JEE Alcohols, Phenols & Ethers Questions

An “unsaturated” carbon is any carbon that is part of a multiple bond (C=C or C≡C).
Therefore, to compare P, Q, R and S we must first identify the ring produced in each sequence and then count the sp2/sp hybridised carbons present in that ring.

Case 1 : Formation of P
The dihydric alcohol supplied in sequence (1) undergoes intramolecular dehydration with conc. $$H_2SO_4$$ to give a saturated cyclic ether (tetrahydrofuran type).
All ring carbons remain $$sp^3$$-hybridised, so

Number of unsaturated carbons in P = 0

Case 2 : Formation of Q
In sequence (2) the diol first forms a leaving-group derivative, followed by a single E2 elimination, producing a cyclohexene ring with one C=C bond.
Only the two carbons of that double bond are $$sp^2$$; hence

Number of unsaturated carbons in Q = 2

Case 3 : Formation of R
Sequence (3) involves two eliminations that furnish a conjugated diene ring, cyclohexa-1,3-diene. Two C=C bonds ⟹ four $$sp^2$$ carbons, so

Number of unsaturated carbons in R = 4

Case 4 : Formation of S
The reagents in sequence (4) carry out stepwise dehydrogenation until an aromatic ring is obtained. The monocyclic product is benzene (or a benzene derivative), which contains six $$sp^2$$ carbons.
Therefore,

Number of unsaturated carbons in S = 6

Comparing: $$P:0 \lt Q:2 \lt R:4 \lt S:6$$

Hence, the product possessing the largest number of unsaturated carbon atoms is S.

Option D which is: S

The given sequence of reagents fits very well with the following set of transformations:

$$\text{Anisole (methoxy-benzene)} \xrightarrow[\text{reflux}]{\text{excess HI}} \underset{X}{\text{Phenol}} \xrightarrow[\text{heat}]{CHCl_3/KOH} \underset{Y}{\text{Salicylaldehyde}} \xrightarrow[\;]{\text{alk. } KMnO_4\; \text{or } K_2Cr_2O_7/H^+} \underset{Z}{\text{Salicylic acid}}$$

Step-1 (Formation of X)
HI cleaves the C-O bond in anisole by an $$S_N1/S_N2$$ mechanism. The aromatic ring retains the -OH group while the alkyl part leaves as $$CH_3I$$. Hence $$X = C_6H_5OH$$ (phenol), an oxygen-containing compound.

Step-2 (Formation of Y)
Phenol on heating with $$CHCl_3$$ in strongly basic medium (Reimer-Tiemann reaction) gives an ortho-formyl derivative, salicylaldehyde ($$2\!-\!HO\!-\!C_6H_4CHO$$). Therefore $$Y$$ is also an oxygen-containing compound (it has both -OH and -CHO groups).

Step-3 (Formation of Z)
The -CHO group of salicylaldehyde is readily oxidised by alkaline $$KMnO_4$$ or $$K_2Cr_2O_7/H^+$$ to the -COOH group, giving salicylic acid ($$2\!-\!HO\!-\!C_6H_4COOH$$). $$Z$$ is a carboxylic acid, not an amine.

Checking the given statements

• Option A: “Both X and Y are oxygen containing compounds.”   Phenol (X) and salicylaldehyde (Y) each contain oxygen, so this statement is true.

• Option B: “Y on heating with $$CHCl_3/KOH$$ forms isocyanide.”   The carbylamine test (formation of isocyanide) requires a primary amine.   Y is an aldehyde, not an amine, so no isocyanide is produced.   Statement B is false.

• Option C: “Z reacts with Hinsberg’s reagent.”   Hinsberg’s test is specific for primary and secondary amines.   Z is a carboxylic acid, hence it does not react with benzenesulfonyl chloride.   Statement C is false.

• Option D: “Z is an aromatic primary amine.”   Z is salicylic acid, not an amine at all.   Statement D is false.

Thus, the only correct statement is:

Option A which is: Both X and Y are oxygen containing compounds.

We begin by writing the four named reactions listed in List-I along with the specific substance that must be taken as the starting material (reactant) in each case.

• P : Kolbe-Schmitt (Kolbe) reaction - it is carried out with $$C_6H_5ONa$$ (sodium phenoxide).
• Q : Reimer-Tiemann reaction - it starts from free phenol $$C_6H_5OH$$ in strongly alkaline medium.
• R : Williamson ether synthesis - it is performed with an alkoxide ion such as $$RO^-\,Na^+$$ (sodium alkoxide).
• S : Etard reaction - the oxidation is applied to toluene $$C_6H_5CH_3$$ with $$CrO_2Cl_2$$ to give benzaldehyde.

Next, let us look at the transformations given in List-II and note the major product formed in each of them.

Case 1 (List-II entry 1) : Decarboxylation of benzoic acid with soda-lime $$C_6H_5COONa \xrightarrow[\text{soda-lime}]{\Delta } C_6H_5CH_3$$ Major product = $$C_6H_5CH_3$$ (toluene).

Case 2 (List-II entry 2) : Dow process / high-temperature hydrolysis of chlorobenzene $$C_6H_5Cl + 2\,NaOH \xrightarrow[623\,\text{K}]{320\,\text{atm}} C_6H_5ONa + NaCl + H_2O$$ Major product before acidification = $$C_6H_5ONa$$ (sodium phenoxide).

Case 3 (List-II entry 3) : Cumene (isopropylbenzene) oxidation followed by acid hydrolysis $$\text{Cumene} \xrightarrow[\text{air}]{\text{O}_2} \text{cumene hydro-peroxide} \xrightarrow[\text{dil.}~H_2SO_4]{} C_6H_5OH + (CH_3)_2CO$$ Major product = $$C_6H_5OH$$ (phenol).

Case 4 (List-II entry 4) : Reaction of ethanol with sodium metal $$2\,C_2H_5OH + 2\,Na \rightarrow 2\,C_2H_5ONa + H_2 \uparrow$$ Major product = $$C_2H_5ONa$$ (sodium ethoxide, a typical alkoxide).

Now we simply link each named reaction from List-I with the entry in List-II that furnishes the required reactant:

• P (Kolbe, needs $$C_6H_5ONa$$)  ← produced in Case 2 ⇒ P→2.
• Q (Reimer-Tiemann, needs $$C_6H_5OH$$)  ← produced in Case 3 ⇒ Q→3.
• R (Williamson ether synthesis, needs an alkoxide such as $$C_2H_5ONa$$)  ← produced in Case 4 ⇒ R→4.
• S (Etard reaction, needs toluene $$C_6H_5CH_3$$)  ← produced in Case 1 ⇒ S→1.

The correct matching is therefore P→2; Q→3; R→4; S→1.

Comparing this sequence with the options supplied, we find that it corresponds to

Option B which is: P→2; Q→3; R→4; S→1.

To determine the correct IUPAC name of the given cyclic compound, we apply the standard IUPAC nomenclature rules by identifying the principal functional group and assigning the appropriate numbering.

The molecule contains both a hydroxyl group $$(-OH)$$ and a carbon-carbon double bond $$\left(C=C\right)$$ in a five-membered ring. According to IUPAC priority rules, the hydroxyl group has higher priority than the double bond. Therefore, the suffix of the compound is “-ol”, and the carbon bearing the $$-OH$$ group is assigned position 1.

After fixing the hydroxyl-bearing carbon as carbon 1, the ring is numbered in the direction that gives the double bond the lowest possible locant. With this numbering, the double bond begins at carbon 2, and the ethyl substituent is located at carbon 4.

Thus, the numbering is:

• Hydroxyl group at C1
• Double bond between C2 and C3
• Ethyl substituent at C4

The name is constructed as follows:

• Substituent: 4-ethyl
• Parent ring: cyclopent
• Double bond: 2-en
• Principal suffix: 1-ol

Hence, the correct IUPAC name of the compound is

4-ethylcyclopent-2-en-1-ol.

We compare the boiling points of four organic compounds with similar molecular weights.

A) $$CH_3CH_2CH_2CH_3$$ (Butane): Non-polar alkane. Only weak London dispersion forces. Boiling point: $$-0.5°C$$.

B) $$CH_3CH_2CH_2CH_2OH$$ (1-Butanol): Contains an $$-OH$$ group capable of strong hydrogen bonding. Boiling point: $$117.7°C$$.

C) $$CH_3CH_2CH_2CHO$$ (Butanal): Aldehyde with a polar $$C=O$$ group, but no H-bonding with itself (no O-H or N-H). Only dipole-dipole interactions. Boiling point: $$75°C$$.

D) $$C_2H_5-O-C_2H_5$$ (Diethyl ether): Polar but cannot form hydrogen bonds with itself. Boiling point: $$34.6°C$$.

The order of boiling points is: Butanol > Butanal > Diethyl ether > Butane.

1-Butanol has the highest boiling point due to strong intermolecular hydrogen bonding. The answer is Option B) $$CH_3CH_2CH_2CH_2OH$$.

• (A) Bu-OH (Butanol): An aliphatic alcohol. Lacks resonance stabilization for its conjugate base, making it the least acidic in the group.
• (C) 4-Methoxyphenol: The $$-\text{OCH}_3$$ group at the para-position exerts a strong electron-donating mesomeric ($$+M$$) effect, which destabilizes the phenoxide ion. It is less acidic than unsubstituted phenol.
• (D) Phenol: Standard reference with resonance-stabilized phenoxide ion.
• (B) 4-Nitrophenol: The $$-\text{NO}_2$$ group at the para-position acts as a strong electron-withdrawing group ($$-M, -I$$), strongly stabilizing the negative charge.
• (E) 2,4-Dinitrophenol: Contains two strong electron-withdrawing nitro groups ($$-\text{NO}_2$$) stabilizing the phenoxide ion via resonance and induction, making it the most acidic.

Final Ascending Order:

$$\text{(A)} < \text{(C)} < \text{(D)} < \text{(B)} < \text{(E)}$$

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Step 1: Identification of Compound P

• Molecular Formula: $$\text{C}_6\text{H}_6\text{O}_3$$
• Ferric Chloride ($$\text{FeCl}_3$$) Test: Indicates the presence of phenolic or enolic hydroxyl ($$-\text{OH}$$) groups.
• No Intramolecular Hydrogen Bonding: This rules out adjacent arrangements like pyrogallol ($$1,2,3$$-trihydroxybenzene).

Consequently, compound $$P$$ is phloroglucinol ($$1,3,5$$-trihydroxybenzene). Phloroglucinol readily exhibits keto-enol tautomerism to exist in its keto form, cyclohexane-$$1,3,5$$-trione. Because it has $$3$$ carbonyl groups in this form, it reacts perfectly with $$3$$ equivalents of hydroxylamine ($$\text{NH}_2\text{OH}$$) to yield the trioxime Q.

Step 2: Formation of Compound R

When phloroglucinol ($$P$$) is treated with excess methyl iodide ($$\text{CH}_3\text{I}$$) in the presence of a strong base like $$\text{KOH}$$, it undergoes exhaustive C-alkylation at its active methylene positions rather than O-alkylation.

Each of the three $$-\text{CH}_2-$$ positions in the cyclohexane-$$1,3,5$$-trione tautomer is methylated twice, leading to $$2,2,4,4,6,6$$-hexamethylcyclohexane-$$1,3,5$$-trione as compound $$R$$.

• Methyl groups in $$R$$: $$6$$ methyl groups attached directly to the ring carbons.

Step 3: Formation of Compound S

Compound $$R$$ contains $$3$$ ketone carbonyl groups. When treated with excess iso-butylmagnesium bromide ($$\text{i-BuMgBr}$$, a Grignard reagent) followed by acid hydrolysis ($$\text{H}_3\text{O}^+$$), each carbonyl carbon undergoes a nucleophilic addition reaction to form a tertiary alcohol.

• Each ketone group adds one iso-butyl group ($$-\text{CH}_2-\text{CH}(\text{CH}_3)_2$$).
• Since there are $$3$$ ketone groups, a total of $$3$$ iso-butyl groups are added.

Step 4: Total Methyl Group Count in S

Let's add up all the methyl ($$-\text{CH}_3$$) groups present in the final structure $$S$$:

• From the core structure of $$R$$: $$6 \text{ methyl groups}$$
• From the $$3$$ attached iso-butyl groups: $$3 \times 2 = 6 \text{ methyl groups}$$

$$\text{Total number of methyl groups} = 6 + 6 = 12$$

We need to evaluate the Assertion (A) and Reason (R) about the acidity of phenol and ethanol.

Assertion A: "$$pK_a$$ value of phenol is 10.0 while that of ethanol is 15.9."

This is a factual statement and is TRUE. A lower $$pK_a$$ value indicates a stronger acid. Phenol ($$pK_a = 10.0$$) is a stronger acid than ethanol ($$pK_a = 15.9$$) because the phenoxide ion ($$C_6H_5O^-$$) is stabilized by resonance with the aromatic ring, while the ethoxide ion ($$C_2H_5O^-$$) has no such stabilization.

Reason R: "Ethanol is stronger acid than phenol."

This is FALSE. As explained above, phenol is a much stronger acid than ethanol (lower $$pK_a$$). The statement has the comparison reversed.

The correct answer is Option (1): A is true but R is false.

Assertion A: Alcohols react both as nucleophiles and electrophiles. This is true. The oxygen atom with its lone pairs allows alcohols to act as nucleophiles (e.g., in Williamson ether synthesis). After protonation of the -OH group, the carbon becomes electrophilic as water becomes a good leaving group.

Reason R: Alcohols react with active metals such as sodium, potassium, and aluminum to yield corresponding alkoxides and liberate hydrogen. This is true: $$2ROH + 2Na \rightarrow 2RONa + H_2\uparrow$$.

However, the reaction with active metals demonstrates the acidic (proton-donating) nature of alcohols, not their dual nucleophilic/electrophilic behavior. Therefore, R is not the correct explanation of A.

The answer is Option D: Both A and R are true but R is NOT the correct explanation of A.

Reaction (A): Conversion of Phenol to Salicylic acid (2-hydroxybenzoic acid). This transformation is achieved via the Kolbe-Schmitt reaction. When phenol is treated with sodium hydroxide ($$NaOH$$) to form a phenoxide ion, followed by reaction with carbon dioxide ($$CO_2$$) under pressure and subsequent acidification with $$HCl$$, salicylic acid is produced. This matches with reagent (IV). Thus, (A) matches with (IV).

Reaction (B): Conversion of Phenol to Salicylaldehyde (2-hydroxybenzaldehyde). This is the Reimer-Tiemann reaction. Phenol reacts with chloroform ($$CHCl_3$$) in the presence of aqueous sodium hydroxide ($$NaOH$$), introducing a formyl group ($$\text{-}CHO$$) at the ortho-position, followed by hydrolysis and acidification ($$HCl$$). This matches with reagent (III). Thus, (B) matches with (III).

Reaction (C): Conversion of Phenol to p-Benzoquinone. Phenol undergoes oxidation when treated with a strong oxidizing agent like acidified sodium dichromate ($$Na_2Cr_2O_7 / H_2SO_4$$) to yield para-benzoquinone. This matches with reagent (I). Thus, (C) matches with (I).

Reaction (D): Conversion of Phenol to Anisole (Methoxybenzene). This is an etherification process (Williamson ether synthesis). Phenol is first converted into sodium phenoxide using $$NaOH$$, which then performs a nucleophilic substitution reaction with methyl chloride ($$CH_3Cl$$) to introduce a methoxy group ($$\text{-}OCH_3$$). This matches with reagent (II). Thus, (D) matches with (II).

Combining all the pairs, the correct matching sequence is: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

Answer: Option D — (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Statement I: p-Nitrophenol is more acidic than m-nitrophenol and o-nitrophenol.

The $$-NO_2$$ group is electron-withdrawing. At the para position, it stabilizes the phenoxide ion through resonance (direct conjugation), making p-nitrophenol the most acidic among the three isomers. This is true.

(Note: While o-nitrophenol has intramolecular hydrogen bonding which reduces its effective acidity in solution, p-nitrophenol has the strongest -R effect stabilization.)

Statement II: Ethanol will give immediate turbidity with Lucas reagent.

Lucas reagent (conc. HCl + ZnCl$$_2$$) is used to distinguish between primary, secondary, and tertiary alcohols. Primary alcohols like ethanol do not give immediate turbidity - they show no reaction at room temperature. Only tertiary alcohols give immediate turbidity. This is false.

Statement I is true but Statement II is false. The answer corresponds to Option (1).

The reaction mentioned in A) is reaction of Aniline with Nitrous acid at low temperature first forming Diazonium salt which then gets substituted by OH- ion from water. Resulting in Phenol-(II)

The Reaction mentioned in B) is a standard oxidation reaction specifically mentioned in NCERT. Phenol on reaction with sodium di-chromate in presence of sulphuric acid produces Benzoquinone-(IV)

The Reaction mentioned in C) is a named reaction i.e. Reimer-Tiemann reaction. The product is Salicylaldehyde-(I)

The Reaction mentioned in D) is a named reaction i.e. Kolbe's reaction. The product formed is Salicylic acid-(III)

First write down the complete data of the question.

List-I (reaction sequences):
P. $$C_6H_5OCH_3 \;\xrightarrow[\text{excess}]{\,HI\,}$$ (anisole is heated with excess hydroiodic acid)
Q. $$C_6H_5OH \;\xrightarrow[]{\,conc.\;HNO_3/\,conc.\;H_2SO_4\,}$$ (phenol treated with conc. nitrating mixture)
R. $$C_6H_5OH \;\xrightarrow[]{\,CH_3COCl/pyridine\,}$$ (phenol acetylated with acetyl chloride in pyridine)
S. $$C_6H_5OH \;\xrightarrow[]{\,NaOH,\;140^{\circ}C,\;7\,atm\,}\;NaO\!C_6H_4O^-$$ $$\xrightarrow[]{\,CO_2\,}\xrightarrow[]{\,H^+\,}$$ (Kolbe-Schmitt reaction followed by acidification)

List-II (probable products):
1. $$o\text{-}HO\!C_6H_4COOH$$  (salicylic acid)
2. $$p\text{-}HO\!C_6H_4NO_2$$  (p-nitrophenol)
3. $$C_6H_5OH + CH_3I$$
4. $$C_6H_5OCOCH_3$$  (phenyl acetate)
5. $$2,4,6\text{-}trinitrophenol$$  (picric acid)

Now analyse every reaction one by one.

Anisole $$\left(C_6H_5OCH_3\right)$$ when heated with excess $$HI$$ undergoes cleavage of the C-O bond attached to the alkyl group, forming phenol and methyl iodide: $$C_6H_5OCH_3 + HI \;\longrightarrow\; C_6H_5OH + CH_3I$$ This matches product 3.

Phenol reacts with concentrated $$HNO_3/H_2SO_4$$ at low temperature to give picric acid (2,4,6-trinitrophenol) because -OH is a strong activator and three nitro groups finally enter the ring: $$C_6H_5OH \xrightarrow[]{conc.\;HNO_3/H_2SO_4} 2,4,6\text{-}trinitrophenol$$ Hence Q corresponds to product 5.

Acetylation of phenol with acetyl chloride in pyridine gives phenyl acetate: $$C_6H_5OH + CH_3COCl \;\xrightarrow[]{pyridine}\; C_6H_5OCOCH_3 + HCl$$ Therefore R matches product 4.

The Kolbe-Schmitt reaction converts phenoxide ion to o-hydroxybenzoate ion which after acidification yields salicylic acid: $$C_6H_5OH \xrightarrow[]{NaOH} C_6H_5O^-Na^+ \xrightarrow[7\,atm]{CO_2,\;140^{\circ}C} o\text{-}HO\!C_6H_4COONa \xrightarrow[]{H^+} o\text{-}HO\!C_6H_4COOH$$ So S corresponds to product 1.

Collecting all matches:
P → 3, Q → 5, R → 4, S → 1

The option that contains this combination is Option A.

Final answer: Option A which is: P-3, Q-5, R-4, S-1.

In reaction A, phenol on heating with $$Zn$$ dust undergoes reduction. Zinc removes the oxygen atom as $$ZnO$$, converting phenol into benzene. Therefore, A matches with Benzene, i.e., $$A \rightarrow (III)$$.

