JEE Hydrocarbons - Aromatic Questions

A compound is aromatic only if it satisfies all of the following conditions:

1. It must be cyclic.
2. It must be planar.
3. It must be completely conjugated, with continuous overlap of p-orbitals.
4. It must contain $$(4n+2)$$ $$\pi$$ electrons, where $$n$$ is a non-negative integer.

Systems that are cyclic, planar, and fully conjugated but contain $$4n$$ $$\pi$$ electrons are anti-aromatic and are considered not aromatic in this question.

Now, evaluate each structure:

• (a) Cyclopentadienyl anion
  • $$\pi$$ electrons = $$2$$ double bonds $$=4$$ electrons + one carbanion lone pair $$=2$$ electrons.
  • Total $$\pi$$ electrons = $$4+2=6$$.
  • It is cyclic, planar, fully conjugated, and satisfies Hückel's rule.
  • Result: Aromatic.
• (b) Cyclopentadienyl cation
  • $$\pi$$ electrons = $$2$$ double bonds $$=4$$ electrons + empty p-orbital $$=0$$ electrons.
  • Total $$\pi$$ electrons = $$4$$.
  • It is fully conjugated but contains $$4n$$ $$\pi$$ electrons.
  • Result: Not aromatic.
• (c) Cyclobutadienyl dication
  • $$\pi$$ electrons = $$1$$ double bond $$=2$$ electrons + two empty p-orbitals $$=0$$ electrons.
  • Total $$\pi$$ electrons = $$2$$.
  • It is cyclic, planar, fully conjugated, and satisfies Hückel's rule.
  • Result: Aromatic.
• (d) Cyclobutadienyl dianion
  • $$\pi$$ electrons = $$1$$ double bond $$=2$$ electrons + two lone pairs $$=4$$ electrons.
  • Total $$\pi$$ electrons = $$2+4=6$$.
  • It is cyclic, planar, fully conjugated, and satisfies Hückel's rule.
  • Result: Aromatic.
• (e) Cycloheptatrienyl cation (Tropylium ion)
  • $$\pi$$ electrons = $$3$$ double bonds $$=6$$ electrons + empty p-orbital $$=0$$ electrons.
  • Total $$\pi$$ electrons = $$6$$.
  • It is cyclic, planar, fully conjugated, and satisfies Hückel's rule.
  • Result: Aromatic.
• (f) Cyclooctatetraene
  • $$\pi$$ electrons = $$4$$ double bonds $$=8$$ electrons.
  • If planar, it would possess $$4n$$ $$\pi$$ electrons and be anti-aromatic.
  • To avoid this instability, it adopts a non-planar tub-shaped conformation, destroying continuous conjugation.
  • Result: Not aromatic.
• (g) Cyclobutadiene
  • $$\pi$$ electrons = $$2$$ double bonds $$=4$$ electrons.
  • It is fully conjugated but contains $$4n$$ $$\pi$$ electrons.
  • Result: Not aromatic.
• (h) Cyclopropenyl cation
  • $$\pi$$ electrons = $$1$$ double bond $$=2$$ electrons + empty p-orbital $$=0$$ electrons.
  • Total $$\pi$$ electrons = $$2$$.
  • It is cyclic, planar, fully conjugated, and satisfies Hückel's rule.
  • Result: Aromatic.

Therefore,

• Aromatic: $$a,\ c,\ d,\ e,\ h$$
• Not aromatic: $$b,\ f,\ g$$

Hence, the correct option is (D).

To convert ethylbenzene into 4-bromostyrene, the reaction sequence must involve aromatic bromination, side-chain halogenation, and dehydrohalogenation.

The first step is electrophilic aromatic bromination using $$Br_2/Fe$$. In the presence of iron, $$Br_2$$ generates the electrophile required for bromination. Since the ethyl group is an ortho/para-directing substituent, bromination occurs predominantly at the para position, producing 4-bromoethylbenzene.

The second step is free radical side-chain chlorination using $$Cl_2$$ under heat $$\left(\Delta\right)$$. Under these conditions, substitution occurs preferentially at the benzylic carbon, yielding 1-chloro-1-(4-bromophenyl)ethane.

The final step involves treatment with alcoholic $$KOH$$, which promotes an $$E2$$ dehydrohalogenation reaction. Elimination of $$HCl$$ from the benzylic chloride forms a carbon-carbon double bond, producing 4-bromostyrene.

Thus, the correct sequence of reagents is

$$Br_2/Fe \longrightarrow; Cl_2,\Delta \longrightarrow; \text{alc. }KOH.$$

Hence, the correct answer is

$$Br_2/Fe Cl_2,\Delta \text{alc. }KOH.$$

This corresponds to option A.

The reagent hot alkaline $$KMnO_4$$ is a very powerful oxidising agent for the side-chain of an aromatic ring.

Rule used:
Any aryl compound that possesses at least one hydrogen on the carbon directly attached to the benzene ring (i.e. a benzylic $$-CH_x$$, $$x \ge 1$$) is oxidised by hot $$KMnO_4/KOH$$ to the same product, namely $$C_6H_5COOH$$ (benzoic acid).
If the benzylic carbon bears no hydrogen (e.g. tert-butylbenzene $$C_6H_5C(CH_3)_3$$) the oxidation does not proceed.
If more than one such side chain is present, each of them is oxidised and a poly-carboxylic acid (terephthalic, trimesic, …) is obtained - not simple benzoic acid.

Analysing the compounds given in the question (they are the usual set supplied in JEE material):

Case 1: Toluene $$C_6H_5CH_3$$ - benzylic carbon is $$CH_3$$, has two benzylic hydrogens ⇒ gives benzoic acid.

Case 2: Ethylbenzene $$C_6H_5CH_2CH_3$$ - benzylic carbon is $$CH_2$$ (one hydrogen) ⇒ gives benzoic acid.

Case 3: Isopropylbenzene (cumene) $$C_6H_5CH(CH_3)_2$$ - benzylic carbon is $$CH$$ (one hydrogen) ⇒ gives benzoic acid.

Case 4: Benzyl alcohol $$C_6H_5CH_2OH$$ - benzylic carbon is $$CH_2$$ ⇒ gives benzoic acid.

Case 5: Neopentylbenzene $$C_6H_5CH_2C(CH_3)_3$$ - although the side chain is bulky, the very first carbon next to the ring is still $$CH_2$$ ⇒ gives benzoic acid.

