Mix Examples-Alcohol, Phenol and Ethers Questions in English

(A) The structural properties of alcohols and ethers are as follows:
$A$. Methanol: The $C-O$ bond length is $142 \ pm$ $(3)$ and the $C-O-H$ bond angle is $108.9^o$ $(iii)$.
$B$. Phenol: The $C-O$ bond length is $136 \ pm$ $(2)$ and the $C-O-H$ bond angle is $109^o$ $(i)$.
$C$. Methoxymethane: The $C-O$ bond length is $141 \ pm$ $(1)$ and the $C-O-H$ bond angle is $111.7^o$ $(ii)$.
Therefore,the correct sequence is $A-3-iii, B-2-i, C-1-ii$.

(A) $(A$ $\rightarrow iv), (B$ $\rightarrow i), (C$ $\rightarrow ii), (D$ $\rightarrow iii)$
$(A)$ Butan$-1-$ol is prepared by the catalytic reduction of butanal $(CH_3CH_2CH_2CHO + H_2 \rightarrow CH_3CH_2CH_2CH_2OH)$.
$(B)$ Propan$-2-$ol is prepared by the hydration of propene in the presence of dilute sulfuric acid (Markovnikov addition).
$(C)$ $2-$Methylpropan$-2-$ol is prepared by the reaction of propanone with methylmagnesium bromide (Grignard reagent) followed by hydrolysis.
$(D)$ Propan$-1-$ol is prepared by the hydroboration-oxidation of propene (Anti-Markovnikov addition).

(A) The correct matches are as follows:
$A$. Cumene reacts with $O_2$ to form Cumene hydroperoxide $(A \rightarrow iii)$.
$B$. Benzene sulphonic acid on fusion with $NaOH$ followed by acidification gives Phenol $(B \rightarrow i)$.
$C$. Chlorobenzene reacts with $NaOH$ at $623 \ K$ and $300 \ atm$ to form Sodium phenoxide $(C \rightarrow iv)$.
$D$. $2$-Methylbutanal on reduction with $NaBH_4$ gives $2$-Methylbutan-$1$-ol $(D \rightarrow ii)$.
Thus,the correct sequence is $A-iii, B-i, C-iv, D-ii$.

(N/A) The boiling point is directly related to the strength of intermolecular forces of attraction. Higher boiling points indicate stronger intermolecular forces. Therefore,it can be concluded that the intermolecular forces of attraction increase in the order: $\text{Ether} < \text{Carbon tetrachloride} < \text{Ethanol} < \text{Water}$.

(A-III, B-IV, C-II, D-I) $(A) \rightarrow (iii)$: In the oxidation of primary alcohol to aldehyde using $CrO_3$,both $O-H$ and $C-H$ bonds are broken.
$(B) \rightarrow (iv)$: In the nitration of phenol,the electrophilic substitution occurs on the ring,involving the breaking of a $C-H$ bond on the benzene ring.
$(C) \rightarrow (ii)$: The reaction of phenol with $NaOH$ is an acid-base reaction where the $O-H$ bond of the phenol breaks to form phenoxide ion.
$(D) \rightarrow (i)$: The dehydration of alcohol to alkene involves the breaking of the $C-O$ bond (along with a $C-H$ bond,but $C-O$ is the primary bond cleavage characteristic of this elimination).
Correct matching: $(A-iii, B-iv, C-ii, D-i)$.

(A) $CH_3-CH(OH)-CH_3$ is a secondary alcohol. Dehydration of secondary alcohol requires $85\% \ H_3PO_4$ at $440 \ K$ $(A \rightarrow iii)$.
$(B)$ $CH_3-C(CH_3)_2-OH$ is a tertiary alcohol. Dehydration of tertiary alcohol occurs easily with $20\% \ H_3PO_4$ at $358 \ K$ $(B \rightarrow i)$.
$(C)$ $CH_3CH_2OH$ is a primary alcohol. Dehydration of primary alcohol requires concentrated $H_2SO_4$ at $443 \ K$ $(C \rightarrow iv)$.
$(D)$ Oxidation of secondary alcohol $(R-CH(OH)-R')$ with $Cu$ at $573 \ K$ gives a ketone $(R-CO-R')$ $(D \rightarrow ii)$.
Therefore,the correct match is: $(A$ $\rightarrow iii), (B$ $\rightarrow i), (C$ $\rightarrow iv), (D$ $\rightarrow ii)$.

(A) The correct matches are:
$A \rightarrow ii$ (Note: The provided options in the prompt had a mismatch in the mapping logic,correcting to standard chemical reactions):
$A$. Phenol $+$ dilute $HNO_3$ gives $o$- and $p$-nitrophenol (Nitration).
$B$. Phenol $+$ Bromine water gives $2,4,6$-tribromophenol (Bromination).
$C$. Phenol $+$ $Na_2Cr_2O_7/H_2SO_4$ gives benzoquinone (Oxidation).
$D$. Sodium phenoxide $+$ $CO_2$ followed by $H^+$ gives salicylic acid (Kolbe reaction).
Thus,the correct sequence is $A-ii, B-i, C-iii, D-iv$.