In reaction B, phenol reacts with $$CHCl_3/NaOH$$ followed by acidification. This is the Reimer-Tiemann reaction, which introduces a formyl group $$(-CHO)$$ at the ortho position of phenol to form salicylaldehyde. Therefore, $$B \rightarrow (I)$$.

In reaction C, phenol reacts with $$CO_2/NaOH$$ followed by acidification. This is the Kolbe-Schmitt reaction, in which a carboxyl group $$(-COOH)$$ is introduced at the ortho position to form salicylic acid. Therefore, $$C \rightarrow (II)$$.

In reaction D, phenol reacts with concentrated $$HNO_3$$. Due to the strongly activating nature of the $$-OH$$ group, nitration occurs at all available ortho and para positions, producing 2,4,6-trinitrophenol (picric acid). Therefore, $$D \rightarrow (IV)$$.

Thus, the correct matching is:

$$A \rightarrow (III)$$

$$B \rightarrow (I)$$

$$C \rightarrow (II)$$

$$D \rightarrow (IV)$$

Hence, the correct answer is:

$$A-(III),\ B-(I),\ C-(II),\ D-(IV)$$

The reagents mentioned in List-I are the classical name reactions of phenolic chemistry. Before matching, recall what each reaction sequence gives:

P  Dow’s process $$C_6H_5Cl \xrightarrow[\;300\;{\rm atm}\;]{NaOH,\;623\;{\rm K}}\;C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$$ → product is simple phenol.

Q  Kolbe-Schmitt reaction $$C_6H_5ONa \xrightarrow{CO_2,\;400\;{\rm K},\;4\!-\!7\;{\rm atm}} \;o\!-\!HO\!-\!C_6H_4COONa \xrightarrow{H^+} o\!-\!HO\!-\!C_6H_4COOH$$ → product is o-hydroxybenzoic acid (salicylic acid).

R  Reimer-Tiemann reaction $$C_6H_5OH \xrightarrow{CHCl_3/NaOH,\;343\;{\rm K}} o\!-\!HO\!-\!C_6H_4CHO$$ → product is o-hydroxybenzaldehyde (salicylaldehyde).

S  Nitration of phenol with conc. $$HNO_3/H_2SO_4$$ (333 K) $$C_6H_5OH \xrightarrow{conc\;HNO_3/conc\;H_2SO_4} 2,4,6\text{-trinitrophenol}$$ → product is picric acid.

Now read List-II (numbered phenolic compounds):

(1) 2,4,6-trinitrophenol (picric acid)
(2) …
(3) phenol
(4) o-hydroxybenzaldehyde (salicylaldehyde)
(5) o-hydroxybenzoic acid (salicylic acid)

Matching each entry:

P (Dow) → phenol → 3
Q (Kolbe-Schmitt) → o-hydroxybenzoic acid → 5
R (Reimer-Tiemann) → o-hydroxybenzaldehyde → 4
S (nitration) → picric acid → 1

Thus P-3, Q-5, R-4, S-1.

The option containing this combination is
Option C which is: P-3, Q-5, R-4, S-1.

The starting compound contains both a phenolic $$-OH$$ group and an aldehyde $$-CHO$$ group.

$$PhMgBr$$ is not only a strong nucleophile but also a very strong base. In the presence of an acidic proton, acid-base reaction occurs preferentially over nucleophilic addition.

The phenolic proton is acidic and reacts immediately with $$PhMgBr$$. As a result, the Grignard reagent is consumed in deprotonating the phenol to form a phenoxide salt and benzene.

Since only one equivalent of $$PhMgBr$$ is present, the entire reagent is consumed in this acid-base reaction. No Grignard reagent remains available to attack the aldehyde group.

During the subsequent treatment with aqueous $$NH_4Cl$$, the phenoxide ion is protonated back to the original phenol.

Therefore, the aldehyde group remains unchanged throughout the reaction sequence, and the final product is identical to the starting material.

The product contains one phenolic hydroxyl group.

Hence, the number of hydroxyl groups present in product $$P$$ is

$$1$$

Assertion A: Butan-1-ol has higher boiling point than ethoxyethane (diethyl ether).

Reason R: Extensive hydrogen bonding leads to stronger association of molecules.

Analysis of Assertion A:

Butan-1-ol (CH₃CH₂CH₂CH₂OH) has a boiling point of approximately 117.7°C.

Ethoxyethane (CH₃CH₂OCH₂CH₃, diethyl ether) has a boiling point of approximately 34.6°C.

Both have the same molecular formula C₄H₁₀O, but butan-1-ol has a significantly higher boiling point.

Assertion A is True.

Analysis of Reason R:

Butan-1-ol has an -OH group capable of forming extensive intermolecular hydrogen bonds, leading to stronger molecular association and higher boiling point. Ethoxyethane cannot form H-bonds with itself (no O-H or N-H groups).

Reason R is True and correctly explains Assertion A.

The correct answer is Option B: Both A and R are true and R is the correct explanation of A.

The sequence of reactions (reagents were supplied in the original problem) converts the same four-carbon skeleton into four different compounds, labelled P, Q, R and S. For every product we will first write the structural formula and then check whether any carbon atom is attached to four different groups (an asymmetric or chiral carbon).

Product P
The first reaction is acid-catalysed hydration of an alkene, and it gives $$\mathbf{2\text{-}butanol}$$ $$CH_3-CH(OH)-CH_2-CH_3$$

The carbon marked with an asterisk bears $$OH$$, $$H$$, $$CH_3$$ and $$CH_2CH_3$$, four different groups:
$$CH_3-\mathbf{C^*}(OH)(H)-CH_2-CH_3$$ Hence P contains one asymmetric carbon and is optically active.

Product Q
In the second step P is treated with Lucas reagent $$HCl/ZnCl_2$$, replacing $$OH$$ by $$Cl$$ to give $$\mathbf{2\text{-}chlorobutane}$$ $$CH_3-CH(Cl)-CH_2-CH_3$$

Again the starred carbon is attached to $$Cl$$, $$H$$, $$CH_3$$ and $$CH_2CH_3$$ (all different):
$$CH_3-\mathbf{C^*}(Cl)(H)-CH_2-CH_3$$ Therefore Q has one asymmetric carbon and is optically active.

Product R
On heating P with concentrated $$H_2SO_4$$, dehydration occurs and the major alkene formed is $$\mathbf{trans\text{-}2\text{-}butene}$$ $$CH_3-CH=CH-CH_3$$

Every carbon in an alkene is $$sp^2$$-hybridised and, at most, bonded to three different groups. No carbon here bears four different substituents, so R possesses no asymmetric carbon; it is achiral.

Product S
Oxidation of P with chromic reagent (or PCC) gives $$\mathbf{butan\text{-}2\text{-}one}$$ $$CH_3-CO-CH_2-CH_3$$

The carbonyl carbon is $$sp^2$$-hybridised and attached to two identical oxygen-linked entities; again, no carbon in the molecule has four different groups. Thus S is also achiral.

Summary of chirality:
• P - chiral (one asymmetric carbon)
• Q - chiral (one asymmetric carbon)
• R - achiral
• S - achiral

Therefore the correct statement is:
Option A which is: Both P and Q have asymmetric carbon(s).

The four compounds are

a : $$CH_3-CH(OH)-CH_3$$  (isopropyl alcohol, 2° alcohol)
b : $$(CH_3)_3C-OH$$  (tert-butyl alcohol, 3° alcohol)
c : $$C_6H_5OH$$  (phenol)
d : $$CH_3CH_2OH$$  (ethanol, 1° alcohol)

Acid strength is judged by the stability of the conjugate base. A smaller $$pK_a$$ means a stronger acid; a larger $$pK_a$$ means a weaker acid.

Step 1 — Alcohols (a, b, d)
Removal of $$H^+$$ from an alcohol gives an alkoxide ion $$(RO^-)$$.  Alkyl groups are electron-releasing (+I) and therefore destabilise the negatively charged oxygen.  The more alkyl groups attached to the carbon bearing the $$OH$$, the greater the +I effect and the weaker the acid.

Hence for simple aliphatic alcohols:
$$3^\circ \text{ alcohol (b)} \; \lt \; 2^\circ \text{ alcohol (a)} \; \lt \; 1^\circ \text{ alcohol (d)}$$ in acidity, or, in terms of $$pK_a$$ (remember, weaker acid ⇒ larger $$pK_a$$):
$$pK_a(b) \; \gt \; pK_a(a) \; \gt \; pK_a(d)$$

Step 2 — Phenol (c)
For phenol, deprotonation gives the phenoxide ion which is resonance-stabilised:

$$C_6H_5OH \;\rightarrow\; C_6H_5O^- + H^+$$

The negative charge can delocalise over the aromatic ring, greatly stabilising the anion. This resonance stabilisation outweighs the +I effect of the ring, so phenol is far more acidic than normal alcohols. Consequently its $$pK_a$$ is much smaller.

Step 3 — Combining the two trends
Putting phenol together with the alcohols, the overall acidity order is

phenol (c) > ethanol (d) > isopropanol (a) > tert-butanol (b)

Converting to $$pK_a$$ values (largest value for the weakest acid):

$$pK_a(b) \; \gt \; pK_a(a) \; \gt \; pK_a(d) \; \gt \; pK_a(c)$$

This matches Option C:

Option C which is: b > a > d > c

We need to arrange the given hydroxyl compounds in order of decreasing acidity.

The acidity of a hydroxyl compound ($$R-OH$$) depends on the stability of the conjugate base ($$R-O^-$$) formed after loss of the proton. The more stable the conjugate base, the stronger the acid.

(A) $$CH_3OH$$ - Methanol (aliphatic alcohol)

(B) $$(CH_3)_3COH$$ - tert-Butanol (aliphatic alcohol)

(C) Phenol ($$C_6H_5OH$$)

(D) p-MeO-Phenol ($$4-CH_3O-C_6H_4-OH$$)

(E) p-$$NO_2$$-Phenol ($$4-NO_2-C_6H_4-OH$$)

Phenols are significantly more acidic than aliphatic alcohols because the phenoxide ion ($$C_6H_5O^-$$) is stabilized by resonance with the aromatic ring. The negative charge is delocalized over the ring. Aliphatic alkoxide ions have no such resonance stabilization. Therefore: All phenols > All alcohols in acidity.

Order among the phenols (E, C, D).

- p-$$NO_2$$-Phenol (E): The $$NO_2$$ group is a strong electron-withdrawing group (both by inductive and resonance effects). It stabilizes the phenoxide ion by pulling electron density away from the oxygen, delocalizing the negative charge further. This makes it the most acidic phenol.

- Phenol (C): The unsubstituted phenol, with moderate acidity.

- p-MeO-Phenol (D): The $$OCH_3$$ group is an electron-donating group by resonance (lone pairs on oxygen donate into the ring). This destabilizes the phenoxide ion by increasing electron density, making it less acidic than unsubstituted phenol. Therefore: E > C > D.

Order among the alcohols (A, B).

- $$CH_3OH$$ (A): Methanol has one methyl group, which is weakly electron-donating by hyperconjugation/induction.

- $$(CH_3)_3COH$$ (B): tert-Butanol has three methyl groups donating electron density to the carbon bearing the $$OH$$. This destabilizes the alkoxide ion more than in methanol. Therefore: A > B in acidity.

$$E > C > D > A > B$$

p-$$NO_2$$-Phenol > Phenol > p-MeO-Phenol > $$CH_3OH$$ > $$(CH_3)_3COH$$

The correct answer is Option 4: E > C > D > A > B.

Following the reaction sequence with propene, the first stage involves hydroboration with $$B_2H_6$$, leading to anti‐Markovnikov addition of $$BH_3$$.
Subsequent oxidation with $$H_2O_2, NaOH$$ converts the intermediate to the anti‐Markovnikov alcohol $$CH_3CH_2CH_2OH$$ (1-propanol).
The primary alcohol is then oxidized by pyridinium chlorochromate ($$PCC$$) to afford the aldehyde $$CH_3CH_2CHO$$ (propanal).
In the final step, reaction with the Grignard reagent $$CH_3MgBr$$ followed by hydrolysis yields the secondary alcohol $$CH_3CH_2CH(OH)CH_3$$ (2-butanol).

The overall product is $$CH_3CH_2CH(OH)CH_3$$ (butan-2-ol), which corresponds to option 3.

When one side of an ether is aryl and the other side is alkyl, hot concentrated $$HI$$ selectively cleaves the alkyl-oxygen bond (because the aryl-oxygen bond has partial double-bond character). Therefore:

$$\text{Phenyl-O-R} \xrightarrow{HI} \text{Phenol }(Q)\;+\;RI$$

Thus $$Q$$ is phenol, $$C_6H_5OH$$.

The Kolbe-Schmitt reaction is the standard industrial route for converting phenol into salicylic acid. The three-step sequence is:

$$C_6H_5OH\; \xrightarrow[\text{(ii) }CO_2]{\text{(i) NaOH, 140-160 °C, 4-7 atm}} \xrightarrow[\;]{\text{(iii) }H_3O^+} o\text{-hydroxybenzoic acid}$$

Hence $$R$$ is salicylic acid, commonly written as $$o\text{-}HO-C_6H_4-COOH$$.

Acetic anhydride acetylates the phenolic -OH of salicylic acid much faster than the -COOH group. The two-step sequence

$$o\text{-}HO-C_6H_4-COOH \xrightarrow[\text{(ii) }H_3O^+]{\text{(i) }(CH_3CO)_2O} o\text{-}AcO-C_6H_4-COOH$$

gives $$S$$ = acetylsalicylic acid, better known as aspirin.

Pharmacologically, aspirin irreversibly acetylates and inactivates cyclo-oxygenase (COX-1 and COX-2). These enzymes catalyse the conversion of arachidonic acid into prostaglandins. By blocking this step, aspirin inhibits the biosynthesis of prostaglandins, which are mediators of pain, fever and inflammation.

Therefore, the correct statement about $$S$$ is:

Option B which is: It inhibits the synthesis of prostaglandin.

The industrial preparation of phenol from cumene occurs in two steps.

Cumene (isopropylbenzene) undergoes aerial oxidation, in which oxygen attacks the benzylic carbon to form a hydroperoxide intermediate:

$$\mathrm{C_6H_5CH(CH_3)_2 \xrightarrow[\text{air}]{O_2} C_6H_5C(CH_3)_2OOH}$$

The intermediate formed is cumene hydroperoxide.

On treatment with dilute acid, cumene hydroperoxide undergoes cleavage to give phenol and acetone:

$$\mathrm{C_6H_5C(CH_3)_2OOH \xrightarrow{H^+} C_6H_5OH + (CH_3)_2CO}$$

Among the given structures:

• Option A represents phenyl acetate.

• Option B represents isopropyl phenyl ether.

• Option C represents cumene (starting material).

• Option D represents cumene hydroperoxide, containing the characteristic $$-OOH$$ group.

Hence, the intermediate formed during the oxidation of cumene is

$$\boxed{\text{Cumene hydroperoxide}}$$

Therefore, the correct answer is $$\boxed{\text{Option D}}$$.

Under acidic conditions, the hydroxyl group is first protonated to form a good leaving group ((-OH_2^+)). Loss of water then generates a secondary carbocation at the carbon bearing the hydroxyl group.

Since this carbocation is adjacent to a four-membered ring, ring expansion occurs to relieve ring strain. The cyclobutane ring expands to form a more stable six-membered cyclohexane ring, generating a rearranged carbocation.

The rearranged carbocation can further undergo hydride or alkyl shifts to produce the most stable tertiary carbocation. Elimination of a proton from an adjacent carbon then gives the alkene product.

According to Zaitsev’s rule, the major product is the most substituted alkene. Among the given options, Option B corresponds to a tetrasubstituted alkene

Since tetrasubstituted alkenes are thermodynamically more stable than trisubstituted alkenes, the rearrangement and elimination sequence favors the formation of the product shown in Option B.

Hence, the major product is

$${\text{Option B}}.$$

The reaction is an acid-catalyzed dehydration of an alcohol using concentrated (H_2SO_4).

Under acidic conditions, the hydroxyl group is first protonated to form a good leaving group ((-OH_2^+)). Loss of water then generates a secondary carbocation at the carbon originally bearing the hydroxyl group.

Since a more stable carbocation can be formed, the secondary carbocation undergoes a (1,2)-hydride shift from the adjacent tertiary carbon, producing a tertiary carbocation.

From this rearranged carbocation, elimination of a (\beta)-hydrogen takes place to form an alkene. According to Zaitsev’s rule, the major product is the most substituted and therefore the most thermodynamically stable alkene.

Among the given options, Option A corresponds to the tetrasubstituted alkene formed after carbocation rearrangement and subsequent elimination. This is more stable than the other possible alkene products and is therefore the major product.

Hence, the correct answer is

$$\boxed{\text{Option A}}.$$

The answer key indicating the last option is incorrect because it does not account for the carbocation rearrangement that occurs before elimination.

The empirical formula of compound $$A$$ is $$C_6H_6O$$. A molecular formula with six carbons and a single oxygen that burns with a sooty flame suggests an aromatic ring containing one oxygen-bearing group. The two most common possibilities are phenol $$(C_6H_5OH)$$ and benzaldehyde $$(C_6H_5CHO)$$.

Benzaldehyde is deactivating and meta-directing; it hardly undergoes bromination in the absence of a Lewis acid. By contrast, phenol has a strongly activating $$-OH$$ group that makes bromination easy even without a catalyst. Hence $$A$$ is phenol.

Next, bromination is carried out “in a low-polarity solvent” (e.g. $$CS_2$$, $$CCl_4$$). In such solvents the bromine remains molecular $$Br_2$$ and the medium contains no water. Therefore:

• Only one bromine atom is introduced (no poly-bromination as in aqueous medium).
• Because the $$-OH$$ group is ortho-/para-directing, the substitution can occur at the ortho or para positions.
• Steric hindrance near the $$-OH$$ group suppresses the ortho product, so the para product is formed in the highest yield.

Thus compound $$B$$ is para-bromophenol (4-bromophenol).

According to the given options, para-bromophenol corresponds to Option A.

Option A which is: para-bromophenol

1. Protonation of the Alcohol

The lone pair of electrons on the oxygen atom of the secondary alcohol $$(-\text{OH}$$) attacks a hydronium ion $$(\text{H}^{+}$$).

The hydroxyl group is converted into an excellent leaving group, a protonated water molecule $$(-\text{OH}_{2}^{+}$$).

2. Loss of Water (Carbocation Formation)

The $$\text{C}-\text{O}$$ bond breaks, and water $$(\text{H}_2\text{O}$$) leaves.

A secondary $$(2^{\circ }$$) carbocation is formed on the carbon chain adjacent to the bulky tert-butyl group.

3. 1,2-Methyl $$(\text{CH}_{3}^{-}$$) Shift

A secondary carbocation is relatively unstable. To gain stability, a methyl group from the adjacent tert-butyl group migrates with its electron pair (a 1,2-methyl shift) to the carbocation carbon.

This rearranges the less stable $$2^{\circ }$$ carbocation into a highly stable tertiary $$(3^{\circ })$$ carbocation at the terminal branch.

4. Intramolecular Cyclization (Ring Closure) 

The $$\pi$$-electrons of the cyclohexene double bond act as an intramolecular nucleophile, attacking the newly formed tertiary carbocation.

This nucleophilic attack closes the chain into a second 6-membered ring (forming a fused decalin system). A new tertiary carbocation is generated at the ring junction (bridgehead position).

5. Deprotonation $$(-\text{H}^{+})$$ to Form the Major Product

A weak base (such as water) removes a proton $$(\text{H}^{+})$$ from the adjacent carbon atom at the ring junction.

Elimination of the proton forms a highly stable, tetrasubstituted double bond directly at the shared bridgehead position between the two fused 6-membered rings.

The given substrate is a β-amino alcohol. Denote it as $$HO\,-\,CH_2\,-\,CH_2\,-\,NH_2$$ (ethanolamine).

Step 1 : protonation with $$HCl$$
The basic $$NH_2$$ group gets protonated much faster than the -OH group, giving the ammonium salt

$$HO\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+}\;Cl^{-} \qquad -(1)$$

In the ammonium form, nitrogen becomes a very good leaving group (it will depart as neutral $$NH_3$$).