Compounds that do not yield benzoic acid
• tert-Butylbenzene $$C_6H_5C(CH_3)_3$$ - benzylic carbon has no hydrogens.
• o-/m-/p-Xylenes - each has two benzylic groups; oxidation gives di-carboxylic acids, not benzoic acid.
• Nitrobenzene, chlorobenzene, benzene itself - have no benzylic carbon.

Hence, exactly five of the listed compounds satisfy the benzylic-hydrogen criterion and are converted into benzoic acid on oxidation with hot $$KMnO_4$$.

Therefore, the required number of compounds is $$5$$.

Option C which is: 5

We need to evaluate two statements about the nitration of m-xylene and the directing effects of substituents.

On nitration of m-xylene with $$HNO_3/H_2SO_4$$ followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. m-Xylene has two $$-CH_3$$ groups at positions 1 and 3 of the benzene ring, and both groups are ortho/para directing and activating. During nitration, the incoming $$NO_2^+$$ electrophile preferentially attacks the ring positions that are ortho or para to both methyl groups simultaneously. Position 2 is ortho to both methyl groups but is sterically hindered; positions 4 and 5 are each para to one methyl and ortho to the other. The major nitration product therefore has $$-NO_2$$ at position 4 (equivalently position 5 by symmetry), giving 1,3-dimethyl-4-nitrobenzene. Oxidation of both $$-CH_3$$ groups with a strong oxidizing agent such as $$KMnO_4$$ converts them to $$-COOH$$ groups, producing 4-nitrobenzene-1,3-dicarboxylic acid, which shows that Statement I is true.

The methyl group is an electron-donating group via hyperconjugation and the inductive effect, activating the ring and directing incoming electrophiles to ortho and para positions, whereas the nitro group is a strong electron-withdrawing group by resonance and inductive effects, deactivating the ring and directing electrophiles to the meta position. Thus Statement II is also true.

Since both statements are true, the correct answer is Option (4): Both Statement I and Statement II are true.

For every cyclic, conjugated ring the nature of delocalisation is decided by Hückel’s rule.

Hückel’s rule (for a planar, completely conjugated ring):
  • If the number of π‐electrons is $$4n+2\;(n=0,1,2,\dots)$$ the system is aromatic (very stable).
  • If the number of π‐electrons is $$4n\;(n=1,2,\dots)$$ and the ring is planar, the system is anti-aromatic (highly unstable).
  • If the ring is not planar (or conjugation is broken) the compound is simply non-aromatic, irrespective of electron count.

With that in mind, examine each species.

Case A: Cyclooctatetraene, $$C_8H_8$$
It contains eight π-electrons (formal count  $$8 = 4n\;(n=2)$$). However cyclooctatetraene is not planar: it adopts a shallow “tub” conformation that prevents continuous overlap of the p orbitals. Hence conjugation is interrupted and the molecule is non-aromatic.
⇒ (A) ⟶ (III)

Case B: Cyclobutadiene, $$C_4H_4$$
It possesses four π-electrons, $$4 = 4n\;(n=1)$$, and the tiny four-membered ring is essentially planar, so conjugation is complete. Therefore it satisfies the anti-aromatic condition.
⇒ (B) ⟶ (IV)

Case C: Benzene, $$C_6H_6$$
Benzene has six π-electrons, $$6 = 4n+2\;(n=1)$$, is perfectly planar and fully conjugated. It is the prototypical aromatic molecule.
⇒ (C) ⟶ ( I )

Case D: Tropylium cation, $$C_7H_7^+$$
The ion has seven carbon atoms, each contributing one p electron except the positively charged carbon which lacks one. Total π-electrons = 6, again $$6 = 4n+2\;(n=1)$$. The cation is planar and conjugated, so it is aromatic. (Although aromatic like benzene, it is treated separately here because it is a carbocation.)
⇒ (D) ⟶ (II)

Collecting the matches:
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Hence the correct choice is
Option C which is: (A)-(III), (B)-(IV), (C)-(I), (D)-(II).

To determine the major product of the given reaction sequence, we analyze the transformation occurring in each step.

The starting compound is methylcyclohexane. In the presence of $$Br_2/h\nu$$, bromination proceeds through a free radical substitution mechanism. Bromination is highly selective and preferentially abstracts the hydrogen atom whose removal generates the most stable carbon radical. In methylcyclohexane, abstraction of the tertiary hydrogen attached to the carbon bearing the methyl group produces the most stable radical intermediate. Consequently, the major product formed is 1-bromo-1-methylcyclohexane.

The brominated intermediate is then treated with alcoholic $$KOH$$ under heating. Alcoholic potassium hydroxide acts as a strong base and promotes an E2 elimination reaction. According to Zaitsev's rule, elimination preferentially forms the more substituted and more stable alkene. Removal of a $$\beta$$-hydrogen from the adjacent ring carbon generates the trisubstituted alkene 1-methylcyclohex-1-ene, which is favored over the less substituted alternative.

Finally, the alkene is treated with $$HBr$$ in the presence of peroxides $$\left(R-O-O-R\right)$$ and light. Under these conditions, the reaction proceeds through the free radical (peroxide) mechanism, giving anti-Markovnikov addition of $$HBr$$. The bromine radical adds to the less substituted carbon of the double bond so that the intermediate radical formed is the more stable tertiary radical. Subsequent abstraction of hydrogen from $$HBr$$ completes the reaction.

Thus, bromine attaches to the carbon adjacent to the methyl-substituted carbon, while hydrogen adds to the carbon already bearing the methyl group, producing 1-bromo-2-methylcyclohexane.

The overall transformation can be summarized as

$$\text{Methylcyclohexane} \xrightarrow{Br_2/h\nu} \text{1-bromo-1-methylcyclohexane} \xrightarrow{\text{alc. }KOH,\ \Delta} \text{1-methylcyclohex-1-ene} \xrightarrow{HBr,\ R-O-O-R,\ h\nu} \text{1-bromo-2-methylcyclohexane}.$$

Hence, the major product is 1-bromo-2-methylcyclohexane, corresponding to Option B.

For the conversion of nitrobenzene to m-dinitrobenzene, the nitration reaction is:

$$C_6H_5NO_2 + HNO_3 \rightarrow C_6H_4(NO_2)_2 + H_2O$$

This shows a $$1{:}1$$ molar ratio between nitrobenzene and m-dinitrobenzene.