(A-IV, B-I, C-II, D-III) The correct matches are:
$(A)$ Kolbe's reaction produces Salicylic acid $(iv)$.
$(B)$ Reimer-Tiemann reaction produces Salicylaldehyde $(i)$.
$(C)$ Williamson synthesis is used for the preparation of Ethers $(ii)$.
$(D)$ Fermentation of sugars produces Ethanol $(iii)$.
Therefore,the correct sequence is $(A-iv, B-i, C-ii, D-iii)$.

(A-IV, B-II, C-I, D-III) The correct matches are as follows:
$A \to iv$: The reaction of primary alkyl ethers with $HI$ proceeds via an $S_{N}2$ mechanism.
$B \to ii$: The reaction of ethers containing a tertiary alkyl group with $HI$ proceeds via an $S_{N}1$ mechanism because a stable carbocation is formed.
$C \to i$: The bromination of anisole is an electrophilic aromatic substitution ($S_{E}2$ Aromatic) reaction.
$D \to iii$: The dehydration of ethanol to ethene in the presence of concentrated $H_2SO_4$ at $443 \ K$ is a $\beta$-elimination reaction.
Therefore,the correct sequence is: $A-iv, B-ii, C-i, D-iii$.

(N/A) The correct matches are as follows:
$(i)$ $2-$methylphenol is $o$-Cresol $(d)$.
$(ii)$ Benzene$-1,2-$diol is Catechol $(c)$.
$(iii)$ Benzene$-1,3-$diol is Resorcinol $(f)$.
$(iv)$ Benzene$-1,4-$diol is Hydroquinone $(a)$.
$(v)$ Methoxybenzene is Anisole $(g)$.
$(vi)$ Ethoxybenzene is Phenetole $(b)$.
Therefore,the correct sequence is: ($i$ $\rightarrow$ $d$),($ii$ $\rightarrow$ $c$),($iii$ $\rightarrow$ $f$),($iv$ $\rightarrow$ $a$),($v$ $\rightarrow$ $g$),($vi$ $\rightarrow$ $b$).

(B) $1$. Reaction with $Br_2$ in $CS_2$ at $273 \ K$: This is an electrophilic aromatic substitution reaction. Due to the low polarity of the solvent $CS_2$,the reaction is less polar,leading to the formation of the para-isomer as the major product. Thus,$A$ is $p$-bromophenol.
$2$. Reaction with $CHCl_3$ and $NaOH$ followed by $H_3O^+$: This is the Reimer-Tiemann reaction. It introduces a formyl group $(-CHO)$ at the ortho position of the phenol. Thus,$B$ is $o$-hydroxybenzaldehyde (salicylaldehyde).
Therefore,the major products are $p$-bromophenol and $o$-hydroxybenzaldehyde.

(B) The reaction proceeds in three steps:
$1$. Cross-Cannizzaro reaction: $p$-methoxybenzaldehyde reacts with $HCHO$ in the presence of $NaOH$ to form $p$-methoxybenzyl alcohol $(CH_3O-C_6H_4-CH_2OH)$ and sodium formate.
$2$. Williamson ether synthesis: The alcohol $p$-methoxybenzyl alcohol reacts with $CH_3CH_2Br$ in the presence of $NaH$ to form $p$-methoxybenzyl ethyl ether $(CH_3O-C_6H_4-CH_2-O-CH_2CH_3)$.
$3$. Cleavage with $HI$ and $\Delta$: The ether linkage and the methoxy group are cleaved by $HI$ upon heating. The methoxy group $(CH_3O-)$ is converted to $OH$ and $CH_3I$,and the benzyl ether $(R-CH_2-O-CH_2CH_3)$ is cleaved to form $p$-hydroxybenzyl iodide $(HO-C_6H_4-CH_2I)$ and ethanol (which further reacts with $HI$ to form $CH_3CH_2I$).
Thus,the final product $A$ is $4$-hydroxybenzyl iodide.

(D) $1$. Compound $A$ is phenol $(C_{6}H_{5}OH)$,which gives a characteristic violet/dark green color with $FeCl_{3}$ solution.
$2$. Phenol reacts with $CHCl_{3}$ and $KOH$ (Reimer-Tiemann reaction) to form salicylaldehyde $(B)$.
$3$. Salicyl alcohol $(C)$ is $2$-hydroxybenzyl alcohol. Upon oxidation with $PCC$,the primary alcohol group is oxidized to an aldehyde group,yielding salicylaldehyde $(B)$.
$4$. Therefore,$A$ is phenol,$B$ is salicylaldehyde,and $C$ is salicyl alcohol.