Step 2 : treatment with strong base $$KOH$$
$$KOH$$ removes the proton from the alcoholic -OH to generate an alkoxide:

$$^{-}O\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+} \qquad -(2)$$

Step 3 : intramolecular $$S_N2$$ displacement
The alkoxide oxygen (a strong nucleophile) attacks the adjacent carbon bearing $$NH_3^{+}$$, expelling neutral $$NH_3$$ and closing a three-membered ring.

$$^{-}O\,-\,CH_2\,-\,CH_2\,-\,NH_3^{+} \;\xrightarrow[\textit{intramolecular }S_N2]{}\; \begin{matrix} & O & \\ / & & \backslash \\ CH_2 &\; & CH_2 \end{matrix} \;+\;NH_3 \qquad -(3)$$

The three-membered ring containing oxygen is called oxirane (ethylene oxide).

Thus the major product ‘B’ is oxirane. Among the given choices this corresponds to Option A.

Answer: Option A which is: ethylene oxide (oxirane).

Reaction for Product A

In the first starting material, we have two different types of $-OH$ groups:

1. Phenolic -OH (Top): Attached directly to the benzene ring.
2. Benzylic -OH (Bottom): Attached to a side-chain CH2 group.

The Mechanism:

• The Phenol is unreactive: The C-O bond in phenol has partial double-bond character due to resonance. Breaking this bond to replace it with Br is extremely difficult and does not happen under these conditions.
• The Benzylic Alcohol reacts: The side-chain -OH gets protonated by HBr to form a good leaving group H2O. It then undergoes a substitution reaction (likely SN1 because the benzylic carbocation is very stable) where the Br- replaces the -OH.
• Result A: The phenolic group remains, and the benzylic alcohol becomes a benzylic bromide -CH2Br.

Reaction for Product B

The reagent used in the given transformation first converts the -OH group into a good leaving group, generates a carbocation adjacent to oxygen and then triggers a 1,2-alkyl shift. Such 1,2-shifts are nothing but the well-known pinacol / semipinacol type rearrangements that enlarge the size of a cyclic skeleton by one carbon at every step.

Key facts about pinacol (or semipinacol) ring expansion that one must always remember:

1. The migration occurs only from the carbon directly bonded to the positively charged centre (a strict 1,2-shift).
2. Each individual shift enlarges the ring by exactly one member: $$n \rightarrow n+1$$.
3. The thermodynamic driving force is release of angle (Baeyer) strain. Three-, four- and even five-membered rings are quite strained, six is nearly strain-free, while seven is also comfortably strain-free.
4. Beyond seven members, two new problems appear: (i) there is hardly any extra angle strain left to be relieved, and (ii) trans-annular (across-the-ring) interactions begin to build up. Hence further expansion (to eight, nine …) is not favoured under ordinary laboratory conditions.

Because of points 3 and 4, a sequence of pinacol-type ring expansions will proceed stepwise until a seven-membered ring is produced and then automatically stop. The same logic is valid for every ring present in the substrate. Therefore, both rings marked ‘A’ and ‘B’ can keep enlarging, but none of them will ever grow beyond seven.

Now examine the statements:

A. Ring expansion can go up to seven-membered rings — TRUE, exactly as argued above.
B. Both rings will finally become six-membered — FALSE; the process normally continues one step further to the seven-membered stage.
C. Both rings will finally become five-membered — FALSE for the same reason.
D. Only A will become six-membered — FALSE; nothing restricts ring B from expanding, and “six” is not the limit anyway.

Hence the only correct statement is:

Option A which is: Ring expansion can go upto seven membered rings

We need to identify which method does NOT correctly prepare alcohols.

Option 1:

This is a correct method. Grignard reagents ($$RMgBr$$) are powerful nucleophiles that attack the carbonyl group of ketones. The mechanism involves nucleophilic addition of $$R^-$$ to the electrophilic carbonyl carbon, forming an alkoxide intermediate. Subsequent hydrolysis (treatment with dilute acid) gives a tertiary alcohol:

$$R'_2C=O + RMgBr \rightarrow R'_2(R)C-OMgBr \xrightarrow{H_3O^+} R'_2(R)C-OH$$

This is one of the most important methods for synthesizing alcohols.

Option 2:

This is a correct method. The hydroxide ion ($$OH^-$$) acts as a nucleophile and attacks the carbon bearing the halogen in an $$S_N2$$ (or $$S_N1$$) reaction, displacing the halide ion:

$$R-X + NaOH(aq) \rightarrow R-OH + NaX$$

This nucleophilic substitution reaction produces an alcohol.

Option 3:

This is a correct method. In this two-step process, the alkene first reacts with borane ($$BH_3$$) to form a trialkylborane (hydroboration), which is then oxidized with alkaline hydrogen peroxide ($$H_2O_2/NaOH$$) to give an alcohol. The reaction gives anti-Markovnikov addition and syn-addition of water across the double bond:

$$R-CH=CH_2 \xrightarrow{BH_3/THF} (RCH_2CH_2)_3B \xrightarrow{H_2O_2/NaOH} RCH_2CH_2OH$$

Option 4:

This is an INCORRECT method for preparing alcohols. Ozonolysis involves treating an alkene with ozone ($$O_3$$) followed by a reductive workup (using $$Zn/H_2O$$ or $$(CH_3)_2S$$). This reaction cleaves the carbon-carbon double bond and produces aldehydes and/or ketones, NOT alcohols:

$$R_2C=CR_2 \xrightarrow{O_3} \text{ozonide} \xrightarrow{Zn/H_2O} R_2C=O + O=CR_2$$

The products are carbonyl compounds (aldehydes or ketones depending on the substitution), not hydroxyl compounds.

The correct answer is Option 4: Ozonolysis of alkene is an incorrect method for preparing alcohols.

Methyl phenyl ether (anisole, PhOCH₃) is prepared by Williamson ether synthesis.

In Williamson synthesis, a phenoxide ion (or alkoxide) reacts with a primary alkyl halide via S_N2 mechanism.

The best combination is: sodium phenoxide (PhO⁻Na⁺) + methyl bromide (MeBr)

$$PhO^-Na^+ + CH_3Br \to PhOCH_3 + NaBr$$

Note: Option 3 (PhBr + MeO⁻Na⁺) would not work well because aryl halides do not undergo S_N2 reactions easily.

The correct answer is Option 4: PhO⁻Na⁺, MeBr.

We need to arrange the following phenols in increasing order of pK$$_a$$ (from most acidic to least acidic):

(A) 2,4-Dinitrophenol

(B) 4-Nitrophenol

(C) 2,4,5-Trimethylphenol

(D) Phenol

(E) 3-Chlorophenol

Lower pK$$_a$$ = stronger acid (more acidic). Electron-withdrawing groups (EWG) increase acidity (lower pK$$_a$$). Electron-donating groups (EDG) decrease acidity (higher pK$$_a$$).

(A) 2,4-Dinitrophenol: Two strong EWG ($$-\text{NO}_2$$) at ortho and para positions strongly stabilize the phenoxide ion. pK$$_a \approx 4.09$$ — most acidic.

(B) 4-Nitrophenol: One EWG at para position. pK$$_a \approx 7.15$$.

(E) 3-Chlorophenol: Chlorine is weakly electron-withdrawing at meta position. pK$$_a \approx 9.02$$.

(D) Phenol: No substituents. pK$$_a \approx 10.0$$.

(C) 2,4,5-Trimethylphenol: Three methyl groups (EDG) destabilize phenoxide ion. pK$$_a \approx 10.6$$ — least acidic.

$$ \text{(A)} < \text{(B)} < \text{(E)} < \text{(D)} < \text{(C)} $$

$$ (4.09) < (7.15) < (9.02) < (10.0) < (10.6) $$

The correct answer is Option B: (A), (B), (E), (D), (C).

We are told that compound X, when treated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$, yields compound Y. Y is used as an acid/base indicator.

First, identify compound Y.

The reaction of a phenolic compound with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ (a Friedel-Crafts type condensation reaction) produces phenolphthalein, which is a well-known acid/base indicator.

Therefore, Y = Phenolphthalein.

First, identify compound X.

Phenolphthalein is synthesized by the condensation of two molecules of phenol with one molecule of phthalic anhydride in the presence of concentrated $$H_2SO_4$$:

$$2 \, C_6H_5OH + C_8H_4O_3 \xrightarrow{H_2SO_4} C_{20}H_{14}O_4 + H_2O$$

Therefore, X = Phenol.

Now, match with options.

Phenol is also known as carbolic acid. Looking at the options:

Option A: Anisole, methyl orange — Incorrect. Anisole is methyl phenyl ether, not phenol.

Option B: Salicylaldehyde, Phenolphthalein — Incorrect. Salicylaldehyde is not used to make phenolphthalein.

Option C: Toludine, Phenolphthalein — Incorrect. Toludine (methylaniline) cannot form phenolphthalein.

Option D: Carbolic acid, Phenolphthalein — Correct. Carbolic acid is phenol, and its reaction with phthalic anhydride gives phenolphthalein.

The correct answer is Option D.

A trisubstituted compound A with formula $$C_{10}H_{12}O_{2}$$ gives a positive $$FeCl_{3}$$ test and decolourises $$KMnO_{4}$$. We wish to determine the number of pi bonds in this molecule.

First, the degree of unsaturation is calculated by the formula

$$DBE = \frac{2(10) + 2 - 12}{2} = 5$$

From the positive $$FeCl_{3}$$ test, a phenolic -OH group is indicated. Treatment with $$NaOH/CH_{3}Br$$ produces $$C_{11}H_{14}O_{2}$$, showing that methylation of the phenol occurs. Further reaction with HI yields $$CH_{3}I$$, confirming the presence of a methoxy (-OCH₃) group. Finally, decolourisation of alkaline $$KMnO_{4}$$ reveals an unsaturation (C=C double bond) outside the benzene ring.

These observations point to a benzene ring (accounting for four degrees of unsaturation: three pi bonds plus one ring) substituted by a phenolic -OH, an -OCH₃ group, and an allyl side chain containing a C=C double bond (one additional degree of unsaturation). The resulting structure corresponds to eugenol, namely 4-allyl-2-methoxyphenol, which can be written as $$\text{HO-}C_{6}H_{3}(OCH_{3})(CH_{2}CH=CH_{2})$$.

In this structure, the benzene ring contributes three pi bonds and the allylic C=C double bond contributes one pi bond, giving a total of 4 pi bonds. Therefore, the answer is 4.

A compound must satisfy both of the following conditions:

• It should give a positive Ceric Ammonium Nitrate (CAN) test, indicating the presence of an alcohol group.

• It should give a positive iodoform test, which requires either a methyl ketone group or the $$CH_3-CH(OH)-R$$ substructure.

Therefore, the compound must be an alcohol containing the $$CH_3-CH(OH)-R$$ moiety.

Evaluation of the given compounds:

• Acetophenone $$\left(PhCOCH_3\right)$$ gives a positive iodoform test but does not contain an alcohol group. Hence, it fails the CAN test.

• 2-Phenylethanol $$\left(PhCH_2CH_2OH\right)$$ gives a positive CAN test but lacks the $$CH_3-CH(OH)-$$ unit required for the iodoform test.

• Acetaldehyde $$\left(CH_3CHO\right)$$ gives a positive iodoform test but is not an alcohol and therefore fails the CAN test.

• 2-Butanol $$\left(CH_3CH(OH)CH_2CH_3\right)$$ is an alcohol and contains the $$CH_3-CH(OH)-$$ group. It gives positive results in both tests.

• Diethyl ether $$\left(CH_3CH_2OCH_2CH_3\right)$$ is an ether and gives neither test.

• 3-Phenylpropan-1-ol $$\left(PhCH_2CH_2CH_2OH\right)$$ gives a positive CAN test but lacks the required $$CH_3-CH(OH)-$$ unit.

• 1-Phenylpropan-2-ol $$\left(PhCH_2CH(OH)CH_3\right)$$ is an alcohol and contains the $$CH_3-CH(OH)-$$ group. It gives positive results in both tests.

• 2-Phenylpropan-2-ol $$\left(PhC(CH_3)_2OH\right)$$ is a tertiary alcohol. Although it gives a positive CAN test, it does not give the iodoform test.

Thus, only the following compounds satisfy both tests:

• 2-Butanol

• 1-Phenylpropan-2-ol

Hence, the number of compounds giving both a positive CAN test and a positive iodoform test is

$$\boxed{2}$$

The starting compound is 1,3,5-tris(4-nitrophenyl)benzene.

Step-1 Sn/HCl reduces each $$\!NO_2$$ group to $$\!NH_2$$, giving 1,3,5-tris(4-aminophenyl)benzene.

Step-2 NaNO$$\_2$$/HCl at 0 °C converts each $$\!NH_2$$ into a diazonium group (-$$N_2^+Cl^-$$).  The product obtained is denoted as $$P$$.

Step-3 Pure water at room temperature hydrolyses every diazonium group to an $$OH$$ group (Sandmeyer hydrolysis), giving $$Q =$$ 1,3,5-tris(4-hydroxyphenyl)benzene.

Step-4 Bromination of $$Q$$ in aqueous medium produces $$R$$.
Phenol rings undergo electrophilic substitution chiefly at the two ortho positions (positions 2 and 6 with respect to the $$OH$$) because the para position (4) is already occupied by the C-C bond that connects the side ring to the central benzene.

Mass-difference data supplied in the problem: $$\text{molar-mass}(R)\;-\;\text{molar-mass}(Q)=474\;\text{g mol}^{-1}$$

Each bromination step replaces an H atom (1 g mol$$^{-1}$$) by a Br atom (80 g mol$$^{-1}$$).
Mass increase per Br introduced = $$80-1=79\;\text{g mol}^{-1}$$.

Number of Br atoms introduced = $$\frac{474}{79}=6$$.

There are three phenolic rings in $$Q$$, so every ring receives $$\frac{6}{3}=2$$ bromine atoms, exactly the two ortho positions expected. Hence $$R$$ is 1,3,5-tris(2,6-dibromo-4-hydroxyphenyl)benzene.

Counting hetero-atoms in one molecule of $$R$$ (all atoms other than C and H):
• Oxygen: 3 (one $$O$$ in each $$OH$$ group)
• Bromine: 6 (two per phenolic ring × three rings)
Total hetero-atoms = $$3+6=9$$.

Therefore, the number of hetero-atoms present in one molecule of $$R$$ is 9.

To determine the decreasing order of acidic strength for the given substituted phenols, we must evaluate how the different substituents affect the stability of the phenoxide ion formed after the loss of a proton ($$H^+$$). As a general rule, electron-withdrawing groups stabilize the negative charge on the phenoxide ion through inductive ($$-I$$) or resonance ($$-R$$) effects, thereby increasing acidity. Conversely, electron-donating groups destabilize the ion by pushing electron density toward the oxygen through positive inductive ($$+I$$) or resonance ($$+R$$) effects, which decreases acidity.

The nitro group ($$-NO_2$$) is a powerful electron-withdrawing group that can exert both $$-I$$ and $$-R$$ effects. In compound (A), p-nitrophenol, the nitro group is located at the para position, allowing both the $$-I$$ and strong $$-R$$effects to heavily stabilize the phenoxide ion, making it the most acidic. In compound (B), m-nitrophenol, the nitro group is at the meta position. Because resonance effects cannot operate from the meta position, only the $$-I$$ effect stabilizes the anion. This makes it highly acidic, but slightly less so than the para isomer.

The methoxy group ($$-OCH_3$$), on the other hand, exerts an electron-withdrawing $$-I$$ effect but a much stronger electron-donating $$+R$$ effect. In compound (C), m-methoxyphenol, the methoxy group is at the meta position where its $$+R$$ effect cannot operate. Therefore, only its mild $$-I$$ effect influences the ring, slightly stabilizing the phenoxide ion. In compound (D), p-methoxyphenol, the methoxy group is at the para position. Here, its dominant electron-donating $$+R$$ effect actively pushes electron density into the ring, highly destabilizing the phenoxide ion and making it the least acidic of all four options.

By comparing the relative stabilization and destabilization caused by these groups at their respective positions, we can conclude the final sequence. The strongest stabilization occurs in compound (A), followed by (B), then a much weaker stabilization in (C), and finally an active destabilization in (D). Therefore, the correct decreasing order of acidic strength is (A) > (B) > (C) > (D).

Thus, D is the right option.

In the presence of acid, the hydroxyl group is first protonated to convert it into a good leaving group.

$$R-OH + H^+ \rightarrow R-OH_2^+$$

The protonated alcohol then loses a water molecule to form a secondary carbocation.

$$R-OH_2^+ \rightarrow R^+ + H_2O$$

The secondary carbocation undergoes a $$1,2$$-methyl shift because it is adjacent to a quaternary carbon. Migration of one methyl group generates a more stable tertiary carbocation.

Carbocation rearrangement:

$$2^\circ\ \text{carbocation} \xrightarrow{\text{1,2-methyl shift}} 3^\circ\ \text{carbocation}$$

The tertiary carbocation then undergoes elimination. A base removes a proton from an adjacent carbon, leading to the formation of an alkene.

According to Zaitsev's rule, the major product is the most substituted and therefore the most stable alkene.

Removal of the proton from the carbon bearing the migrated methyl group gives a tetrasubstituted alkene, which is the most stable product.

Hence, the major product is the rearranged alkene having the double bond between the two most substituted carbon atoms.

Therefore, the correct answer is the structure shown in Option (C).

We need to determine the reason for the difference in bromination products of phenol in chloroform versus water medium.

When phenol is treated with $$Br_2$$ in chloroform ($$CHCl_3$$), we get monobromination—specifically a mixture of ortho-bromophenol and para-bromophenol because in a non-polar solvent, $$Br_2$$ remains as a molecular species and acts as a weaker electrophile. The reaction proceeds through a typical electrophilic aromatic substitution requiring a Lewis acid catalyst, and only one hydrogen is replaced.

When phenol is treated with $$Br_2$$ in water ($$Br_2$$/aq), we get 2,4,6-tribromophenol as a white precipitate. In water, bromine partially dissociates as follows:

The polar solvent facilitates the generation of a more reactive electrophile ($$Br^+$$ or its equivalent from $$HOBr$$), and the phenoxide ion ($$C_6H_5O^-$$) formed partially in water is even more reactive than phenol. This leads to tribromination at all available ortho and para positions.

The key difference arises due to the polarity of the solvent:

- In a polar solvent (water), $$Br_2$$ generates a stronger electrophilic species, and phenol is more activated, leading to tribromination.

- In a non-polar solvent (chloroform), $$Br_2$$ remains molecular and is a weaker electrophile, leading to only monobromination.

Option A: Hyperconjugation — Not relevant here.

Option B: Polarity of solvent — Correct. The solvent polarity determines the strength of the electrophile generated.

Option C: Free radical formation — The reaction is electrophilic, not free radical.

Option D: Electromeric effect of the substrate — The electromeric effect of phenol is the same regardless of solvent; it is the solvent that changes the outcome.

Hence, the correct answer is Option B: Polarity of solvent.

We need to determine which compound yields carbolic acid upon hydrolysis. Carbolic acid is the common name for phenol ($$C_6H_5OH$$). To this end, each option is examined for its hydrolysis products.

In the case of Option A: Cumene (isopropylbenzene), simple hydrolysis does not occur. Instead, cumene undergoes oxidation by $$O_2$$ in the industrial cumene process to afford phenol and acetone, but this transformation is an oxidation rather than hydrolysis.

When Option B: Benzenediazonium chloride ($$C_6H_5N_2^+Cl^-$$) is warmed with water, hydrolysis proceeds to give phenol:

$$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\text{warm}} C_6H_5OH + N_2 + HCl$$

This reaction directly furnishes phenol (carbolic acid).

On hydrolysis, Option C: Benzal chloride ($$C_6H_5CHCl_2$$) yields benzaldehyde ($$C_6H_5CHO$$) rather than phenol:

$$C_6H_5CHCl_2 + H_2O \rightarrow C_6H_5CHO + 2HCl$$

Furthermore, hydrolysis of Option D: Ethylene glycol ketal regenerates the corresponding carbonyl compound along with ethylene glycol, without producing phenol.