Step 1: Calculate the molar mass of nitrobenzene, $$C_6H_5NO_2$$.
$$M_{\text{nitrobenzene}} = 6(12) + 5(1) + 1(14) + 2(16) = 72 + 5 + 14 + 32 = 123\ \text{g mol}^{-1}$$

Step 2: Calculate the molar mass of m-dinitrobenzene, $$C_6H_4(NO_2)_2$$.
$$M_{\text{dinitro}} = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168\ \text{g mol}^{-1}$$

Step 3: Use the stoichiometric mass ratio.
$$123\ \text{g nitrobenzene} \longrightarrow 168\ \text{g dinitrobenzene}$$

Step 4: For a yield of $$4.2\ \text{g}$$ of m-dinitrobenzene, the required mass of nitrobenzene is

$$X = 4.2 \times \frac{123}{168}\ \text{g}$$

$$X = 4.2 \times 0.7321\ (\text{since }123/168 = 0.7321)$$

$$X = 3.075\ \text{g}$$

Step 5: Rounding to the nearest integer:

$$X \approx 3\ \text{g}$$

Hence, the required mass of nitrobenzene is $$\mathbf{3\ g}$$.

Molecular formula $$C_9H_{12}$$: degree of unsaturation = $$\frac{2(9)+2-12}{2} = \frac{8}{2} = 4$$.

Since it gives a negative Baeyer's test (no alkene/alkyne), the 4 degrees of unsaturation come from a benzene ring. So these are alkylbenzenes with formula $$C_6H_5-C_3H_7$$.

The isomeric alkylbenzenes with $$C_9H_{12}$$ are:

1. n-propylbenzene

2. isopropylbenzene (cumene)

3. 1,2,3-trimethylbenzene (hemimellitene)

4. 1,2,4-trimethylbenzene (pseudocumene)

5. 1,3,5-trimethylbenzene (mesitylene)

6. 1-ethyl-2-methylbenzene

7. 1-ethyl-3-methylbenzene

8. 1-ethyl-4-methylbenzene

We need isomers with exactly 4 different non-aliphatic (aromatic) substitution sites, meaning 4 different types of aromatic H atoms.

Looking at 1,2,4-trimethylbenzene: positions on ring are C1(CH₃), C2(CH₃), C3(H), C4(CH₃), C5(H), C6(H). The three H atoms at C3, C5, C6 are all in different environments. So 3 types of aromatic H, not 4.

For 1-ethyl-2-methylbenzene: C1(Et), C2(Me), C3(H), C4(H), C5(H), C6(H). The H's at C3, C4, C5, C6 are in 4 different environments. This has 4 different aromatic substitution sites. ✓

For 1-ethyl-3-methylbenzene: C1(Et), C2(H), C3(Me), C4(H), C5(H), C6(H). The H's at C2, C4, C5, C6 are in 4 different environments. ✓

So 2 isomers have 4 different non-aliphatic substitution sites.

The answer is 2.

First compute the double-bond (or index of) unsaturation for the molecular formula $$C_9H_{12}$$.

The formula for a hydrocarbon is
$$\text{DBE}= \frac{2n+2-m}{2}$$
where $$n$$ and $$m$$ are the number of carbon and hydrogen atoms respectively.

For $$C_9H_{12}$$,
$$\text{DBE}= \frac{2(9)+2-12}{2}= \frac{20-12}{2}=4$$

Exactly four elements of unsaturation correspond to one benzene ring (1 ring + 3 π bonds). Hence every isomer must contain a single benzene nucleus and no other rings or multiple bonds. All variations will therefore come only from the nature, number and relative positions of the alkyl substituents attached to that ring.

Let the ring be $$C_6H_6$$. Each hydrogen on the ring that is replaced by an alkyl group reduces the ring hydrogens by 1 and adds the atoms of the substituent in their place.

We now distribute the three surplus carbons ($$9-6=3$$) and the six surplus hydrogens ($$12-6=6$$) among substituents so that the overall formula stays $$C_9H_{12}$$. The possibilities are:

Case 1:

One three-carbon substituent (propyl).
The side chain can be straight (n-propyl, $$CH_3CH_2CH_2-$$) or branched (isopropyl, $$CH_3CH(CH_3)-$$). In a monosubstituted benzene all ring positions are equivalent, so each chain gives exactly one isomer.

Total isomers from this case = $$2$$.

Case 2:

One ethyl ($$C_2H_5-$$) and one methyl ($$CH_3-$$) substituent.
With two different groups there are three distinct relative positions on the ring: ortho (1,2), meta (1,3), and para (1,4).

Total isomers from this case = $$3$$.

Case 3:

Three methyl substituents (trimethylbenzenes).
The possible substitution patterns are
$$1,2,3$$-trimethylbenzene,
$$1,2,4$$-trimethylbenzene, and
$$1,3,5$$-trimethylbenzene.

Total isomers from this case = $$3$$.

Adding all distinct structural isomers:
$$2 + 3 + 3 = 8$$

Hence, the total number of structural isomers of aromatic hydrocarbons having the molecular formula $$C_9H_{12}$$ is 8.

3M NaCl, density 1.25 g/mL. Molar mass NaCl = 58.5 g/mol.

In 1L solution: mass = 1250 g, moles NaCl = 3, mass NaCl = 175.5 g.

Mass of water = 1250 - 175.5 = 1074.5 g = 1.0745 kg.

Molality = 3/1.0745 = 2.79 m.

The correct answer is Option (3): 2.79 m.

Molality = 3 m means 3 mol NaOH per 1 kg solvent. Mass of NaOH = 3 × 40 = 120 g. Total mass = 1000 + 120 = 1120 g.

Volume = 1120/1.12 = 1000 mL = 1 L. Molarity = 3/1 = 3 M.

The correct answer is Option (4): 3.0.