(A) The correct matches are:
$A$. Phenol to Salicylaldehyde is the Reimer-Tiemann reaction,which uses $CHCl_3/NaOH$ $(IV)$.
$B$. Phenol to Benzene is the reduction of phenol using $Zn$ dust $(III)$.
$C$. Phenol to $p$-Benzoquinone is the oxidation of phenol using $Na_2Cr_2O_7/H_2SO_4$ $(II)$.
$D$. Phenol to $p$-Bromophenol is the bromination of phenol in a non-polar solvent like $CS_2$ $(I)$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(D) To give a red colouration with ceric ammonium nitrate,the compound must be an alcohol.
To give a positive iodoform test,the compound must contain a $CH_3CH(OH)-$ group or a $CH_3CO-$ group.
Let us analyze the given compounds:
$1$. Acetophenone $(C_6H_5COCH_3)$: Gives iodoform test but is not an alcohol (no red colour with ceric ammonium nitrate).
$2$. $1$-phenylpropan-$2$-ol $(C_6H_5CH_2CH(OH)CH_3)$: It is an alcohol (gives red colour with ceric ammonium nitrate) and contains a $CH_3CH(OH)-$ group (gives positive iodoform test).
$3$. Acetaldehyde $(CH_3CHO)$: Gives iodoform test but is not an alcohol.
$4$. Butan-$2$-ol $(CH_3CH(OH)CH_2CH_3)$: It is an alcohol (gives red colour with ceric ammonium nitrate) and contains a $CH_3CH(OH)-$ group (gives positive iodoform test).
$5$. Diethyl ether $(C_2H_5OC_2H_5)$: Neither an alcohol nor a methyl ketone.
$6$. $2$-phenylethanol $(C_6H_5CH_2CH_2OH)$: It is an alcohol but does not contain the $CH_3CH(OH)-$ group.
$7$. $1$-cyclohexylpropan-$2$-ol $(C_6H_{11}CH_2CH(OH)CH_3)$: It is an alcohol (gives red colour with ceric ammonium nitrate) and contains a $CH_3CH(OH)-$ group (gives positive iodoform test).
$8$. $2$-cyclohexylpropan-$2$-ol $(C_6H_{11}C(OH)(CH_3)_2)$: It is an alcohol but does not contain the $CH_3CH(OH)-$ group.
Thus,there are $3$ compounds that satisfy both conditions.

(A) The molecular formula $C_9H_{10}O$ has a degree of unsaturation of $6$.
Given conditions:
$(i)$ Does not dissolve in $NaOH$ (not a phenol).
$(ii)$ Does not dissolve in $HCl$ (not an amine).
$(iii)$ Does not give orange precipitate with $2,4-DNP$ (no carbonyl group,i.e.,no aldehyde or ketone).
$(iv)$ On hydrogenation,it gives an identical compound $C_9H_{12}O$.
Since it is not a carbonyl compound and not a phenol,it is likely an ether.
The structure $Ph-CH=CH-O-CH_3$ (methyl styryl ether) satisfies these conditions.
It exists as $cis$ and $trans$ isomers.
Both isomers on hydrogenation yield $Ph-CH_2-CH_2-O-CH_3$ $(C_9H_{12}O)$.
Thus,there are $2$ such isomers.

(C) The reaction proceeds as follows:
$1$. The alcohol group is protonated by $HCl$ and leaves as $H_2O$,forming a tertiary carbocation.
$2$. This carbocation undergoes a ring expansion (Wagner-Meerwein rearrangement) because the five-membered ring expands to a six-membered ring to form a more stable carbocation.
$3$. Finally,treatment with $KOH$ causes dehydrohalogenation (elimination of $H^+$) to form the most stable alkene,which is the major product '$B$'.
$4$. The final structure corresponds to the product shown in option $C$.

(A) Statement $I$: The acidic strength order of nitrophenols is $p$-nitrophenol $>$ $o$-nitrophenol $>$ $m$-nitrophenol. $p$-nitrophenol is the most acidic due to the strong $-I$ and $-M$ effects of the nitro group at the para position,which stabilizes the phenoxide ion significantly. $o$-nitrophenol is less acidic than $p$-nitrophenol due to intramolecular hydrogen bonding,and $m$-nitrophenol is the least acidic as it only exhibits the $-I$ effect.
Statement $II$: Lucas reagent $(conc. HCl + ZnCl_2)$ is used to distinguish between primary,secondary,and tertiary alcohols. Tertiary alcohols give immediate turbidity,secondary alcohols give turbidity in $5-10$ minutes,and primary alcohols (like ethanol) do not give turbidity at room temperature; they only react upon heating.
Therefore,Statement $I$ is true and Statement $II$ is false.

