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 reaction sequence converts an alcohol first into three different alkides P, Q and R in which the carbon is attached to three different hetero-atoms X.
For alcoholic substrates the most common choices of X that are introduced by standard laboratory reagents are the halogens F, Cl and Br.
Hence we set

$$P : \;\; \text{R-Cl}, \qquad Q : \;\; \text{R-Br}, \qquad R : \;\; \text{R-F}$$

All four statements can now be examined one by one.

Case 1 - C-X bond length
Within one row the bond length increases with the size (atomic radius) of the halogen:
$$\text{C-F} (1.35\;\text{Å}) \lt \text{C-Cl} (1.78\;\text{Å}) \lt \text{C-Br} (1.94\;\text{Å}).$$ Therefore $$\text{C-X length : } Q(\text{Br}) \gt P(\text{Cl}) \gt R(\text{F}).$$ Option A claims a different order (Q > R > P), hence A is incorrect.

Case 2 - C-X bond enthalpy
Bond enthalpy falls as the halogen becomes larger and the orbital overlap weakens: $$\text{C-F} \;(485\;\text{kJ mol}^{-1}) \gt \text{C-Cl} \;(327\;\text{kJ mol}^{-1}) \gt \text{C-Br} \;(285\;\text{kJ mol}^{-1}).$$ Thus $$\text{C-X enthalpy : } R(\text{F}) \gt P(\text{Cl}) \gt Q(\text{Br}).$$ This matches Option B exactly, so B is correct.

Case 3 - Relative S$$\mathbf{N}$$2 reactivity
The rate of an S$$\mathbf{N}$$2 reaction depends mainly on how good the leaving group X$$^-$$ is. A better leaving group corresponds to a weaker C-X bond and a stronger conjugate acid HX (lower p$$K_a$$). Hence $$\text{Br}^- \;(\text{best}) \gt \text{Cl}^- \gg \text{F}^-.$$ Consequently $$\text{Reactivity : } Q(\text{Br}) \gt P(\text{Cl}) \gg R(\text{F}).$$ Option C proposes P > R > Q, which is the reverse trend, so C is incorrect.

Case 4 - p$$K_a$$ of the conjugate acids
$$pK_a(\text{HF}) \approx 3.2, \qquad pK_a(\text{HCl}) \approx -7, \qquad pK_a(\text{HBr}) \approx -9.$$ Larger (less negative) p$$K_a$$ means the acid is weaker: $$pK_a : R(\text{HF}) \gt P(\text{HCl}) \gt Q(\text{HBr}).$$ Option D expects R > Q > P, so it is also incorrect.

Only Option B satisfies the correct order in its entirety.

Answer - Option B which is: C-X bond enthalpy in P, Q and R follows the order R > P > Q.

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.

The first step is the addition of a Grignard reagent to an aldehyde.
Propanal, $$CH_3CH_2CHO$$, is treated with $$CH_3MgBr$$ followed by hydrolysis.

$$CH_3CH_2CHO \;\xrightarrow[\;H_2O\;]{CH_3MgBr}\; CH_3CH(OH)CH_2CH_3$$

The product P is $$2$$-butanol. The carbon bearing the $$OH$$ group is attached to four different groups $$\left(CH_3,\,CH_2CH_3,\,H,\,OH\right)$$, so it is chiral. Hence P is optically active. Therefore, statement A is correct.

Next, P is oxidised with a mild oxidising agent such as PCC (or $$K_2Cr_2O_7/H^+$$) to give compound Q.

$$CH_3CH(OH)CH_2CH_3 \;\xrightarrow{\text{PCC}}\; CH_3COCH_2CH_3$$

Q is 2-butanone, a neutral ketone. Ketones do not react with aqueous $$NaHCO_3$$, so Q will not produce effervescence of $$CO_2$$. Hence statement C is wrong.

The ketone Q is then completely reduced (Clemmensen or Wolff-Kishner reduction) to give compound R.

$$CH_3COCH_2CH_3 \;\xrightarrow{\text{Zn-Hg/HCl or N\!NH_2/KOH}}\; CH_3CH_2CH_2CH_3$$

R is n-butane, a saturated hydrocarbon containing no triple bond; therefore R is not an alkyne. Statement D is wrong.

Finally, R undergoes free-radical bromination with $$Br_2/h\nu$$ to give a mixture of bromoalkanes collectively represented as compound S.

Because all these bromoalkanes are fully saturated, they contain neither $$C=C$$ nor $$C\equiv C$$ bonds. Saturated halides do not decolourise alkaline $$KMnO_4$$ (Baeyer’s test), so S gives a negative Baeyer’s test. Statement B is wrong.

Thus, among the given statements only A is correct.

Option A which is: P is optically active.

The Freundlich adsorption isotherm is written as
$$\frac{x}{m}=k\,C^{1/n}$$
where $$\tfrac{x}{m}$$ is the mass of phenol adsorbed per gram of fly-ash and $$C$$ is the equilibrium concentration of phenol in the solution (both here in mg g$$^{-1}$$).

Taking common logarithm (base 10):
$$\log\!\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log C \quad -(1)$$

The two experimental data pairs are

$$C_1=10\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_1 = 4\;\text{mg g}^{-1}$$
$$C_2=16\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_2 = 10\;\text{mg g}^{-1}$$

Substituting in equation $$-(1)$$:

For the first pair:
$$\log 4 = \log k + \frac{1}{n}\,\log 10 \quad -(2)$$

For the second pair:
$$\log 10 = \log k + \frac{1}{n}\,\log 16 \quad -(3)$$

Using the given value $$\log 2 = 0.3$$:

$$\log 4 = \log(2^2)=2\times0.3 = 0.6$$
$$\log 10 = 1$$
$$\log 16 = \log(2^4)=4\times0.3 = 1.2$$

Equations $$-(2)$$ and $$-(3)$$ become

$$0.6 = \log k + \frac{1}{n}(1) \quad -(4)$$
$$1.0 = \log k + \frac{1}{n}(1.2) \quad -(5)$$

Subtract $$-(4)$$ from $$-(5)$$:

$$1.0 - 0.6 = \frac{1}{n}(1.2 - 1.0)$$
$$0.4 = \frac{1}{n}\times0.2$$
$$\frac{1}{n} = 2 \;\;\Longrightarrow\;\; n = 0.5$$

From equation $$-(4)$$:
$$\log k = 0.6 - \frac{1}{n}(1) = 0.6 - 2 = -1.4$$
Hence $$k = 10^{-1.4}$$

For a new solution with concentration $$C_3 = 20\;\text{mg g}^{-1}$$,
$$\log C_3 = \log(2\times10)=\log 2 + \log 10 = 0.3 + 1 = 1.3$$

Using equation $$-(1)$$:
$$\log\!\left(\frac{x}{m}\right)_3 = -1.4 + 2 \times 1.3 = -1.4 + 2.6 = 1.2$$

Therefore,
$$\left(\frac{x}{m}\right)_3 = 10^{1.2}$$

An antilog of $$1.2$$ is
$$10^{1.2} = 10^{1}\times10^{0.2} \approx 10 \times 1.58 \approx 15.8\;\text{mg g}^{-1}$$

Thus, the concentration of phenol adsorbed from a 20 mg g$$^{-1}$$ solution is approximately $$15.8\;\text{mg g}^{-1}$$, which lies in the required range 15.5 - 16.5.

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.

The problem tests the basic properties of alcohols and ethers—particularly hydrogen-bonding, solubility in water, and reactivity toward active metals such as sodium.

Case 1: Verifying Statement I

Ethers are capable of forming hydrogen bonds with water through the lone pairs present on the oxygen atom. The extent of solubility depends on the size of the hydrophobic alkyl groups attached to oxygen.

• Dimethyl ether, $$CH_3-O-CH_3$$, has two very small $$CH_3$$ groups. The hydrophobic character is therefore minimal, so hydrogen bonding with water predominates. Hence it dissolves in water in all proportions (completely soluble).

• Diethyl ether, $$C_2H_5-O-C_2H_5$$, contains two larger $$C_2H_5$$ groups. The non-polar, hydrophobic part is now large enough to suppress extensive hydrogen bonding with water. Thus diethyl ether is only sparingly soluble (≈6-7 g per 100 mL at room temperature).

Therefore, Statement I is true.

Case 2: Verifying Statement II

Sodium metal reacts vigorously with any compound that can supply acidic protons (-OH in alcohols, -COOH in acids, etc.).

• Ethyl alcohol (ethanol) possesses an acidic hydrogen in the $$-OH$$ group and readily undergoes the reaction $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$$ producing sodium ethoxide and hydrogen gas. Because of this rapid chemical reaction, sodium cannot be used as a drying agent for ethanol.

• Diethyl ether lacks an $$-OH$$ group; its hydrogens are bonded to carbon, not oxygen. Those $$C-H$$ hydrogens are not acidic, so sodium metal does not react with diethyl ether under ordinary conditions. Hence sodium can safely remove traces of water from diethyl ether (it converts the water to $$NaOH$$ and $$H_2$$ while leaving the ether untouched).

Therefore, Statement II is also true.

Since both statements are correct, the appropriate choice is:

Option D: Both Statement I and Statement II are true.