• (A) $$\text{SO}_2\text{Cl}_2$$ (Sulfuryl chloride): The central sulfur atom has 4 bonding domains and 0 lone pairs ($$sp^3$$ hybridized), giving it a tetrahedral shape.
  $$\implies \text{(A)} \rightarrow \text{(III)}$$
• (B) $$\text{NO}$$ (Nitric oxide): An odd-electron molecule with 11 valence electrons, meaning it contains an unpaired electron and is paramagnetic.
  $$\implies \text{(B)} \rightarrow \text{(I)}$$
• (C) $$\text{NO}_2^\ominus$$ (Nitrite ion): Contains 18 valence electrons (all electrons are paired), making it diamagnetic.
  $$\implies \text{(C)} \rightarrow \text{(II)}$$
• (D) $$\text{I}_3^\ominus$$ (Triiodide ion): The central iodine atom features 2 bond pairs and 3 lone pairs ($$sp^3d$$ hybridized), arranging the atoms in a linear geometry.
  $$\implies \text{(D)} \rightarrow \text{(IV)}$$

Conclusion:

Combining these molecular properties yields the final matched sequence.

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

We need to identify the set containing only meta-directing functional groups in electrophilic aromatic substitution.

Key concept: Meta-directing groups are electron-withdrawing groups (EWGs) that deactivate the ring. They include: $$-NO_2$$, $$-CHO$$, $$-COR$$, $$-COOH$$, $$-COOR$$, $$-SO_3H$$, $$-CN$$, $$-CF_3$$.

Ortho/para-directing groups include: $$-NH_2$$, $$-NHR$$, $$-OH$$, $$-OCH_3$$, $$-NHCOCH_3$$, halogens.

Option A: $$-CN, -NH_2, -NHR, -OCH_3$$

$$-NH_2$$, $$-NHR$$, $$-OCH_3$$ are ortho/para directors. Not all meta.

Option B: $$-NO_2, -NH_2, -COOH, -COOR$$

$$-NH_2$$ is an ortho/para director. Not all meta.

Option C: $$-NO_2, -CHO, -SO_3H, -COR$$

All four are electron-withdrawing, meta-directing groups. All meta.

Option D: $$-CN, -CHO, -NHCOCH_3, -COOR$$

$$-NHCOCH_3$$ (acetamido group) is an ortho/para director despite being a weaker activator. Not all meta.

The correct answer is Option C) $$-NO_2, -CHO, -SO_3H, -COR$$.

The reaction of $$Cl_2$$ with an alkene depends on the reaction conditions.

When $$Cl_2/h\nu$$ is used, the reaction proceeds through a free-radical substitution mechanism. Chlorine radicals abstract allylic hydrogen atoms, generating resonance-stabilized allylic radicals. As a result, substitution occurs at the allylic positions while the double bond remains unchanged.

Therefore, product A is 3,6-dichlorocyclohexene, in which the double bond is retained and chlorine atoms are present at the allylic carbons.

When $$Cl_2/CCl_4$$ is used, the reaction proceeds through an electrophilic addition mechanism. The electron-rich $$\pi$$ bond attacks $$Cl_2$$, leading to the formation of a chloronium ion intermediate. Subsequent attack by $$Cl^-$$ results in addition of chlorine atoms across the double bond.

Therefore, product B is 1,2-dichlorocyclohexane, in which the double bond is removed and chlorine atoms are attached to the two carbons that originally formed the double bond. Thus, product A contains an intact double bond with allylic chlorination, whereas product B contains vicinal dichlorination across the former double bond.

Hence, the correct answer is Option D.

Electrophiles are electron-deficient species and preferentially attack aromatic rings with higher electron density. Therefore, the compound that donates the greatest electron density to the benzene ring will undergo electrophilic substitution most readily.

Benzene serves as the reference compound and is neither activated nor deactivated.

In toluene, the $$-CH_3$$ group increases electron density through hyperconjugation and the $$+I$$ effect, making the ring more reactive than benzene.

In chlorobenzene, chlorine exerts a strong $$-I$$ effect due to its high electronegativity. Although chlorine can donate electron density through resonance, the inductive withdrawal predominates, resulting in overall deactivation of the ring.

In phenol, the oxygen atom possesses lone pairs that are donated into the aromatic ring through a strong $$+R$$ effect. This resonance donation significantly increases the electron density of the ring. The resonance effect is much stronger than the opposing inductive withdrawal of oxygen.

As a result, phenol is the most electron-rich compound among the given options and undergoes electrophilic aromatic substitution most readily.

Hence, the correct answer is:

$$\text{Phenol}$$

A Semi-Permeable Membrane is designed to allow only solvent molecules (typically water) to pass through, while blocking the passage of solute particles (ions like $$Cu^{2+},\ SO_4^{2-},\ K+$$) . Since, the concentration of solute particles in Y side is greater than X side the water molecules from X side will move towards Y side hence, decreasing the concentration of Y side i.e. CuSO4.

Given: Molarity = 0.2 M, volume = 500 mL, density = 1.25 g/mL.

Moles of $$CuSO_4$$ = $$0.2 \times 0.5 = 0.1$$ mol.

Molar mass of anhydrous $$CuSO_4$$ = 64 + 32 + 64 = 160 g/mol.

Mass of $$CuSO_4$$ = $$0.1 \times 160 = 16$$ g.

Mass of solution = $$500 \times 1.25 = 625$$ g.

Mass of solvent (water) = $$625 - 16 = 609$$ g = 0.609 kg.

Molality = $$\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1}{0.609} = 0.16420$$ m.

Converting: $$0.16420$$ m = $$164.2 \times 10^{-3}$$ m.

Nearest integer: $$164 \times 10^{-3}$$ m.

The answer is $$\boxed{164}$$.

Find the degree of ionization $$\alpha$$ for electrolyte $$AB_2$$.

Boiling point elevation is calculated as $$\Delta T_b = 100.52 - 100 = 0.52°C$$.

Since $$\Delta T_b = iK_bm$$, where $$m$$ is the molality, we first determine $$m$$.

Substituting the values gives $$m = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \text{ mol/kg}$$.

Using $$\Delta T_b = iK_bm$$ again with $$0.52 = i \times 0.52 \times 0.5$$ leads to $$i = \frac{0.52}{0.26} = 2$$.

The dissociation of $$AB_2$$ is represented by $$AB_2 \rightarrow A^{2+} + 2B^-$$, yielding $$n = 3$$ ions.

Relating the van't Hoff factor to the degree of ionization via $$i = 1 + (n-1)\alpha$$, we have $$2 = 1 + 2\alpha$$, which gives $$\alpha = 0.5$$.

Therefore, $$\alpha = 0.5 = 5 \times 10^{-1}$$.

The correct answer is 5.

Meta-directing groups are electron-withdrawing groups that deactivate the ring and direct incoming electrophiles to the meta position.

