Alcohol, Phenols and Ethers

(i) Oxidation of isopentane

$ \begin{array}{r} \mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{CH}_2-\mathrm{CH}_3 \xrightarrow[358 \mathrm{~K}]{20 \% \mathrm{H}_3 \mathrm{PO}_4} \\ \mathrm{CH}_3-\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{OH}-\mathrm{CH}_2 \mathrm{CH}_3 \\ \text { Product } X: 2 \text { methvl-2-butanol } \end{array} $

(ii) Dehydration with given acid;

$ \begin{array}{r} \mathrm{CH}_3-\mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{OH}-\mathrm{CH}_2-\mathrm{CH}_3 \xrightarrow{\mathrm{H}_2 \mathrm{PO}_4} \\ \mathrm{CH}_3-\mathrm{C}\left(\mathrm{CH}_3\right)=\mathrm{CH}-\mathrm{CH}_3+\mathrm{H}_2 \mathrm{O} \end{array} $

$ \text { Product } Y: 2 \text {-methyl-2-butene. } $

Since, the alcohol $(X)$ gives turbidity immediately with Lucas reagent, thus the alcohol is tertiary. So, the reaction is as follows

Thus, $X$ and $Y$ respectively are 2 -methyl-2-butanol i.e.

The number of $\sigma$ bonds is 12 and $\pi$ bonds is 4 . The ratio of $\sigma$ and $\pi$ bonds is $3: 1$.

The complete reaction sequence are as follows

$ \text { The complete reaction is as follows } $

• Thus, $B$ and $C$ are functional isomers.

• With $\mathrm{H}_2$ /catalyst $B$ gives $1^{\circ}$ amine and $C$ gives $2^{\circ}$ amine.

• $C$ forms isocyanate with HgO . So, statement I, II and IV are correct.

• Ethyl cyanide on hydrolysis gives propionic acid while ethyl isocyanide on hydrolysis in acidic medium gives formic acid.

Among the given options, intramolecular hydrogen bonding is present in catechol.

Incorrect match with respect to compounds to be distinguished and reagent used is given in option (d). It is because 2,4 -dinitrophenylhydrazine confirms the presence of carbonyl group ( $\mathrm{C}=\mathrm{O}$ ) which is present in both aldehydes and ketones. Thus, this reagent cannot be used to distinguish given compounds.

$ \text { The complete Reimer-Tiemann reaction is as follows, } $

$ \text { Thus, } X, Y \text { and } Z \text { are } $

The correct increasing order of boiling points is

It is because of strength of intermolecular forces present in each molecule, with hydrogen bonding in alcohols being the strongest and van der Waals' forces in alkanes being the weakest.

The complete reaction sequence is as follows

$ \text { Thus the compound } X \text { and } Y \text { are } $

The complete reaction sequence is as follows

1. Understand pKa and Acidity: A lower pKa value indicates a stronger acid.

2. Analyze each compound:

   • (C) Ethanol: Alcohols are generally weak acids. Their acidity is similar to water. The pKa of ethanol is known to be around 16.

     • Looking at List-II, (I) 15.9 is a perfect match for Ethanol.

     • So, C-I.

   • (B) Phenol: Phenol is significantly more acidic than alcohols due to the resonance stabilization of its conjugate base (phenoxide ion). Its pKa is typically around 10.

     • Looking at List-II, (III) 10.0 is a direct match for Phenol.

     • So, B-III.

   • (A) p-nitrophenol: The nitro group (-NO2) is a strong electron-withdrawing group (EWG) by both inductive (-I) and resonance (-R) effects. When an EWG is at the para position, it strongly stabilizes the phenoxide ion through resonance, making p-nitrophenol much more acidic than phenol. Therefore, its pKa should be significantly lower than 10.

     • Looking at List-II, (II) 7.1 and (V) 8.3 are options below 10.0. The nitro group typically lowers the pKa of phenol by about 2-3 units. So, 10.0 - 2.9 = 7.1.

     • Thus, (II) 7.1 is an excellent match for p-nitrophenol.

     • So, A-II.

   • (D) p-cresol: p-cresol is 4-methylphenol. The methyl group (-CH3) is an electron-donating group (EDG) due to hyperconjugation and inductive effect (+I). EDGs destabilize the phenoxide ion, making p-cresol slightly less acidic than phenol. Therefore, its pKa should be slightly higher than phenol's pKa (10.0).

     • Looking at List-II, we have (IV) 10.2 left. This value is slightly higher than 10.0 (phenol), which is consistent with the presence of an EDG.

     • So, D-IV.

3. Consolidate the matches:

   • (A) p-nitrophenol - (II) 7.1

   • (B) Phenol - (III) 10.0

   • (C) Ethanol - (I) 15.9

   • (D) p-cresol - (IV) 10.2

4. Check the options:

   • Option A: A-II, B-V, C-I, D-III (Incorrect, B is 8.3, D is 10.0)

   • Option B: A-II, B-III, C-I, D-IV (This perfectly matches our derived pairs.)

