUPAC name of the following compounds are given below

(a) 3-Ethyl-5-methylhexane-2, 4-diol.

(b)

IUPAC name $\rightarrow$ 1-Methoxy-3-nitrocyclohexane.

Answer

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

Answer

Solubility of alcohols in water depends upon the following two factors :

(i) Hydrogen bonding: $\mathrm{O}-\mathrm{H}$ group in alcohol being polar forms H -bonds with $\mathrm{H}_2 \mathrm{O}$. Greater the extent of H-bonding, higher is the solubility. For alcohols having comparable molecular masses, solubility increases as the number of OH groups increases, i.e.,

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHOHCH}_2 \mathrm{OH}<\mathrm{HOCH}_2 \mathrm{CHOHCH}_2 \mathrm{OH} $

(ii) Size of the alkyl/aryl group: Higher the hydrocarbon part, lower is the extent of H-bonding and hence lower is the solubility. For example, solubility decreases in the order :

$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$

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.

Answer

Pyridinium chlorochromate, $\mathrm{C}_5 \mathrm{H}_5 \stackrel{+}{\mathrm{N}} \mathrm{HCrO}_3 \mathrm{Cl}^{-}\left(\mathrm{CrO}_3\right.$, pyridine and $HCl$ ) oxidises

$>\mathrm{CHOH} \longrightarrow \quad \mathrm {>} {C}=\mathrm{O}$ without causing oxidation at the double bond.

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.

Answer

Pyridinium chlorochromate in $CH_2 Cl_2$

$ C H_3\underset{\text { Ethanol }}{CH_2 OH}\xrightarrow [CH_2 Cl_2]{C_5 H_5 NH^+ CrO_3 Cl^-} \underset{\text { Ethanal }}{CH_3 CHO} $

Answer

Ethanol can be converted into ethanoic acid by using acidified $KMnO_4$ or $K_2 Cr_2 O_7$. Both $KMnO_4$ and $K_2 Cr_2 O_7$ are strong oxidising agents.

$ \underset{\text{Ethanol}}{\mathrm{CH_3CH_2OH}} \xrightarrow[\text{or} \quad K_2 Cr_2 O_7 / H_2 SO_4 ]{\substack{\text \\ \mathrm{KMnO_4 /H_2SO_4 }}} \underset{\text{Ethanoic acid}}{\mathrm{CH_3COOH}} $

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.

Answer

Due to - I and - R-effect of the $\mathrm{NO}_2$ group, electron density in the $\mathrm{O}-\mathrm{H}$ bond decreases while due to +I or hyperconjugation effect, of the $\mathrm{CH}_3$ group, the electron density in the $\mathrm{O}-\mathrm{H}$ bond increases.

As a result, $\mathrm{O}-\mathrm{H}$ bond is $o$-nitrophenol is much weaker than the $\mathrm{O}-\mathrm{H}$ bond is $o$-cresol and hence $o$-nitrophenol is much more acidic than $o$-cresol.

Answer

When phenol is treated with bromine water, white ppt. of 2, 4,6-tribromophenol is obtained.

Answer

Due to the-I-and-R-effect of o-nitrophenol, it is a stronger acid than phenol thus, making the release of $H^{+}$ ion from $\mathrm{O}-\mathrm{H}$ bond easier. -I - and -R- effect increases the acidic strength by increasing the polarity of -OH bond.

On the other hand, $-\mathrm{CH}_3$ group in o-cresol produces +I -effect which 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 < o-nitrophenol.

Answer

Alkyl groups have +I -effect. As the number of alkyl groups increases from 1 to 2 to 3 in going from $1^{\circ}$ to $2^{\circ}$ to $3^{\circ}$ alcohols, the electron density in the $\mathrm{O}-\mathrm{H}$ bond increases and the bond becomes stronger and stronger.

In other words, acidic character of alcohols decreases from $1^{\circ}$ to $2^{\circ}>3^{\circ}$.

Since Na metal is basic and alcohols are acidic in nature, therefore, reactivity of Na metal towards alcohols decreases as the acidic strength of alcohols decreases, i.e., in the order : $1^{\circ}>2^{\circ}>3^{\circ}$.