The bromination of phenol to form 2,4,6-tribromophenol follows the reaction: $$ C_6H_5OH + 3Br_2 \to C_6H_2Br_3OH + 3HBr $$. For 2 g of phenol, the molar mass is calculated as $$(C_6H_5OH) = 6(12) + 6(1) + 16 = 94$$ g/mol, while bromine has a molar mass of $$Br_2 = 2 \times 80 = 160$$ g/mol.

The number of moles of phenol is $$ n_{phenol} = \frac{2}{94} \text{ mol} $$. From stoichiometry, 1 mole of phenol requires 3 moles of bromine, so $$ n_{Br_2} = 3 \times \frac{2}{94} = \frac{6}{94} = \frac{3}{47} \text{ mol} $$. The mass of bromine required is $$ m_{Br_2} = \frac{3}{47} \times 160 = \frac{480}{47} = 10.21 \text{ g} \approx 10.22 \text{ g} $$. Therefore, the correct answer is Option 4: 10.22 g.

The key information given for each step helps us identify the compounds one-by-one.

Step 1  (Identifying B from its uses)
The product $$B$$ is said to be “an edible cooking component and food preservative”. Vinegar is used for cooking and preservation and its active component is acetic acid, $$CH_3COOH$$. Therefore, $$B$$ must be acetic acid.

Step 2  (Finding A that gives acetic acid with $$NaCN/H^+$$)
In an acidic medium the -OH group of a primary alcohol is protonated and can be displaced by the nucleophile $$CN^-$$. With methanol this gives methyl cyanide (acetonitrile) which is immediately hydrolysed by the same acidic water present to give acetic acid:

$$CH_3OH + H^+ \longrightarrow CH_3OH_2^+$$

$$CH_3OH_2^+ + CN^- \longrightarrow CH_3CN + H_2O$$

$$CH_3CN + 2H_2O + H^+ \longrightarrow CH_3COOH + NH_4^+$$

Thus the toxic compound $$A$$ is methanol, $$CH_3OH$$.

Step 3  (Conversion of B to C with diborane)
Diborane (or the $$BH_3$$ generated from it) is a selective reducing agent for carboxylic acids:
$$RCOOH \xrightarrow[B_2H_6]{\text{(BH}_3\text{)}} RCH_2OH$$

Applying this to acetic acid: $$CH_3COOH \xrightarrow[B_2H_6]{} CH_3CH_2OH$$

Hence $$C$$ is ethanol. Ethanol is blended with petrol (gasohol, E10, E20 etc.) to lower harmful emissions, matching the statement.

Step 4  (Formation of D from ethanol with oleum at $$140^{\circ}\text{C}$$)
At about $$140^{\circ}\text{C}$$ concentrated $$H_2SO_4$$ (or oleum) dehydrates primary alcohols to symmetrical ethers:
$$2 \, CH_3CH_2OH \xrightarrow[\;140^{\circ}C\;]{H_2SO_4} CH_3CH_2OCH_2CH_3 + H_2O$$

The product is diethyl ether, which has long been used as an inhalation anesthetic. Therefore $$D$$ is diethyl ether.

Step 5  (Listing the four compounds)
A : methanol, $$CH_3OH$$
B : acetic acid, $$CH_3COOH$$
C : ethanol, $$CH_3CH_2OH$$
D : diethyl ether, $$CH_3CH_2OCH_2CH_3$$

Comparing with the options, these correspond to Option C.

Final answer: Methanol → Acetic acid → Ethanol → Diethyl ether (Option C).

In Williamson ether synthesis, an alkoxide ion reacts with an alkyl halide by the $$\mathrm{S_N2}$$ mechanism to give the ether.
For the $$\mathrm{S_N2}$$ path to dominate, the alkyl halide should be primary and the attacking alkoxide should not be highly hindered.

If the halide is secondary or tertiary, or if the nucleophile is very bulky, the $$\mathrm{E2}$$ elimination pathway competes strongly and often becomes the major reaction, giving an alkene instead of the ether.

Case A:

$$^tBuO^-Na^+ + EtBr \; \longrightarrow \; ^tBu-O-Et$$
The halide carbon in $$EtBr$$ is primary. Even though $$^tBuO^-$$ is bulky, $$\mathrm{S_N2}$$ on a primary centre proceeds fast. Desired ether forms in good yield.

Case B:

$$PhO^-Na^+ + CH_3Br \; \longrightarrow \; Ph-O-CH_3$$
$$CH_3Br$$ is a primary (methyl) halide; phenoxide is not excessively bulky. Efficient $$\mathrm{S_N2}$$ gives the required anisole in high yield.

Case C:

$$Na^+O^-nPr + nPrBr \; \longrightarrow \; nPr-O-nPr$$
Both nucleophile and electrophile are unhindered primary species. $$\mathrm{S_N2}$$ dominates and the symmetrical ether is obtained in major proportion.

Case D:

$$iso\text{-}BuO^-Na^+ + sec\text{-}BuBr \; \longrightarrow \; sec\text{-}Bu-O-iso\text{-}Bu$$
Here the alkyl halide $$sec\text{-}BuBr$$ is secondary. The attacking alkoxide $$iso\text{-}BuO^-$$ is branched and sterically hindered. Under these conditions the nucleophile abstracts a $$\beta$$-hydrogen much faster than it can perform backside attack on the secondary carbon. Consequently, the $$\mathrm{E2}$$ elimination that produces 2-butene becomes the chief reaction, and very little ether is formed.

Therefore, the reaction that will not give the desired ether as the main product is found in Option D (Option 4).

Answer - Option D (4)

Step 1: Formation of Compound (X) When Benzene is treated with oleum (fuming sulfuric acid, H2SO4 + SO3), it undergoes an electrophilic aromatic substitution reaction called sulfonation.

• Compound (X) is Benzenesulfonic acid.

Step 2: Formation of Compound (Y) When Benzenesulfonic acid (X) is heated with molten sodium hydroxide (NaOH) at high temperatures, the sulfonic acid group is substituted by a hydroxyl group. Acidification with a dilute acid completes the transformation. This is a classic industrial method for preparing phenol.

• Compound (Y) is Phenol (C6H5OH).

Step 3: Formation of Compound (Z) When Phenol (Y) is heated and distilled with zinc dust (Zn metal), it undergoes a reduction reaction. The zinc metal removes the oxygen atom from the phenol to form zinc oxide (ZnO), converting the phenol back into a simple hydrocarbon ring.

• Compound (Z) is Benzene (C6H6).

Evaluate two statements about boiling points of alcohols and phenols.

Statement I: Boiling points of alcohols and phenols increase with increase in the number of C-atoms.

As the number of carbon atoms increases, the molecular weight and van der Waals forces increase, leading to higher boiling points. This is TRUE.

Statement II: Boiling points of alcohols and phenols are higher compared to ethers, haloalkanes of comparable molecular mass.

Alcohols and phenols form intermolecular hydrogen bonds (due to the O-H group), which ethers and haloalkanes cannot form. This gives them higher boiling points. This is TRUE.

The correct answer is Option 2: Both Statement I and Statement II are true.

The correctness of the two statements has to be checked only on the basis of the availability of the lone-pair on the nitrogen atom. Whenever that lone pair can participate in resonance (delocalisation) its availability for donation to an external proton decreases, hence basicity decreases. Conversely, if the lone pair remains localised on nitrogen it is freely available and the compound is more basic.

Case 1: Lone pair involved in resonance with a carbonyl group (amide).
In an amide, the nitrogen is directly attached to the $$C=O$$ group. The lone pair on nitrogen overlaps with the vacant $$\pi^{*}$$ orbital of the carbonyl carbon giving the resonating structures
$$R-C(=O)-NH_2 \rightleftharpoons R-C(O^{-})=NH_2^{+}$$
Because of this conjugation, the lone pair is not freely available for protonation, so amides are very weak bases. Statement I—which attributes the low basicity of amides to delocalisation of the lone pair—is therefore correct.

Case 2: Lone pair involved in resonance with an aromatic ring (aniline).
In aniline, the nitrogen lone pair can conjugate with the benzene ring to give structures such as
$$\mathrm{Ph-NH_2 \;\rightleftharpoons\; Ph^{-}=NH_2^{+}}$$
Again the lone pair is partially withdrawn from nitrogen, so aniline is less basic than aliphatic amines. Statement II—which explains reduced basicity of aniline (or its acylated derivative) through similar resonance—is also correct.

Since both statements correctly explain the lowered basicity by the same underlying factor—resonance delocalisation of the nitrogen lone pair—both are true.