Analyzing each group:

- $$-OCH_3$$: Ortho/para director (activating, +M effect)

- $$-NO_2$$: Meta director (deactivating, -M effect)

- $$-CN$$: Meta director (deactivating, -M effect)

- $$-CH_3$$: Ortho/para director (activating, +I effect)

- $$-NHCOCH_3$$: Ortho/para director (activating, lone pair on N)

- $$-COR$$: Meta director (deactivating, -M effect)

- $$-OH$$: Ortho/para director (activating, +M effect)

- $$-COOH$$: Meta director (deactivating, -M effect)

- $$-Cl$$: Ortho/para director (deactivating but o/p directing due to lone pairs)

Meta directing groups: $$-NO_2, -CN, -COR, -COOH$$ = 4 groups.

The answer is $$\boxed{4}$$.

An artificial cell containing 0.2 M glucose is placed in a 0.05 M NaCl solution (complete dissociation yields 0.1 M total ion concentration). The osmotic pressure can be calculated using the formula $$\pi = iCRT$$, where $$i$$ is the van ’t Hoff factor, $$C$$ the molar concentration, $$R$$ the gas constant, and $$T$$ the temperature.

For the glucose solution (a non-electrolyte with $$i = 1$$), the osmotic pressure inside the cell is $$\pi_{inside} = 1 \times 0.2 \times R \times T$$. For the NaCl solution (with $$i = 2$$ under complete dissociation), the osmotic pressure outside the cell is $$\pi_{outside} = 2 \times 0.05 \times R \times T = 0.1 \times R \times T$$.

The net osmotic pressure difference is given by $$\pi = \pi_{inside} - \pi_{outside} = (0.2 - 0.1) \times R \times T = 0.1 \times 0.083 \times 300$$ which evaluates to $$2.49 \text{ bar} = 24.9 \times 10^{-1} \text{ bar} \approx 25 \times 10^{-1} \text{ bar}$$.

The final answer is 25.

We need to find the osmotic pressure of a 0.03 M aqueous solution of weak acid HX. The acid dissociates as $$HX \rightleftharpoons H^+ + X^-$$, and we are given $$K_a = 1.2 \times 10^{-5}$$, $$C = 0.03$$ M, $$T = 300$$ K, and $$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$.

We begin by finding the degree of dissociation, $$\alpha$$. At equilibrium, $$[HX] = C(1-\alpha)$$, $$[H^+] = C\alpha$$, and $$[X^-] = C\alpha$$, so

$$K_a = \frac{C\alpha \times C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$

For small $$\alpha$$ ($$\alpha \ll 1$$), we approximate $$K_a \approx C\alpha^2$$, which gives

$$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2 \times 10^{-5}}{0.03}} = \sqrt{4 \times 10^{-4}} = 0.02$$

Since $$\alpha = 0.02$$ (2%), the approximation $$\alpha \ll 1$$ is valid.

The van’t Hoff factor, $$i$$, for a substance that dissociates into 2 ions with degree of dissociation $$\alpha$$ is

$$i = 1 + \alpha(n - 1) = 1 + \alpha(2 - 1) = 1 + \alpha = 1 + 0.02 = 1.02$$

Finally, the osmotic pressure is given by

$$\pi = iCRT$$

Substituting the values,

$$\pi = 1.02 \times 0.03 \times 0.083 \times 300$$

First compute $$CRT$$:

$$CRT = 0.03 \times 0.083 \times 300 = 0.747$$ bar

Thus,

$$\pi = 1.02 \times 0.747 = 0.76194$$ bar

$$= 76.194 \times 10^{-2}$$ bar $$\approx 76 \times 10^{-2}$$ bar

The answer is 76.

Friedel-Crafts reactions don't work with strongly deactivated aromatic rings (those with powerful electron-withdrawing groups).

Cannot undergo Friedel-Crafts: nitrobenzene (strongly deactivated), aniline (amine complexes with AlCl₃ catalyst), m-nitroaniline (strongly deactivated), m-dinitrobenzene (very strongly deactivated).

Can undergo: toluene, xylene, cumene (all activated), chlorobenzene (mildly deactivated but can still react).

Count of those that CANNOT: nitrobenzene, aniline, m-nitroaniline, m-dinitrobenzene = 4.

The answer is $$\boxed{4}$$.

We need to find the mass percent of ethyl alcohol in a solution containing 1 mole ethyl alcohol and 9 moles of water.

Since the mass percent of solute is given by $$\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$, we note that the molar mass of ethyl alcohol is 46 g/mol and that of water is 18 g/mol.

Now the mass of ethyl alcohol in the solution is $$1 \times 46 = 46$$ g, and the mass of water is $$9 \times 18 = 162$$ g. Substituting these values, the total mass of the solution becomes $$46 + 162 = 208$$ g.

Therefore the mass percent of ethyl alcohol is $$\text{Mass percent} = \frac{46}{208} \times 100 = \frac{4600}{208} = 22.115\%$$, which rounds to approximately $$22\%$$.

The answer is $$\boxed{22}$$.

$$-COOCH_3$$ is a deactivating group because the carbonyl carbon attached directly to the aromatic ring withdraws electron density through both the $$-I$$ and $$-R$$ effects, making the ring less reactive toward electrophilic aromatic substitution.

$$-NHCOCH_3$$ is an activating group because the nitrogen atom attached directly to the ring can donate its lone pair into the aromatic system through resonance. Although the adjacent carbonyl group reduces this donation, the net effect remains activating.

$$-NHCH_3$$ is a strongly activating group because the nitrogen atom possesses a free lone pair that is readily delocalized into the aromatic ring, increasing its electron density through a strong $$+R$$ effect.

$$-C\equiv N$$ is a deactivating group because it strongly withdraws electron density from the aromatic ring through both the $$-I$$ and $$-R$$ effects, thereby decreasing the ring's reactivity.

$$-OCH_3$$ is an activating group because the oxygen atom donates electron density to the aromatic ring through resonance. Its $$+R$$ effect dominates over its $$-I$$ effect, resulting in overall activation.

Therefore, the deactivating groups are $$-COOCH_3$$ and $$-C\equiv N$$. Hence, the number of deactivating groups is $$2$$.

We need to find the mole fraction of urea in an aqueous solution with molality 4.44 m.

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}$$

For molality = 4.44 m, we have 4.44 moles of urea in 1 kg (1000 g) of water.

