Alcohol, Phenols and Ethers

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

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.

$\mathrm{CO}+\mathrm{H}_2 \xrightarrow[\text { High pressure }(673 \mathrm{~K}, 300 \mathrm{~atm}) \mathrm{ZnO}-\mathrm{Cr}_2 \mathrm{O}_3]{\text { Migh temperature }} \underset{\text { Methanol }}{\mathrm{CH}_3 \mathrm{OH}}$

Methanol is industrially prepared when mixture of carbon monoxide and hydrogen is subjected to high temperature and pressure in presence of $\mathrm{ZnO}-\mathrm{Cr}_2 \mathrm{O}_3$ as catalyst.

Acidity $\propto$ stability of the anion formed. The anions formed are

Chlorine is an ortho-para directing group. Chlorine is electronegative species but it has lone pair of electrons which has tendency to added in resonance structure of benzene ring which develop negative charge on ortho and para position and minimum electron density at meta position.

Alkyl bromide in presence of $\mathrm{Mg} / \mathrm{ether}$ gives $R \mathrm{Mg}^{+} \mathrm{Br}^{-}$. Here, alkyl group is nucleophilic and attacks $\mathrm{CO}_2$ to form carboxylic acid.

$\mathrm{SOCl}_2$ is used to replace $-\mathrm{OH}$ group to $-\mathrm{Cl}$ group and evolved $\mathrm{SO}_2$ and $\mathrm{HCl}$ gas.