Therefore, the correct choice is Option B: Benzenediazonium chloride.

Explanation: Benzene first undergoes Friedel-Crafts alkylation with isopropyl chloride in the presence of $$AlCl_3$$ to form cumene.

Cumene is oxidized by air to form cumene hydroperoxide, which on acidic hydrolysis gives phenol.

Therefore, intermediate $$P$$ is phenol.

Phenol reacts with:

$$Na_2Cr_2O_7/H_2SO_4$$

to undergo oxidation and form $$p$$-benzoquinone.

Phenol also reacts with bromine in non-polar solvent $$CS_2$$ to undergo monobromination.

Since the $$-OH$$ group is ortho/para-directing and steric hindrance favors substitution at the para position, the major product formed is $$p$$-bromophenol.

Thus, $$A = p\text{-benzoquinone}$$, $$B = p\text{-bromophenol}$$. The right option is B.

This question involves the cumene process, an industrial method for producing phenol and acetone.

Cumene is isopropylbenzene: $$C_6H_5CH(CH_3)_2$$. It reacts with $$O_2$$ to form cumene hydroperoxide: $$C_6H_5CH(CH_3)_2 + O_2 \rightarrow C_6H_5C(CH_3)_2{-}OOH$$, which under acid-catalyzed conditions ($$H^+/H_2O$$) decomposes to phenol and acetone: $$C_6H_5C(CH_3)_2{-}OOH \xrightarrow{H^+/H_2O} C_6H_5OH + CH_3COCH_3$$.

This process yields: $$\bullet$$ Compound A = Phenol ($$C_6H_5OH$$), $$\bullet$$ Compound B = Acetone ($$CH_3COCH_3$$).

Option A: Cresol and $$CH_3COOH$$ — Incorrect, since cresol (methylphenol) is not formed. Option B: Catechol and $$CH_3COCH_3$$ — Incorrect, as catechol is 1,2-dihydroxybenzene. Option C: Phenol and $$CH_3COCH_3$$ — Correct, these are the standard products of the cumene process. Option D: Hydroquinone and $$CH_3COCH_3$$ — Incorrect, as hydroquinone is 1,4-dihydroxybenzene.

The correct answer is Option C: Phenol and $$CH_3COCH_3$$.

We have a 1.84 mg sample of a polyhydric alcohol 'X' with molar mass 92.0 g/mol, and it gives 1.344 mL of $$H_2$$ gas at STP. We need to find the number of alcoholic (-OH) hydrogens in 'X'.

We begin by finding the number of moles of compound 'X'. The mass is 1.84 mg = $$1.84 \times 10^{-3}$$ g, and the molar mass is 92.0 g/mol, so:

$$n_X = \frac{1.84 \times 10^{-3}}{92.0} = 2.0 \times 10^{-5} \text{ mol}$$

Now we find the moles of $$H_2$$ produced. At STP, 1 mole of any gas occupies 22400 mL. The volume of $$H_2$$ is 1.344 mL, so:

$$n_{H_2} = \frac{1.344}{22400} = 6.0 \times 10^{-5} \text{ mol}$$

When a polyhydric alcohol reacts with an active metal like sodium, each -OH group releases half a mole of $$H_2$$. If the compound has $$k$$ alcoholic hydrogen atoms, then one mole of the compound produces $$\frac{k}{2}$$ moles of $$H_2$$:

$$R(OH)_k + kNa \rightarrow R(ONa)_k + \frac{k}{2}H_2$$

We can write the ratio:

$$\frac{n_{H_2}}{n_X} = \frac{k}{2}$$

$$\frac{6.0 \times 10^{-5}}{2.0 \times 10^{-5}} = \frac{k}{2}$$

$$3 = \frac{k}{2}$$

$$k = 6$$

We can verify: a compound with molar mass 92 and 6 -OH groups would be consistent with a highly hydroxylated small molecule. Indeed, the moles and volumes check out perfectly.

Hence, the correct answer is 6.

We need to identify compound P, its nitration products, and find the number of oxygen atoms in compound C.

Identify compound P

The key clues are:

- P on mild nitration (dilute $$HNO_3$$) gives two isomers A and B

- A and B can be separated by steam distillation

- A shows intramolecular hydrogen bonding

- B shows intermolecular hydrogen bonding

This is a classic description of phenol ($$C_6H_5OH$$). When phenol is nitrated with dilute $$HNO_3$$:

- It gives ortho-nitrophenol (A) and para-nitrophenol (B)

- ortho-nitrophenol has intramolecular hydrogen bonding (between -OH and -$$NO_2$$ groups on adjacent carbons), making it steam volatile

- para-nitrophenol has intermolecular hydrogen bonding, making it less volatile

- They can be separated by steam distillation because ortho-isomer is steam volatile while para-isomer is not

Identify compound C

When phenol reacts with concentrated $$HNO_3$$, it gives 2,4,6-trinitrophenol (picric acid).

$$C_6H_2(NO_2)_3OH$$

Picric acid is a yellow compound and is a strong acid, matching the description of compound C.

Count oxygen atoms in picric acid

The molecular formula of picric acid is $$C_6H_3N_3O_7$$:

- 3 nitro groups ($$-NO_2$$): each has 2 oxygen atoms = $$3 \times 2 = 6$$ oxygen atoms

- 1 hydroxyl group ($$-OH$$): 1 oxygen atom

- Total oxygen atoms = 6 + 1 = 7

The correct answer is 7.

The balanced chemical equation for the Zerewitinoff determination is:

$$\mathrm{R-OH + CH_3MgI \longrightarrow CH_4\uparrow + R-OMgI}$$

From the stoichiometry of the reaction:

$$\mathrm{1\ mol\ Alcohol \longrightarrow 1\ mol\ CH_4}$$

Volume of methane evolved:

$$\mathrm{3.1\ mL}$$

Molar volume of gas:

$$\mathrm{22700\ mL\ mol^{-1}}$$

Moles of methane evolved:

$$\mathrm{Moles\ of\ CH_4 = \frac{3.1}{22700}}$$

Since the molar ratio is $$\mathrm{1:1}$$:

$$\mathrm{Moles\ of\ R-OH = \frac{3.1}{22700}\ mol}$$

Mass of alcohol given:

$$\mathrm{4.5\ mg = 4.5\times10^{-3}\ g}$$

Molecular weight of alcohol:

$$\mathrm{Molecular\ Weight = \frac{Mass}{Moles}}$$

$$\mathrm{= \frac{4.5\times10^{-3}}{\frac{3.1}{22700}}}$$

$$\mathrm{= \frac{4.5\times10^{-3}\times22700}{3.1}}$$

$$\mathrm{= \frac{102.15}{3.1}}$$

$$\mathrm{\approx 32.95\ g\ mol^{-1}}$$

Rounding to the nearest whole number:

$$\boxed{\mathrm{33\ g\ mol^{-1}}}$$

Step 1: Identify product A from the fermentation of glucose.

$$\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$$

Product A is ethanol ($$\text{C}_2\text{H}_5\text{OH}$$).

Step 2: Identify product B from the iodoform reaction.

NaOI is sodium hypoiodite, the reagent for the iodoform reaction. Ethanol undergoes the iodoform reaction:

$$\text{CH}_3\text{CH}_2\text{OH} + 4\text{NaOI} \xrightarrow{\Delta} \text{CHI}_3 + \text{HCOONa} + 3\text{NaOH}$$

The products are iodoform ($$\text{CHI}_3$$) and sodium formate ($$\text{HCOONa}$$).

Therefore, product B is sodium formate ($$\text{HCOONa}$$), which contains 1 carbon atom.

The number of carbon atoms in product B is 1.

We need to determine the number of chlorine atoms in bithionol. Bithionol is a well-known antiseptic and disinfectant used in medicated soaps. Its IUPAC name is 2,2'-thiobis(4,6-dichlorophenol), and its chemical formula is $$\text{C}_{12}\text{H}_{6}\text{Cl}_{4}\text{O}_{2}\text{S}$$.

We have two phenol rings connected by a sulfur bridge (thio linkage). Each phenol ring carries two chlorine atoms at the 4 and 6 positions. Since there are two such rings, the total number of chlorine atoms is $$2 \times 2 = 4$$.

Hence, the correct answer is $$\boxed{4}$$.

We begin by recalling what a Novolac resin is. Novolac is the product obtained when phenol reacts with formaldehyde ($$\text{HCHO}$$) in an acidic medium. The first step of this condensation reaction is the electrophilic substitution of formaldehyde at the ortho and para positions of the phenolic ring.

Using the general mechanism of electrophilic aromatic substitution, we can state:

$$\text{Phenol} + \text{HCHO} \;\xrightarrow[\text{acid}]{\text{condensation}}\; o\text{-Hydroxymethyl-phenol} \;+\; p\text{-Hydroxymethyl-phenol}$$

Here we clearly see that the immediate product formed—before the polymer chain starts growing—is the substituted phenol that carries a -CH2OH group at the ortho or para position with respect to the -OH already present on the ring. These substituted phenols then link through methylene (-CH2-) bridges to give the linear Novolac polymer. Therefore, each repeating unit (monomeric unit) of the polymer chain is precisely this $$o$$-hydroxymethylphenol (and its para analogue).

Among the options offered:

Option D states “o-Hydroxymethylphenol,” which matches exactly the compound produced in the first condensation step and which ultimately acts as the monomeric unit that repeats to form Novolac.

Hence, the correct answer is Option 4.

The reaction described here is the formation of a phenolphthalein-type dye. When a phenolic compound is heated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ (which acts as a dehydrating agent), a condensation reaction occurs. The resulting product, when dissolved in NaOH, gives a characteristic pink colour.

For this condensation to happen, the phenol must have a free $$-OH$$ group attached directly to the aromatic ring. The electrophilic substitution during the condensation takes place at the ortho or para positions relative to the $$-OH$$ group.

Now, 4-methylphenol (p-cresol) has the structure where the $$-OH$$ group is at position 1 and the $$-CH_3$$ group is at position 4. The para position (position 4) is occupied by $$-CH_3$$, but both ortho positions (positions 2 and 6) are free. The condensation with phthalic anhydride requires two available positions on the ring, and the ortho positions serve this purpose.

Among the other options, option 2 contains an alcoholic $$-OH$$ (not directly on the ring) which cannot undergo this phenol-specific condensation. Options 3 and 4 contain dihydroxy compounds (resorcinol-type), which would give fluorescein (a green fluorescent dye) rather than the pink phenolphthalein-type dye.

Hence, 4-methylphenol gives the pink colour with phthalic anhydride in concentrated $$H_2SO_4$$ followed by NaOH treatment.

Hence, the correct answer is Option 1.

The phenolic -OH group (the one attached directly to the benzene ring) does not react with thionyl chloride ($SOCl_2$) due to resonance stabilization. The lone pairs of electrons on the phenolic oxygen atom delocalize into the $\pi$-system of the adjacent benzene ring. This resonance interaction grants the carbon-oxygen (C-O) bond a partial double-bond character. Because of this partial double bond, the C-O bond is significantly stronger and much harder to cleave than a standard single bond. Additionally, $SOCl_2$ replaces hydroxyl groups via a nucleophilic substitution mechanism that requires an $sp^3$ hybridized carbon; it cannot successfully attack the $sp^2$ hybridized carbon of an aromatic ring. Therefore, the phenolic -OH remains untouched, while the other two aliphatic -OH groups (which are on $sp^3$ carbons and have standard, easily broken single bonds) will be successfully replaced by chlorine atoms.

Halogenation: On treating phenol with bromine, different reaction products are formed under different experimental conditions.

(a) When the reaction is carried out in solvents of low polarity such as $$\mathrm{CHCl_3}$$ or $$\mathrm{CS_2}$$ and at low temperature, monobromophenols are formed.

$$\mathrm{C_6H_5OH + Br_2 \xrightarrow[\ 273\ K\ ]{CS_2} o\text{-}Bromophenol + p\text{-}Bromophenol}$$

The usual halogenation of benzene takes place in the presence of a Lewis acid such as $$\mathrm{FeBr_3}$$, which polarises the halogen molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid due to the highly activating effect of the $$\mathrm{-OH}$$ group attached to the benzene ring.

(b) When phenol is treated with bromine water, $$\mathrm{2,4,6}$$-tribromophenol is formed as a white precipitate.

$$\mathrm{C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr}$$

$$\mathrm{2,4,6\text{-}Tribromophenol}$$

Thus b, c, d give the right options. a gives a different product. Thus, the right option is C.

We begin by recalling the general principle of the Williamson ether synthesis. The statement of the method is: An ether can be prepared by the reaction of an alkoxide ion with a primary alkyl halide under S}_\mathrm{N}2 \text{ conditions. Symbolically, $$\mathrm{R'O^-Na^+ \;+\; R''-X \;\longrightarrow\; R'O\!\!-\!R'' \;+\; NaX}$$ where $$X$$ is generally $$\mathrm{Cl,\, Br}$$ or $$\mathrm{I}$$ and the $$R''$$ group should be such that it can undergo the back-side attack required for the $$S_\mathrm{N}2$$ mechanism.

Now we look at the target molecule ethyl phenyl ether, also called phenetole, having the structure $$\mathrm{C_6H_5-O-CH_2CH_3}.$$ To obtain this by Williamson synthesis, we choose the pair of reactants so that one provides the $$\mathrm{O^-}$$ and the other provides the alkyl group. A convenient combination is:

$$\mathrm{C_6H_5O^-Na^+ \;+\; BrCH_2CH_3 \;\longrightarrow\; C_6H_5OCH_2CH_3 \;+\; NaBr}$$

Here $$\mathrm{C_6H_5O^-}$$ is the phenoxide ion, and $$\mathrm{BrCH_2CH_3}$$ (ethyl bromide) is a primary alkyl halide, perfectly suited for the $$S_\mathrm{N}2$$ reaction. Every requirement of the Williamson synthesis is satisfied, so the preparation is indeed feasible. Thus, the Assertion (A) that ethyl phenyl ether may be achieved by Williamson synthesis is correct.

Next, we examine the Reason (R): “Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether.” Writing the suggested reaction, we would have

$$\mathrm{C_6H_5Br \;+\; Na^+O^-CH_2CH_3 \;\longrightarrow\; ?}$$

For this to give the desired ether, the ethoxide ion would have to displace the bromide ion directly from the aromatic ring of bromobenzene. However, an aryl halide such as $$\mathrm{C_6H_5Br}$$ cannot undergo the $$S_\mathrm{N}2$$ mechanism because the carbon bearing the halogen is part of a rigid, conjugated, planar $$\mathrm{sp^2}$$ system. The back-side attack required in $$S_\mathrm{N}2$$ is sterically and electronically impossible on such a substrate. Therefore the ethoxide ion does not replace bromide, and no ether is formed under ordinary Williamson conditions. Hence the Reason (R) is false.

Summarising, we have:

• Assertion (A): correct.
• Reason (R): incorrect.

Therefore the option that matches this evaluation is Option B.

Hence, the correct answer is Option B.

Thus, the correct option is B.

Chlorobenzene reacts with sodium hydroxide only under very harsh conditions because the $$\mathrm{C-Cl}$$ bond in aryl halides has partial double bond character due to resonance, making nucleophilic substitution difficult.

When chlorobenzene is fused with aqueous $$\mathrm{NaOH}$$ at high temperature and high pressure, it forms sodium phenoxide.

$$\mathrm{C_6H_5Cl + NaOH \xrightarrow[300\ atm]{623\ K} C_6H_5ONa + NaCl}$$

On subsequent acidification, sodium phenoxide is converted into phenol.

$$\mathrm{C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl}$$

Therefore, the required reaction conditions for the conversion of chlorobenzene to sodium phenoxide are:

• Temperature = $$623\ \mathrm{K}$$
• Pressure = $$300\ \mathrm{atm}$$

Hence, the correct answer is Option (D): $$623\ \mathrm{K},\ 300\ \mathrm{atm}$$.

The reaction of this secondary alcohol with HCl proceeds via an SN1 mechanism involving a carbocation rearrangement.

1. Formation of Carbocation: The hydroxyl (-OH) group is first protonated by the acid to form a good leaving group (-OH₂⁺), which then departs as a water molecule. This leaves behind a secondary (2°) carbocation on the side chain.
2. 1,2-Hydride Shift: Carbocations always rearrange to gain stability. The positive charge is currently on a secondary carbon, but the adjacent carbon in the cyclohexane ring is tertiary (3°). A hydrogen atom moves from the ring carbon to the side chain (a 1,2-hydride shift), moving the positive charge onto the more stable tertiary position within the ring.
3. Nucleophilic Attack: Finally, the chloride ion (Cl⁻) attacks this newly formed, stable tertiary carbocation.

The final major product 'A' is 1-chloro-1-(1-methylethyl)cyclohexane, where the chlorine is attached directly to the tertiary carbon of the ring.

The reaction begins with a crossed Cannizzaro reaction between $$\mathrm{4\text{-}Methoxybenzaldehyde}$$ and $$\mathrm{HCHO}$$ in presence of $$\mathrm{NaOH}$$.

Since neither aldehyde contains $$\mathrm{\alpha}$$-hydrogens, disproportionation occurs.

Formaldehyde is preferentially oxidised to formate, while the aromatic aldehyde is reduced to:

$$\mathrm{4\text{-}Methoxybenzyl\ Alcohol}$$

In the second step, $$\mathrm{NaH}$$ deprotonates the alcohol to form an alkoxide ion.

The alkoxide attacks $$\mathrm{CH_3CH_2Br}$$ through an $$\mathrm{S_N2}$$ mechanism in DMF, forming:

$$\mathrm{4\text{-}Methoxybenzyl\ Ethyl\ Ether}$$

In the final step, concentrated $$\mathrm{HI}$$ and heat cleave both ether linkages.

In the aryl methyl ether, iodide attacks the methyl group, forming a phenol group $$\mathrm{(-OH)}$$.

In the benzylic ether, iodide attacks the benzylic carbon, replacing the ethoxy group with:

$$\mathrm{(-CH_2I)}$$

The final major product contains a phenol group and an iodomethyl group at para positions on the benzene ring.

Correct Option: $$\mathrm{A}$$

The reaction involves the Grignard reagent $$\mathrm{CH_3MgBr}$$, which acts as a strong nucleophile and strong base.

Compound $$\mathrm{A}$$ is an ester, $$\mathrm{Phenyl\ Acetate}$$. Esters react with excess Grignard reagent to form tertiary alcohols.

In the first step, one equivalent of $$\mathrm{CH_3MgBr}$$ attacks the ester carbonyl carbon.

Nucleophilic acyl substitution occurs, expelling the phenoxy group $$\mathrm{(-OC_6H_5)}$$ and forming $$\mathrm{Acetone\ (CH_3COCH_3)}$$.

Since the Grignard reagent is present in excess, a second equivalent attacks the ketone carbonyl carbon.

Acidic hydrolysis protonates the alkoxide intermediate to form $$\mathrm{2\text{-}Methyl\text{-}2\text{-}propanol}$$ with structure $$\mathrm{(CH_3)_3C-OH}$$, which is a tertiary alcohol.

In compound $$\mathrm{B}$$, the terminal alkyne proton is highly acidic.

Hence, $$\mathrm{CH_3MgBr}$$ first undergoes an acid-base reaction, deprotonating the alkyne instead of forming a tertiary alcohol.

Therefore, the correct option is $$\mathrm{A}$$.

When a phenol is heated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ the first step is an electrophilic acylation at a position that is ortho (or, more rarely, para) to the -OH group. After two such substitutions and intramolecular ring closure, a phthalein dye (e.g. phenolphthalein, fluorescein, etc.) is obtained. The reaction therefore has two essential requirements:

1. The aromatic ring must be activated toward electrophilic substitution (-OH is an ortho/para-directing activator).
2. At least one of the positions ortho or para to the -OH must be unsubstituted so that the acylation can occur.

Let us check the four phenols given in the question.

Phenol itself has both ortho and the para positions free; it readily forms the coloured dye phenolphthalein.

p-Cresol has both ortho positions free; it also condenses with phthalic anhydride to give a coloured p-tolyl-phthalein.