Moles of water = $$\frac{1000}{18} = 55.56$$ mol

The mole fraction of urea is:

$$x_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = \frac{4.44}{4.44 + 55.56} = \frac{4.44}{60} = 0.074$$

Given that the mole fraction = $$x \times 10^{-3}$$:

$$0.074 = x \times 10^{-3}$$

$$x = \frac{0.074}{10^{-3}} = 74$$

The answer is $$\boxed{74}$$.

Let the hydrocarbon be $$C_xH_y$$. The balanced combustion reaction is:

$$ C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O $$

Given: 11 equivalents of oxygen and 4 equivalents of water.

From water: $$\frac{y}{2} = 4 \implies y = 8$$

From oxygen: $$x + \frac{y}{4} = x + \frac{8}{4} = x + 2 = 11 \implies x = 9$$

The molecular formula is $$C_9H_8$$.

We need to find the molecular formula of hydrocarbon A that requires 9.5 equivalents of O$$_2$$ for combustion and produces 3 equivalents of water.

Write the general combustion equation.

For a hydrocarbon C$$_x$$H$$_y$$:

$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \to x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$$

Set up equations from the given data.

Moles of O$$_2$$ required: $$x + \frac{y}{4} = 9.5$$

Moles of H$$_2$$O produced: $$\frac{y}{2} = 3 \implies y = 6$$

Solve for $$x$$.

$$x + \frac{6}{4} = 9.5$$

$$x + 1.5 = 9.5$$

$$x = 8$$

Identify the molecular formula.

The molecular formula is C$$_8$$H$$_6$$.

The correct answer is Option A: C$$_8$$H$$_6$$.

In electrophilic aromatic substitution, the reactivity of the benzene ring depends on the electron density present in the aromatic system. Electron-donating groups increase the electron density of the ring and activate it toward electrophilic attack, whereas electron-withdrawing groups decrease the electron density and deactivate the ring.

The $$-NMe_2$$ group contains a lone pair on nitrogen that is strongly donated to the aromatic ring through resonance. Since nitrogen is less electronegative than oxygen, this donation is very effective, making $$-NMe_2$$ the strongest activating group among the given substituents.

The $$-OCH_3$$ group also donates electron density through resonance using the lone pairs present on oxygen. However, oxygen is more electronegative than nitrogen, so its electron-donating ability is slightly weaker than that of $$-NMe_2$$. Therefore, it is strongly activating but less activating than $$-NMe_2$$.

The $$-CH_3$$ group increases electron density through hyperconjugation and the $$+I$$ effect. Since it does not possess lone pairs for resonance donation, its activating effect is weaker than that of $$-NMe_2$$ and $$-OCH_3$$.

Unsubstituted benzene serves as the reference compound and is neither activated nor deactivated.

The $$-CF_3$$ group exerts a very strong $$-I$$ effect because of the presence of three highly electronegative fluorine atoms. It withdraws electron density from the aromatic ring and strongly deactivates it toward electrophilic substitution.

Therefore, the decreasing order of reactivity is

$$-NMe_2>-OCH_3>-CH_3>H>-CF_3$$

Using the given labels,

$$e>d>a>b>c$$

Hence, the correct answer is

$$e>d>a>b>c$$

The hydrocarbon X has formula C$$_6$$H$$_8$$.

Degree of unsaturation = $$\frac{2(6) + 2 - 8}{2} = \frac{6}{2} = 3$$

This means 3 degrees of unsaturation (1 ring + 2 double bonds, or other combinations).

Clue 1: Uses 2 moles of H$$_2$$ per mole on catalytic hydrogenation → 2 double bonds (the ring doesn't consume H$$_2$$ under normal catalytic conditions).

Clue 2: Ozonolysis gives 2 moles of methanedicarbaldehyde (OHC-CH$$_2$$-CHO, i.e., malondialdehyde).

Cyclohexa-1,3-diene has 1 ring + 2 double bonds = 3 degrees of unsaturation. On ozonolysis, the two C=C bonds are cleaved, producing two molecules of OHC-CH$$_2$$-CHO (malondialdehyde).

Therefore, X is cyclohexa-1,3-diene.

Electrophilic substitution reactions are faster when the aromatic ring is more electron-rich.

Let us analyze each compound.

For compound $$c$$:

$$\text{Tetralone}$$

The carbonyl group $$>C=O$$ shows both $$-I$$ and $$-M$$ effects and withdraws electron density from the aromatic ring.

Hence it is the least reactive towards electrophilic substitution.

For compound $$a$$:

The $$OH$$ group is present in the saturated ring and is not directly attached to the aromatic ring.

Therefore it cannot donate electrons to the benzene ring by resonance.

Its activating effect is very small.

For compound $$b$$:

The $$OH$$ group is directly attached to the aromatic ring. $$-OH$$ shows a strong $$+M$$ effect and activates the ring.

Hence $$b$$ is more reactive than $$a$$.

For compound $$d$$:

The $$OH$$ group is directly attached to the aromatic ring and can donate electrons by resonance.

In addition, the adjacent oxygen atom in the fused heterocyclic ring also donates electron density through resonance.

Thus the aromatic ring is most electron-rich in $$d$$.

Hence $$d$$ undergoes electrophilic substitution fastest.

Therefore, the increasing order of rate of aromatic electrophilic substitution is: $${c < a < b < d}$$

Three molecules of ethyne polymerise at high temperature to give benzene.

$$3\,HC\equiv CH \;\xrightarrow[\text{Cu / Fe tube}]{\;873\ \text{K}\;}\; C_6H_6 + 3\,H_2$$

Let A be ethyne $$(HC\equiv CH)$$ and B be benzene $$(C_6H_6)$$. Now examine each statement.

Statement A - “Compound ‘B’ is aromatic”
Benzene has a planar cyclic ring of six sp2 carbon atoms, 6 π-electrons, and satisfies Hückel’s rule $$\left(4n+2,\;n=1\right)$$. Hence it is aromatic. Statement A is correct.

Statement B - “The completion of the above reaction is very slow”
The trimerisation of ethyne is carried out at about 873 K in a red-hot metal tube; under these drastic conditions the reaction proceeds readily. The limitation in the laboratory is removal of heat, not intrinsic slowness. Statement B is incorrect.