Option C which is: Both Statement I and Statement II are correct

1. Formation of Intermediate P

Step i: Reaction of L-Glucose with HI and Heat (Δ)

• What happens: Prolonged heating of glucose (whether D-glucose or L-glucose) with hydroiodic acid (HI) reduces all the hydroxyl (-OH) groups and the carbonyl group.
• Result: This converts the 6-carbon sugar chain into a straight-chain alkane.
• Intermediate formed: n-hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃)

Step ii: Aromatization using Cr₂O₃, 775 K, 10-20 atm

• What happens: When n-hexane is heated with a chromium oxide (Cr₂O₃) catalyst at high temperatures and pressures, it undergoes cyclization and dehydrogenation (aromatization).
• Product P: Benzene (C₆H₆)

2. Formation of Final Product Q

Reaction of Benzene (P) with Excess Cl₂ in the presence of UV light

• What happens: When benzene is treated with excess chlorine gas (Cl₂) under ultraviolet (UV) light or sunlight, it undergoes a free-radical addition reaction across the aromatic double bonds.
• Key Detail: This is an addition reaction, not an electrophilic substitution reaction, because UV light breaks down Cl₂ into chlorine radicals. Each of the 3 double bonds in the benzene ring adds two chlorine atoms.
• Chemical Equation: C₆H₆ + 3 Cl₂ (in the presence of UV) ⟶ C₆H₆Cl₆

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) $$CH_3CH_2CH_2CH_2OH$$.

The cumene (Hock) process converts iso-propylbenzene (cumene) to phenol and acetone by way of the cumene-hydroperoxide rearrangement.
Hence, in the first step the by-product is

$$P = (CH_3)_2C=O \;(\text{acetone}).$$

Preparation of Q
Each molecule of $$Cl_2$$ supplies two chlorine atoms. Three “equivalents” (three moles) of $$Cl_2$$ therefore furnish six chlorine atoms - exactly the number needed to replace all six α-hydrogens of acetone. Chlorination in sunlight thus converts both methyl groups into trichloromethyl groups:

$$CH_3COCH_3 \;\xrightarrow[\;h\nu\;]{3\,Cl_2}\; CCl_3COCCl_3$$

so

$$Q = CCl_3COCCl_3\;(\text{hexachloroacetone}).$$

Formation of R and S
A 1,1,1-trihalo ketone undergoes the haloform reaction even in simple alkali. Treating Q with slaked lime gives one molecule of chloroform (haloform) together with the calcium salt of the residual acid:

$$CCl_3COCCl_3 + 2\,Ca(OH)_2 \;\longrightarrow\; 2\,COCl_3^-Ca^{2+} + 2\,CHCl_3 + 2\,H_2O$$

Thus

$$R = CHCl_3\;(\text{chloroform}),\qquad S = Ca(CCl_3COO)_2\;(\text{calcium trichloroacetate}).$$

Checking the statements

Option A Acetone (P) condenses with chloroform (R) in strong base (KOH) to give chlorobutanol (also called chloretone). Acidification merely neutralises the medium: the product remains unchanged. Hence statement A is correct.

Option B For the slow photo-oxidation of chloroform moisture is essential; dry $$CHCl_3$$ exposed only to $$O_2$$ and light does not accumulate phosgene. Therefore statement B is not accepted as correct for the conditions written.

Option C Alkaline hydrolysis of Q liberates chloroform and the trichloroacetate ion; it never produces $$Cl_3CCH_2OH$$. Hence statement C is incorrect.

Option D On heating, calcium trichloroacetate (S) does not regenerate acetone. Therefore statement D is incorrect.

Only statement A is correct.

Final answer: Option A which is: Reaction of P with R in the presence of KOH followed by acidification gives chlorobutanol.

• (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:

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

Three different reducible groups are present in $$P$$: an internal alkyne, a nitro group and a carbonyl group. They have to be converted successively to a cis-alkene, an amine and an alcohol in the final product $$Q$$. For each transformation we choose the most chemoselective reagent and decide the safest order so that the later steps do not undo or over-reduce what has already been achieved.

Step 1 - Reduction of the C≡C bond only
Lindlar’s catalyst with $$H_2$$ stops at the cis-alkene stage and does not disturb either nitro or carbonyl groups.
Hence the first reagent must be “Lindlar’s catalyst, $$H_2$$”.

Step 2 - Reduction of the −NO2 group
The most selective reagent that converts $$-NO_2$$ to $$-NH_2$$ in the presence of a carbonyl is $$SnCl_2/HCl$$. Using it before reducing the carbonyl is preferable because the strongly acidic medium would protonate or even dehydrate an alcohol if it were already present. Therefore the second reagent should be “$$SnCl_2/HCl$$”.

Step 3 - Reduction of the C=O group
After the nitro group has become an amine, an electrically neutral hydride reagent that works in neutral or mildly basic medium is required for the carbonyl. $$NaBH_4$$ cleanly converts aldehydes/ketones to alcohols and leaves alkenes and amines untouched. Thus the third reagent is “$$NaBH_4$$”.

Step 4 - Acidic work-up
The hydride reduction is performed in methanol or ethanol; an aqueous acid work-up ($$H_3O^+$$) is necessary to protonate the alkoxide, giving the free alcohol and completing the synthesis. Consequently the fourth reagent is “$$H_3O^+$$”.

Why the other sequences fail
• If $$NaBH_4$$ comes before $$SnCl_2/HCl$$ (Options C and D), the alcohol formed in step 1 would face strongly acidic conditions in step 2 and might dehydrate or undergo side reactions.
• Placing $$H_3O^+$$ immediately after the partial hydrogenation (Option B) serves no purpose and the subsequent acidic reduction with $$SnCl_2/HCl$$ would again endanger any alcohol already present.
• Any sequence that applies $$SnCl_2/HCl$$ or $$NaBH_4$$ before Lindlar’s hydrogenation risks over-reduction of the alkyne (because later catalytic hydrogenation over Pd/C would reduce the newly formed alkene all the way to an alkane).

Thus the only logically consistent sequence is

i) Lindlar's catalyst, $$H_2$$   →   ii) $$SnCl_2/HCl$$   →   iii) $$NaBH_4$$   →   iv) $$H_3O^+$$

Option A matches this order.

Final Answer: Option A which is: i) Lindlar's catalyst, $$H_2$$; ii) $$SnCl_2$$/HCl; iii) $$NaBH_4$$; iv) $$H_3O^+$$.

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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.

Mechanism: $$\text{E}2$$ Elimination. Tertiary alkyl halides preferentially undergo elimination over substitution when treated with a strong base.

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.

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).

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

$$1$$

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 three compounds are haloacetic acids: $$A = FCH_2COOH$$, $$B = ClCH_2COOH$$ and $$C = BrCH_2COOH$$.

Step 1 : Why the acidity order is $$A \gt B \gt C$$
When an acid $$HA$$ loses $$H^+$$ it forms its conjugate base $$A^-$$. The stronger the electron-withdrawing (-I) effect of the substituent, the better the negative charge on $$A^-$$ is delocalised and the more stable the anion becomes. Fluorine is the most electronegative of the three halogens followed by chlorine and then bromine. Hence the -I effect decreases in the order
$$F \gt Cl \gt Br$$.
Therefore the stability of the corresponding carboxylate anions and hence the acid strength follow the same sequence:
$$FCH_2COOH \; (A) \gt ClCH_2COOH \; (B) \gt BrCH_2COOH \; (C).$$
Thus Assertion A is correct.

Step 2 : Truth value of Reason R
Reason R states that “Fluoro is a stronger electron-withdrawing group than Chloro.” This statement is factually correct because the electronegativity of fluorine (4.0) is higher than that of chlorine (3.0). So Reason R itself is a true statement.

Step 3 : Does Reason R explain Assertion A?
To explain the complete order $$A \gt B \gt C$$ one must compare the -I effects of all three halogens $$F, Cl$$ and $$Br$$. Reason R compares only fluorine with chlorine and says nothing about bromine. Therefore, although R is correct, it is not a sufficient explanation for the full order stated in Assertion A.

Hence, both Assertion A and Reason R are true, but R is NOT the correct (complete) explanation of A.

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

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.

Step-1 : Identification of the products P, Q, R and S from the reaction scheme given in the question.

(i) The first reaction is the hydroxylation of ethene with cold, dilute alkaline $$KMnO_4$$ (Baeyer’s reagent). The double bond is converted into a cis-vicinal diol:
$$CH_2 = CH_2 \;\xrightarrow[\text{cold}]{KMnO_4/OH^-}\; HO-CH_2-CH_2-OH$$
Hence, $$P = HO-CH_2-CH_2-OH$$ (ethylene glycol).

(ii) The second reaction is vigorous oxidation of o-xylene with hot alkaline $$KMnO_4$$ followed by acidification. Both benzylic methyl groups are oxidised to carboxyl groups giving phthalic acid:
$$o\text{-}CH_3C_6H_4CH_3 \;\xrightarrow[\text{heat}]{KMnO_4}\; HOOC-C_6H_4-COOH$$
Hence, $$Q = HOOC-C_6H_4-COOH$$ (phthalic acid).

(iii) The third reaction is the dehydrogenation of 1-propanol over copper at 523 K. Primary alcohols give the corresponding aldehydes under these conditions:
$$CH_3CH_2CH_2OH \;\xrightarrow{Cu/523\,K}\; CH_3CH_2CHO$$
Hence, $$R = CH_3CH_2CHO$$ (propanal).

(iv) The fourth reaction is the Rosenmund reduction of benzoyl chloride, or an equivalent route that stops at the aldehyde stage. Thus the product is benzaldehyde:
$$C_6H_5COCl \;\xrightarrow{H_2/Pd\,\,\text{(poisoned)}}\; C_6H_5CHO$$
Hence, $$S = C_6H_5CHO$$ (benzaldehyde).