Answer

When benzene diazonium chloride is heated with water then phenol is formed.

Answer

A stronger acid displaces a weaker acid from its salt. Since $\mathrm{H}_2 \mathrm{O}$ displaces ROH from RONa and both $\mathrm{H}_2 \mathrm{O}$ and alcohol displace acetylene from sodium acetylide, therefore, water is the strongest, followed by alcohol while acetylene is the weakest acid. In other words, acidity decreases in the order :

$\mathrm{H}_2 \mathrm{O}>\mathrm{ROH}>\mathrm{HC} \equiv \mathrm{CH}$

$\underset{\text { Stronger acid }}{\mathrm{H}_2 \mathrm{O}}+\underset{\text { Sod. alkoxide }}{\mathrm{RONa}} \longrightarrow \underset{\text { Weaker acid }}{\mathrm{R}-\mathrm{OH}}+\mathrm{NaOH}$

$\underset{\text { Stronger acid }}{\mathrm{H}_2 \mathrm{O}}+ \underset{\text { Sod. acetylide }}{\mathrm{HC} \equiv \mathrm{CNa}} \longrightarrow \underset{\text { Weaker acid }}{\mathrm{HC} \equiv \mathrm{CH}}+\mathrm{NaOH}$

$\underset{\text { Stronger acid }}{\mathrm{ROH}} + \underset{\text { Sod. acetylide }} {\mathrm{HC} \equiv \mathrm{CNa}} \longrightarrow \underset{\text { Weaker acid }}{\mathrm{HC} \equiv \mathrm{CH}}+\mathrm{RONa}$,

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.

$ C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text { Invertase }} \underset{\text { Glucose }}{C_6 H_{12}O_6} + \underset{\text { Fructose }}{C_6H_{12}O_6} $

$ C_6H_{12}O_6 \xrightarrow{\text { Zymase }} 2 C_2H_5OH + 2CO_2 $

Answer

Propan-2-one can be converted into tert-butyl alcohol by reacting with $\mathrm{CH}_3 \mathrm{MgBr} /$ ether followed by acid hydrolysis.

Answer

Some compounds can rotate the plane polarised light, when it is passed through their solution. Such compounds are called optically active compounds. The structures of the isomers of alcohols with molecular formula $ \mathrm{C} _{4} \mathrm{H} _{10} \mathrm{O}$ are as follows

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.

Answer

Due to donation of the lone pair of electrons on the oxygen atom to the benzene ring, phenol can be written as a resonance hybrid of five structures, $\mathrm{I}-\mathrm{V}$ as given in image.

Out of these, structures II, III and IV contain a carbon-oxygen double bond. In other words, carbon-oxygen bond in phenols has some double bond character while in alcohols it is a pure single bond. Therefore, OH group in phenols is more strongly held as compared to OH group in alcohols.

Answer

OH group in phenols is a strong electron donating group. As a result, electron density on the benzene ring is quite high and hence it repels nucleophiles. In other words, nucleophiles cannot attack the benzene ring and hence phenols usually do not give nucleophilic substitution reactions.

Answer

Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule.

$ \mathrm{CH} _{2}=\mathrm{CH} _{2}+\mathrm{H} _{2} \mathrm{O} \stackrel {H^+} { \leftrightharpoons } \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH} $

$ \mathrm{CH} _{3}\mathrm{CH}= \mathrm{CH_2} +\mathrm{H} _{2} \mathrm{O} \stackrel {H^+} { \leftrightharpoons } \mathrm{CH} _{3} - \mathrm{CH OH} - \mathrm{CH_3} $

This addition reaction takes place in accordance with Markownikoff’s rule.

Mechanism The mechanism of the reaction involves the following three steps:

Step 1 Protonation of alkene to form carbocation by electrophilic attack of $ \mathrm{H} _{3} \mathrm{O}^{+}$

$\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}$

Step 2 Nucleophilic attack of water on carbocation.

Step 3 Deprotonation to form an alcohol.

