Properties of Phenols Questions in English

(D) The phthalein dye test is a characteristic chemical test used to identify the presence of phenolic groups.
In this test, a phenol reacts with phthalic anhydride in the presence of concentrated sulphuric acid $(H_2SO_4)$ as a dehydrating agent.
This reaction results in the formation of a phthalein dye (such as phenolphthalein), which exhibits a distinct color change in an alkaline medium.
Therefore, the correct functional group identified by this test is the phenolic group.

(C) $1$. Acidic strength of phenols depends on the stability of the phenoxide ion formed after the loss of a proton.
$2$. Electron-withdrawing groups $(-EWG)$ increase acidic strength by stabilizing the phenoxide ion, while electron-donating groups $(-EDG)$ decrease it.
$3$. The substituents at the $p$-position are:
- $p$-nitrophenol $(III)$: $-NO_2$ is a strong $-I$ and $-M$ group (strongest acid).
- $p$-chlorophenol $(I)$: $-Cl$ is $-I$ (electron-withdrawing) but $+M$ (electron-donating). The $-I$ effect dominates, making it more acidic than phenol.
- $p$-Cresol $(II)$: $-CH_3$ is $+I$ and hyperconjugation (weakly electron-donating).
- $p$-methoxyphenol $(IV)$: $-OCH_3$ is $-I$ but strong $+M$ (strongly electron-donating).
$4$. The order of electron-donating/withdrawing ability is: $-NO_2$ $(-M)$ > $-Cl$ $(-I > +M)$ > $-CH_3$ $(+I)$ > $-OCH_3$ $(+M)$.
$5$. Thus, the decreasing order of acidic strength is: $III > I > II > IV$.

(D) $1$. $A$ trihydric phenol is a benzene ring substituted with three hydroxyl $(-OH)$ groups.
$2$. Hydroquinone is a dihydric phenol ($1,4$-dihydroxybenzene).
$3$. Catechol is a dihydric phenol ($1,2$-dihydroxybenzene).
$4$. $o$-Cresol is a monohydric phenol ($2$-methylphenol).
$5$. Pyrogallol is a trihydric phenol ($1,2,3$-trihydroxybenzene).
$6$. Therefore, the correct option is $D$.

(D) $1$. Solubility in water depends on the ability of the compound to form hydrogen bonds with water molecules.
$2$. $o-Nitrophenol$ exhibits intramolecular hydrogen bonding, which reduces its ability to form intermolecular hydrogen bonds with water.
$3$. $p-Nitrophenol$ exhibits intermolecular hydrogen bonding with water molecules due to the presence of the polar $-NO_2$ group and the $-OH$ group, making it more soluble than the others.
$4$. Phenol and $p-Cresol$ have hydrophobic hydrocarbon parts that limit their solubility compared to $p-Nitrophenol$.

(C) $1$. $o-Nitrophenol$ exhibits intramolecular hydrogen bonding, which reduces the intermolecular forces of attraction.
$2$. $p-Nitrophenol$ exhibits strong intermolecular hydrogen bonding, leading to the association of molecules.
$3$. Due to this strong intermolecular association, $p-Nitrophenol$ requires more energy to break the lattice structure, resulting in a significantly higher melting point compared to $Phenol$, $o-Nitrophenol$, and $p-Cresol$.

(C) When $phenol$ reacts with bromine water, it undergoes electrophilic substitution at all ortho and para positions simultaneously.
This results in the formation of $2,4,6-tribromophenol$, which appears as a white precipitate.
The chemical equation is: $C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH + 3HBr$.

(D) $1$. The molecular formula $C_6H_6O$ corresponds to the degree of unsaturation $U = 6 - \frac{6}{2} + 1 = 4$, which suggests an aromatic ring.
$2$. The compound dissolves in $NaOH$, indicating it is acidic in nature.
$3$. It gives a characteristic violet colour with neutral $FeCl_3$, which is a standard test for phenolic groups.
$4$. It reacts with bromine water to form a tri-bromo derivative $(2,4,6-\text{tribromophenol})$, which is characteristic of phenol due to the strong activating effect of the $-OH$ group.
$5$. Therefore, the compound is phenol.

(B) The given reaction sequence is the Reimer-Tiemann reaction followed by oxidation:
$1$. $Phenol$ reacts with $CHCl_3$ and $NaOH$ to form $o-hydroxybenzaldehyde$ (Salicylaldehyde) as the major product $(A)$.
$2$. The intermediate $A$ is already $Salicylaldehyde$. The step $\xrightarrow[H_3O^+]{} B$ typically refers to the acidification of the phenoxide intermediate to obtain the final aldehyde product $B$.
$3$. Oxidation of $Salicylaldehyde$ $(B)$ using an oxidizing agent converts the $-CHO$ group into a $-COOH$ group, resulting in $Salicylic \ acid$ $(C)$.
Therefore, the final product $C$ is $Salicylic \ acid$.

(B) Step $1$: The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ at $340 \text{ K}$ followed by hydrolysis leads to the formation of salicylaldehyde (o-hydroxybenzaldehyde).
Step $2$: This specific chemical reaction is known as the Reimer-Tiemann reaction.
Step $3$: The electrophile involved in this reaction is dichlorocarbene $(:CCl_2)$.

(B) Step $1$: The reaction between a phenol $(Ar-OH)$ and an acid chloride $(R-COCl)$ in the presence of a base like pyridine is known as the Schotten-Baumann reaction.
Step $2$: The lone pair on the oxygen atom of the phenol attacks the electrophilic carbonyl carbon of the acid chloride.
Step $3$: The chloride ion $(Cl^-)$ acts as a leaving group, and the pyridine neutralizes the $HCl$ produced to drive the reaction forward.
Step $4$: The product formed is an ester, specifically an aryl ester, with the structure $Ar-O-C(=O)-R$.

(D) The reaction between phenol $(C_6H_5OH)$ and formaldehyde $(HCHO)$ in the presence of an acid catalyst is an electrophilic aromatic substitution reaction.
Phenol is an ortho/para-directing group due to the electron-donating $-OH$ group.
Formaldehyde acts as an electrophile in the presence of acid.
The reaction leads to the formation of hydroxybenzyl alcohol (also known as saligenin) as the major product, specifically the ortho-isomer ($o$-hydroxybenzyl alcohol).
Thus, the product $P$ is $o$-hydroxybenzyl alcohol.

(D) The reaction shows the oxidation of phenol to $p$-benzoquinone.
Phenol undergoes oxidation in the presence of chromic acid ($CrO_3$ in acidic medium) to form a conjugated diketone known as $p$-benzoquinone.
Therefore, the correct condition '$A$' is oxidation with $CrO_3$.

(C) $1$. Acid strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$2$. $p-Nitrophenol$ $(d)$ is the strongest acid because the $-NO_2$ group exerts a strong $-I$ and $-M$ effect, which stabilizes the phenoxide ion significantly.
$3$. $Phenol$ $(c)$ is more acidic than alcohols because the phenoxide ion is stabilized by resonance.
$4$. Among alcohols, $Ethanol$ $(a)$ is more acidic than $2-Methylpropan-2-ol$ $(b)$ because the electron-donating inductive effect $(+I)$ of the three methyl groups in $2-Methylpropan-2-ol$ destabilizes the alkoxide ion more than the single methyl group in $Ethanol$.
$5$. Thus, the decreasing order of acid strength is $d > c > a > b$.