Resorcinol (1,3-dihydroxybenzene) is even more activated because it contains two -OH groups. It reacts smoothly, giving fluorescein (intense yellow-green fluorescence).

2,4,6-Trinitrophenol (picric acid) carries three -NO2 groups at all ortho and para positions relative to the -OH group. Two consequences follow:
(i) Every required entry position is already occupied; sterically the ring offers no vacant ortho/para site for the incoming acylium ion.
(ii) The strongly -M and -I nitro groups make the ring highly deactivated toward electrophilic substitution.
Because of these two factors, picric acid cannot undergo the initial acylation step, so no phthalein framework is formed and, hence, no colour is produced.

Therefore the only phenol that fails to give a coloured product with phthalic anhydride and conc. $$H_2SO_4$$ is the one in Option D, 2,4,6-trinitrophenol (picric acid).

Option D which is: 2,4,6-trinitrophenol (picric acid)

The compound $$\mathrm{C_6H_6O}$$ gives a dark green colour with neutral $$\mathrm{FeCl_3}$$, indicating the presence of a phenolic $$\mathrm{-OH}$$ group. Hence, compound $$\mathrm{A}$$ is $$\mathrm{Phenol}$$.

Phenol reacts with $$\mathrm{CHCl_3/KOH}$$ followed by acidic workup through the $$\mathrm{Reimer\text{-}Tiemann\ Reaction}$$. This introduces a formyl group $$\mathrm{(-CHO)}$$ mainly at the ortho position, forming $$\mathrm{o\text{-}Hydroxybenzaldehyde\ (Salicylaldehyde)}$$.

Thus, compound $$\mathrm{B}$$ is salicylaldehyde.

Compound $$\mathrm{B}$$ is also formed by oxidation using $$\mathrm{PCC}$$. Since PCC oxidises primary alcohols into aldehydes, compound $$\mathrm{C}$$ must contain a $$\mathrm{-CH_2OH}$$ group at the ortho position.

Therefore, compound $$\mathrm{C}$$ is $$\mathrm{o\text{-}Hydroxybenzyl\ Alcohol}$$.

Correct Option: $$\mathrm{A}$$

First the alcohol is replaced and a carbocation is formed, which rearranges itself for stability. Then an alkene is formed. 

Then addition of H and OH takes place per anti-markonikoff's rule and we get option D. Thus, option D is the right answer.

Compound $$\mathrm{A}$$ has molecular formula $$\mathrm{C_6H_{12}O_2}$$, corresponding to a saturated ester or carboxylic acid.

Reduction with $$\mathrm{LiAlH_4/H_3O^+}$$ converts an ester:

$$\mathrm{RCOOR' \rightarrow RCH_2OH + R'OH}$$

The resulting primary alcohols on oxidation form corresponding carboxylic acids.

Option $$\mathrm{C}$$ is:

$$\mathrm{CH_3CH_2CH_2COOCH_2CH_3}$$

which is ethyl butanoate.

Reduction produces:

$$\mathrm{1\text{-}Butanol\ (C_4)\ +\ Ethanol\ (C_2)}$$

Oxidation of $$\mathrm{1\text{-}Butanol}$$ gives:

$$\mathrm{Butanoic\ Acid\ (C_4)}$$

matching the condition given.

Option $$\mathrm{D}$$ produces only $$\mathrm{C_3}$$ alcohols and hence forms $$\mathrm{C_3}$$ carboxylic acid on oxidation.

Therefore, the correct compound is:

$$\mathrm{CH_3CH_2CH_2COOCH_2CH_3}$$

Correct Option: $$\mathrm{C}$$

The acid-catalyzed dehydration of 3,3-dimethylbutan-2-ol proceeds through a detailed four-step mechanism:

Step 1: Protonation of the alcohol

The reaction begins with the strong acid, concentrated $$H_2SO_4$$, dissociating to provide a proton ($$H^+$$). The oxygen atom of the hydroxyl group ($$-OH$$) in 3,3-dimethylbutan-2-ol possesses lone electron pairs. One of these pairs acts as a base and accepts the proton. This step is crucial because the neutral hydroxyl group is a poor leaving group, but protonating it converts it into an oxonium ion ($$-OH_2^+$$), which is an excellent leaving group because it can depart as a stable, neutral water molecule.

Step 2: Formation of the secondary carbocation

Once the leaving group is formed, the carbon-oxygen bond breaks. The water molecule departs, taking both electrons from the bond with it. This leaves the carbon atom deficient in electrons, resulting in a full positive charge. Because the positively charged carbon is directly bonded to two other carbon atoms (one methyl group and one tert-butyl group), this intermediate is classified as a secondary carbocation.

Step 3: Carbocation rearrangement via a 1,2-methyl shift

Secondary carbocations are moderately stable, but the molecule will spontaneously rearrange if a more stable intermediate can be formed. Directly adjacent to the positively charged secondary carbon is a quaternary carbon atom (bonded entirely to other carbons, specifically three methyl groups). To achieve a lower overall energy state, an entire methyl group ($-CH_3$) migrates from the quaternary carbon to the adjacent positively charged carbon, taking its bonding pair of electrons with it. This specific rearrangement is called a 1,2-methyl shift. As a result, the positive charge moves to the carbon that just lost the methyl group. This new positively charged carbon is now bonded to three other carbons, making it a highly stable tertiary carbocation.

Step 4: Elimination to form the alkene

The final step is the elimination of a proton to form the carbon-carbon double bond, restoring neutrality to the molecule. A weak base present in the reaction mixture, such as water or the bisulfate ion ($$HSO_4^-$$), will remove a proton ($$H^+$$) from a carbon atom immediately adjacent to the positive charge.

The molecule has two main pathways for elimination:

Removing a proton from one of the outer methyl groups would yield a disubstituted alkene with very few alpha hydrogens.

Removing a proton from the adjacent carbon in the main chain yields a double bond right in the middle of the molecule.

Following Zaitsev's rule, elimination strongly favors the formation of the most highly substituted alkene because it is thermodynamically more stable. The proton is eliminated from the internal carbon, forming a double bond between the two central carbons. This results in a tetrasubstituted alkene, 2,3-dimethylbut-2-ene. This specific product is overwhelmingly favored because it is surrounded by four methyl groups, providing a massive total of 12 alpha hydrogens that stabilize the double bond through strong hyperconjugation.

The target molecule is 2-methylpropan-2-ol, which is (CH₃)₃COH — a tertiary alcohol.

The starting material is ethyl ethanoate (ethyl acetate): CH₃COOC₂H₅.

The Grignard reagent is CH₃MgBr (methylmagnesium bromide).

In the first step, CH₃MgBr attacks the carbonyl carbon of ethyl ethanoate. One equivalent of Grignard adds to give a tetrahedral intermediate (a magnesium alkoxide): CH₃-C(CH₃)(OMgBr)-OC₂H₅. This intermediate is a hemiketal-like species that collapses: the ethoxide (C₂H₅O⁻) is expelled as a leaving group, generating a ketone intermediate — in this case acetone (propan-2-one): CH₃COCH₃.

In the second step, a second equivalent of CH₃MgBr adds to the acetone formed. The Grignard adds to the carbonyl of acetone to give the magnesium alkoxide: (CH₃)₃COMgBr. Aqueous workup gives the tertiary alcohol: (CH₃)₃COH = 2-methylpropan-2-ol.

Thus, 2 equivalents of CH₃MgBr are required in total: 1 equivalent to react with ethyl ethanoate to produce acetone, and 1 equivalent to add to the acetone to give 2-methylpropan-2-ol.

Therefore, the answer is $$\boxed{2}$$ equivalents.

We start by recalling what Bakelite is. Bakelite is the very first synthetic polymer, technically called phenol-formaldehyde resin. It is obtained when phenol $$\left(C_{6}H_{5}OH\right)$$ reacts with formaldehyde $$\left(HCHO\right)$$ under either acidic or basic conditions and then undergoes further cross-linking.

The first stage of the process is the introduction of $$-CH_{2}OH$$ (hydroxymethyl) groups into the aromatic ring of phenol. In organic chemistry, whenever an aromatic ring such as benzene or phenol is attacked by an electron-deficient species, the mechanism is called electrophilic substitution. Here the electrophile is the carbon of formaldehyde, which carries a partial positive charge $$\left(\delta ^+\right)$$ because oxygen is more electronegative than carbon.

So, step one is an electrophilic substitution:

$$C_{6}H_{5}OH + HCHO \;\xrightarrow{acid / base\; C6H4(OH)(CH2OH)}$$

Because more than one $$-CH_{2}OH$$ group can enter the ring, we often get mixtures such as $$p$$- and $$o$$-hydroxymethyl phenol.

Now, once two phenol molecules each carry $$-CH_{2}OH$$ groups, they can link together. Two $$-CH_{2}OH$$ groups condense by eliminating a molecule of water $$\left(H_{2}O\right)$$ and form a methylene bridge $$\left(-CH_{2}-\right)$$ between two aromatic rings. This removal of water is literally a dehydration.

The dehydration step can be shown schematically as:

$$\bigl[C_{6}H_{4}(OH)(CH_{2}OH)\bigr]_2 \;\xrightarrow[-H_{2}O]{\; C6H4(OH)-CH2-C6H4(OH)}$$

So we clearly have two fundamental types of reactions taking place:

1. Electrophilic substitution (introduction of $$-CH_{2}OH$$ on the aromatic ring).

2. Dehydration (elimination of water to form methylene bridges, leading to polymer growth).

Looking at the options, Option C exactly states “Electrophilic substitution and dehydration,” which matches our mechanistic description.

Hence, the correct answer is Option C.

To determine the increasing order of boiling points of the given substituted phenols, we compare the intermolecular forces present in each compound, particularly hydrogen bonding and dipole-dipole interactions.

All four compounds contain a phenolic $$-OH$$ group capable of forming intermolecular hydrogen bonds. Therefore, differences in boiling point arise mainly from the nature of the substituent attached to the benzene ring, which influences the strength of these intermolecular interactions.

In $$p$$-cresol (I), the $$-CH_3$$ group is an electron-donating substituent through the $$+I$$ effect. It slightly decreases the acidity of the phenolic $$-OH$$ group and weakens intermolecular hydrogen bonding, resulting in the lowest boiling point among the given compounds.

In $$p$$-methoxyphenol (IV), the $$-OCH_3$$ group donates electrons through the $$+R$$ effect. Although it enhances molecular polarity to some extent, its overall contribution to intermolecular attraction is greater than that of the methyl group, giving it a boiling point slightly higher than $$p$$-cresol.

In $$p$$-nitrophenol (II), the strongly electron-withdrawing $$-NO_2$$ group exerts both $$-I$$ and $$-R$$ effects, increasing the acidity of the phenolic $$-OH$$ group and strengthening intermolecular hydrogen bonding. The high polarity of the nitro group also enhances dipole-dipole interactions, leading to a comparatively higher boiling point.

In $$p$$-aminophenol (III), the $$-NH_2$$ group itself can participate in intermolecular hydrogen bonding in addition to the phenolic $$-OH$$ group. The presence of both hydrogen bond donor and acceptor sites creates an extensive hydrogen-bonding network, resulting in the highest boiling point among the four compounds.

Therefore, the increasing order of boiling points is

$$\boxed{\text{I} < \text{IV} < \text{II} < \text{III}}$$

which corresponds to the option B.

We have to identify an organic compound (A) whose molecular formula is $$C_6H_{12}O_2$$. On hydrolysis with dilute $$H_2SO_4$$ it gives a carboxylic acid (B) and an alcohol (C). This description exactly fits the behaviour of an ester, because an ester undergoes acid-catalysed hydrolysis according to the general equation

$$\text{RCOOR'}\;+\;H_2O\;\xrightarrow[\;]{\;H_2SO_4\;}\;\text{RCOOH}\;+\;\text{R'OH}$$

So, compound (A) must be an ester made up of an acid part $$\text{RCO-}$$ and an alcohol part $$\text{-OR'}$$.

Now, alcohol (C) gives white turbidity immediately with anhydrous $$ZnCl_2$$ and conc. HCl. This is the well-known Lucas test for classifying alcohols:

• Tertiary alcohol → turbidity within seconds (immediate).
• Secondary alcohol → turbidity in 5-10 min.
• Primary alcohol → no turbidity at room temperature.

Because turbidity is obtained at once, alcohol (C) must be a tertiary alcohol.

Let us search for the smallest tertiary alcohol that can arise from an ester whose total formula is $$C_6H_{12}O_2$$. The simplest tertiary alcohol is tert-butyl alcohol (2-methyl-2-propanol):

$$\text{(CH}_3)_3\text{C-OH}$$

Its molecular formula is $$C_4H_{10}O$$ (four carbons, one oxygen, ten hydrogens).

If alcohol (C) contributes four carbon atoms, the remaining two carbon atoms (because the ester has six carbons in all) must belong to the acid (B). The only common two-carbon carboxylic acid is acetic acid (ethanoic acid):

$$\text{CH}_3\text{COOH}$$

Acetic acid has the formula $$C_2H_4O_2$$.

Now we must check that these two fragments really fit the molecular formula of (A) when they combine to form an ester. Combining an acid and an alcohol produces an ester with the loss of one molecule of water ($$H_2O$$). Let us add their formulas algebraically and then subtract the elements of water.

• Acid part: $$C_2H_4O_2$$
• Alcohol part: $$C_4H_{10}O$$
• Sum: $$C_{2+4}H_{4+10}O_{2+1}=C_6H_{14}O_3$$
• Subtract water $$\left(H_2O\right)$$: $$C_6H_{14-2}O_{3-1}=C_6H_{12}O_2$$

Exactly the required molecular formula. Therefore our choice of fragments is correct.

Hence compound (A) must be the ester formed from acetic acid and tert-butyl alcohol, i.e. tert-butyl acetate:

$$\text{CH}_3\text{COO-C(CH}_3)_3$$

Among the given structural options, this structure corresponds to Option (1).

Hence, the correct answer is Option (1).

To determine the major product of the given reaction sequence, we first analyze the cleavage of the ether with $$HI$$ and then examine the dehydration of the alcohol formed in the first step.

The starting compound is the asymmetric ether 1-ethoxy-2-methylbutane, having the structure

$$CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3.$$

When ethers are treated with excess $$HI$$ under heating, the oxygen atom is protonated and the $$C-O$$ bond undergoes cleavage. The iodide ion preferentially attacks the less sterically hindered alkyl group via an $$S_N2$$ mechanism. Consequently, the ethyl group forms ethyl iodide, while the larger alkyl fragment is converted into the corresponding alcohol.

Thus, the intermediate formed is 2-methylbutan-1-ol, having the structure

$$CH_3-CH_2-CH(CH_3)-CH_2OH.$$

On treatment with concentrated $$H_2SO_4$$ and heat, this alcohol undergoes acid-catalyzed dehydration. Initially, the hydroxyl group is protonated and leaves as water, generating a primary carbocation. This unstable carbocation undergoes a hydride shift to produce a more stable tertiary carbocation.

The rearranged tertiary carbocation then eliminates a proton according to Zaitsev's rule, leading to the formation of the most substituted and thermodynamically most stable alkene.

Hence, the major product obtained is 2-methylbut-2-ene, whose structure is

$$CH_3-CH=C(CH_3)-CH_3.$$

Therefore, the correct answer is

$$CH_3-CH=C(CH_3)-CH_3.$$

To determine which alcohol derivative is unstable in an aqueous base, we evaluate the behavior of each functional group under basic conditions and identify the one that readily undergoes hydrolysis.

An ester $$ROCOCH_3$$ contains a reactive carbonyl group and undergoes base-catalyzed hydrolysis (saponification) in the presence of aqueous alkali. During this process, the hydroxide ion attacks the electrophilic carbonyl carbon, resulting in cleavage of the ester bond to form a carboxylate salt and an alcohol. Hence, esters are unstable in aqueous base.

A benzyl ether $$ROCH_2Ph$$ is generally resistant to hydrolysis under basic conditions because the $$C-O$$ bond is not sufficiently electrophilic for nucleophilic attack by $$OH^-$$.

A tetrahydropyranyl (THP) ether is an acetal-type protecting group. Acetals are stable in basic media and are hydrolyzed only under acidic conditions.

A tert-butyl ether $$ROC(CH_3)_3$$ is also stable in aqueous base and does not undergo hydrolysis under these conditions.

Therefore, among the given alcohol derivatives, only the ester (Option A) is unstable in an aqueous base because it readily undergoes saponification.

Hence, the correct answer is Option A.

1. Step 1: Epoxide Ring Opening under Acidic Conditions ($$\text{HBr}$$, $$\text{SN}_2$$)

   When 1-methyl ethylene oxide is treated with $$\text{HBr}$$, the oxygen atom of the epoxide ring is first protonated by the strong acid to form a highly reactive oxonium ion intermediate. This protonation increases the electrophilicity of both ring carbon atoms.

   Although the reaction follows an $$\text{SN}_2$$-like pathway where the bromide ion ($$\text{Br}^\ominus$$) attacks as a nucleophile, the bond-breaking in a protonated epoxide precedes bond-making. This builds up significant partial positive charge ($$\delta^+$$) on the ring carbons. Because the more substituted secondary ($$2^\circ$$) carbon stabilizes this partial positive charge much better than the primary ($1^\circ$$) carbon, the bromide ion selectively attacks the more substituted secondary carbon:

   $$\text{CH}_3\text{--(CH--O--CH}_2\text{)} \xrightarrow{\text{H}^\oplus} \text{CH}_3\text{--(CH--OH}^\oplus\text{--CH}_2\text{)} \xrightarrow{\text{Br}^\ominus} \text{CH}_3\text{--CH(Br)--CH}_2\text{OH}$$

   This regioselective ring opening yields 1-bromopropan-2-ol as the intermediate.




2. Step 2: Substitution of the Hydroxyl Group with Excess $$\text{HBr}$$

   Because the reaction mixture contains an excess of $$\text{HBr}$$, the remaining secondary alcohol group ($$\text{--OH}$$) in 1-bromopropan-2-ol undergoes a subsequent nucleophilic substitution reaction:

   • The hydroxyl group is protonated by another equivalent of $$\text{HBr}$$ to form an excellent leaving group ($$\text{--OH}_2^\oplus$$).
   • A second bromide ion ($$\text{Br}^\ominus$$) then attacks the carbon, displacing the water molecule.
   $$\text{CH}_3\text{--CH(Br)--CH}_2\text{OH} \xrightarrow{\text{H}^\oplus} \text{CH}_3\text{--CH(Br)--CH}_2\text{OH}_2^\oplus \xrightarrow{\text{Br}^\ominus} \text{CH}_3\text{--CH(Br)--CH}_2\text{Br} + \text{H}_2\text{O}$$

   This converts the secondary alcohol into a vicinal dibromide: 1,2-dibromopropane.

Why Other Options Are Incorrect:

• Halohydrin Intermediates (e.g., 1-bromopropan-2-ol or 2-bromopropan-1-ol):

  Structures that contain a remaining hydroxyl group ($$\text{--OH}$$) alongside a single bromine atom only represent the intermediate species formed after the initial ring opening. They fail to account for the secondary substitution step driven by the specified excess reagent conditions.




• Geminal Dibromides (e.g., 1,1-dibromopropane or 2,2-dibromopropane):

  Geminal dibromides require both halogens to be added to the exact same carbon atom. The epoxide ring opening naturally distributes functional groups across adjacent carbons (carbons 1 and 2), making geminal placement chemically impossible here.

Conclusion:

The combination of regioselective acidic epoxide cleavage followed by acid-catalyzed substitution of the remaining alcohol transforms 1-methyl ethylene oxide completely into 1,2-dibromopropane.

Answer: Option B (1,2-dibromopropane)

1. Electrophilic Addition:

   Dilute sulfuric acid acts as a source of hydronium ions ($$\text{H}_3\text{O}^\oplus$$). The electron-rich carbon-carbon double bonds ($$\text{C=C}$$) undergo electrophilic addition by attacking the protons ($$\text{H}^\oplus$$).