   • Option C: A-V, B-IV, C-II, D-III (Incorrect, C is 7.1)

   • Option D: A-IV, B-III, C-I, D-V (Incorrect, A is 10.2, D is 8.3)

Therefore, Option B is the correct answer.

The final answer is $\boxed{B}$

The complete mechanism of reaction is,

$ \text { Bromination of alkene } $

$ X=\text { Butan-2-ol, } Y=\mathrm{KOH} $

$ \text { The complete reaction involved is } $

$ \text { Thus, } X \text { undergoes substitution with water in two steps. } $

The correct order of boiling point is,

$ \mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A} $

Alcohol has highest boiling point due to hydrogen bonding.

Order is,

Alcohol $>$ Ketone $>$ Aldehyde $>$ Ether

There are 4 primary alcohol for $\mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$ they are pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol and 2,2-dimethylpropan-1-ol There are 3 secondary alcohol possible for $\mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$ which are pentan-2-ol, pentan-3-ol and methylbutan-2-ol. Only one tertiary alcohol possible for $\mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$ which is 2-methylbutan-2-ol.

In quinol, intramolecular hydrogen bonding is absent

Williamson ether synthesis An $\mathrm{S}_{\mathrm{N}} 2$ reaction between an alkoxide and an alkyl halide. Williamson synthesis work best with $2^{\circ}$ and $3^{\circ}$. If often gives E2 elimination product.

The complete reaction is as follows,

$ \begin{array}{l} \underset{\substack{\text { Lauryl alcohol } \\\\ (X)}}{\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{10} \mathrm{CH}_{2} \mathrm{OH} \xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}}} \\\\ \underset{\substack{\text { Lauryl hydrogen sulphate } \\\\ (Y)}}{\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{10} \mathrm{CH}_{2} \mathrm{OSO}_{3} \mathrm{H}} \xrightarrow{\mathrm{NaOH}} \\\\ \underset{\substack{\text { Sodium lauryl sulphate product (Z) }}}{\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{10} \mathrm{CH}_{2} \mathrm{OSO}_{2} \mathrm{Na}} \end{array} $

The correct match is

A-II, B-I, C-IV, D-III

The complete reaction is as follow

Bithionol $\left(\mathrm{C}_{12} \mathrm{H}_6 \mathrm{Cl}_4 \mathrm{O}_2 \mathrm{~S}\right)$ Terpineol $\left(\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}\right)$

Chloroxylenol $\left(\mathrm{C}_8 \mathrm{H}_9 \mathrm{ClO}\right)$

Thus, the number of -OH groups present in bithionol, terpineol and chloroxylenol is $2,1,1$.

The complete reaction is as follows.

The correct of boiling point is

$ \text { II }>\text { III }>\text { IV }>\text { I } $

The number of H -bonding guides the boiling point. In case of alcohol, the trend of boiling point is, $3^{\circ}$ alcohol $>2^{\circ}$ alcohol $>1^{\circ}$ alcohol. Hence, the correct order is

$ >\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{CH}_3>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O} $

The alcohol $(X) \mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$ can be prepared by acid catalyst hydration of 2-methyl-1-butene.

$\Rightarrow X$ is a $3^{\circ}$-alcohol, as it produces turbidity, instantly with Lucas reagent ( $\mathrm{Conc} . \mathrm{HCl}+$ anh. $\mathrm{ZnCl}_2$ ).

Chlorobenzene is not formed in the reaction of phenol with thionyl chloride because the nucleophilic substitution reaction required to substitute the hydroxyl group in phenol with a chlorine atom does not readily occur. The reason for this is that the carbon-oxygen bond in phenol has partial double bond character due to resonance. This double bond character makes the C-O bond strong and less reactive towards nucleophilic substitution reactions, which typically involve the cleavage of a single C-O bond.

Hence, the assertion (A) is correct. The reason (R) provided is also correct, and it is indeed the correct explanation for why phenol does not react in the proposed manner. The carbon-oxygen bond's partial double bond character reduces the tendency of the C-O bond to participate in substitution reactions required for chlorobenzene formation.

Thus, the correct answer is:

Option A: Both (A) and (R) are correct and (R) is the correct explanation of (A).

Since the alcohol $X$ does not give turbidity with conc. HCl and $\mathrm{ZnCl}_2$ at room temperature. Thus, the $X$ is primary alcohol. The required reaction is as follows.

Thus, the compounds $Y$ and $Z$ are 2, 4, 6-trinitrophenol (picric acid) and 2, 4, 6-tribromophenol.

The reaction involved is as follows.

In o-nitrophenol, hydrogen of $ \mathrm{O}-\mathrm{H} $ group and oxygen of nitro group form intramolecular hydrogen bonding producing a six membered ring structure while in $ m $-nitrophenol, $ p $-nitrophenol and phenol, intermolecular hydrogen bonding are present.

The structure is,

2-bromopropane undergoes dehydrohalogenation reaction in presence of alcoholic KOH to give unsaturated hydrocarbon i.e. propene ( $X$ ) which further undergoes hydroboration oxidation to give alcohol i.e. propanol.

The common name of benzene 1,3-diol is resorcinol.