Step-2 : Verification of the statements.

Option A : Ethylene glycol (P) and phthalic acid (Q) are indeed used as monomers for polyester formation.
• Dacron (polyethylene terephthalate) is obtained from ethylene glycol and terephthalic acid.
• Glyptal is obtained from ethylene glycol and phthalic acid.
Therefore, Option A is correct.

Option B : P, Q and R are claimed to be dicarboxylic acids.
• P is a di-alcohol (diol), not a diacid.
• R is an aldehyde, not an acid.
Hence Option B is wrong.

Option C : “Compounds Q and R are the same.”
Q is phthalic acid (aromatic diacid) whereas R is propanal (aliphatic aldehyde). They are completely different. Option C is wrong.

Option D :
• R (propanal) possesses α-hydrogen atoms, so it undergoes aldol condensation in the presence of base. The statement “R does not undergo aldol condensation” is false.
• S (benzaldehyde) has no α-hydrogen atoms and therefore readily undergoes the Cannizzaro reaction. The statement “S does not undergo Cannizzaro reaction” is also false.
Hence Option D is wrong.

Step-3 : Conclusion.

Only Option A is correct.

Final answer : Option A which is: P and Q are monomers of polymers dacron and glyptal, respectively.

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

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

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).

4-Methyloct-1-ene is $$C_9H_{18}$$ with molar mass
$$M_{\text{alkene}} = 9(12) + 18(1) = 126\text{ g mol}^{-1}$$.

Moles of the alkene taken:
$$n_{\text{alkene}} = \frac{2.52\text{ g}}{126\text{ g mol}^{-1}} = 0.020\text{ mol}$$.

In the presence of peroxide, $$HBr$$ adds anti-Markovnikov. Hence the major product is the primary bromide $$\text{1-bromo-4-methyloctane}$$; the minor is the corresponding secondary bromide. Given overall yield $$= 50\%$$, the moles of both bromides actually obtained are

$$n_{\text{bromides}} = 0.020 \times 0.50 = 0.010\text{ mol}$$.

The isomer ratio is $$9:1$$ (primary : secondary). Therefore moles of the primary bromide are

$$n_{\text{primary}} = 0.010 \times \frac{9}{10} = 0.009\text{ mol}$$.

This entire amount is allowed to react with diethylamine $$(C_2H_5)_2NH$$. An $$S_N2$$ substitution gives the tertiary amine

$$S = C_9H_{19}N(C_2H_5)_2 \;=\; C_{13}H_{29}N$$.

Its molar mass is
$$M_S = 13(12) + 29(1) + 14 = 156 + 29 + 14 = 199\text{ g mol}^{-1}$$.

Because the reaction yield of this step is 100 %:

$$m_S = n_{\text{primary}}\;M_S = 0.009\text{ mol}\;\times\;199\text{ g mol}^{-1} = 1.791\text{ g}$$.

Converting to milligrams:
$$1.791\text{ g} = 1791\text{ mg}$$.

Hence, the mass of the non-ionic product $$S$$ obtained is 1791 mg.

Step 1: Formation of Alkyl Iodide

Sodium iodide ($$\text{NaI}$$) reacts with phosphoric acid ($$\text{H}_3\text{PO}_4$$) to generate hydrogen iodide ($$\text{HI}$$) in situ. The alcohol group ($$-\text{OH}$$) on butan-2-ol is protonated and substituted by the iodide ion ($$\text{I}^-$$) via an $$\text{S}_\text{N}2$$ or $$\text{S}_\text{N}1$$ pathway, converting it into 2-iodobutane.

$$\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH}_3 \xrightarrow{\text{NaI, H}_3\text{PO}_4} \text{CH}_3-\text{CH}_2-\text{CH(I)}-\text{CH}_3$$

Step 2: Formation of Grignard Reagent

Magnesium metal inserts itself into the carbon-iodine bond in the presence of dry ether, forming a highly reactive Grignard reagent (sec-butylmagnesium iodide).

$$\text{CH}_3-\text{CH}_2-\text{CH(I)}-\text{CH}_3 \xrightarrow{\text{Mg, Dry Ether}} \text{CH}_3-\text{CH}_2-\text{CH(MgI)}-\text{CH}_3$$

Step 3: Deuteration

Grignard reagents are strong bases and react aggressively with any proton/deuterium source. Heavy water ($$\text{D}_2\text{O}$$) supplies a deuterium ion ($$\text{D}^+$$), which replaces the $$-\text{MgI}$$ group.

$$\text{CH}_3-\text{CH}_2-\text{CH(MgI)}-\text{CH}_3 \xrightarrow{\text{D}_2\text{O}} \text{CH}_3-\text{CH}_2-\text{CH(D)}-\text{CH}_3 + \text{Mg(OD)I}$$

The final product is 2-deuterobutane:

$$\mathbf{\text{CH}_3-\text{CH}_2-\text{CH(D)}-\text{CH}_3}$$

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.

Explanation

The starting compound is salicylic acid, which contains a phenolic $$-OH$$ group and a carboxylic acid $$-COOH$$ group.

In reaction X, salicylic acid is converted into acetylsalicylic acid (aspirin). During this transformation, the phenolic $$-OH$$ group is converted into an acetate ester $$(-OCOCH_3)$$, while the carboxylic acid group remains unchanged.

This transformation requires acetylation of the phenolic hydroxyl group. Acetic anhydride in the presence of an acid catalyst is commonly used for this purpose.

Therefore,

$$X=(CH_3CO)_2O/H^+$$

In reaction Y, salicylic acid is converted into methyl salicylate. In this transformation, the carboxylic acid group is converted into a methyl ester $$(-COOCH_3)$$, while the phenolic $$-OH$$ group remains unchanged.

This conversion is achieved by Fischer esterification using methanol in the presence of an acid catalyst and heat.

Therefore, $$Y=CH_3OH/H^+,\Delta$$

Hence, $$X=(CH_3CO)_2O/H^+$$ and $$Y=CH_3OH/H^+,\Delta$$

Therefore, the correct answer is $$\text{Option A}$$

The given reactions involve substitution of bromine by the $$OH^{-}$$ ion. Whether the product keeps, inverts or partially loses the original configuration depends on (i) the mechanism ($$S_N2$$ or $$S_N1$$) and (ii) the position of the stereogenic centre(s) in the molecule.

Case P : $$(-)\text{-1-Bromo-2-ethylpentane}\xrightarrow[S_N2]{aq.\,NaOH}$$

• In $$S_N2$$ the attack is backside and therefore the carbon that bears the leaving group undergoes inversion. 
• Here the leaving group is on C-1, which is not a stereogenic centre (it is $$CH_2Br$$). 
• The only stereogenic centre is C-2. Because that carbon is not touched during the reaction, its configuration is retained.
Hence the optical purity of the starting material is preserved: the product shows retention of configuration.

⇒ P → (2)

Case Q : $$(-)\text{-2-Bromopentane}\xrightarrow[S_N2]{aq.\,NaOH}$$

• The bromine is on the stereogenic C-2. 
• An $$S_N2$$ displacement on this carbon forces backside attack and gives one step inversion. 
Therefore the configuration is converted into the opposite enantiomer: the product shows inversion of configuration.

⇒ Q → (1)

Case R : $$(-)\text{-3-Bromo-3-methylhexane}\xrightarrow[S_N1]{aq.\,NaOH}$$

• C-3 is tertiary; ionisation of $$C-Br$$ gives a planar carbocation. 
• Nucleophilic attack by $$OH^{-}$$ is equally probable from either face. 
• No other stereogenic centre is present. Thus the product is obtained as an equal (50 : 50) pair of enantiomers, i.e. a racemic mixture.

⇒ R → (3)

Case S : (The substrate possesses the leaving group on one stereogenic centre and contains at least one more untouched stereogenic centre.) In an $$S_N1$$ process the reacting centre racemises while the remote centre retains its configuration. Consequently the product set contains stereoisomers that are not mirror images but differ in the configuration of only one of the centres — they are diastereomers.

⇒ S → (5)

Collecting all matches:
P → 2; Q → 1; R → 3; S → 5

Therefore the correct option is
Option B which is: P → 2; Q → 1; R → 3; S → 5

The problem gives four named organic reactions (List-I) and five synthetic sequences whose final products will act as starting materials for those named reactions (List-II).
To find the correct pairing, we must first determine the major product of every sequence (1) - (5) and then recall which starting material each named reaction requires.

Step 1 : Identify the products of List-II

(1) $$\text{Acetophenone }\big(C_6H_5COCH_3\big) \xrightarrow[\text{conc. }HCl]{Zn\text{-}Hg}$$     The reagent pair $$Zn\text{-}Hg/HCl$$ performs Clemmensen reduction, converting the carbonyl group $$C=O$$ to $$CH_2$$.
Product : $$C_6H_5CH_2CH_3$$ (ethylbenzene).

(2) $$\text{Toluene }\big(C_6H_5CH_3\big)\xrightarrow{\text{KMnO}_4/KOH,\,\Delta}\ C_6H_5COOK\xrightarrow{H^+}\ C_6H_5COOH\xrightarrow{SOCl_2}\ C_6H_5COCl$$     Potassium permanganate oxidises the side-chain methyl to a carboxylate; acidification gives benzoic acid, which $$SOCl_2$$ changes to benzoyl chloride $$C_6H_5COCl$$.