2. Carbocation Stability (Markovnikov's Rule):

   Protonation occurs selectively to form the more stable tertiary ($$3^\circ$$) carbocations rather than less stable secondary carbocations at both alkene sites.




3. Nucleophilic Attack by Water:

   Water molecules ($$\text{H}_2\text{O}$$) present in the dilute medium act as nucleophiles, attacking both of the tertiary carbocations. After deprotonation, hydroxyl ($$-\text{OH}$$) groups are successfully attached to those positions.

Neopentyl alcohol is $$HO{-}CH_2{-}C(CH_3)_3$$ (IUPAC : 2,2-dimethyl-1-propanol). On heating with a protonic acid (conc. $$H_2SO_4$$, $$H_3PO_4$$ etc.) it undergoes acid-catalysed dehydration by the $$E_1$$ mechanism.

Step 1 : Protonation and loss of water
$$HO{-}CH_2{-}C(CH_3)_3 + H^+ \;\longrightarrow\; HOH^+-CH_2{-}C(CH_3)_3$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\longrightarrow\; CH_2^{+}{-}C(CH_3)_3 + H_2O$$
The leaving of water gives a highly unstable primary carbocation.

Step 2 : 1,2-Methyl shift
A neighbouring methyl group migrates with its bonding pair (1,2-alkyl shift) to stabilise the cation:

$$CH_2^{+}{-}C(CH_3)_3 \;\xrightarrow[\text{methyl shift}]{}\; (CH_3)_2C^{+}{-}CH(CH_3)$$
The rearranged carbocation is tertiary, hence far more stable.

Step 3 : Elimination ($$E_1$$) from the tertiary carbocation
The positive carbon is adjacent to three kinds of $$\beta$$-hydrogen:

Case A - Removal of the single hydrogen on the $$CH(CH_3)$$ group
$$\;\;(CH_3)_2C^{+}{-}CH(CH_3) \; \xrightarrow[-H^+]{E_1}\; (CH_3)_2C=CHCH_3$$
This gives the tri-substituted alkene 2-methyl-2-butene.

Case B - Removal of a hydrogen from either of the two equivalent methyl groups attached directly to the carbocation centre
$$\;\;(CH_3)_2C^{+}{-}CH(CH_3) \; \xrightarrow[-H^+]{E_1}\; CH_2=C(CH_3)CH_2CH_3$$
This furnishes the di-substituted alkene 2-methyl-1-butene.

Step 4 : Product distribution
According to Zaitsev’s rule, the more substituted alkene (2-methyl-2-butene) is formed preferentially. Experimentally the ratio is about 85 % (A) : 15 % (B).

Hence
A (85 %) = 2-methyl-2-butene $$[(CH_3)_2C=CHCH_3]$$
B (15 %) = 2-methyl-1-butene $$[CH_2=C(CH_3)CH_2CH_3]$$

Option B which is: 2-methyl-2-butene and 2-methyl-1-butene.

To determine the major product of the given reaction, we analyze the acid-catalyzed dehydration of the substituted cyclohexanone containing a tertiary alcohol group.

When treated with concentrated $$H_2SO_4$$, the reaction proceeds through an E1 elimination mechanism. Initially, the hydroxyl group is protonated by the acid, converting the poor leaving group $$-OH$$ into the better leaving group $$-OH_2^+$$. Loss of water then generates a stable tertiary carbocation at the carbon bearing the hydroxyl group.

The resulting carbocation can undergo elimination by the removal of a proton from either of the adjacent carbon atoms. According to Zaitsev's rule, the preferred pathway is the one that forms the more substituted and thermodynamically more stable alkene.

In this case, elimination can produce either an exocyclic double bond or an endocyclic double bond. Formation of the endocyclic double bond is favored because it generates an $$\alpha,\beta$$-unsaturated ketone, allowing conjugation between the newly formed $$C=C$$ bond and the existing carbonyl group. This conjugated enone system is significantly more stable than the alternative non-conjugated product.

Therefore, the major product is 3-ethylcyclohex-2-en-1-one, which consists of a cyclohexenone ring bearing an ethyl substituent at the $$3$$-position and a conjugated double bond adjacent to the carbonyl group.

Hence, the correct answer is the conjugated enone structure, i.e., 3-ethylcyclohex-2-en-1-one.

Step 1: Functional Groups in Novestrol

Novestrol contains three primary reactive functional groups:

1. A Phenolic Group (-OH attached directly to a benzene ring): Found on the aromatic ring A.
2. An Alcohol Group (Secondary/Tertiary -OH): A tertiary hydroxyl group located on ring D.
3. A Terminal Alkyne Group (-C triple bond CH): Also located on ring D, adjacent to the tertiary alcohol.

Step 2: Testing Reagents Against the Functional Groups

Let's look at the reagents given in Option B and match them to these functional groups:

• Br2 / water (Bromine Water): * The phenolic ring is highly activated due to the resonance donation from the -OH group.
  • When treated with bromine water, it readily undergoes electrophilic aromatic substitution to form a brominated derivative (it can also react with the terminal alkyne group). Therefore, Novestrol gives a positive test with Br2/water.
• ZnCl2 / HCl (Lucas Reagent):
  • This reagent is used to distinguish between primary, secondary, and tertiary alcohols based on how fast they form an insoluble alkyl chloride.
  • Since Novestrol contains an alcohol functional group, it will react with the Lucas reagent.
• FeCl3 (Neutral Ferric Chloride):
  • This is the characteristic diagnostic test for compounds containing a phenolic group.
  • Because Novestrol possesses a phenol ring, adding FeCl3 results in a characteristic colored complex formation (usually violet or green/blue), indicating a successful reaction.

Why the Other Options are Incorrect

• Options A and C (Alcoholic HCN): Novestrol does not contain a carbonyl group (like an aldehyde or ketone) or an unhindered electrophilic center that typically undergoes nucleophilic addition with HCN. Thus, it will not react with alcoholic HCN.
• Options C and D (NaOCl): Sodium hypochlorite (NaOCl) is commonly used in haloform reactions for methyl ketones or specific oxidation reactions that do not selectively or characteristically target the key functional groups of Novestrol in this standard testing context.

We have phenol, $$\mathrm{C_6H_5OH}$$, dissolved in chloroform, $$\mathrm{CHCl_3}$$, and the mixture is treated with aqueous $$\mathrm{NaOH}$$. This set of reagents is characteristic of the Reimer-Tiemann reaction. In this reaction, phenol first forms the phenoxide ion under the basic conditions, and the chloroform generates the dichlorocarbene species $$\mathrm{:CCl_2}$$, which subsequently undergoes electrophilic substitution at the ortho position of the phenoxide ring. On hydrolysis of the intermediate, we obtain o-hydroxybenzaldehyde as the major product.

So, compound $$P$$ is o-hydroxybenzaldehyde (also called salicylaldehyde). Its molecular formula is

$$\mathrm{C_6H_4(OH)(CHO)} \;=\; \mathrm{C_7H_6O_2}.$$

Now we calculate the mass percentage of carbon in this compound. First, we write down the molar masses of all the constituent atoms:

$$\text{Atomic mass of C} = 12 \text{ u}, \qquad \text{Atomic mass of H} = 1 \text{ u}, \qquad \text{Atomic mass of O} = 16 \text{ u}.$$

The total molar mass of $$\mathrm{C_7H_6O_2}$$ is obtained by adding the contributions from every atom:

$$ \begin{aligned} M &= \bigl(7 \times 12\bigr) + \bigl(6 \times 1\bigr) + \bigl(2 \times 16\bigr) \\ &= 84 + 6 + 32 \\ &= 122 \text{ u}. \end{aligned} $$

The mass contributed by carbon alone is

$$m_C = 7 \times 12 = 84 \text{ u}.$$

Using the formula for percentage composition,

$$ \text{Percentage of C} = \frac{m_C}{M} \times 100 = \frac{84}{122} \times 100 = 68.85\% \approx 69\%. $$

So, the answer is $$69\%$$.

We have to select that compound which fulfils both of the following conditions:

(i) it must react with the Grignard reagent $$\mathrm{C_2H_5MgBr}$$ (ethyl-magnesium bromide), and

(ii) it must decolourize an aqueous bromine solution (bromine water).

First we recall the general behaviour of a Grignard reagent. Ethyl-magnesium bromide is represented as $$\mathrm{C_2H_5^- \,\,Mg^{2+}Br^-}$$, so the carbon attached to magnesium behaves as a very strong base and nucleophile. Therefore, any compound containing an acidic hydrogen (-OH, -NH, -SH, etc.) instantly reacts with a Grignard reagent according to

$$\mathrm{R{-}H \;+\; C_2H_5MgBr \;\longrightarrow\; R^-\,MgBr^+ \;+\; C_2H_6}\,,$$

where $$\mathrm{R{-}H}$$ denotes the acidic compound and the ethyl group becomes $$\mathrm{C_2H_6}$$ (ethane).

Now we recall the test with bromine water. Aqueous $$\mathrm{Br_2}$$ is ruddy brown; it gets decolourised by

  • compounds containing an electron-rich double bond $$\mathrm{C=C}$$ (alkenes, enols), and

  • compounds whose aromatic ring is strongly activated, for example phenol, which undergoes electrophilic bromination to give 2,4,6-tribromophenol:

$$\mathrm{C_6H_5OH + 3\,Br_2(aq) \;\longrightarrow\; C_6H_2Br_3OH + 3\,HBr}$$

and the reddish colour of $$\mathrm{Br_2}$$ disappears.

Let us check each option qualitatively:

• Option A carries a $$\mathrm{C\equiv N}$$ and a carbonyl oxygen. It does react with a Grignard reagent (nitrile and carbonyl are both reactive), but neither group can decolourise bromine water because there is no strongly activated aromatic ring nor an alkenic double bond able to add bromine in water. So Option A fails condition (ii).

• Option B contains a nitrile $$\mathrm{-C\!\equiv N}$$ and an ester $$\mathrm{-CO_2CH_3}$$. Again, it reacts with the Grignard reagent, but an ester/nitrile does not affect bromine water. Hence Option B also fails condition (ii).

• Option C shows an ether $$\mathrm{-OCH_3}$$ and a carbon-carbon double bond $$\mathrm{CH=CH_2}$$. The double bond definitely decolourises bromine water, satisfying condition (ii). However, simple alkenes do not possess an acidic hydrogen and therefore do not react with ethyl-magnesium bromide. Thus Option C fails condition (i).

• Option D bears a hydroxyl group $$\mathrm{-OH}$$, and from the context of typical JEE questions this -OH is almost always on an aromatic ring (phenol). We analyse it quantitatively:

Reaction with Grignard reagent

Applying the general acid-base formula written earlier, we let $$\mathrm{ArOH}$$ represent phenol:

$$\mathrm{ArOH + C_2H_5MgBr \;\longrightarrow\; ArO^-\,MgBr^+ + C_2H_6}.$$

The hydrogen of the hydroxyl group is sufficiently acidic (pKₐ ≈ 10), so it is abstracted, giving magnesium phenoxide and ethane gas. Condition (i) is thus satisfied.

Decolourisation of bromine water

The aromatic ring of phenol is strongly activated by the $$\mathrm{-OH}$$ group. When bromine water is added, electrophilic substitution occurs at the ortho and para positions:

$$\mathrm{ArOH + 3\,Br_2 (aq) \;\longrightarrow\; Ar(Br)_3OH + 3\,HBr},$$

and the brown colour of $$\mathrm{Br_2}$$ vanishes. Hence condition (ii) is also satisfied.

Therefore only the compound possessing the hydroxyl (phenolic) group fulfils both required properties.

Hence, the correct answer is Option 4.

More Pka means more basic nature of the compound. The +M and +I effect increase the value of PKa.

This reaction involves Concentrated HBr (excess) and heat, which will attack two different functional groups on the molecule: the ether and the alkene.

Concentrated HBr and heat will break the aryl-alkyl ether bond.

• The Oxygen gets protonated, and the Bromide ion Br- attacks the smaller methyl group via an SN2 mechanism.
• Result: The group is converted into a Phenol, and methyl bromide is released as a by-product.

The alkene group undergoes an electrophilic addition reaction following Markovnikov's rule.

• The H+ from HBr adds to the terminal carbon to form the most stable carbocation (the benzylic carbocation).
• The Br- then attacks this stable carbocation.
• Result: The -CH=CH2 group is converted into a -CH(Br)CH3 group.

Firstly an E2 elimination takes place leading to the formation of an alkene with C1-C2 carbons. Then oxymercurataion and de-mercuration reaction adds an -OH group at C1 giving the desired product mentioned in option A.

1. Ether Cleavage ($$\text{S}_{\text{N}}2$$):

   The oxygen atom of the aryl-alkyl ether ($$-\text{O--CH}_3$$) is protonated by $$\text{HBr}$$. The bromide ion ($$\text{Br}^\ominus$$) attacks the smaller alkyl group ($$-\text{CH}_3$$) via an $$\text{S}_{\text{N}}2$$ pathway, breaking the $$\text{O--CH}_3$$ bond to yield a phenol ($$-\text{OH}$$) and bromomethane ($$\text{CH}_3\text{Br}$$). The sturdier sp2 aryl $$\text{C--O}$$ bond remains completely intact.




2. Electrophilic Addition (Markovnikov's Rule):

   The alkene double bond ($$-\text{CH}=\text{CH}--\text{CH}_3$$) undergoes protonation to generate the more stable secondary carbocation adjacent to the benzene ring, which is heavily stabilized by benzylic resonance. Subsequent attack by the bromide ion ($$\text{Br}^\ominus$$) completes the Markovnikov addition.

We need to determine the mechanism and major product for the reaction of sodium phenoxide with methyl iodide ($$\text{CH}_3\text{I}$$).

Reaction Mechanism: Williamson Ether Synthesis

1. Nucleophilic Attack via $$\text{S}_\text{N}2$$ Pathway:

   The reaction between sodium phenoxide ($$\text{Ph--O}^\ominus\text{Na}^\oplus$$) and methyl iodide ($$\text{CH}_3\text{I}$$) is a classic example of a Williamson ether synthesis. The strongly nucleophilic phenoxide oxygen atom attacks the relatively unhindered primary carbon atom of the methyl iodide electrophile:

   $$\text{Ph--O}^\ominus + \text{CH}_3\text{--I} \xrightarrow{\text{S}_\text{N}2} \text{Ph--O--CH}_3 + \text{I}^\ominus$$



2. Transition State and Displacement:

   As the nucleophilic oxygen atom forms a new covalent bond with the methyl group, the iodide ion simultaneously departs as an excellent leaving group ($$-\text{NaI}$$), yielding 2-methoxy-1-methylbenzene (or anisole depending on ring substitutions).

Conclusion:

The reaction cleanly proceeds via a bimolecular nucleophilic substitution ($$\text{S}_\text{N}2$$) process to give the corresponding alkyl aryl ether.

Answer: Option B (2-methoxy-1-methylbenzene formed via S$$_\text{N}$$2 reaction)

1. Electrophilic Aromatic Substitution:

   The hydroxyl group ($$-\text{OH}$$) is a powerful activating group that directs incoming electrophiles to its ortho and para positions. Both positions ortho to the $$-\text{OH}$$ group are open and readily undergo bromination.




2. Ipso-Bromodebenzylation (Ipso-Substitution):

   The para position is already occupied by a sulfonic acid group ($$-\text{SO}_3\text{H}$$). Because the sulfonic acid group is an excellent leaving group, a third bromine electrophile attacks this occupied position (ipso-attack), displacing the sulfur trioxide ($$-\text{SO}_3$$) group.

Conclusion:

The combined ortho-bromination and para-desulfonation results in the complete conversion of the substrate into a symmetrically trisubstituted phenol.

Answer: Option A — 2,4,6-Tribromophenol

1. Step 1: Nucleophilic Substitution with $$\text{HCl}$$

   The aliphatic hydroxyl group ($$-\text{OH}$$) undergoes protonation followed by nucleophilic substitution by the chloride ion ($$\text{Cl}^\ominus$$). This replaces the alcohol with a chlorine atom, forming an alkyl chloride intermediate.




2. Step 2: Intramolecular Friedel-Crafts Alkylation

   Anhydrous $$\text{AlCl}_3$$ abstracts the chloride ion from the intermediate to generate a stable carbocation. This carbocation acts as an electrophile and undergoes a rapid intramolecular electrophilic substitution at the highly activated position ortho to the phenolic hydroxyl group, closing the chain to form a fused five-membered ring (indane core).

1. Step 1: Base-Mediated Dehydrohalogenation ($$\alpha,\beta$$-Elimination)

   The reaction begins with the elimination of hydrogen bromide ($$-\text{HBr}$$) from the cyclic bromo-ketone precursor. A base abstracts a proton from the $$\beta$$-position relative to the leaving group ($$\text{--Br}$$), triggering an $$\alpha,\beta$$-elimination that forms a stable, conjugated diene intermediate (a cyclohexadienone derivative):

   $$\text{Bromo-ketone Precursor} \xrightarrow{-\text{HBr}} \text{Cyclohexadienone Intermediate [Conjugated diene, stable]}$$



2. Step 2: Tautomerization (Enolisation to Attain Aromaticity)

   The non-aromatic cyclohexadienone intermediate contains a highly acidic sp3 hybridized carbon bearing two hydrogen atoms adjacent to the carbonyl group ($$\text{C=O}$$). To relieve ring strain and gain immense thermodynamic stability, the compound undergoes rapid enolisation (keto-enol tautomerization). This restores the fully delocalized $$\pi$$-system of the benzene ring:

   $$\text{Cyclohexadienone (Keto form)} \rightleftharpoons \text{Phenol Derivative (Enol form) [Very stable, aromatic]}$$

Conclusion:

The driving force of the entire transformation is the immense resonance energy gained through aromatization, shifting the equilibrium completely toward the formation of the phenol derivative.

Answer: The option displaying the fully aromatic phenol derivative as the major product (Option B).

1. Step 1: Radical Oxidation to Cumene Hydroperoxide

   When cumene is reacted with oxygen ($$\text{O}_2$$) in the presence of air, a free-radical oxidation process occurs selectively at the highly reactive tertiary benzylic carbon atom. This leads to the formation of cumene hydroperoxide as a stable intermediate:

   $$\text{C}_6\text{H}_5\text{--CH(CH}_3)_2 + \text{O}_2 \rightarrow \text{C}_6\text{H}_5\text{--C(CH}_3)_2\text{--O--O--H}$$



2. Step 2: Acid-Catalyzed Cleavage (Hock Rearrangement)

   Subsequent treatment of cumene hydroperoxide with dilute hydrochloric acid ($$\text{dil. HCl}$$ or $$\text{H}^\oplus$$) protonates the terminal oxygen atom. This triggers a simultaneous 1,2-phenyl migration to the electron-deficient oxygen atom with the loss of a water molecule.

   The resulting oxacarbocation intermediate is then attacked by water, undergoing proton transfer and bond cleavage to yield two highly valuable chemical industrial products: Phenol and Acetone (propan-2-one).

   $$\text{C}_6\text{H}_5\text{--C(CH}_3)_2\text{--O--O--H} \xrightarrow{\text{H}^\oplus, \, \text{H}_2\text{O}} \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{--CO--CH}_3$$

Conclusion:

The industrial preparation of phenol from cumene cleanly delivers phenol as the principal product along with acetone as a commercially significant co-product.

Answer: Option C — Phenol and Acetone

1. Aryl-Alkyl Ether Cleavage ($$\text{S}_{\text{N}}2$$):

   The methoxy oxygen ($$-\text{OCH}_3$$) attached to the benzene ring is protonated. Iodide ($$\text{I}^\ominus$$) attacks the methyl group via an $$\text{S}_{\text{N}}2$$ pathway, cleaving the alkyl $$\text{C--O}$$ bond to produce a phenol ($$-\text{OH}$$) group and iodomethane ($$\text{CH}_3\text{I}$$).