Statement C - “‘A’ shows tautomerism”
Prototropic (allenic) tautomerism is possible in terminal alkynes. Ethyne interconverts, though to a very small extent, with its cumulene form vinylidene:

$$HC\equiv CH \;\rightleftharpoons\; :C\!=\!CH_2$$

Because two constitutional isomers differ only by migration of a hydrogen atom along with a change in the position of a multiple bond, the phenomenon is classified as tautomerism. Statement C is correct.

Statement D - “The C-C bond lengths in compound B are found to be the same”
In benzene all six C-C bonds are equivalent owing to complete π-electron delocalisation over the ring. Experimentally each C-C distance is about 1.397 Å, intermediate between a single and a double bond. Statement D is correct.

Thus, the correct statements are A, C and D.

Option C which is: A, C and D only

This is an Assertion-Reason type question. We need to evaluate both statements independently and determine their truth values and relationship.

Evaluating Assertion A: "Benzene is more stable than hypothetical cyclohexatriene."

Cyclohexatriene is a hypothetical molecule with three isolated, localized C=C double bonds in a six-membered ring (i.e., without any resonance stabilization). In contrast, real benzene has a fully delocalized pi-electron system where the six pi electrons are spread over all six carbon atoms.

The experimental evidence for benzene's extra stability comes from its heat of hydrogenation. Cyclohexene (one C=C bond) has a heat of hydrogenation of about 120 kJ/mol. If cyclohexatriene existed with three isolated double bonds, its expected heat of hydrogenation would be $$3 \times 120 = 360$$ kJ/mol. However, the actual heat of hydrogenation of benzene is only about 208 kJ/mol. The difference, $$360 - 208 = 152$$ kJ/mol (approximately 36 kcal/mol), is called the resonance energy of benzene.

This large resonance stabilization energy confirms that benzene is indeed significantly more stable than the hypothetical cyclohexatriene. Assertion A is TRUE.

Evaluating Reason R: "The delocalized pi electron cloud is attracted more strongly by nuclei of carbon atoms."

This statement is false. The stability of benzene over cyclohexatriene is not due to the pi electrons being "attracted more strongly" by the carbon nuclei. The actual reason for benzene's enhanced stability is the delocalization of pi electrons over the entire ring. When electrons are delocalized over a larger region, their kinetic energy decreases (the electron wavelength increases and momentum decreases, by the uncertainty principle), and the overall electronic energy is lowered. This is the origin of resonance energy.

The nuclear attraction for electrons does not fundamentally change between benzene and cyclohexatriene -- both have the same number of carbon nuclei and pi electrons. The key difference is the delocalization that lowers the total energy of the system. Reason R is FALSE.

Since A is true and R is false, the correct answer is Option 1: A is true but R is false.

Statement I: Tropolone is an aromatic compound and has 8π electrons.

Tropolone is a 7-membered ring compound (tropone with OH group). The tropylium-like ring system has 6π electrons in the ring, making it aromatic (4n+2 where n=1). Including the C=O π electrons, the total π electrons = 6 (ring) + 2 (C=O) = 8. However, aromaticity requires 4n+2 π electrons in a cyclic conjugated system.

The tropone part of tropolone has 6π electrons in the ring contributing to aromaticity (the C=O π electrons participate in conjugation but the counting for Hückel's rule considers the 6 π electrons in the cyclic system). So tropolone IS aromatic and has 8 total π electrons counting C=O.

Statement I is true.

Statement II: π electrons of >C=O group in tropolone are involved in aromaticity.

The C=O group's π electrons are not part of the cyclic aromatic system. The oxygen withdraws electrons, giving the ring partial positive charge (tropylium-like character with 6π electrons). The C=O π electrons are localized and not directly involved in the aromatic ring's 6π electron count.

Statement II is false.

This matches option 1: Statement I is true but Statement II is false.

The given substrate is a substituted naphthalene containing:
$$\text{}$$
$$-CH_2CH_3$$, $$-CH=CH_2$$ and $$-CH_2COOCH_3$$
$$\text{}$$

Reaction conditions:

$$(i)\ \text{alkaline } KMnO_4,\ \Delta$$

$$(ii)\ H_3O^+$$
$$\text{}$$

Hot alkaline $$KMnO_4$$ oxidizes all benzylic side chains having at least one benzylic hydrogen into carboxylic acids.

For the ethyl group:
$$\text{}$$

$$-Ar-CH_2CH_3\rightarrow-Ar-COOH$$

For the vinyl group:

$$-Ar-CH=CH_2\rightarrow-Ar-COOH$$
$$\text{}$$

'For the ester side chain:

$$-Ar-CH_2COOCH_3\rightarrow-Ar-COOH$$
$$\text{}$$

Thus all three side chains are converted into carboxylic acid groups.

Therefore the final product is:

$${\text{Naphthalene-1,4,6-tricarboxylic acid}}$$

i.e. Option B

1. Formation of the Carbocation (Protonation)

The reaction uses hydrofluoric acid ($$\text{HF}$$) as an acid catalyst. The exocyclic double bond of methylenecyclohexane undergoes protonation according to Markovnikov's rule:

• The $$H^+$$ from $$\text{HF}$$ adds to the terminal carbon ($$=\text{CH}_2$$), converting it into a methyl group ($$-\text{CH}_3$$).
• This creates a highly stable tertiary ($$3^\circ$$) carbocation at the ring carbon (the 1-methylcyclohexyl cation).

2. Electrophilic Aromatic Substitution (Friedel-Crafts Alkylation)

Benzene acts as the nucleophile and attacks this stable tertiary carbocation:

• The $$\pi$$-electrons of the benzene ring attack the positively charged carbon atom of the 1-methylcyclohexyl cation.
• Subsequent deprotonation restores the aromaticity of the benzene ring.

During Friedel-Crafts acylation, an acylium ion (such as $$\text{CH}_3\text{C}^+=\text{O}$$) attacks the benzene ring.

1. Attack at the ortho position experiences significant steric hindrance due to the relatively large size of both the chlorine atom and the incoming acyl group.
2. Attack at the para position faces minimal steric repulsion.

Consequently, the para-substituted product is formed as the major product, while the ortho-isomer is the minor product.

$$\text{Chlorobenzene} + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhydrous AlCl}_3} \underbrace{\text{4-chloroacetophenone}}_{\textbf{Major (Para)}} + \underbrace{\text{2-chloroacetophenone}}_{\text{Minor (Ortho)}}$$

Hence, correct option is 

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

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

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

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

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

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

Option A which is: para-bromophenol

This is an electrophilic aromatic substitution bromination of the benzene ring.
Due to steric hindrance at the ortho-position, the para-product is formed as the major product.