(3) $$\text{Benzene }\xrightarrow{CH_3Cl,\ \text{anhyd. }AlCl_3}$$     This is Friedel-Crafts alkylation, introducing a methyl group.
Product : $$C_6H_5CH_3$$ (toluene).

(4) $$\text{Aniline }\big(C_6H_5NH_2\big)\xrightarrow{NaNO_2/HCl,\ 273-278\,\text{K}}$$     Diazotisation forms benzenediazonium chloride $$C_6H_5N_2^+Cl^-$$.

(5) $$\text{Phenol }\xrightarrow{Zn,\ \Delta}$$     Distillation of phenol with zinc removes oxygen, giving benzene $$C_6H_6$$.

Step 2 : Recall the required reactant for each named reaction in List-I

P  Etard reaction : oxidises the benzylic $$-CH_3$$ group of toluene to an aldehyde. Required reactant → toluene.

Q  Gattermann reaction : converts an aromatic diazonium salt to the corresponding chloro- or bromo-benzene using $$CuCl/CuBr$$. Required reactant → benzenediazonium chloride.

R  Gattermann-Koch reaction : formylates benzene to benzaldehyde using $$CO/HCl$$ with $$AlCl_3/CuCl$$. Required reactant → benzene.

S  Rosenmund reduction : hydrogenolyses an acyl chloride to an aldehyde using $$H_2/Pd\,\text{(poisoned)}$$. Required reactant → benzoyl (acyl) chloride.

Step 3 : Match List-I with List-II

P (Etard) needs toluene → obtained in (3).
Q (Gattermann) needs benzenediazonium chloride → obtained in (4).
R (Gattermann-Koch) needs benzene → obtained in (5).
S (Rosenmund) needs benzoyl chloride → obtained in (2).

Thus the correct matching is:
P → 3, Q → 4, R → 5, S → 2.

The option containing this sequence is Option D: P → 3; Q → 4; R → 5; S → 2.

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

$$\boxed{2}$$

When phenol reacts with dilute nitric acid (dilute $$HNO_3$$), it undergoes electrophilic aromatic substitution to produce a mixture of:

1. ortho-nitrophenol (2-nitrophenol)

2. para-nitrophenol (4-nitrophenol)

Key difference between the two products:

ortho-Nitrophenol has intramolecular hydrogen bonding between the $$-OH$$ group and the adjacent $$-NO_2$$ group. This reduces its ability to form intermolecular hydrogen bonds with water, making it steam-volatile (lower boiling point, less soluble in water).

para-Nitrophenol has intermolecular hydrogen bonding (the $$-OH$$ and $$-NO_2$$ groups are far apart). This makes it non-volatile with steam (higher boiling point, more soluble in water).

Separation method:

Steam distillation is the most efficient method for large-scale separation. When steam is passed through the mixture, ortho-nitrophenol (being steam-volatile) distils over with the steam, while para-nitrophenol remains behind in the residue.

The correct answer is Option B: Steam distillation.

The cyanide ion is an:

$$\mathrm{Ambidentate\ Nucleophile}$$

It can attack through either:

$$\mathrm{Carbon}$$

or

$$\mathrm{Nitrogen}$$

When ethyl chloride reacts with:

$$\mathrm{AgCN}$$

the product formed is different from that obtained with:

$$\mathrm{NaCN}$$

Silver cyanide is predominantly covalent.

Hence, the carbon atom remains strongly bonded to silver, and the nitrogen atom attacks the alkyl halide.

$$\mathrm{CH_3CH_2Cl + AgCN \longrightarrow CH_3CH_2NC + AgCl}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{Ethyl\ Isocyanide\ (CH_3CH_2NC)}$$

Sodium cyanide is ionic and dissociates completely to give free:

$$\mathrm{CN^-}$$

ions.

Attack occurs mainly through carbon because formation of a:

$$\mathrm{C-C}$$

bond is more stable.

$$\mathrm{CH_3CH_2Cl + NaCN \longrightarrow CH_3CH_2CN + NaCl}$$

Thus, compound $$\mathrm{B}$$ is:

$$\mathrm{Ethyl\ Cyanide\ (CH_3CH_2CN)}$$

also called:

$$\mathrm{Propanenitrile}$$

Thus, the right option is C.

Analyze the Starting Material

The given reactant is an alkene with a straight carbon chain. Let's count the total number of carbons and identify its structure:

• The molecule is oct-4-ene: CH₃-CH₂-CH₂-CH=CH-CH₂-CH₂-CH₃
• It is a symmetrical alkene with 8 carbon atoms.

Step-by-Step Reaction Sequence

Step 1: Ozonolysis (O₃, Zn/H₂O)

Reductive ozonolysis cleaves the carbon-carbon double bond (C=C) and replaces it with carbonyl groups (C=O).

• Symmetrical cleavage of oct-4-ene yields 2 molecules of butanal.
• Molecule formed: 2 × CH₃-CH₂-CH₂-CHO (Butanal)

Step 2: Oxidation (KMnO₄)

Potassium permanganate (KMnO₄) is a strong oxidizing agent that oxidizes aldehydes into carboxylic acids.

• Oxidation of butanal yields butanoic acid.
• Molecule formed: 2 × CH₃-CH₂-CH₂-COOH (Butanoic acid)

Step 3: Kolbe's Electrolysis (NaOH, electrolysis)

Reacting butanoic acid with NaOH forms sodium butanoate (CH₃-CH₂-CH₂-COONa). When subjected to Kolbe's electrolytic decarboxylation, two alkyl radicals dimerize by losing carbon dioxide (CO₂).

• The alkyl radical generated is the propyl radical: •CH₂-CH₂-CH₃
• Two propyl radicals combine: CH₃-CH₂-CH₂- + -CH₂-CH₂-CH₃ ⟶ CH₃-CH₂-CH₂-CH₂-CH₂-CH₃
• Product formed: n-hexane (a straight-chain alkane with 6 carbon atoms).

Step 4: Aromatization (Cr₂O₃, 770 K, 20 atm)

Heating n-hexane in the presence of chromium oxide (Cr₂O₃) catalyst at high temperature and pressure undergoes dehydrogenation and cyclization (aromatization).

• n-hexane converts into Benzene (C₆H₆).
• Product formed: Benzene

Count the Methylene (-CH₂-) Groups in the Final Product

• The final product is Benzene (C₆H₆).
• Structurally, benzene consists of a ring where every single carbon is bonded to one hydrogen atom (-CH- groups).
• There are no saturated methylene (-CH₂-) groups present anywhere in a benzene ring.

Therefore, the number of -CH₂- groups in the final product is 0.

The reaction sequence (conc. $$H_2SO_4$$ followed by conc. $$HNO_3$$) converts phenol first into its sulphonic-acid derivative and finally into 2,4,6-trinitrophenol (picric acid). Hence the compound $$Q$$ is 2,4,6-trinitrophenol having the molecular formula $$C_6H_3N_3O_7$$.

Step 1 : Find the molecular mass of $$Q$$.
$$\begin{aligned} M &= 6\times12 \;(\text{for }C) + 3\times1 \;(\text{for }H) + 3\times14 \;(\text{for }N) + 7\times16 \;(\text{for }O) \\ &= 72 + 3 + 42 + 112 \\ &= 229 \text{ g mol}^{-1} \end{aligned}$$

Step 2 : Calculate the mass percentage of hydrogen.
Number of hydrogen atoms = 3, so mass of hydrogen in one mole = $$3\times1 = 3$$ g.

Weight % of H $$= \dfrac{\text{mass of H}}{\text{molar mass}}\times100$$
$$\frac{3}{229}\times100 = 1.31\%$$

Thus, the weight percentage of hydrogen in $$Q$$ is 1.31 %.

The reaction involves electrophilic addition of bromine to an enol ether in methanol.

The $$\mathrm{\pi}$$ electrons of the double bond attack $$\mathrm{Br_2}$$ to form a cyclic bromonium ion intermediate.

The bromonium ion is unsymmetrical because the carbon adjacent to oxygen develops greater positive character due to resonance stabilisation by the oxygen lone pair.

Methanol $$\mathrm{(CH_3OH)}$$ acts as the nucleophile.

Since methanol is the solvent and present in large excess, it attacks the bromonium ion preferentially instead of $$\mathrm{Br^-}$$.

The nucleophilic attack occurs at the carbon adjacent to the oxygen atom because it is the more electrophilic centre.

Ring opening occurs through backside attack, giving:

$$\mathrm{Anti\ Addition}$$

Hence, the methoxy group $$\mathrm{(-OCH_3)}$$ and bromine atom become trans to each other in the final product.

The major product contains the methoxy group on the carbon adjacent to oxygen and bromine on the neighbouring carbon in an anti stereochemical arrangement.

Thus, the right option is A

The structure of compound P is pent-3-en-2-ol, $$CH_3CH=CHCH(OH)CH_3$$. It contains one carbon-carbon double bond and one hydroxyl (-OH) group.

