Alcohol, Phenols and Ethers

Metamerism arises due to different alkyl chains on either side of the functional group in the molecule pair of molecules $\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OCH}_2 \mathrm{CH}_3$ are metamers of each other.

$ \text { } \begin{aligned}So,\,\, \mathrm{X} & =\mathrm{POCl}_3 \\ \mathrm{Z} & =\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \end{aligned} $

Phthalein Dye test: Phenol on heating with phthalic anhydride in presence of concentrated sulphuric acid forms a colourless condensation compound called phenolphthalein. On further reaction with NaOH it gives pink colour.

So phenolic group is correct answer.

On the basis of polarity the correct order is

$ \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5<\mathrm{CH}_3 \mathrm{COCH}_3<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}<\mathrm{CH}_3 \mathrm{COOH} $

So correct order is

$ A< C< B< D $

Tertiary alcohols react instantaneously with Lucas reagent and gives immediate turbidity. In case of tertiary alcohols, they form halides easily with Lucas reagent (conc. $\mathrm{HCl}$ and $\mathrm{ZnCl}_2$)

Let's analyze the given reactions and determine the products $\mathrm{A}$ and $\mathrm{B}$.

Reaction with PCl₃:

When an alcohol ($\mathrm{ROH}$) reacts with phosphorus trichloride ($\mathrm{PCl}_3$), the hydroxyl group ($\mathrm{OH}$) of the alcohol is replaced by a chloro group ($\mathrm{Cl}$), resulting in the formation of an alkyl chloride ($\mathrm{RCl}$). Additionally, the reaction produces phosphorous acid as a byproduct. This can be represented as:

$3 \mathrm{ROH} + \mathrm{PCl}_3 \rightarrow 3 \mathrm{RCl} + \mathrm{H}_3\mathrm{PO}_3$

Here, $\mathrm{A}$ is $\mathrm{H}_3\mathrm{PO}_3$ (phosphorous acid).

Reaction with PCl₅:

When an alcohol reacts with phosphorus pentachloride ($\mathrm{PCl}_5$), it also replaces the hydroxyl group with a chloro group. The reaction differs slightly in the byproducts; phosphorus oxychloride ($\mathrm{POCl}_3$) and hydrogen chloride ($\mathrm{HCl}$) are produced:

$\mathrm{ROH} + \mathrm{PCl}_5 \rightarrow \mathrm{RCl} + \mathrm{HCl} + \mathrm{POCl}_3$

Here, $\mathrm{B}$ is $\mathrm{POCl}_3$ (phosphorus oxychloride).

Thus, in the reactions provided:

• $\mathrm{A} = \mathrm{H}_3\mathrm{PO}_3$
• $\mathrm{B} = \mathrm{POCl}_3$

Matching these findings against the provided options, we see that Option D correctly identifies $\mathrm{A}$ as $\mathrm{H}_3\mathrm{PO}_3$ and $\mathrm{B}$ as $\mathrm{POCl}_3$:

Option D: $\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{POCl}_3$

This is the correct answer.

Due to presence of conjugation in product.

At STP condition substitution at sp² carbon atom is not feasible.

$\bullet$ Nitro group has electron withdrawing tendency. It can withdraw electrons both by $-$I effect and $-$R effect. Thus the acidic strength of monosubstituted nitrophenol is higher than phenol.

$\bullet$ Nitro group present at o- and p-positions will have strong $-$R effect while nitro group present at m-position will influence only $-$I effect hence acidity or o/p isomer will be more meta isomer.

Primary, secondary and tertiary alcohols can be differentiated by their reaction with (HCl + anhy ZnCl₂) Lucas reagent

$\bullet$ 3$^\circ$ alcohol $\buildrel {ZnC{l_2} + HCl} \over \longrightarrow $ Immediate turbidity at room temperature

$\bullet$ 2$^\circ$ alcohol $\buildrel {ZnC{l_2} + HCl} \over \longrightarrow $ Turbidity after 5 minutes at room temperature

$\bullet$ 1$^\circ$ alcohol $\buildrel {ZnC{l_2} + HCl} \over \longrightarrow $ Do not gives turbidity at room temperature