Answer

$CO_2$ is a linear molecule. The dipole moments of the two $C = O$ bonds being equal and opposite cancel out each other and hence $CO_2$ is a non-polar molecule

In contrast, in $\mathrm{R}-\mathrm{O}-\mathrm{R}$ molecule, the two dipoles of the $\mathrm{R}-\mathrm{O}$ bonds are inclined to each other at angle of $110^{\circ}$, i.e., the two dipoles do not cancel out and hence have a finite resultant. In other words, $\mathrm{R}-\mathrm{O}-\mathrm{R}$ is a polar molcule.

Answer

The reaction of alcohols with Lucas reagent (conc. $ \mathrm{HCl}$ and $ \mathrm{ZnCl} _{2}$ ) follow $ \mathrm{S} _{\mathrm{N}}{ }{1}$ mechanism. $ \mathrm{S} _{\mathrm{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.

Answer

Aspirin can be prepared by the reaction of salicylic acid with acetic anhydride. Salicylic acid is prepared by the reaction of phenol with $ \mathrm{CO} _{2}$ and $ \mathrm{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 $-\mathrm{OH}$ group i.e., acetylation occurs at $-\mathrm{OH}$ group of salicylic acid.

Reaction is as follows:

Answer

Nitration of benzene and phenol is an electrophilic substitution reaction. The rate of any electrophilic substitution reaction depends upon the electron density in the aromatic ring. Obviously, higher the electron density in the aromatic ring, higher is the rate of electrophilic substitution reaction.

Now the presence of OH group in phenol, increases the electron density at ortho and para position by +R effect. Since the electron density is more in phenol than in benzene, therefore, phenol is more easily nitrated than benzene.

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 $ \mathrm{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.

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, $ \mathrm{C}-\mathrm{O}$ bond in phenol is less polar than $ \mathrm{C}-\mathrm{O}$ bond in methanol and therefore, the dipole moment of phenol is smaller than that of methanol.

Answer

Williamson’s synthesis occurs by $\mathrm{S}_{\mathrm{N}} 2$ mechanism in which sodium alkoxide reacts with an alkyl halide. Now to prepare di-tert-butyl ether, sodium tert-butoxide must be reacted with tert-butyl bromide.

Since $3^{\circ}$ alkyl halides prefer to undergo elimination rather than substitution, therefore, sodium tert-butoxide reacts with tert-butyl bromide and favours elimination to form isobutylene rather than substitution to form di-tertbutyl ether.

Answer

The bond angle in $ \mathrm{C}-{{\mathrm{O}}}-\mathrm{H}$ in alcohols is slightly less than tetrahedral angle $\left(109^{\circ} 28^{\prime}\right)$. 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 $ \mathrm{C}-\mathrm{O}-\mathrm{C}$ bond angle in ether is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky $(-R)$ groups.

Answer

Solubility of alcohols in $\mathrm{H}_2 \mathrm{O}$ is due to H -bonding between the $\mathrm{O}-\mathrm{H}$ group of alcohols and $\mathrm{H}_2 \mathrm{O}$ molecules. The hydrocarbon part (i.e., R group) on the other hand, tends to prevent the formation of H -bonds and thus tends to decrease the solubility of alcohols in $\mathrm{H}_2 \mathrm{O}$.

Since in lower molecular mass alcohols, the hydrocarbon part is small, therefore, lower molecular mass alcohols are soluble in water.

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

Answer

Nitro group of phenol produces $-I$ and $-R$ effect. Because of these two effects $-\mathrm{NO} _{2}$ group is electron withdrawing in nature. So, the electron density in the $ \mathrm{O}-\mathrm{H}$ bond of $p$-nitrophenol decreases relative to the $ \mathrm{O}-\mathrm{H}$ bond of phenol.

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

Answer

Alcohols undergo intermolecular H-bonding. So the molecules of alcohols are held together by strong intermolecular forces of attraction and hence exist as associated molecules.

Ethers, on the other hand, do not have H atom on the O -atom and hence do not form H-bonds. As a result, ethers exist as discrete molecules which are held only by weak dipole-dipole attractions.