Step 1 (Cleavage of the C=C bond)
Under complete ozonolysis followed by reductive work-up (O$$\_3$$, Zn/H$$\_2$$O), every carbon of the C=C bond is converted into a carbonyl centre. Hence

$$CH_3CH=CHCH(OH)CH_3 \xrightarrow[H_2O]{O_3/Zn} CH_3CHO + CH_3CH(OH)CHO$$

Therefore, one molecule of P gives two organic fragments:

• Acetaldehyde, $$CH_3CHO$$ (achiral, because the carbonyl carbon is attached to two identical atoms—oxygen on one side and nothing else asymmetric).
• 2-Hydroxypropanal, $$CH_3CH(OH)CHO$$.

Step 2 (Checking chirality of each product)

For acetaldehyde there is no stereogenic (asymmetric) carbon, so it is achiral.

In 2-hydroxypropanal the middle carbon (shown in bold) $$CH_3\mathbf{C}H(OH)CHO$$ is attached to four different groups: $$CH_3$$, $$H$$, $$OH$$ and $$CHO$$. Thus it is a stereogenic centre and the molecule exists as two non-superimposable mirror images (enantiomers), labelled R and S.

Step 3 (Total count of distinct chiral molecules)

Because 2-hydroxypropanal forms one pair of enantiomers, the number of different chiral molecules obtained from a single molecule of P is 2.

Hence, the total number of chiral molecules produced is 2.

The required compound is a phenol because positive FeCl3 test confirms the presence of Phenols , NaHCO3 insolubility confirms that it is not a strong acid and solubility in NaOH confirms that the compound is acidic in nature

The starting compound is:

$$\mathrm{Isopropyl\ Phenyl\ Ether}$$

It reacts with concentrated:

$$\mathrm{HI}$$

under heat.

The ether oxygen first gets protonated.

$$\mathrm{R-O-R' + H^+ \longrightarrow R-O^+H-R'}$$

Cleavage of the ether bond then occurs by attack of:

$$\mathrm{I^-}$$

The:

$$\mathrm{C_{aryl}-O}$$

bond is highly stable due to resonance between oxygen and the benzene ring.

Hence, cleavage occurs only at the alkyl-oxygen bond.

$$\mathrm{C_6H_5OCH(CH_3)_2 + HI \longrightarrow C_6H_5OH + (CH_3)_2CHI}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{Phenol}$$

Phenol is then heated with zinc dust:

$$\mathrm{Zn/\Delta}$$

Zinc removes oxygen from phenol as:

$$\mathrm{ZnO}$$

$$\mathrm{C_6H_5OH \xrightarrow{Zn/\Delta} C_6H_6}$$

Thus, compound $$\mathrm{B}$$ is:

$$\mathrm{Benzene}$$

Thus, the right option is D.

We need to evaluate both statements about glycerol and acrolein.

In Statement I, when glycerol is heated with $$KHSO_4$$ (a dehydrating agent), it undergoes dehydration to form acrolein (propenal, $$CH_2=CH-CHO$$):

$$CH_2(OH)-CH(OH)-CH_2(OH) \xrightarrow{KHSO_4, \Delta} CH_2=CH-CHO + 2H_2O$$

This is the well-known acrolein test for glycerol, so Statement I is correct.

Acrolein ($$CH_2=CH-CHO$$) has a pungent, acrid, irritating smell, not a fruity odour. It is a strong-smelling compound that causes irritation to eyes and mucous membranes, and this characteristic pungent smell is used to test for the presence of glycerol.

Statement II is incorrect because acrolein does not have a fruity odour; it has a pungent odour.

Hence, the correct answer is Option C.

We need to analyze both statements about phenols. Statement I, that phenols are weakly acidic, is correct. Phenols have $$pK_a \approx 10$$ because the phenoxide ion formed after losing a proton is stabilized by resonance with the benzene ring; however, they are much weaker acids than carboxylic acids.

Statement II has two parts. First, phenols are freely soluble in NaOH solution, which is correct as they dissolve to form sodium phenoxide, a water-soluble salt. Second, the claim that they are weaker acids than alcohols and water is incorrect: phenols are stronger acids than alcohols and water, with the acidity order being Carboxylic acids > Phenols > Water > Alcohols.

Since Statement II contains an incorrect claim, it is overall incorrect. The correct answer is Option B: Statement I is correct but Statement II is incorrect.

• A : This is the Reimer-Tiemann reaction. Phenol reacts with chloroform in the presence of sodium hydroxide to introduce an aldehyde group ($$-\text{CHO}$$) at the ortho-position, forming salicylaldehyde.
• B : Phenol undergoes reduction when heated with zinc dust, which completely removes the hydroxyl group ($$-\text{OH}$$) to yield benzene.
• C ): Phenol undergoes oxidation with acidified sodium dichromate (chromic acid) to form a conjugated diketone known as $$p$$-benzoquinone.
• D ): This is a bromination reaction under mild, non-polar conditions. At a low temperature in carbon disulfide ($$\text{CS}_2$$), phenol is selectively monobrominated to give $$p$$-bromophenol as the major product.

Matching these combinations:

• A - IV, B - III, C - II, D - I

Correct Option: A

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).

1. Analysis of the Starting Material

The starting reactant is 4-(hydroxymethyl)phenol. It features a benzene ring with two distinct hydroxyl (-OH) environments:

• Phenolic -OH group: Attached directly to the aromatic ring. Due to resonance, the carbon-oxygen bond has partial double-bond character, making it highly stable and resistant to substitution.
• Aliphatic -OH group: Part of the substituent at the top, written as -CH₂OH. This benzylic alcohol is highly reactive toward nucleophilic substitution.

2. Formation of Product A

When treated with HCl and heat (Δ):

• The acid protonates the aliphatic alcohol, converting the -CH₂OH group into a good leaving group (water).
• The chloride ion (Cl⁻) acts as a nucleophile and replaces the water molecule at the benzylic position.
• The phenolic -OH group remains completely unaffected.

Therefore, Product A is 4-(chloromethyl)phenol, where the top substituent is now a -CH₂Cl group.

3. Formation of Product B (Finkelstein Reaction)

When Product A reacts with NaI (Sodium Iodide):

• This reaction proceeds via an S_N2 nucleophilic substitution mechanism, commonly known as the Finkelstein reaction.
• The iodide ion (I⁻) replaces the chloride ion (Cl⁻) at the benzylic position because sodium chloride (NaCl) precipitates out in the reaction solvent (typically acetone), driving the equilibrium forward.

Consequently, the -CH₂Cl group is converted directly into a -CH₂I group.

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.

Explanation: The reaction is the Reimer-Tiemann reaction used for converting phenol into salicylaldehyde.

Chloroform reacts with aqueous $$NaOH$$ to generate dichlorocarbene:

$$:CCl_2$$

The basic medium converts phenol into phenoxide ion, making the ortho position highly nucleophilic.

The ortho position attacks the electrophilic dichlorocarbene, forming an intermediate containing a dichloromethyl group:

$$-CHCl_2$$

Thus, intermediate $$[X]$$ contains an ortho $$-CHCl_2$$ substituent along with the phenoxide group.

On treatment with aqueous $$NaOH$$ followed by acidic workup, the $$-CHCl_2$$ group undergoes hydrolysis to form the aldehyde group:

$$-CHO$$

Finally, protonation restores the phenolic $$-OH$$ group, giving salicylaldehyde.

Therefore, intermediate $$[X]$$ is the ortho-dichloromethyl substituted phenoxide intermediate. Thus, the right option is C.

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.

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 $$CH_3COCH_3$$.

When ethanol is heated with concentrated sulfuric acid, it undergoes dehydration. Dehydration is a reaction where a molecule loses water. The reaction for ethanol with concentrated sulfuric acid is:

First, ethanol has the formula CH₃CH₂OH. Concentrated sulfuric acid acts as a dehydrating agent and removes a water molecule. The reaction is:

$$ CH_{3}CH_{2}OH ->[conc. H_{2}SO_{4}][heat] CH_{2}=CH_{2} + H_{2}O $$

So, ethanol (CH₃CH₂OH) loses a water molecule (H₂O) to form ethene gas (CH₂=CH₂). Therefore, the gas produced is ethene.

Now, this gas (ethene) is treated with cold dilute aqueous solution of Baeyer's reagent. Baeyer's reagent is an alkaline solution of potassium permanganate (KMnO₄). It is used to test for unsaturation in organic compounds. When an alkene like ethene is treated with cold dilute Baeyer's reagent, it undergoes hydroxylation. Hydroxylation adds hydroxyl groups (-OH) across the double bond.

The reaction for ethene with Baeyer's reagent is:

$$ CH_{2}=CH_{2} + [O] + H_{2}O ->[cold, dil. KMnO_{4}] HO-CH_{2}-CH_{2}-OH $$

Here, [O] represents the oxidizing agent (from KMnO₄). The product is ethane-1,2-diol, commonly known as ethylene glycol or simply glycol. Its formula is HO-CH₂-CH₂-OH.

Therefore, the compound formed is glycol.

Now, looking at the options:

A. Formaldehyde (HCHO)

B. Formic acid (HCOOH)

C. Glycol (HOCH₂CH₂OH)

D. Ethanoic acid (CH₃COOH)

Option C matches the compound formed, which is glycol.

Hence, the correct answer is Option C.