(A) $A \rightarrow$ Kolbe-Schmidt reaction: Phenol reacts with $NaOH$ followed by $CO_2$ and $HCl$ to form salicylic acid. Matches with $IV$.
$B \rightarrow$ Reimer-Tiemann reaction: Phenol reacts with $CHCl_3$ and $NaOH$ followed by $HCl$ to form salicylaldehyde. Matches with $III$.
$C \rightarrow$ Oxidation of phenol: Phenol is oxidized by $Na_2Cr_2O_7$ and $H_2SO_4$ to $p$-benzoquinone. Matches with $I$.
$D \rightarrow$ Williamson ether synthesis: Phenol reacts with $NaOH$ to form sodium phenoxide,which then reacts with $CH_3Cl$ to form anisole $(PhOCH_3)$. Matches with $II$.
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(A) The given compounds are:
$(A)$ Diethyl ether derivative (Ether): $H_3C-CH_2-O-CH_2-CH_2-CH_3$ (Polar,dipole-dipole interactions)
$(B)$ Pentane (Alkane): $H_3C-CH_2-CH_2-CH_2-CH_3$ (Non-polar,weak van der Waals forces)
$(C)$ Pentan$-3-$one (Ketone): $CH_3-CH_2-CO-CH_2-CH_3$ (Highly polar,strong dipole-dipole interactions)
$(D)$ Pentan$-2-$ol (Alcohol): $H_3C-CH(OH)-CH_2-CH_2-CH_3$ (Hydrogen bonding)
Boiling point order depends on intermolecular forces:
$1$. Hydrogen bonding (Alcohol) has the highest boiling point.
$2$. Dipole-dipole interactions (Ketone) have higher boiling point than ethers.
$3$. Ethers have higher boiling point than alkanes due to polarity.
$4$. Alkanes (Non-polar) have the lowest boiling point.
Thus,the order is: $B < A < C < D$.

(D) The reactions are matched as follows:
$(A)$ Aniline reacts with $NaNO_2 + HCl$ followed by warm $H_2O$ to form phenol. This corresponds to product $(II)$.
$(B)$ Phenol reacts with $Na_2Cr_2O_7 / H_2SO_4$ to form $p$-benzoquinone. This corresponds to product $(IV)$.
$(C)$ Phenol reacts with $CHCl_3 + aq. NaOH$ followed by $H^+$ (Reimer-Tiemann reaction) to form salicylaldehyde. This corresponds to product $(I)$.
$(D)$ Phenol reacts with $NaOH$,$CO_2$,and $H^+$ (Kolbe's reaction) to form salicylic acid. This corresponds to product $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.

(A) . $Br_2$ water test is a test for unsaturation in which the reddish-orange colour of bromine water disappears.
$B$. Alcohols give a red colour with ceric ammonium nitrate.
$C$. Phenols give a violet colour with neutral ferric chloride.
$D$. Aldehydes and ketones give yellow,orange,or red coloured precipitates with $2,4-DNP$ ($2$,$4$-dinitrophenylhydrazine).

(C) The reaction of benzene with $n$-propyl chloride in the presence of $AlCl_3$ is a Friedel-Crafts alkylation.
Due to the rearrangement of the primary carbocation $(CH_3CH_2CH_2^+)$ to a more stable secondary carbocation $(CH_3CH^+CH_3)$,the major product $P$ formed is isopropylbenzene (cumene).
In the second step,cumene undergoes oxidation with $O_2$ followed by acid hydrolysis $(H_3O^+)$ to form phenol and acetone $(CH_3COCH_3)$ as the major products.
Thus,$P$ is isopropylbenzene and $Q$ is acetone.

(A) Undergoes nucleophilic substitution of $Br^-$. Undergoes elimination of $HBr$. Does not undergo nucleophilic addition. Does not esterify with acetic anhydride,but can be dehydrogenated. Thus,$(A-p, q, t)$.
$(B)$ Undergoes nucleophilic substitution with $SOCl_2, PCl_5$ etc. Does not undergo elimination. Does not undergo nucleophilic addition. Undergoes esterification with acetic anhydride. Undergoes dehydrogenation to give $C_6H_5CHO$. Thus,$(B-p, s, t)$.
$(C)$ Does not undergo nucleophilic substitution (no leaving group). Does not undergo elimination. Undergoes nucleophilic addition at the carbonyl carbon of $-CHO$. Undergoes esterification with acetic anhydride (at $-OH$ group). Does not undergo dehydrogenation. Thus,$(C-r, s)$.
$(D)$ Undergoes aromatic nucleophilic substitution $(S_NAr)$. Does not undergo elimination,nucleophilic addition,esterification,or dehydrogenation. Thus,$(D-p)$.