2. Aliphatic Ether Cleavage ($$\text{S}_{\text{N}}1$$):

   The secondary aliphatic ether nitrogen-stabilized carbon center ($$-\text{CH}(\text{CN})--\text{O}--\text{CH}(\text{CH}_3)_2$$) undergoes protonation. Because it can form a highly stable, bulky tertiary carbocation ($$-\text{C}^\oplus(\text{CH}_3)_2$$) on the isopropyl side, the bond breaks via an $$\text{S}_{\text{N}}1$$ pathway to yield 2-iodopropane ($$\text{I--CH}(\text{CH}_3)_2$$) and a cyanohydrin intermediate ($$-\text{CH}(\text{CN})\text{OH}$$).

Iodoform test would oxidise the alcohol to carboxylic acid.

Williamson Ether Synthesis

This reaction follows the mechanism of a classic Williamson Ether Synthesis, which proceeds via an $$\text{S}_\text{N}2$$ nucleophilic substitution pathway:

1. Deprotonation: Potassium carbonate ($$\text{K}_2\text{CO}_3$$) acts as a mild base. It selectively removes the acidic phenolic proton ($$-\text{OH}$$) from m-cresol to generate a highly nucleophilic m-cresolate (phenoxide) anion.
2. Nucleophilic Attack: The phenoxide oxygen atom attacks the electrophilic $$\text{sp}^3$$ carbon attached to the bromine atom in propargyl bromide ($$\text{HC}\equiv\text{C--CH}_2\text{Br}$$).
3. Leaving Group Departure: The bromide ion ($$\text{Br}^-$$) leaves simultaneously, forming a new carbon-oxygen ($$\text{C--O}$$) ether linkage.

Detailed Evaluation of Each Option:

• Option A: Electrophilic Ring Substitution Products

  This statement is incorrect. Under these conditions, the base deprotonates the oxygen atom, making it the most nucleophilic site. Ring alkylation (Friedel-Crafts style) does not compete with the rapid $$\text{S}_\text{N}2$$ displacement on oxygen.




• Option B (Correct Matrix): Propargyl m-Tolyl Ether

  This statement is correct. The nucleophilic phenoxide ion directly bonds with the propargyl group ($$-\text{CH}_2\text{--C}\equiv\text{CH}$$), giving an ether product where the intact m-cresol ring skeleton is preserved with a newly appended propargyloxy side chain: $$\text{CH}_3\text{--C}_6\text{H}_4\text{--O--CH}_2\text{--C}\equiv\text{CH}$$.




• Option C & D: Elimination or Alkyne Addition Products

  These statements are incorrect. Mild bases like $$\text{K}_2\text{CO}_3$$ in acetone do not cause elimination reactions on terminal propargyl halides, nor is there any active reagent or catalyst present to add across the triple bond ($$\text{C}\equiv\text{C}$$).

Conclusion:

The reaction cleanly transforms the phenolic group into an ether linkage through an $$\text{S}_\text{N}2$$ substitution without altering the aromatic ring or the terminal alkyne structure.

Answer: Option B — Propargyl m-tolyl ether

When an alcohol is dehydrated under acidic conditions, the $$-OH$$ group is protonated and lost as water, forming a carbocation intermediate which then loses a proton to give an alkene. The ease of dehydration depends on (i) the stability of the carbocation intermediate or (ii) whether a special driving force (like conjugation) stabilises the product.

Let us analyse each compound:

Option A: 4-Hydroxypentan-2-one

Structure: $$CH_3 - CO - CH_2 - CH(OH) - CH_3$$

The carbonyl is at C-2 and the $$-OH$$ is at C-4. The C-4 is the $$\beta$$-carbon relative to the carbonyl group (C-2 is carbonyl, C-3 is $$\alpha$$, C-4 is $$\beta$$).

This is a $$\beta$$-hydroxy ketone. Under acidic conditions, $$\beta$$-hydroxy carbonyl compounds undergo very facile dehydration because the product is an $$\alpha,\beta$$-unsaturated ketone (a conjugated enone):

$$CH_3 - CO - CH = CH - CH_3$$

This product (pent-3-en-2-one) is stabilised by conjugation between the $$C=C$$ double bond and the $$C=O$$ group. The extended conjugation provides a strong thermodynamic driving force, making this dehydration very easy.

Option B: 3-Hydroxypentan-2-one

Structure: $$CH_3 - CO - CH(OH) - CH_2 - CH_3$$

Here the $$-OH$$ is at C-3, which is the $$\alpha$$-carbon relative to the carbonyl at C-2. This is an $$\alpha$$-hydroxy ketone. Dehydration would need to remove $$-OH$$ from C-3 and $$-H$$ from either C-2 (which is the carbonyl carbon — not possible as it has no H) or C-4. Eliminating between C-3 and C-4 gives $$CH_3-CO-C(=CH-CH_3)$$, which would place a double bond at C-3=C-4. This is not conjugated with the carbonyl group in the favourable way that the $$\beta$$-hydroxy ketone product is, so the driving force is weaker.

Option C: 1-Pentanol

Structure: $$CH_3CH_2CH_2CH_2CH_2OH$$

This is a simple primary alcohol. Dehydration of primary alcohols is the most difficult among 1°, 2°, and 3° alcohols because the resulting primary carbocation is very unstable. The reaction requires harsh conditions (high temperature, strong acid). No special stabilisation of the product occurs.

Option D: 2-Hydroxycyclopentanone

This is an $$\alpha$$-hydroxy ketone ($$-OH$$ on the carbon adjacent to the carbonyl). Dehydration would lead to a cyclopentenone, but the $$\alpha$$-hydroxy position makes it less facile than $$\beta$$-elimination because the geometry in the five-membered ring adds strain, and $$\alpha$$-elimination is less favourable than $$\beta$$-elimination.

Among all the options, 4-hydroxypentan-2-one (Option A) is a $$\beta$$-hydroxy ketone which undergoes the easiest acid-catalysed dehydration due to the formation of a conjugated $$\alpha,\beta$$-unsaturated ketone product.

The correct answer is Option A.

The given compound contains a phenolic$$\ \ -OH$$ and an aliphatic$$\ \ -OH$$

$$K_2CO_3$$ is a weak base. It will selectively deprotonate the phenolic$$\ \ -OH$$ as the resulting phenoxide ion $$Ar\ -\ O^{-\ }$$ is stabilized by resonance, while the aliphatic$$\ \ -OH$$ remains mostly undissociated.

The phenoxide ion then acts as a nucleophile and performs an $$S_N2$$ reaction with $$CH_3I$$ to form $$Ar-OCH_3$$, which is the compound given in option A.

We begin by recalling what the Bouveault-Blanc reduction is. This named reaction is classically carried out when an ester is treated with metallic sodium in the presence of an alcohol, most commonly ethanol. The overall transformation converts the ester functional group into the corresponding primary alcohols.

In other words, if we take a generic ester written as $$RCOOR'$$ and subject it to the reagents $$Na$$ (metallic sodium) and $$C_2H_5OH$$ (ethanol), the reaction proceeds as

$$RCOOR' \;+\; 4\,Na \;+\; 4\,C_2H_5OH \;\rightarrow\; RCH_2OH \;+\; R'OH \;+\; 4\,C_2H_5ONa.$$

So, the key points we must match with the options are:

1. The substrate must be an ester.

2. The reducing system must be $$Na/C_2H_5OH$$ (sodium in an alcoholic solvent).

Now we look at the given options one by one:

Option A mentions the reduction of an acyl halide with $$H_2/Pd$$. This is hydrogenation and is not the Bouveault-Blanc reduction.

Option B mentions the reduction of an anhydride with $$LiAlH_4$$. That is a lithium aluminium hydride reduction, again unrelated to Bouveault-Blanc.

Option C states the reduction of an ester with $$Na/C_2H_5OH$$. This matches exactly with the reaction we have just described.

Option D talks about reducing a carbonyl compound with $$Na/Hg$$ and $$HCl$$. That is the Clemmensen reduction, not Bouveault-Blanc.

Only Option C aligns perfectly with the definition and reagents of the Bouveault-Blanc reduction.

Hence, the correct answer is Option C.

To prepare allyl phenyl ether, which has the structure CH2=CH-CH2-O-C6H5, we use the Williamson ether synthesis. This method involves the reaction of an alkoxide ion with an alkyl halide to form an ether. The general reaction is R-ONa + R'-X → R-O-R' + NaX, where R and R' are alkyl or aryl groups.

In this case, allyl phenyl ether has two parts: the allyl group (CH2=CH-CH2-) and the phenyl group (C6H5-). We need to choose the correct combination of reactants from the options.

Let's evaluate each option:

Option A: CH2=CH-Br + C6H5-CH2-ONa
Here, CH2=CH-Br is vinyl bromide. Vinyl halides are unreactive in nucleophilic substitution reactions because the carbon is sp2 hybridized, and the C-Br bond has partial double bond character due to resonance. This makes it difficult for nucleophiles to attack. The product would be CH2=CH-O-CH2-C6H5 (vinyl benzyl ether), not allyl phenyl ether. So, this option is incorrect.

Option B: C6H5-CH=CH-Br + CH3-ONa
Here, C6H5-CH=CH-Br is styryl bromide (a vinyl halide). Vinyl halides are unreactive for the same reasons as in option A. The product would be C6H5-CH=CH-O-CH3 (styryl methyl ether), not allyl phenyl ether. So, this option is incorrect.

Option C: CH2=CH-CH2-Br + C6H5ONa
Here, CH2=CH-CH2-Br is allyl bromide, and C6H5ONa is sodium phenoxide. Allyl bromide is a primary alkyl halide but is highly reactive due to resonance stabilization of the intermediate carbocation in SN1 or the transition state in SN2 reactions. Sodium phenoxide is a strong nucleophile. The reaction proceeds as:
CH2=CH-CH2-Br + C6H5ONa → CH2=CH-CH2-O-C6H5 + NaBr
The product is allyl phenyl ether (CH2=CH-CH2-O-C6H5). This reaction is feasible under heating conditions.

Option D: C6H5Br + CH2=CH-CH2-ONa
Here, C6H5Br is bromobenzene (an aryl halide), and CH2=CH-CH2-ONa is sodium allyloxide. Aryl halides like bromobenzene do not undergo nucleophilic substitution easily under normal heating conditions because the carbon is sp2 hybridized, and the C-Br bond is strong due to resonance. Special conditions (e.g., high temperature and pressure) are required for substitution. The product would be C6H5-O-CH2-CH=CH2 (allyl phenyl ether), but the reaction does not occur readily. So, this option is incorrect.

Therefore, only option C provides a feasible route to synthesize allyl phenyl ether via Williamson ether synthesis.

Hence, the correct answer is Option C.

To determine which compound will not be soluble in sodium bicarbonate (NaHCO₃), we must recall that sodium bicarbonate is a weak base and reacts only with acids stronger than carbonic acid (H₂CO₃). Carbonic acid has a pKa value of approximately 6.3. Therefore, any acid with a pKa less than or equal to 6.3 will react with NaHCO₃, producing carbon dioxide gas and dissolving in the solution. Acids with pKa greater than 6.3 are weaker than carbonic acid and will not react with NaHCO₃, hence they are insoluble.

Now, let's evaluate each option:

Option A: 2,4,6-Trinitrophenol (Picric acid)
This compound has three nitro groups attached to the phenol ring, which are strong electron-withdrawing groups. These groups significantly increase the acidity. The pKa of picric acid is about 0.4, which is much less than 6.3. Therefore, it is a strong acid and will react with NaHCO₃, making it soluble.

Option B: Benzene sulphonic acid
Sulphonic acids are very strong acids. The pKa of benzene sulphonic acid is approximately -6.5, which is far less than 6.3. Hence, it will readily react with NaHCO₃ and dissolve.

Option C: o-Nitrophenol
This is a phenol derivative with a nitro group at the ortho position. While the nitro group is electron-withdrawing and increases acidity compared to phenol (pKa ≈ 10), the pKa of o-nitrophenol is about 7.23. Since 7.23 is greater than 6.3, o-nitrophenol is a weaker acid than carbonic acid. Thus, it will not react with NaHCO₃ and will not dissolve.

Option D: Benzoic acid
Benzoic acid is a carboxylic acid. Its pKa is approximately 4.2, which is less than 6.3. Therefore, it is stronger than carbonic acid and will react with NaHCO₃, leading to solubility.

Based on the pKa values, o-Nitrophenol (Option C) is the only compound that will not dissolve in sodium bicarbonate because it is too weak an acid to react with it.

Hence, the correct answer is Option C.

$$-CH_3$$ (Option A): This is an electron-donating group through the $$+I$$ (inductive) effect and hyperconjugation. It destabilizes the phenoxide ion.

$$-OCH_3$$ (Option B): It can donate into the ring via the $$+M$$ (mesomeric) effect. This makes it a strong electron-donating group that destabilizes the ion.

$$-COCH_3$$ (Option C): This acetyl group is a strong electron-withdrawing group. It exerts both a $$-I$$ effect and a powerful $$-M$$ (resonance withdrawing) effect.

Because it is at the para-position, the negative charge from the phenoxide oxygen can be delocalized through the ring directly onto the oxygen of the carbonyl group. This is the most effective form of stabilization.

We begin with the primary alcohol $$R - CH_2 - OH$$ and we desire to oxidise it only up to the aldehyde stage, $$R - CHO$$. In other words, we want

$$R - CH_2 - OH \;\;\longrightarrow\;\; R - CHO$$

without further oxidation to the corresponding carboxylic acid $$R - COOH$$.

Whenever a primary alcohol is treated with a strong oxidising agent under aqueous or strongly acidic conditions, the reaction generally proceeds beyond the aldehyde and gives the acid. Therefore, we must analyse the strength and selectivity of each oxidising reagent listed in the options.

Option A: $$KMnO_4$$

The permanganate ion $$MnO_4^-$$ in acidic, neutral, or alkaline medium is a very strong oxidant. The accepted oxidation pathway for a primary alcohol is

$$R - CH_2 - OH \;\;\xrightarrow{KMnO_4 / \text{heat}} \;\; R - COOH$$

Thus, $$KMnO_4$$ would over-oxidise the aldehyde further to the acid. Hence it is unsuitable when we intend to stop at the aldehyde.

Option B: $$K_2Cr_2O_7$$ (acidic dichromate)

The chromate-dichromate system $$Cr_2O_7^{2-}$$ in acidic medium is also a powerful oxidising agent. The general result is

$$R - CH_2 - OH \;\;\xrightarrow{K_2Cr_2O_7 / H^+}\;\; R - COOH$$

Again, the aldehyde intermediate is not isolated because it is further oxidised in the same reaction mixture. Therefore $$K_2Cr_2O_7$$ is not selective enough for our requirement.

Option C: $$CrO_3$$ (chromic anhydride, Jones oxidation)

When $$CrO_3$$ is used in aqueous sulphuric acid (Jones reagent), the medium remains strongly acidic and water is present. Under these conditions, a primary alcohol undergoes

$$R - CH_2 - OH \;\;\xrightarrow{CrO_3 / H_2SO_4 / H_2O}\;\; R - COOH$$

Again, the aldehyde stage is not the final product. So $$CrO_3$$ in its common form is also unsuitable.

Option D: PCC (Pyridinium Chlorochromate)

PCC is a complex of $$CrO_3$$ with hydrochloric acid and pyridine. The crucial feature is that the reagent is used in anhydrous organic solvents such as dichloromethane, so there is no bulk water present. The absence of water prevents the further oxidation of the aldehyde to the carboxylic acid. The sequence is therefore limited to

$$R - CH_2 - OH \;\;\xrightarrow{\text{PCC, CH}_2Cl_2}\;\; R - CHO$$

The aldehyde does not hydrate (no water), and the reaction stops at the required oxidation level. Thus PCC is regarded as the most selective reagent for converting a primary alcohol into an aldehyde.

After examining all four reagents, only PCC (Option D) satisfies the criterion of stopping the oxidation at the aldehyde stage.

Hence, the correct answer is Option D.

The Williamson synthesis of ether is a method used to prepare ethers. It involves the reaction between an alkoxide ion and an alkyl halide. Let me explain this step by step.

Consider the general reaction: an alkoxide ion, represented as $$ RO- $$, reacts with an alkyl halide, represented as $$ R'X $$, where R and R' are alkyl groups (which can be the same or different), and X is a halogen atom like chlorine, bromine, or iodine. The reaction produces an ether and a salt. The chemical equation is:

$$ RO- + R'X -> ROR' + X- $$

Now, let's analyze the mechanism. The alkoxide ion $$ RO- $$ has a negatively charged oxygen atom, making it electron-rich. This means it acts as a nucleophile (a species that donates an electron pair). The alkyl halide $$ R'X $$ has a polar carbon-halogen bond, where the carbon atom is partially positive (electrophilic) due to the higher electronegativity of the halogen. The nucleophile ($$ RO- $$) attacks the electrophilic carbon atom of the alkyl halide. This leads to the breaking of the carbon-halogen bond and the formation of a new carbon-oxygen bond. As a result, the halogen atom (X) is displaced as a halide ion.

This process involves the substitution of the halogen atom in the alkyl halide by the alkoxy group (OR). Since the attacking species is a nucleophile, and it replaces another group (the halide), this reaction is classified as a nucleophilic substitution reaction.

Now, comparing this with the given options:

Option A: Nucleophilic substitution - This matches the mechanism described above.

Option B: Electrophilic addition - This involves an electrophile adding to a double bond, which does not occur here as there is no unsaturated bond in the reactants.

Option C: Nucleophilic addition - This involves a nucleophile adding across a double bond, typically seen in carbonyl compounds, but here we have substitution, not addition.

Option D: Electrophilic substitution - This is common in aromatic compounds where an electrophile replaces a hydrogen atom, but Williamson synthesis involves aliphatic compounds and nucleophilic attack.

Therefore, Williamson synthesis is a nucleophilic substitution reaction.

Hence, the correct answer is Option A.

$$\text{C}_6\text{H}_5\text{ONa} + \text{CO}_2 \xrightarrow[5\text{ Atm}]{125^\circ\text{C}} \text{C}_6\text{H}_4(\text{OH})(\text{COONa}) \quad \text{[Sodium Salicylate (B)]}$$

$$\text{C}_6\text{H}_4(\text{OH})(\text{COONa}) \xrightarrow{\text{Ac}_2\text{O } / \text{ H}^+} \text{C}_6\text{H}_4(\text{OCOCH}_3)(\text{COOH}) + \text{CH}_3\text{COOH} + \text{Na}^+$$

We start by testing each of the four statements one by one, because the question asks us to find the single statement that is not correct.

Statement A says: “Alcohols are weaker acids than water.”

To compare acidities we use the definition $$pK_a = -\log K_a.$$ A larger $$pK_a$$ means a smaller $$K_a$$ and therefore a weaker acid. Experimentally we have

$$pK_a(\text{H}_2\text{O}) \approx 15.7,$$ $$pK_a(\text{ROH})$$ (for a simple primary alcohol such as ethanol) $$\approx 16\text{--}18.$$

Because $$16-18 \gt 15.7,$$ the $$pK_a$$ of an alcohol is larger, so its $$K_a$$ is smaller, hence an alcohol is indeed a weaker acid than water. Therefore Statement A is correct.

Statement B claims: “Acid strength of alcohols decreases in the order $$\text{RCH}_2\text{OH} \gt \text{R}_2\text{CHOH} \gt \text{R}_3\text{COH}.$$”

Acidity depends on the stability of the conjugate base. After deprotonation an alkoxide ion $$\text{RO}^-$$ is formed. Alkyl groups are electron-donating by the inductive effect $$\left(+I\right).$$ More alkyl groups push electron density toward oxygen, destabilising the negative charge. Hence:

• A primary alkoxide $$(\text{RCH}_2\text{O}^-)$$ has one $$+I$$ group.
• A secondary alkoxide $$(\text{R}_2\text{CHO}^-)$$ has two $$+I$$ groups and is less stable.
• A tertiary alkoxide $$(\text{R}_3\text{CO}^-)$$ has three $$+I$$ groups and is the least stable.

Lower stability of the conjugate base means lower acidity of the conjugate acid, so the acid strength indeed falls in the sequence primary > secondary > tertiary. Hence Statement B is correct.

Statement C says: “Carbon-oxygen bond length in methanol, $$\text{CH}_3\text{OH},$$ is shorter than that of the C-O bond length in phenol.”