Since lesser energy is required to break weak dipole-dipole attractions than to break H -bonds in associated molecules of alcohols, therefore, boiling points of alcohols are much higher than those of ethers of comparable molecular mass.

Answer

Positive resonance effect is observed in phenols. Due to this $+R$ effect, lone pair of electrons of $-\mathrm{OH}$ 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 $ \mathrm{C}-\mathrm{O}$ bond of phenol acquires some partial double bond character while the $ \mathrm{C}-\mathrm{O}$ bond of methanol is purely single bond.

Therefore, the carbon-oxygen bond in phenol is slightly stronger than that in methanol.

Answer

The phenoxide ion obtained after the removal of a proton is stabilised by resonance whereas the ethoxide ion obtained after the removal of a proton is destabilised by ’ +I ’ effect of $-\mathrm{C}_2 \mathrm{H}_5$ group.

Therefore phenol is stronger acid than ethanol. On the other hand ethanol is weaker acid than water because electron releasing $-\mathrm{C}_2 \mathrm{H}_5$ group in ethanol inreases the electron density on oxygen and hence the polarity of $\mathrm{O}-\mathrm{H}$ bond in ethanol decreases which results in the decreasing acidic strength.

Increasing order of acidity is ethanol < water < phenol.

Answer

A. $\rightarrow(4)$

B. $\rightarrow (3)$

C. $\rightarrow(6)$

D. $\rightarrow(1)$

E. $\rightarrow(7)$

F. $\rightarrow (2)$

(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.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(5)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

(A) $ \mathrm{CH} _{3}-\mathrm{O}-\mathrm{CH} _{3}$ is a symmetrical ether so the products are $ \mathrm{CH} _{3}I $ and $ \mathrm{CH} _{3} \mathrm{OH}$.

(B) In $\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CH}-\mathrm{O}-\mathrm{CH} _{3}$ unsymmetrical ether, one alkyl group is primary while another is secondary.

So, it follows $ \mathrm{S} _{\mathrm{N}}{ }{2}$ mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are

$ \begin{aligned} & \mathrm{C} \mathrm{H} _{3} \\ & \mathrm{C} \mathrm{H} _{3} \end{aligned}>\mathrm{CHOH}+\mathrm{CH} _{3} \mathrm{I} $

(C) In this case, one of the alkyl group is tertiary and the other is primary. It follows $ \mathrm{S} _{\mathrm{N}}{ }{1}$ mechanism and halide ion attacks the tertiary alkyl group and the products are $\left(\mathrm{CH} _{3}\right) _{3} \mathrm{C}-\mathrm{I}$ and $ \mathrm{CH} _{3} \mathrm{OH}$.

(D) Here, the unsymmetrical ether is alkyl aryl ether. In this ether $ \mathrm{O}-\mathrm{CH} _{3}$ bond is weaker than $ \mathrm{O}-\mathrm{C} _{6} \mathrm{H} _{5}$ bond which has partial double bond character due to resonance.

So, the halide ion attacks on alkyl group and the products are $ \mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{OH}$ and $ \mathrm{CH} _{3} \mathrm{I}$.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(6)$

C. $\rightarrow(4)$

D. $\rightarrow(3)$

E. $\rightarrow(1)$

F. $\rightarrow(2)$

(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. $ \mathrm{HNO} _{3}$.

(D) Methanol, $ \mathrm{CH} _{3} \mathrm{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 $ \mathrm{NaOH}$.

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

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(6)$

D. $\rightarrow(5)$

E. $\rightarrow(3)$

F. $\rightarrow(2)$

(A) Methanol is also known as ‘wood spirit’ as it was produced by the destructive distillation of wood.

(B) In Kolbe’s reaction, 2 - hydroxy benzoic acid (salicylic acid) is prepared by the reaction of phenol with $ \mathrm{CO} _{2}$ gas.

(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-\mathrm{X}+\mathrm{R}-\mathrm{ONa} \longrightarrow \mathrm{ROR}+\mathrm{NaX} $

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

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

(E) On treating phenol with chloroform in the presence of $ \mathrm{NaOH}$, an aldehydic group is introduced at ortho position of benzene ring

(F) Ethanol is prepared by the fermentation of sugars.