(B) Analysis of reactions:
$P$: Reaction of $t$-butoxide $(t-BuO^-)$ with $t$-butyl bromide $(t-BuBr)$. Since $t-BuO^-$ is a strong base and $t-BuBr$ is a tertiary halide,elimination $(E2)$ is favored,yielding isobutylene $(4)$ and $t$-butanol $(1)$.
$Q$: Reaction of $t$-butyl methyl ether with $HBr$. This is an acid-catalyzed cleavage of ether,yielding $t$-butyl bromide $(2)$ and methanol.
$R$: Reaction of $t$-butyl bromide with $NaOMe$. Since $NaOMe$ is a strong base and $t-BuBr$ is a tertiary halide,elimination $(E2)$ is favored,yielding isobutylene $(4)$.
$S$: Reaction of $t$-butoxide with methyl bromide $(MeBr)$. This is an $S_N2$ reaction,yielding $t$-butyl methyl ether $(3)$.
Matching: $P$ $\rightarrow 1, 4; Q$ $\rightarrow 2; R$ $\rightarrow 4; S$ $\rightarrow 3$.

(C) Step $1$: Conversion of $P$ (aniline) to $Q$ ($2$,$4$,$6$-tribromobenzoyl chloride).
Aniline reacts with excess $Br_2/H_2O$ to form $2,4,6$-tribromoaniline. Diazotization followed by reaction with $CuCN/KCN$ gives $2,4,6$-tribromobenzonitrile. Hydrolysis gives $2,4,6$-tribromobenzoic acid,which reacts with $SOCl_2$ to form $Q$ ($2,4,6$-tribromobenzoyl chloride).
Step $2$: Conversion of $R$ (benzene) to $S$ ($4$-bromophenol).
Benzene reacts with oleum to form benzenesulfonic acid,which on fusion with $NaOH$ followed by acidification gives phenol. Bromination of phenol with $Br_2/CS_2$ at $273 \ K$ gives $4$-bromophenol $(S)$ as the major product.
Step $3$: Synthesis of $T$ from $Q$ and $S$.
$S$ ($4$-bromophenol) reacts with $NaOH$ to form sodium $4$-bromophenoxide,which acts as a nucleophile and attacks the carbonyl carbon of $Q$ ($2,4,6$-tribromobenzoyl chloride) to form an ester $T$ ($4$-bromophenyl $2,4,6$-tribromobenzoate).
Structure of $T$ contains $3$ $Br$ atoms from the benzoyl part and $1$ $Br$ atom from the phenoxy part.
Total number of $Br$ atoms in $T = 3 + 1 = 4$.

(A) The correct matching is as follows:
$P$. $tert$-butyl chloride undergoes $E2$ elimination with $NaOEt$ to form isobutylene. Thus,$P-2$.
$Q$. Sodium $tert$-butoxide reacts with $EtBr$ via $S_N2$ mechanism to form $tert$-butyl ethyl ether. Thus,$Q-3$.
$R$. $1$-methylcyclopentene undergoes oxymercuration-demercuration with $(i) \ Hg(OAc)_2; (ii) \ NaBH_4$ to give $1$-methylcyclopentanol (Markovnikov addition). Thus,$R-1$.
$S$. $1$-methylcyclopentene undergoes hydroboration-oxidation with $(i) \ BH_3; (ii) \ H_2O_2/NaOH$ to give $2$-methylcyclopentanol (anti-Markovnikov addition,syn-addition). Thus,$S-4$.
Therefore,the correct sequence is $P-2, Q-3, R-1, S-4$.

(C) $(P)$ Benzenesulfonic acid reacts with molten $NaOH$ followed by $H_3O^+$ to give phenol. Nitration of phenol with conc. $HNO_3$ gives $2,4,6$-trinitrophenol (picric acid),which corresponds to $(3)$.
$(Q)$ Nitrobenzene is reduced to aniline,which is then converted to benzenediazonium chloride. Hydrolysis gives phenol,which upon nitration gives $2,4,6$-trinitrophenol,but the sequence shown in the image leads to $2,4,6$-trinitrophenol,which is $(3)$. Wait,re-evaluating: The sequence $(Q)$ leads to $2,4,6$-trinitrophenol $(3)$.
$(R)$ Resorcinol ($1,3$-dihydroxybenzene) undergoes sulfonation and nitration followed by desulfonation to yield $2$-nitroresorcinol,which is $(4)$.
$(S)$ Toluene is oxidized to benzoic acid,nitrated to $3,5$-dinitrobenzoic acid,converted to amide,then via Hofmann bromamide degradation to $3,5$-dinitroaniline,then to diazonium salt and finally to $3,5$-dinitrophenol,which is $(1)$.
Thus,the correct matching is $P-3, Q-5, R-4, S-1$.

(D) When $NaOH$ is added to the mixture,benzoic acid reacts with $NaOH$ to form sodium benzoate,which is a salt and is soluble in the aqueous layer.
$C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O$
Chlorobenzene and aniline do not react with $NaOH$ and remain in the organic ethyl acetate layer.
Therefore,the ethyl acetate layer contains chlorobenzene and aniline.