Since in Williamson ether synthesis, $1^{\circ}$ (alkyl halide) forms ether via $\mathrm{S}_{\mathrm{N}} 2$ mechanism but with $3^{\circ}$ alkylhalide, due to steric hindrance, it undergoes elimination reaction to form alkene.

Among the given options, reaction given in option (b) is feasible pyridinium chlorochromate (PCC) is a complex of chromium trioxide with pyridine and HCl . This reagent is used for the oxidation of primary alcohols to aldehydes.

$ \begin{aligned} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{OH} \xrightarrow{\mathrm{PCC}} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \end{aligned} $

• Reaction (a) does not occur because reaction with $\mathrm{SOCl}_2$ generally follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism, which occurs at $s p^3$-hybridised carbon atom containing a leaving group But in case of phenol, the carbon attached to phenolic OH is $s p^2$-hybridised. Hence, phenols do not react with $\mathrm{SOCl}_2$.

• Reaction (c), is not feasible as oxidation of primary alcohol with acidic $\mathrm{KMnO}_4$ does not stop reaction at aldehyde stage. It will further leads to formation of carboxylic acid.

• Also in (d), for primary alcohol a mild reagent ( $20 \% \mathrm{H}_3 \mathrm{PO}_4 / 273 \mathrm{~K}$ ) is used for dehydration which does not give alkene. These kind of reagents are use for dehydration of secondary and tertiary alcohols.

Lucas test is used to differentiate between primary, secondary and tertiary alcohols using a solution of anhydrous $\mathrm{ZnCl}_2$ in conc. HCl which is a Lucas reagent. Hence, 1-propanol (primary alcohol) can be distinguished from 2-propanol (secondary alcohol) by Lucas test.

Anisoledoesn't react with

sodium metal as it anisole doesn't contain acidic hydrogen.

In Reimer Tiemann reaction, phenol reacts with chloroform $\left(\mathrm{CHCl}_3\right)$ in presence of bases like $\mathrm{NaOH}, \mathrm{KOH}$ etc. it gives 0 -hydroxy benzaldehyde as major product.

Kolbe's reaction is used to prepare o-hydroxy carboxylic acid from phenol in presence of $\mathrm{CO}_2$ and strong base.

The oxidation of cumene (isopropylbenzene) in the presence of air gives cumene hydro-peroxide. The reaction between cumene hydroxide and dilute acid gives phenol. The reaction involved is as follows :

A is true but R is false.

Lucas reagent is a solution of anhydrous $\mathrm{ZnCl}_2$ and conc. HCl . It is used to distinguish between $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ alcohol.

Primary alcohol remain colourless unless it is subjected to heat.

Secondary alcohol turns turbid in 3 to 5 min .

Tertiary alcohol produce turbidity immediate with Lucas reagent.

Primary alcohol and secondary alcohol give aldehyde and ketones respectively when heated with Cu at 573 K as they undergoes dehydrogenation while tertiary alcohol undergoes dehydration.

(i) First phenol is distilled with Zn -dust to give benzene. It is an example of reduction.

(ii) Benzene reacts with chlorine in presence of anhydrous $\mathrm{FeCl}_3$ to give chlorobenzene.

$\mathrm{p} K_a$ value is inversely proportional to the acidity, i.e. the smaller the $\mathrm{pK}_a$ value, the stronger is the acid.

$-\mathrm{NO}_2$ group at ortho and para-position withdraws electron of -OH group towards itself due to their $-R$ effect while - $\mathrm{NO}_2$ group at meta-position withdraw electron from -OH group by

the weaker -1 effect. Thus, ortho and para-nitrophenols are more acidic than meta-nitrophenol. Also, para-nitrophenol is more acidic than ortho-nitrophenol due to intermolecular H -bonding in ortho-nitrophenol which makes loss of proton little more difficult. Phenol is least acidic in nature due to absence of electron-withdrawing group. So, acidity order will be

Hence, the correct $\mathrm{p} K_a$ value order will be

$ A>C>B>D $

Since Williamson's ether synthesis takes place via $\mathrm{S}_{\mathrm{N}} 2$ mechanism. So, alkyl chloride with more steric hindrance will be less reactive in nature. Out of (i), (ii), (iii) and (iv), (i) will have high steric hindrance and will be least reactive. Alkyl chloride (ii) reacts faster than (iii) and (iv). Since $\mathrm{CH}_2=\mathrm{CH}$ - is electron withdrawing which makes $\mathrm{CH}_2$ more electron deficient than (iii) and (iv). In other words, nucleophilic attack occur faster on (ii). Out of (iii) and (iv), (iii) will be more electron deficient than (iv) due to absence of one methyl group, i.e. electron-donating group.

Hence, correct order of reactivity towards Williamson's ether synthesis will be

$ \text { (ii) }>\text { (iii) }>\text { (iv) }>\text { (i) } $

The rate of acid mediated dehydration reaction of alcohol depends on the stability of the carbocation. The order of stability of carbocations for given alcohols is

Hence, the correct order of rate of dehydration is

$ \mathrm{III}>\mathrm{IV}>\mathrm{II}>\mathrm{I} . $