$ \mathrm{C} _{12} \mathrm{H} _{22} \mathrm{O} _{11}+\mathrm{H} _{2} \mathrm{O} \xrightarrow{\text { Invertase }} \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}+\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} $

$ \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} \xrightarrow{\text { Zymase }} 2 \mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}+2 \mathrm{CO} _{2} $

Answer:(b) Assertion and reason both are wrong statements.

Correct Assertion: Addition reaction of water to but-1-ene in acidic medium yields butan-2-ol.

Correct Reason: Addition of water in acidic medium proceeds through the formation of secondary carbocation.

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

$p$-nitrophenol is more acidic than phenol because nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance.

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

Correct Assertion: The IUPAC name of the given compound is 1-(2-propoxy) propane.

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

Correct Assertion: Bond angle in ether is slightly more than the tetrahedral angle.

Answer:(b) Assertion and reason both are wrong statements.

Correct Assertion: Boiling points of alcohols are higher than that of ethers of comparable molecular mass.

Correct Reason: Alcohols can form intermolecular hydrogen bonding while ethers cannot.

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

Correct Assertion: Bromination of benzene but not of phenol is carried out in presence of a Lewis acid.

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

Correct Explanation: Due to the presence of intramolecular hydrogen bonding, o-nitrophenol does not form hydrogen bonds with $ \mathrm{H} _{2} \mathrm{O}$ but $m$ and $p$-nitrophenol form hydrogen bonds with water.

Answer:(c) Assertion is correct statement but reason is wrong statement.

Correct Reason: Phenoxide ion is stabilised by resonance but ethoxide ion is not stabilised by resonance.

Resonance in phenoxide ion

Answer:(b) Assertion and reason both are wrong statements.

Correct Assertion: Phenol form 2, 4, 6-tribromophenol on treatment with bromine in water.

Correct Reason: In water, phenol gives phenoxide ion. This phenoxide ion activates the ring towards electrophilic substitution reaction.

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

Correct Assertion: Phenols give $ \mathrm{o}$ and $p$-nitrophenol on nitration with dil. $ \mathrm{HNO} _{3}$ at $298 \mathrm{~K}$.

Answer

In case of alkyl aryl ethers, the products are always phenol and an alkyl halide because due to resonance $ \mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{O}$ bond has partial double bond character.

The mechanism is given below.

Mechanism Protonation of anisole gives methylphenyl oxonium ion.

In this ion, the bond between $ \mathrm{O}-\mathrm{CH} _{3}$ is weaker than the bond between $ \mathrm{O}-\mathrm{C} _{6} \mathrm{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 $ \mathrm{O}$-atom and the $s p^{2}$ hybridised carbon atom of the phenyl group.

Therefore, attack by $I^{-}$ ion exclusively breaks the weaker $ \mathrm{O}-\mathrm{CH} _{3}$ bond forming methyl iodide and phenol.

Answer

(a) The starting material used in the industrial preparation of phenol is cumene.

(b) Phenols when treated with bromine water gives polyhalogen derivatives in which all the hydrogen atoms present at ortho and para positions with respect to $-\mathrm{OH}$ group are replaced by bromine atoms.

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.

However, in non-aqueous medium such as $ \mathrm{CS} _{2}, \mathrm{CCl} _{4}, \mathrm{CHCl} _{3}$, monobromophenols are obtained.

In non-aqueous solution ionisation of phenol is greatly suppressed. Therefore, ring is activated slightly and hence monosubstitution occur.

(c) In bromination of benzene, Lewis acid is used to polarize $\mathrm{Br}_2$ to form the reactive electrophile, $\mathrm{Br}^{+}$. In case of phenol, Lewis acid is not required because the O -atom of phenol itself polarizes the $\mathrm{Br}_2$ molecule to form $\mathrm{Br}^{+}$ ions.

Further, +R -effect of OH group makes phenol highly activated towards electrophilic substitution reactions.

Answer

Phenol is converted into salicylic acid. This salicylic acid is treated with acetic anhydride. On acetylation of salicylic acid, aspirin (acetyl salicylic acid) is formed.

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

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