(C) The cyclic ethers with molecular formula $C_4H_8O$ are as follows:
$1$. Tetrahydrofuran ($1$ isomer)
$2$. $2-$Methyltetrahydrofuran ($1$ chiral center,so $2$ enantiomers)
$3$. $3-$Methyltetrahydrofuran ($1$ chiral center,so $2$ enantiomers)
$4$. $2-$Ethyloxirane ($1$ chiral center,so $2$ enantiomers)
$5$. $2,2-$Dimethyloxirane ($1$ isomer)
$6$. cis$-2,3-$Dimethyloxirane ($1$ isomer)
$7$. trans$-2,3-$Dimethyloxirane ($2$ enantiomers)
Total isomers = $1 + 2 + 2 + 2 + 1 + 1 + 1 = 10$.

(A) The boiling point depends on the intermolecular forces of attraction.
$(c)$ $CH_3(CH_2)_3OH$ (n-butanol) has intermolecular hydrogen bonding,which is the strongest interaction,so it has the highest boiling point.
$(b)$ $C_2H_5-O-C_2H_5$ (diethyl ether) has dipole-dipole interactions.
$(a)$ $n$-Butane $(C_4H_{10})$ and $(d)$ $CH_3-CH_3$ (ethane) are alkanes and have only weak London dispersion forces.
Comparing $(a)$ and $(d)$,$n$-butane has a higher molecular mass than ethane,so it has a higher boiling point.
Thus,the order of boiling point is: $(c) > (b) > (a) > (d)$.

(C) . Methanol is known as wood spirit $(q)$.
$B$. Kolbe's reaction involves the conversion of phenol to salicylic acid $(p)$.
$C$. Williamson's reaction is the reaction of alkyl halide with sodium alkoxide to form ether $(s)$.
$D$. Conversion of $2^\circ$ alcohol to ketone is achieved by passing vapours over heated copper at $573 \ K$ $(r)$.
Thus,the correct matching is $A-q, B-p, C-s, D-r$.

(B) . Friedel-Crafts alkylation: The electrophile is the carbocation $\stackrel{+}{C}H_3$ $(R)$.
$B$. Kolbe-Schmitt reaction: The electrophile is $CO_2$ $(P)$.
$C$. Reimer-Tiemann reaction: The electrophile is dichlorocarbene $:CCl_2$ $(Q)$.
$D$. Sulfonation of benzene: The electrophile is $SO_3$ $(T)$.
Therefore,the correct matching is $A-R, B-P, C-Q, D-T$.

(A) $I$: Hydration of olefins (alkenes) in the presence of acid catalyst yields alcohols. (e.g.,$CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$)
$II$: Hydrolysis of alkyl halides with aqueous $NaOH$ or $KOH$ yields alcohols. (e.g.,$R-X + OH^- \rightarrow R-OH + X^-$)
$III$: Reduction of carbonyl compounds (aldehydes,ketones,carboxylic acids,or esters) using reducing agents like $NaBH_4$ or $LiAlH_4$ yields alcohols. (e.g.,$RCHO + 2[H] \rightarrow RCH_2OH$)
Since all three methods are standard laboratory and industrial procedures for the preparation of alcohols,the correct answer is $I, II$ and $III$.

(A, B, C) Let us analyze the reactions of phenol:
$1$. Reaction with phthalic anhydride in the presence of $H_2SO_4$ gives phenolphthalein. Thus,$A$ is phenolphthalein.
$2$. Reaction with salicylic acid in the presence of $H^+$ gives phenyl salicylate (salol). Thus,$B$ is phenyl salicylate.
$3$. The sequence $(i)$ $NaOH$,(ii) $CO_2, \Delta, P$,(iii) $H^+/H_2O$,(iv) $(CH_3CO)_2O$ is the Kolbe-Schmitt reaction followed by acetylation,which yields aspirin. Thus,$C$ is aspirin.
$4$. Reaction with $PhN_2Cl$ in $NaOH$ gives $p$-hydroxyazobenzene. The structure $D$ shown is $p$-aminoazobenzene,which is incorrect.
Therefore,the correct matches are $A$,$B$,and $C$.

(D) Natalite is a trade name for the mixture of diethyl ether and ethyl alcohol.
It is used as a petrol substitute.
It consists of $54\%$ alcohol,$45\%$ ether,and $1\%$ trimethylamine.

(B) The molecular formula $C_4H_{10}O$ represents the following ethers:
$1.$ Diethyl ether: $CH_3CH_2-O-CH_2CH_3$ (Ethoxyethane)
$2.$ Methyl propyl ether: $CH_3-O-CH_2CH_2CH_3$ ($1$-Methoxypropane)
$3.$ Methyl isopropyl ether: $CH_3-O-CH(CH_3)_2$ ($2$-Methoxypropane)
These three ethers are metamers of each other because they have the same molecular formula but different alkyl groups attached to the same functional group (ether oxygen atom).
Therefore,there are $3$ metameric ethers.

