Mix Examples-Alcohol, Phenol and Ethers Questions in English

(D) Statement $(A)$ is correct: Phenol is more acidic than alcohols because the phenoxide ion formed after deprotonation is stabilized by resonance with the benzene ring.
Statement $(B)$ is incorrect: Melamine plastic is produced from melamine and formaldehyde,not phenol.
Statement $(C)$ is correct: Phenol reacts with neutral $FeCl_3$ solution to form a violet-colored iron-phenoxide complex.
Statement $(D)$ is incorrect: Phenol reacts with acetyl chloride $(CH_3COCl)$ to form phenyl acetate $(C_6H_5OCOCH_3)$,not phenetole $(C_6H_5OC_2H_5)$.
Therefore,statements $(A)$ and $(C)$ are correct.

(D) Let us analyze each reaction:
$I$. $CH_3CH_2CH=CH_2 \xrightarrow[(ii) NaOH/H_2O_2]{(i) B_2H_6} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). This is hydroboration-oxidation,which follows anti-Markovnikov addition of water.
$II$. $CH_3CH_2CHO \xrightarrow[(ii) H_2O]{(i) CH_3MgI} CH_3CH_2CH(OH)CH_3$ (butan-$2$-ol). This is a Grignard reaction forming a secondary alcohol.
$III$. $CH_3CH_2CH_2CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2, HCl} CH_3CH_2CH_2CHO$ $\xrightarrow{NaBH_4} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). The Stephen reduction converts nitrile to aldehyde,which is then reduced to primary alcohol.
$IV$. $CH_3CH_2CH_2CO_2H \xrightarrow{B_2H_6} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). Diborane specifically reduces carboxylic acids to primary alcohols.
Thus,reactions $I, III$ and $IV$ produce $n$-butanol.

(A) $1$. When salicylic acid is heated with sodalime $(NaOH + CaO)$,it undergoes decarboxylation to form phenol $([A])$.
$2$. When phenol $([A])$ is distilled with zinc dust,it undergoes reduction to form benzene $([B])$.
$3$. When salicylic acid reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution and decarboxylation to form $2,4,6-\text{tribromophenol}$ $([C])$.
$4$. When salicylic acid is heated at $200-230^{\circ}C$,it undergoes decarboxylation to form phenol $([D])$.

(A) The reaction proceeds as follows:
$(i)$ Phenol reacts with $NaOH$ and $CH_3I$ to form anisole (methoxybenzene) via Williamson ether synthesis.
(ii) Anisole undergoes Friedel-Crafts acylation with propionyl chloride $(CH_3CH_2COCl)$ in the presence of anhydrous $AlCl_3$. Since the $-OCH_3$ group is ortho/para-directing,and the para-position is sterically less hindered,the major product is $p$-methoxypropiophenone.
(iii) Reduction of the ketone group with $NaBH_4$ yields the corresponding secondary alcohol,$1$-($4$-methoxyphenyl)propan$-1-$ol.
Therefore,the correct option is $A$.

(C) In reaction $A$,phenol reacts with sodium $(Na)$ to release hydrogen gas $(x = H_2)$.
Molar mass of $H_2 = 2 \ g/mol$.
In reaction $B$,benzoic acid reacts with sodium bicarbonate $(NaHCO_3)$ to release carbon dioxide gas $(y = CO_2)$.
Molar mass of $CO_2 = 44 \ g/mol$.
Sum of molar masses = $2 + 44 = 46$.

(D) The correct matching is as follows:
$A$. $H_{3}C-CH(OH)-CH_{3}$ is formed by the reaction of $CH_{3}COOH$ with $CH_{3}OH$ followed by reduction (Grignard or similar sequence). This matches with $III$.
$B$. $CH_{3}COOH \rightarrow CH_{3}CH_{2}OH$ is a reduction reaction. This matches with $II$.
$C$. $CH_{3}CH_{2}OH \rightarrow H_{3}C-CH(OH)-CH_{3}$ involves dehydration to ethene followed by hydration. This matches with $IV$.
$D$. Benzene $\rightarrow$ Phenol is the Cumene process. This matches with $I$.
Therefore,the correct sequence is $A-III, B-II, C-IV, D-I$. Note: The provided options in the prompt were slightly inconsistent with standard chemical pathways; based on the logic,the correct mapping is $A-III, B-II, C-IV, D-I$.