Let us recall typical experimental bond lengths:

• In methanol (a simple aliphatic alcohol) the $$\text{C-O}$$ bond is a pure single bond; its length is about $$143\ \text{pm}.$$
• In phenol, resonance causes partial double-bond character between the ring carbon and oxygen:

$$\text{Ar-O-H} \;\;\;\rightleftharpoons\;\;\; \text{Ar}=\text{O}^- -H^+$$

Because of this partial double-bond character the $$\text{C-O}$$ bond in phenol is shorter, about $$136\ \text{pm}.$$ Numerically, $$136\ \text{pm} \lt 143\ \text{pm},$$ so the bond in phenol is the shorter one. Statement C claims the opposite, therefore Statement C is not correct.

Statement D states: “The bond angle O-C-H in methanol is 108.9°.”

In methanol the carbon atom is $$sp^3$$-hybridised. An ideal $$sp^3$$ angle is $$109.5^\circ.$$ Slight differences arise because an $$\text{O-C}$$ bond and $$\text{C-H}$$ bonds are not identical in size or repulsion. Experimental gas-phase measurements give $$\angle(\text{O-C-H}) \approx 108.9^\circ,$$ well within normal error of the ideal tetrahedral value. Therefore Statement D is correct.

Summarising our findings:

• A is correct, • B is correct, • C is incorrect, • D is correct.

Hence, the only statement that is not correct is Statement C, which corresponds to Option 3.

Hence, the correct answer is Option C.

In the Victor-Meyer's test, alcohols are classified as primary (1°), secondary (2°), or tertiary (3°) based on the colour they produce. The test involves several steps:

First, the alcohol is converted to an alkyl iodide using iodine and red phosphorus. Then, the alkyl iodide reacts with silver nitrite to form a nitroalkane. The nitroalkane is treated with nitrous acid (generated from sodium nitrite and hydrochloric acid), and finally, the mixture is made alkaline with sodium hydroxide.

The colour observed depends on the type of alcohol:

For a primary alcohol (1°), the nitroalkane formed is primary. This primary nitroalkane reacts with nitrous acid to form a nitrolic acid. When treated with sodium hydroxide, the nitrolic acid gives a red colour.

For a secondary alcohol (2°), the nitroalkane formed is secondary. This secondary nitroalkane reacts with nitrous acid to form a pseudonitrol (a nitroso compound), which appears blue.

For a tertiary alcohol (3°), no reaction occurs in the initial steps because tertiary alcohols undergo dehydration to form alkenes instead of alkyl iodides. Therefore, no nitro compound is formed, and the mixture remains colourless.

So, the colours given by 1°, 2°, and 3° alcohols are red, blue, and colourless, respectively.

Now, comparing with the options:

Option A: Red, colourless, blue → Incorrect, as it assigns colourless to 2° and blue to 3°.

Option B: Red, blue, colourless → Correct, as it matches red for 1°, blue for 2°, and colourless for 3°.

Option C: Colourless, red, blue → Incorrect, as it assigns colourless to 1° and blue to 3°.

Option D: Red, blue, violet → Incorrect, as it assigns violet to 3° instead of colourless.

Hence, the correct answer is Option B.

The rate of dehydration of alcohols to form alkenes depends on the mechanism and the stability of the intermediate or transition state. Dehydration typically occurs via an acid-catalyzed elimination reaction (E1 or E2 mechanism), where the rate-determining step involves the formation of a carbocation for E1 or the stability of the transition state for E2.

First, recall the stability order of carbocations: tertiary (3°) carbocations are more stable than secondary (2°), which are more stable than primary (1°), and methyl carbocation (CH3+) is the least stable. This stability directly affects the rate in E1 mechanisms because the formation of a more stable carbocation is faster.

For tertiary alcohols (3°), dehydration proceeds via E1 mechanism, forming a stable tertiary carbocation, leading to a fast reaction. Secondary alcohols (2°) also undergo E1 but form a less stable secondary carbocation, so the rate is slower than tertiary but faster than primary. Primary alcohols (1°) usually undergo E2 mechanism because primary carbocations are highly unstable. In E2, the reaction rate depends on the stability of the transition state, which is influenced by the ability to remove a β-hydrogen. Primary alcohols have fewer electron-donating groups, making the transition state less stable, hence slower dehydration compared to secondary and tertiary.

Now, consider methanol (CH3OH). It is a primary alcohol but lacks a β-carbon and β-hydrogen. Without a β-hydrogen, intramolecular dehydration to form an alkene is impossible. Methanol can undergo intermolecular dehydration to form dimethyl ether under specific conditions, but this is not the same as alkene formation. Therefore, for alkene production, methanol has no rate or an extremely slow rate.

Comparing the rates:

• Tertiary alcohols dehydrate fastest due to stable carbocation formation.
• Secondary alcohols are next, slower than tertiary but faster than primary.
• Primary alcohols (with β-hydrogens, e.g., ethanol) dehydrate slowly via E2.
• Methanol cannot dehydrate to an alkene, so it is the slowest.

Thus, the order of dehydration rate is: tertiary > secondary > primary > methanol, or $$3° > 2° > 1° > CH_3OH$$.

Evaluating the options:

• Option A: $$2° > 1° > CH_3OH > 3°$$ → Incorrect, as tertiary is fastest.
• Option B: $$3° > 2° > 1° > CH_3OH$$ → Correct, matches the order.
• Option C: $$2° > 3° > 1° > CH_3OH$$ → Incorrect, as tertiary is faster than secondary.
• Option D: $$CH_3OH > 1° > 2° > 3°$$ → Incorrect, as methanol is slowest.

p-Nitrophenol (III): The $$-NO_2$$ group is a powerful electron-withdrawing group. At the para position, it effectively withdraws electron density through resonance ($$-M$$), making it the most acidic among the given options.

p-Chlorophenol (I): Chlorine is electronegative and exerts a $$-I$$ effect. Although it has a $$+M$$ effect, the $$-I$$ effect is more significant in halogens. It is more acidic than alkyl-substituted phenols but less than nitro-substituted ones.

p-Cresol (II): The methyl group is electron-donating through hyperconjugation ($$+H$$) and inductive effect ($$+I$$). This destabilizes the phenoxide ion relative to phenol, making it less acidic than (I) and (III).

p-Methoxyphenol (IV): The methoxy group exhibits a very strong $$+M$$ effect due to the lone pair on the oxygen atom. At the para position, this effect is dominant over its $$-I$$ effect, making it the most electron-donating and thus the least acidic.

The order of decreasing acidity is: III > I > II > IV

The benzylic $$-OH$$ group is protonated and leaves as $$H_2O$$ to form a stable benzylic carbocation, which is then attacked by $$Cl^-$$ to form a $$-CH_2Cl$$ group.

The $$-OH$$ group directly attached to the benzene ring does not react because the $$C-O$$ bond has partial double-bond character due to resonance, making it difficult to break.

The major product is p-hydroxybenzyl chloride, as shown in Option D.

The question asks about the name of the reaction where phenol reacts with benzoyl chloride to form phenyl benzoate.

First, let's recall the reaction: phenol (C6H5OH) reacts with benzoyl chloride (C6H5COCl) to produce phenyl benzoate (C6H5COOC6H5). The reaction can be written as:

$$ C_{6}H_{5}OH + C_{6}H_{5}COCl -> C_{6}H_{5}COOC_{6}H_{5} + HCl $$

This reaction typically requires a base, such as sodium hydroxide (NaOH), to neutralize the hydrochloric acid (HCl) produced and drive the reaction forward. The overall process is known as benzoylation.

Now, let's examine the options:

Option A: Claisen reaction. This refers to the Claisen condensation, which involves the reaction between two ester molecules in the presence of a strong base to form a β-keto ester. For example, ethyl acetate reacts to form acetoacetic ester. It does not involve phenol or benzoyl chloride, so this is incorrect.

Option B: Schotten-Baumann reaction. This reaction specifically describes the benzoylation of compounds with active hydrogen atoms (like phenols or amines) using benzoyl chloride in the presence of an aqueous base (such as NaOH). The base helps by neutralizing the acid formed and making the nucleophile (phenol) more reactive. This matches the given reaction perfectly, as phenol reacts with benzoyl chloride under Schotten-Baumann conditions to yield phenyl benzoate.

Option C: Reimer-Tiemann reaction. This reaction involves phenol reacting with chloroform (CHCl3) in the presence of a strong base (like NaOH) to produce ortho-hydroxybenzaldehyde (salicylaldehyde). It introduces an aldehyde group (-CHO) at the ortho position. Since benzoyl chloride is not used, this option is incorrect.

Option D: Gatterman-Koch reaction. This reaction is used to introduce an aldehyde group (-CHO) into an aromatic ring using carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of a catalyst (like AlCl3 and CuCl). It is typically applied to benzene or its derivatives, not phenol, and does not involve benzoyl chloride. Hence, this is incorrect.

Therefore, the reaction of phenol with benzoyl chloride to form phenyl benzoate is known as the Schotten-Baumann reaction.

Hence, the correct answer is Option B.

Phenol on heating with $$CHCl_3$$ and NaOH gives salicylaldehyde. The reaction is called Reimer - Tiemann reaction.

Phenol reacts with bromine water to form 2,4,6-tribromophenol due to the highly activating nature of the phenoxide ion formed in the aqueous medium.

Lucas reagent is a mixture of concentrated $$HCl$$ and anhydrous $$ZnCl_2$$. It converts an alcohol $$R{-}OH$$ into the corresponding alkyl chloride $$R{-}Cl$$.

First, state the general reaction that takes place in the presence of Lucas reagent:

$$R{-}OH + HCl \;\xrightarrow[\text{conc.}]{ZnCl_2}\; R{-}Cl + H_2O$$

The mechanism proceeds by the $$S_N1$$ pathway for secondary and tertiary alcohols and by a much slower $$S_N2$$ pathway for primary alcohols. In an $$S_N1$$ reaction the rate-determining step is the formation of a carbocation, and the stability of this carbocation directly controls the speed of the reaction. The order of carbocation stability is

$$\text{tertiary} \;>\; \text{secondary} \;>\; \text{primary}$$

Therefore, the order of reactivity of alcohols with Lucas reagent follows the same trend:

$$\text{tertiary alcohol} \;>\; \text{secondary alcohol} \;>\; \text{primary alcohol}$$

Now classify each given alcohol:

1. Pentanol is $$\text{CH}_3\!-\!\text{CH}_2\!-\!\text{CH}_2\!-\!\text{CH}_2\!-\!\text{CH}_2OH$$.   The hydroxyl group is attached to an end carbon, so it is a primary alcohol.

2. 2-Methyl butanol is $$\text{(CH}_3)_2\text{CHCH}_2CH_2OH$$ (also written as 2-methyl-1-butanol).   Again the OH is on a terminal carbon, so this is a primary alcohol.

3. 2-Pentanol is $$\text{CH}_3\!-\!\text{CH}_2\!-\!\text{CH}(OH)\!-\!\text{CH}_2\!-\!\text{CH}_3$$.   The OH is on the second carbon, which is bonded to two other carbons, hence it is a secondary alcohol.

4. 2-Methyl butan-2-ol is $$\text{(CH}_3)_3\text{C}OH$$.   The carbon bearing the OH is attached to three other carbons, so it is a tertiary alcohol.

Using the previously stated reactivity order, the tertiary alcohol will react the fastest with conc. $$HCl$$ and $$ZnCl_2$$. Among the four options, only 2-methyl butan-2-ol is tertiary.

Hence, the correct answer is Option D.

First we recall the basic facts about the Lucas test. The Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride, written in equations as $$ZnCl_2 + conc\,HCl$$. It converts an alcohol $$R{-}OH$$ into the corresponding alkyl chloride $$R{-}Cl$$: $$R{-}OH + HCl \rightarrow R{-}Cl + H_2O$$, with $$ZnCl_2$$ acting as a Lewis-acid catalyst. The turbidity (cloudiness) that appears comes from insoluble alkyl chloride droplets. The time taken for this turbidity tells us the class of the alcohol.

Now we state the empirical observation that governs the test:

$$\text{Rate of reaction: } \text{tertiary} \; > \; \text{secondary} \; > \; \text{primary}.$$

To justify this order we analyse the stepwise mechanism. The first and rate-determining step is the formation of a carbocation via protonation followed by departure of water. Writing every step:

1. Protonation of the alcohol oxygen by concentrated $$HCl$$ gives the oxonium ion:
$$R{-}OH + H^+ \rightarrow R{-}OH_2^+.$$

2. The zinc chloride coordinates with the lone pairs on oxygen, pulling electron density away and making the -OH2+ group an even better leaving group:
$$(R{-}OH_2^+)\,ZnCl_2 \rightarrow R{-}OH_2^+\;ZnCl_2.$$

3. Departure of water produces the carbocation:
$$R{-}OH_2^+ \rightarrow R^+ + H_2O.$$

4. Finally chloride ion attacks the carbocation to give the alkyl chloride:
$$R^+ + Cl^- \rightarrow R{-}Cl.$$

This sequence—slow ionisation to a carbocation followed by fast nucleophilic attack—is the classical $$S_N1$$ mechanism (substitution, nucleophilic, unimolecular). In an $$S_N1$$ process the stability of the carbocation controls the rate: more stable carbocation ⇒ faster reaction.

Carbocation stability follows the rule

$$3^\circ \; > \; 2^\circ \; > \; 1^\circ,$$

because additional alkyl groups donate electron density by both hyperconjugation and the inductive effect, dispersing the positive charge.

Therefore a tertiary alcohol, which forms a highly stabilised tertiary carbocation, loses water almost instantly; the solution becomes cloudy within a few seconds. A secondary alcohol forms a less stable carbocation, so the cloudiness appears more slowly (often in a few minutes). A primary alcohol forms an unstable primary carbocation, so the $$S_N1$$ path is extremely slow; instead it can react by an $$S_N2$$ path only on prolonged heating, which is why primary alcohols usually give no turbidity at room temperature.

Putting everything together, the fastest reacting alcohol in the Lucas test is the tertiary alcohol and it proceeds through the $$S_N1$$ mechanism.

Hence, the correct answer is Option D.

The ether (A) has the molecular formula $$C_5H_{12}O$$. When heated with excess hot concentrated HI, it undergoes cleavage to form two alkyl iodides. The reaction of an ether with HI is:

$$ \text{R-O-R'} + 2\text{HI} \rightarrow \text{R-I} + \text{R'-I} + \text{H}_2\text{O} $$

These alkyl iodides are then treated with NaOH, which hydrolyzes them to alcohols:

$$ \text{R-I} + \text{NaOH} \rightarrow \text{R-OH} + \text{NaI} $$

$$ \text{R'-I} + \text{NaOH} \rightarrow \text{R'-OH} + \text{NaI} $$

So, the compounds (B) and (C) are alcohols. Oxidation of (B) gives propanone ($$CH_3COCH_3$$), which is a ketone. Ketones are formed by the oxidation of secondary alcohols. The only secondary alcohol that oxidizes to propanone is propan-2-ol ($$CH_3CH(OH)CH_3$$):

$$ \text{CH}_3\text{CH(OH)CH}_3 \rightarrow \text{CH}_3\text{COCH}_3 $$

Oxidation of (C) gives ethanoic acid ($$CH_3COOH$$), which is a carboxylic acid. Carboxylic acids are formed by the oxidation of primary alcohols. The primary alcohol that oxidizes to ethanoic acid is ethanol ($$CH_3CH_2OH$$), as ethanol oxidizes first to ethanal and then to ethanoic acid:

$$ \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{COOH} $$

Therefore, (B) is propan-2-ol and (C) is ethanol. The alkyl groups in the ether (A) correspond to these alcohols. Propan-2-ol comes from the isopropyl group ($$(CH_3)_2CH-$$), and ethanol comes from the ethyl group ($$CH_3CH_2-$$). Thus, the ether (A) is isopropyl ethyl ether, with the structure $$(CH_3)_2CH-O-CH_2CH_3$$.

To determine the IUPAC name, identify the longest carbon chain. The isopropyl group has a three-carbon chain (propane), and the ethyl group has two carbons. The parent chain is propane, with the ethoxy group attached to carbon 2. Therefore, the IUPAC name is 2-ethoxypropane.

Verifying with the options:

• Option A: 2-ethoxypropane matches the structure $$(CH_3)_2CH-O-CH_2CH_3$$.
• Option B: ethoxypropane is ambiguous but typically implies 1-ethoxypropane ($$CH_3CH_2CH_2-O-CH_2CH_3$$), which would give propan-1-ol and ethanol upon reaction; oxidation of propan-1-ol gives propanoic acid, not propanone.
• Option C: methoxybutane (e.g., $$CH_3O-CH_2CH_2CH_2CH_3$$) would give methanol and butan-1-ol; oxidation gives formaldehyde/formic acid and butanoic acid, not propanone or ethanoic acid.
• Option D: 2-methoxybutane ($$CH_3O-CH(CH_3)CH_2CH_3$$) would give methanol and butan-2-ol; oxidation gives formaldehyde/formic acid and butanone, not propanone or ethanoic acid.

Hence, only 2-ethoxypropane produces the correct products.

So, the answer is Option A.

We first recall the essence of the Williamson ether synthesis. It is an $$S_N2$$ nucleophilic substitution reaction in which an alkoxide ion $$RO^-$$ attacks an alkyl halide $$R'X$$ to give an ether:

$$RO^- + R'X \;\rightarrow\; ROR' + X^-$$

The fundamental requirement for a smooth $$S_N2$$ reaction is that the carbon atom bearing the leaving group $$X$$ should be unhindered; in other words, a primary alkyl halide is ideal because the nucleophile can approach the carbon from the back side and displace the halide ion in a single concerted step.

Now we analyse the situation given in the question. We wish to prepare a mixed ether that contains one primary alkyl group and one tertiary alkyl group. The chemist has two choices for the halide partner:

1. Take the primary alkyl group as the halide and the tertiary alkyl group as the alkoxide, or

2. Take the tertiary alkyl group as the halide and the primary alkyl group as the alkoxide.

The question specifically says that the tertiary halide is used. So we are in case 2. Let us write the reaction that we would be attempting:

$$R_3C{-}X \;+\; R' O^- \;\longrightarrow\; R_3C{-}O{-}R' \;+\; X^-$$

Here $$R_3C{-}X$$ is a tertiary alkyl halide, and $$R' O^-$$ is a primary alkoxide ion. Because the carbon $$R_3C{-}$$ is tertiary, it is heavily crowded. An $$S_N2$$ attack requires the alkoxide to approach from the back side of the $$C{-}X$$ bond, but steric bulk blocks this route. Therefore the rate of $$S_N2$$ substitution becomes extremely slow.

At the same time, the alkoxide is a strong base. Whenever a strong base meets a tertiary alkyl halide, an elimination pathway is highly favourable. The base abstracts a β-hydrogen, the electrons form a double bond, and the halide ion departs. This is the $$E2$$ mechanism. We can symbolise it as follows (for simplicity a single β-hydrogen is shown):

$$\begin{array}{c} \text{Base }(R'O^-) + R_3C{-}CH_2{-}X \;\;\xrightarrow{E2}\;\; R_3C=CH_2 + R'OH + X^-\\[6pt] \text{(alkene)}\hspace{45pt}\text{(recovered alcohol)} \end{array}$$

Because the $$E2$$ pathway has a much lower steric requirement (the base merely needs to abstract a hydrogen and does not have to approach the crowded carbon atom directly), elimination overwhelmingly dominates over substitution when a tertiary halide is present.

Thus, instead of the desired mixed ether, the chief product that comes out of the reaction mixture is an alkene derived from the tertiary alkyl halide. The ether yield is negligible.

Comparing this conclusion with the options given:

A. “Rate of reaction will be slow due to slow cleavage of carbon-halogen bond.”  →  While the $$S_N2$$ path is slow, this statement ignores the dominant elimination and is therefore not the best choice.

B. “Alkene will be the main product.”  →  This matches our analysis exactly.

C. “Simple ether will form instead of mixed ether.”  →  Not supported; elimination predominates.

D. “Expected mixed ether will be formed.”  →  Contrary to the mechanistic reasoning.

Therefore the correct statement is option B.

Hence, the correct answer is Option B.