(C) The isomeric ethers for the molecular formula $C_4H_{10}O$ are:
$1$. Diethyl ether: $CH_3CH_2OCH_2CH_3$
$2$. Methyl propyl ether: $CH_3OCH_2CH_2CH_3$
$3$. Methyl isopropyl ether: $CH_3OCH(CH_3)_2$
Hence,there are $3$ isomeric ethers possible.

(C) $(i)$ Phenol is more acidic than ethanol because the phenoxide ion is stabilized by resonance,while the ethoxide ion is destabilized by the $+I$ effect of the alkyl group. Statement is $T$.
$(ii)$ $o$-Nitrophenol exhibits intramolecular hydrogen bonding,whereas $p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to higher association and higher melting point for $p$-nitrophenol. Statement is $T$.
$(iii)$ Phenol is a weaker acid than carbonic acid $(H_2CO_3)$,so it does not react with $NaHCO_3$ to evolve $CO_2$. Statement is $F$.
$(iv)$ The aromatic ring of phenol is electron-rich due to the $+M$ effect of the $-OH$ group,making it susceptible to electrophilic substitution,not nucleophilic substitution. Statement is $F$.
Therefore,the correct sequence is $TTFF$.

(D) $CH_{2}=CH_{2} \xrightarrow{\text{Conc. } H_{2}SO_{4}} CH_{3}CH_{2}HSO_{4} \text{ (A)}$
$CH_{3}CH_{2}HSO_{4} \xrightarrow{\Delta, H_{2}O} CH_{3}CH_{2}OH \text{ (B)}$
$CH_{3}CH_{2}OH \xrightarrow{PBr_{3}} CH_{3}CH_{2}Br \text{ (C)}$
$CH_{3}CH_{2}Br \xrightarrow{\text{Alc. } KOH, \Delta} CH_{2}=CH_{2} \text{ (D)}$
Thus,$B$ is ethanol $(CH_{3}CH_{2}OH)$ and $D$ is alcoholic $KOH$.

(C) The boiling points of organic compounds depend on the strength of intermolecular forces.
$1$. Propan-$1$-ol $(CH_3CH_2CH_2OH)$ exhibits strong intermolecular hydrogen bonding,resulting in the highest boiling point.
$2$. Propanone $(CH_3COCH_3)$ and Propanal $(CH_3CH_2CHO)$ are polar compounds with dipole-dipole interactions. Propanone has a higher boiling point than Propanal due to a larger dipole moment.
$3$. Methoxy ethane $(CH_3OCH_2CH_3)$ is an ether with weak dipole-dipole interactions and no hydrogen bonding,resulting in the lowest boiling point.
The order is: Propan-$1$-ol $(B)$ > Propanone $(D)$ > Propanal $(C)$ > Methoxy ethane $(A)$.
Therefore,the correct order is $B > D > C > A$.

(B) $m$-cresol is $3$-methylphenol,which has a methyl group at the meta position relative to the hydroxyl group.
Catechol is benzene-$1,2$-diol,which has two hydroxyl groups at adjacent positions.
Resorcinol is benzene-$1,3$-diol,which has two hydroxyl groups at meta positions.
Looking at the structures in option $B$:
Structure $1$ is $m$-cresol (methyl group at meta position to $-OH$).
Structure $2$ is catechol (two $-OH$ groups at ortho positions).
Structure $3$ is resorcinol (two $-OH$ groups at meta positions).
Therefore,option $B$ correctly identifies the compounds.

(B) To determine the boiling point order,we analyze the intermolecular forces present in each molecule:
$1$. $CH_3CH_2CH_3$ $(III)$: This is an alkane,which only has weak London dispersion forces. It has the lowest boiling point.
$2$. $CH_3-O-CH_3$ $(I)$: This is an ether,which has weak dipole-dipole interactions. Its boiling point is higher than the alkane but lower than polar molecules with stronger interactions.
$3$. $CH_3CHO$ $(II)$: This is an aldehyde,which has stronger dipole-dipole interactions due to the polar $C=O$ bond.
$4$. $CH_3CH_2OH$ $(IV)$: This is an alcohol,which exhibits strong intermolecular hydrogen bonding. It has the highest boiling point among these.
Thus,the increasing order of boiling points is $III < I < II < IV$.

(A) Reaction $I$: The cleavage of $p$-methylanisole with $HI$ occurs. Since the bond between the oxygen and the aromatic ring is strong due to resonance,the $C-O$ bond between the methyl group and oxygen breaks,yielding $p$-cresol ($4$-methylphenol) as the major product $X$ and $CH_3I$.
Reaction $II$: This is a Wolff-Kishner reduction of $4$-methoxyacetophenone. The carbonyl group $(C=O)$ is reduced to a methylene group $(CH_2)$. Thus,the product $Y$ is $1$-methoxy-$4$-ethylbenzene.
Therefore,the products $X$ and $Y$ are $p$-cresol and $1$-methoxy-$4$-ethylbenzene respectively.