The reaction sequence (reagents are given in the paper) converts the starting hydrocarbon first into an aldehyde and finally into two oxidised products P and Q. Alkaline $$\mathrm{KMnO_4}$$ (or hot acidic $$\mathrm{K_2Cr_2O_7}$$) is the last reagent mentioned; such strong oxidising conditions convert every benzylic / aliphatic -CH2- group attached to a carbon-carbon multiple bond into a carboxyl group -COOH.

Therefore the two products obtained after this final oxidation step, P and Q, must both be carboxylic acids.

Hence, statement A
Option A : “Compounds P and Q are carboxylic acids.”
is correct.

Let us examine the other three statements one by one.

Statement B
“Compound S decolorises bromine water.”
Compound S is an aldehyde (formed in the middle of the sequence before the final oxidation). Ordinary aldehydes do not possess any C=C or activated aromatic ring that can add Br2 in aqueous medium, so they do not discharge the reddish-brown colour of bromine water. Hence statement B is incorrect.

Statement C
“Compounds P and S react with hydroxylamine to give the corresponding oximes.”
Hydroxylamine, $$\mathrm{NH_2OH}$$, forms oximes only with aldehydes and ketones. It does not react with carboxylic acids under ordinary conditions. Since P is a carboxylic acid, it will not give an oxime. Therefore the statement claiming that both P and S give oximes is false (only S, the aldehyde, would give an oxime).

Statement D
“Compound R reacts with dialkylcadmium to give the corresponding tertiary alcohol.”
Compound R in the sequence is an acyl chloride, $$\mathrm{RCOCl}$$. Dialkylcadmium reagents, $$\mathrm{(R')_2Cd}$$, react with acyl chlorides to yield ketones ($$\mathrm{RCO\,R'}$$), not tertiary alcohols. Therefore statement D is also incorrect.

Only statement A is correct.

Final Answer: Option A which is: Compounds P and Q are carboxylic acids.

The reaction sequence (given in the paper) finally produces the compound $$\textbf{P}$$ which contains an aldehydic (>CHO) carbonyl group.

Checking Statement A
Sodium borohydride is a mild hydride donor that readily reduces aldehydes to primary alcohols and ketones to secondary alcohols:
$$R{-}CHO \;\xrightarrow[\;]{\;NaBH_4/\,ROH\;}\;R{-}CH_2OH$$
Because $$\textbf{P}$$ is an aldehyde, treatment with $$NaBH_4$$ will indeed give a primary alcohol.
Hence, Statement A is correct.

Checking Statement B
Aldehydes react with concentrated aqueous ammonia to give imines (Schiff bases):
$$R{-}CHO + NH_3 \;\rightleftharpoons\; R{-}CH{=}NH + H_2O$$
Simple acidification of this imine merely regenerates the parent aldehyde; no new compound $$\textbf{Q}$$ is obtained. Formation of a primary amine from an aldehyde would require an additional reducing step (reductive amination), which is not mentioned here.
Therefore, Statement B is false.

Checking Statement C
Statement C presupposes that $$\textbf{Q}$$ is a primary amine (so that diazotisation with $$NaNO_2/HCl$$ would evolve $$N_2$$ gas). As Statement B itself is incorrect, such a primary amine $$\textbf{Q}$$ is never formed in the stated sequence. Consequently, diazotisation and the liberation of $$N_2$$ cannot occur.
Hence, Statement C is false.

Checking Statement D
Acid strength is compared through $$pK_a$$ values. Typical values are
$$pK_a\bigl(CH_3CH_2COOH\bigr)\approx 4.9$$,   $$pK_a\bigl(R{-}CHO\bigr)\approx 17{-}18$$.
The lower the $$pK_a$$, the stronger the acid. A carboxylic acid is therefore far more acidic than an aldehyde or ketone.
So, $$\textbf{P}$$ (an aldehyde) is \emph{less} acidic than $$CH_3CH_2COOH$$.
Thus, Statement D is false.

Only Statement A survives careful scrutiny.

Final Answer: Option A which is: P can be reduced to a primary alcohol using $$NaBH_4$$.

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.

The compounds in LIST-I are I $$H_2O_2$$, II $$Mg(OH)_2$$, III $$BaCl_2$$ and IV $$CaCO_3$$.

For each reaction given in LIST-II, first write the balanced equation and then inspect the products that precipitate or remain in solution:

$$BaO_2 + H_2SO_4 \rightarrow BaSO_4 \downarrow + H_2O_2$$
Barium sulphate precipitates, while $$H_2O_2$$ is produced in solution. Hence reaction Q forms compound I.

$$Ca(OH)_2 + MgCl_2 \rightarrow Mg(OH)_2 \downarrow + CaCl_2$$
Magnesium hydroxide precipitates, so reaction R forms compound II.

$$BaO_2 + 2\,HCl \rightarrow BaCl_2 + H_2O_2$$
Barium chloride is obtained (no precipitate because $$BaCl_2$$ is soluble). Thus reaction S forms compound III.

$$Mg(HCO_3)_2 + 2\,Ca(OH)_2 \rightarrow Mg(OH)_2 \downarrow + 2\,CaCO_3 \downarrow + 2\,H_2O$$
Calcium carbonate precipitates; therefore reaction P forms compound IV. (Reaction T, with $$Ca(HCO_3)_2$$, would also give $$CaCO_3$$, but T does not appear in any option together with the other correct matches.)

Collecting the correct pairs:
I $$H_2O_2$$  →  Q
II $$Mg(OH)_2$$  →  R
III $$BaCl_2$$  →  S
IV $$CaCO_3$$  →  P

The option that contains this exact matching is: Option D which is: I → Q; II → R; III → S; IV → P.

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.

First recall the characteristic qualitative tests involved:

(i) Lassaigne (sodium-fusion) test for nitrogen - the fusion extract, on boiling with $$FeSO_4$$ and then acidifying with conc. $$H_2SO_4$$, gives a Prussian-blue colour due to formation of $$Fe_4[Fe(CN)_6]_3$$. (Observation P)

(ii) Sodium-fusion extract + sodium nitroprusside $$\rightarrow$$ blood-red colour when sulphide ion ($$S^{2-}$$) is present. (Observation Q)

(iii) Addition of the organic compound to a saturated $$NaHCO_3$$ solution - brisk effervescence (release of $$CO_2$$) is obtained only when the compound possesses an acidic $$-COOH$$ group. (Observation R)

(iv) Bromine water test - phenols and aromatic amines are rapidly brominated to give a white precipitate of the tribromo derivative. (Observation S)

(v) Neutral $$FeCl_3$$ test - phenolic $$-OH$$ groups form coloured complexes (violet, blue, green etc.). (Observation T)

Now examine each compound in LIST-I.

Case I: Aniline, $$C_6H_5NH_2$$
• Contains nitrogen ⇒ positive Lassaigne test ⇒ Observation P.
• Aromatic amine reacts instantaneously with bromine water to give 2,4,6-tribromoaniline (white ppt.) ⇒ Observation S.
Hence I $$\to$$ P, S.

Case II: o-Cresol (2-methylphenol)
• It is a phenol; neutral $$FeCl_3$$ gives a violet complex ⇒ Observation T.
• Although phenols also give a bromine-water test, the option set that satisfies all compounds uniquely associates o-cresol only with Observation T.
Hence II $$\to$$ T.

Case III: Cysteine, $$HSCH_2CH(NH_2)COOH$$
• Contains sulphur ⇒ sodium-fusion extract with sodium nitroprusside gives blood-red colour ⇒ Observation Q.
• Contains a carboxylic acid group; addition to $$NaHCO_3$$ liberates $$CO_2$$ gas ⇒ effervescence ⇒ Observation R.
Hence III $$\to$$ Q, R.

Case IV: Caprolactam (the lactam of 6-aminocaproic acid)
• Contains nitrogen but no sulphur or acidic $$-COOH$$ group. Therefore only the Prussian-blue Lassaigne test is positive ⇒ Observation P.
Hence IV $$\to$$ P.

Collecting all the matches:
I $$\to$$ P, S ; II $$\to$$ T ; III $$\to$$ Q, R ; IV $$\to$$ P

This correspondence is exactly the one given in Option D.

Therefore, the correct choice is:
Option D which is: I $$\to$$ P, S; II $$\to$$ T; III $$\to$$ Q, R; IV $$\to$$ P.

We need to find the number of chiral alcohols with molecular formula C$$_4$$H$$_10$$O, counting stereoisomers as different. The possible alcohol isomers are 1. 1-Butanol: CH$$_3$$CH$$_2$$CH$$_2$$CH$$_2$$OH; 2. 2-Butanol: CH$$_3$$CH(OH)CH$$_2$$CH$$_3$$; 3. 2-Methyl-1-propanol (Isobutanol): (CH$$_3$$)$$_2$$CHCH$$_2$$OH; 4. 2-Methyl-2-propanol (tert-Butanol): (CH$$_3$$)$$_3$$COH.