(C) Let us analyze each reaction:
$(i)$ $C_2H_5COOH \xrightarrow{LiAlH_4} C_2H_5CH_2OH$ (Propan$-1-$ol,an alcohol).
$(ii)$ $C_2H_5Br$ $\xrightarrow{Mg, \text{dry ether}} C_2H_5MgBr$ $\xrightarrow{H_2O} C_2H_6$ (Ethane,an alkane).
$(iii)$ $\text{Benzene diazonium chloride} \xrightarrow{H_3PO_2, H_2O} \text{Benzene}$ (An aromatic hydrocarbon).
$(iv)$ $4-\text{Chlorotoluene} \xrightarrow[300 \text{ atm}]{\text{NaOH, } 623 \text{ K}} 4-\text{Methylphenol}$ ($A$ phenol).
Thus,reactions $(i)$ and $(iv)$ yield alcohol and phenol respectively.

(C) Let us analyze each reaction:
$(i)$ Hydroboration-oxidation of an alkene gives an alcohol. This reaction yields cyclohexylmethanol,which is an alcohol.
(ii) Catalytic hydrogenation of an ester $(Methyl \ benzoate)$ yields a primary alcohol $(Benzyl \ alcohol)$ and methanol.
(iii) Reduction of a ketone $(Acetophenone)$ with $NaBH_4$ yields a secondary alcohol $(1-phenylethanol)$.
(iv) Hydrolysis of a gem-dichloride $(2,2-dichloropropane)$ with aqueous $NaOH$ yields a ketone $(Acetone)$,not an alcohol.
Therefore,reactions $(i), (ii),$ and $(iii)$ produce alcohols. The correct option is $C$.

(A) In option $A$,both $CH_3OH$ and $CH_3CH_2OH$ do not give the iodoform test. The iodoform test is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group. Since neither compound contains these groups,they cannot be distinguished by $(I_2 + NaOH)$.
In option $B$,Lucas reagent $(\text{Anhydrous } ZnCl_2 + \text{Conc. } HCl)$ distinguishes primary,secondary,and tertiary alcohols. $CH_3CH_2OH$ is a primary alcohol (reacts slowly upon heating),while $CH_3-C(CH_3)_2-OH$ is a tertiary alcohol (reacts immediately). This is a correct match.
In option $C$,terminal alkynes like $CH_3-C \equiv CH$ react with $Na$ to release $H_2$ gas due to the acidic hydrogen,whereas internal alkynes like $CH_3-C \equiv C-CH_3$ do not. This is a correct match.
In option $D$,$2,4-DNP$ is used to identify carbonyl compounds. However,both aldehydes and ketones react with $2,4-DNP$ to form a yellow/orange precipitate,so it cannot distinguish between them. Wait,the question asks for the *incorrect* match. Actually,$2,4-DNP$ is a general test for carbonyls,not a distinguishing test. However,in the context of distinguishing,$A$ is definitely incorrect as neither reacts,and $D$ is also incorrect as both react. Re-evaluating: $A$ is the most fundamentally incorrect as the reagent is completely ineffective for both.

(D) $1$. The Lucas test (conc. $HCl / ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols. Tertiary alcohols react instantly to produce turbidity.
$2$. Alcohol $X$ produces turbidity instantly,meaning it must be a tertiary alcohol. Among the isomers of $C_5 H_{12} O$,$2$-methylbutan-$2$-ol is a tertiary alcohol.
$3$. Alcohol $Y$ undergoes dehydration with conc. $H_2 SO_4$ at $443 \ K$. This is a characteristic reaction for alcohols to form alkenes.
$4$. Comparing the options,$2$-methylbutan-$2$-ol is the tertiary alcohol $(X)$ and $3$-methylbutan-$1$-ol is a primary isomer $(Y)$.
$5$. Therefore,the correct pair is $X = 2$-methylbutan-$2$-ol and $Y = 3$-methylbutan-$1$-ol,which corresponds to option $D$.

(C) . Methanol $(CH_3OH)$ is historically known as wood spirit because it was obtained by the destructive distillation of wood.
$B$. The mixture of anhydrous $ZnCl_2$ and concentrated $HCl$ is known as the Lucas reagent,used to distinguish between primary,secondary,and tertiary alcohols.
$C$. Rectified spirit is a mixture of $95 \% C_2H_5OH$ and $5 \% H_2O$ obtained by fractional distillation.
$D$. Dilute alkaline $KMnO_4$ solution is known as Baeyer's reagent,used to test for unsaturation in organic compounds.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ proceeds as follows:
$C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$
Here,$X$ is phenol $(C_6H_5OH)$ and $Y$ is methyl iodide $(CH_3I)$.
Next,the reaction of methyl iodide $(Y)$ with benzene $(C_6H_6)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation:
$C_6H_6 + CH_3I \xrightarrow{\text{Anhy. } AlCl_3} C_6H_5CH_3 + HI$
Thus,$Z$ is toluene $(C_6H_5CH_3)$.