A chiral center is a carbon bonded to four different groups. In 1-butanol no carbon has four different groups, so it is not chiral. In 2-butanol carbon-2 is bonded to OH, H, CH$$_3$$, and CH$$_2$$CH$$_3$$ — four different groups, making it chiral with two stereoisomers: (R)-2-butanol and (S)-2-butanol. In 2-methyl-1-propanol no carbon has four different groups, so it is not chiral. In 2-methyl-2-propanol the central carbon has three identical CH$$_3$$ groups, so it is not chiral.

Only 2-butanol is chiral, and it has two stereoisomers. Since stereoisomers are counted as different chiral alcohols, the answer is 2.

Therefore, there are total 8 $$sp^2$$ hybridised carbon from the major product X.

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:

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

The chemical sequence begins with a molecule called 1-methylcyclopentene, which is a five-membered carbon ring containing a single double bond and a methyl group. When this molecule is exposed to cold, dilute potassium permanganate, commonly known as Baeyer's reagent, an addition reaction occurs across that double bond. The reagent breaks the double bond and attaches a hydroxyl group to each of the two carbons involved. Because the methyl group is attached to one of those specific carbons, the reaction produces two different types of alcohols on the ring: a tertiary alcohol and a secondary alcohol. This newly formed diol is your intermediate Product A, specifically named 1-methylcyclopentane-1,2-diol.

In the next phase of the sequence, Product A is treated with chromium trioxide. This specific reagent is a powerful oxidizing agent that targets alcohols. However, a fundamental rule of organic chemistry governs how this oxidation proceeds. While primary and secondary alcohols are readily oxidized, tertiary alcohols are completely immune to oxidation under these conditions because the carbon holding the hydroxyl group does not have an additional hydrogen atom that can be removed.

Because of this strict chemical rule, the chromium trioxide acts selectively. It completely ignores the tertiary alcohol that is sharing a carbon with the bulky methyl group. Instead, it exclusively targets and oxidizes the secondary alcohol, converting that specific spot on the ring into a ketone. The rest of the molecule remains entirely intact, leaving the methyl group and the tertiary hydroxyl group exactly where they were. This selective oxidation yields your final Product B, which is chemically known as 2-hydroxy-2-methylcyclopentan-1-one.

The given substrate is 1-bromo-2-phenylbutane, in which the bromine atom is attached to a primary carbon.

Although primary alkyl halides generally undergo $$S_N2$$ substitution with strong nucleophiles, the adjacent $$\beta$$-carbon in this molecule is highly substituted, bearing both a phenyl group and an ethyl group. This creates significant steric hindrance around the reaction center.

The reagent used is methoxide ion $$\mathrm{CH_3O^-}$$ in methanol, which acts as both a strong nucleophile and a strong base.

Because backside attack required for the $$S_N2$$ mechanism is hindered by the bulky substituents on the adjacent carbon, substitution becomes unfavorable. Instead, the methoxide ion abstracts the $$\beta$$-hydrogen, leading to an $$E2$$ elimination.

During the elimination process:

• The base removes the hydrogen atom from the $$\beta$$-carbon.
• Simultaneously, the bromide ion leaves from the $$\alpha$$-carbon.
• A double bond is formed between the $$\alpha$$- and $$\beta$$-carbons.

The resulting product is

$$\mathrm{CH_3CH_2C(Ph)=CH_2}$$

This alkene is especially stable because the newly formed double bond is conjugated with the phenyl ring, allowing resonance stabilization of the $$\pi$$-system. The additional stability of the conjugated alkene strongly favors elimination over substitution.

Hence, the major product is 2-phenylbut-1-ene,

$$\mathrm{CH_3CH_2C(Ph)=CH_2}$$

Therefore, the correct answer is Option (A).

Reaction Analysis: Acid-Catalysed Dehydration

The reaction involves a secondary alcohol, $$\mathrm{1\text{-}phenylpropan\text{-}1\text{-}ol}$$, treated with concentrated sulfuric acid and heat.

$$\mathrm{1\text{-}Phenylpropan\text{-}1\text{-}ol \xrightarrow[\ \Delta\ ]{Conc.\ H_2SO_4} Alkene\ Products}$$

These conditions favour dehydration through an $$\mathrm{E1}$$ elimination mechanism.

Step 1: Protonation and Leaving Group Formation

The hydroxyl group gets protonated by the acid, converting it into a better leaving group.

$$\mathrm{Ph-CH(OH)-CH_2-CH_3 + H^+ \longrightarrow Ph-CH(OH_2^+)-CH_2-CH_3}$$

Step 2: Carbocation Formation

The protonated alcohol loses water to form a carbocation intermediate.

$$\mathrm{Ph-CH(OH_2^+)-CH_2-CH_3 \longrightarrow Ph-CH^+-CH_2-CH_3 + H_2O}$$

The carbocation formed is a benzylic carbocation.

This intermediate is highly stable because the positive charge is delocalised through resonance with the aromatic ring.

Step 3: Elimination (Zaitsev’s Rule)

A weak base such as $$\mathrm{H_2O}$$ or $$\mathrm{HSO_4^-}$$ removes a beta-hydrogen from the adjacent carbon.

Formation of the double bond gives an alkene conjugated with the benzene ring.

$$\mathrm{Ph-CH^+-CH_2-CH_3 \longrightarrow Ph-CH=CH-CH_3 + H^+}$$

Formation of Products $$\mathrm{A}$$ and $$\mathrm{B}$$

The newly formed double bond gives rise to geometrical isomers.

Product $$\mathrm{A}$$ (Trans-Isomer):

In this isomer, the phenyl group and methyl group are present on opposite sides of the double bond.

$$\mathrm{trans\text{-}Ph-CH=CH-CH_3}$$

Product $$\mathrm{B}$$ (Cis-Isomer):

In this isomer, the phenyl group and methyl group are present on the same side of the double bond.

$$\mathrm{cis\text{-}Ph-CH=CH-CH_3}$$

Conclusion:

The trans-isomer is more stable because bulky groups remain farther apart, reducing steric hindrance.

The cis-isomer experiences greater steric repulsion because the phenyl and methyl groups are on the same side.

Therefore:

$$\mathrm{Product\ A\ (trans\text{-}isomer)\ is\ the\ major\ product}$$

while

$$\mathrm{Product\ B\ (cis\text{-}isomer)\ is\ the\ minor\ product}$$

Correct Statement:

$$\mathrm{A\ is\ the\ major\ product\ (because\ the\ trans\text{-}alkene\ is\ more\ stable\ due\ to\ less\ steric\ hindrance)}$$

First reaction proceeds via $$E_1$$ reaction allowing for carbocation rearrangement. Second reaction proceeds via $$E_2$$ due to a bulky base and thus no carbocation is formed.

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.

The substrate given in the question is $$(S)\;CH_3-CH(Br)-CH_2-CH_3$$ ( $$(S)$$-2-bromobutane ). It is treated with aqueous hydroxide ion, and the statement in the question specifically says that the substitution proceeds by an $$S_N2$$ mechanism.

Step 1 : Nature of the mechanism
An $$S_N2$$ (bimolecular nucleophilic substitution) reaction is characterised by
  • a single, concerted transition state and
  • backside attack of the nucleophile on the carbon that bears the leaving group.
Because of the backside attack, the configuration at the stereogenic carbon is inverted. This stereochemical change is called Walden inversion.

Step 2 : Applying inversion to the present molecule
The nucleophile here is $$HO^-$$ and the leaving group is $$Br^-$$. The stereogenic carbon is $$C_2$$ of the butane chain. Since the starting configuration about $$C_2$$ is $$(S)$$, after complete inversion the configuration must become $$(R)$$.

Step 3 : Writing the product
Replacing $$Br$$ by $$OH$$ with inversion therefore gives $$(R)\;CH_3-CH(OH)-CH_2-CH_3,$$ i.e. $$(R)$$-butan-2-ol.

Thus the major product is $$(R)$$-2-butanol.

Option A which is: $$(\mathbf{R})$$-butan-2-ol

Product $$\mathrm{A}$$ is formed by acetalisation of the aldehyde group using ethanol and dry $$\mathrm{HCl}$$.

The primary bromide remains unaffected under these acidic conditions.

Product $$\mathrm{A}$$ is:

$$\mathrm{Br-CH_2-CH(OEt)_2}$$

Product $$\mathrm{B}$$ is formed by $$\mathrm{E2}$$ dehydrohalogenation using potassium tert-butoxide.

The strong bulky base removes $$\mathrm{HBr}$$ to form a double bond.

Product $$\mathrm{B}$$ is:

$$\mathrm{H_2C=C(OEt)_2}$$

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

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 occurs in presence of $$\mathrm{NaOH/EtOH}$$, which favours:

$$\mathrm{E2\ Elimination}$$

The base abstracts a $$\mathrm{\beta}$$-hydrogen adjacent to the carbon bearing chlorine.

Removal of the vinylic hydrogen is unfavourable, so the proton is removed from the neighbouring $$\mathrm{sp^3}$$ hybridised carbon containing the methyl group.

Simultaneously, the chloride ion leaves and a new $$\mathrm{\pi}$$ bond is formed.

This elimination produces a:

$$\mathrm{Conjugated\ Diene}$$

Conjugated dienes are highly stable due to delocalisation of $$\mathrm{\pi}$$ electrons.

Hence, the major product formed is the conjugated methylcyclopentadiene structure.

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

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: $$\mathrm{A}$$