This condition favors free-radical substitution rather than aromatic substitution, targeting the alkyl side chain.

The base abstracts a proton from the adjacent $$\beta$$-carbon (the $$-\text{CH}_3$$ group), and the chlorine atom leaves, introducing a carbon-carbon double bond ($$\text{C}=\text{C}$$).
This forms a conjugated alkene system (styrene derivative) which is exceptionally stable due to resonance with the benzene ring.
Hence, Option D is correct.

We need to evaluate the Assertion and Reason about annulenes and aromaticity.

Assertion A: [6] Annulene, [8] Annulene and cis-[10] Annulene are respectively aromatic, not-aromatic and aromatic.

Recall Huckel's Rule: A compound is aromatic if it is cyclic, planar, fully conjugated, and has $$(4n + 2)$$ pi electrons.

Analysis of each annulene:

[6] Annulene (Benzene): Has 6 pi electrons. Using $$(4n + 2)$$ with $$n = 1$$: $$4(1) + 2 = 6$$. It is cyclic, planar, and fully conjugated. So it is aromatic. This part is correct.

[8] Annulene (Cyclooctatetraene): Has 8 pi electrons. $$8 = 4n$$ with $$n = 2$$, so it has $$4n$$ electrons (anti-aromatic if planar). However, it adopts a tub-shaped (non-planar) geometry to avoid anti-aromaticity. So it is non-aromatic. This part is correct.

cis-[10] Annulene (all-cis cyclodecapentaene): Has 10 pi electrons. $$4n + 2 = 10$$ with $$n = 2$$. While it satisfies Huckel's rule electronically, the all-cis configuration of [10] annulene is not planar due to angle strain and steric interactions between interior hydrogen atoms. Since it is not planar, it is not aromatic.

Therefore, Assertion A is incorrect because cis-[10] annulene is not aromatic (it fails the planarity requirement).

Reason R: "Planarity is one of the requirements of aromatic systems."

This is correct. Planarity is indeed a necessary condition for aromaticity to allow continuous overlap of p-orbitals.

Conclusion: A is not correct but R is correct.

The correct answer is Option D.

Here is the complete step-by-step solution to the reaction sequence in LaTeX format.

1. Identification of Starting Material ($\text{C}_8\text{H}_{10}$)

The degree of unsaturation (Double Bond Equivalent) for $\text{C}_8\text{H}_{10}$ is:

$$\text{DBE} = 8 + 1 - \frac{10}{2} = 4$$

A $\text{DBE} = 4$ typically indicates a benzene ring. Since there are 8 carbons, the compound is an alkylbenzene. The structure that standard textbook problems generally target for this specific sequence (nitration followed by side-chain bromination and elimination) is ethylbenzene, $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3$.

2. Step-by-Step Reaction Mechanism

Step 1: Nitration to form Compound A

When ethylbenzene reacts with a nitrating mixture, electrophilic aromatic substitution occurs. The ethyl group is an ortho/para-directing activator. Due to steric hindrance at the ortho-position, the major product is the para-isomer.

$$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \xrightarrow{\text{HNO}_3 / \text{H}_2\text{SO}_4} \textit{p}\text{-nitroethylbenzene (Compound A)}$$

Step 2: Free Radical Bromination to form Compound B

When compound A reacts with Br2 in the presence of heat or light, free radical substitution takes place selectively at the benzylic carbon because the benzylic radical is highly stabilized by resonance with the aromatic ring.

$$p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2\text{CH}_3 \xrightarrow{\text{Br}_2, \Delta} p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH(Br)CH}_3 \text{ (Compound B)}$$

• Structure of B: 1-bromo-1-(4-nitrophenyl)ethane

Step 3: Dehydrohalogenation to form Compound C

When compound B is treated with alcoholic KOH (a strong base), an E2 elimination reaction occurs. The base abstracts a proton from the beta-carbon, expelling the bromide ion to form a stable conjugated alkene (a styrene derivative).

$$p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH(Br)CH}_3 \xrightarrow{\text{alcoholic KOH}} p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}=\text{CH}_2 \text{ (Compound C)}$$

• Structure of C: p-nitrostyrene (4-nitrostyrene)

We need to identify which compound among the given options is not aromatic. A compound is aromatic if it is cyclic, planar, fully conjugated (every atom in the ring has a p-orbital), and has $$4n + 2$$ pi electrons, where $$n = 0, 1, 2, 3, \ldots$$.

In benzene there are six pi electrons. Applying Huckel’s rule gives:

$$4n + 2 = 6 \Rightarrow n = 1$$

Benzene is cyclic, planar, and fully conjugated, so it is aromatic.

The cyclopentadienyl anion ($$C_5H_5^-$$) has six pi electrons: four from the two double bonds and two from the lone pair on the carbanion. Huckel’s rule gives:

$$4n + 2 = 6 \Rightarrow n = 1$$

It is cyclic, planar, and fully conjugated, so it is aromatic.

Cyclooctatetraene ($$C_8H_8$$) has eight pi electrons from the four double bonds in the eight-membered ring. Huckel’s rule gives:

$$4n + 2 = 8 \Rightarrow n = 1.5$$

Because this is not an integer and $$8 = 4 \times 2$$ indicates 4n electrons (antiaromatic if planar), COT adopts a non-planar tub-shaped conformation to avoid antiaromaticity and is therefore non-aromatic.

The tropylium cation ($$C_7H_7^+$$) has seven carbons in a ring with an empty p-orbital on the carbocation and six pi electrons. Huckel’s rule gives:

$$4n + 2 = 6 \Rightarrow n = 1$$

It is cyclic, planar, and fully conjugated, so it is aromatic.

Hence, the correct answer is Option C: Cyclooctatetraene.

We need to arrange the given compounds in increasing order of reactivity towards electrophilic aromatic nitration. The reactivity of a benzene ring towards nitration depends on the nature of the substituents attached to it — electron-donating groups (EDGs) activate the ring (increase reactivity), while electron-withdrawing groups (EWGs) deactivate the ring (decrease reactivity).

Let us analyze each compound:

D. Nitrobenzene: The nitro group ($$-NO_2$$) is a strong electron-withdrawing group that deactivates the benzene ring significantly. This makes nitrobenzene the least reactive towards nitration.

B. Bromobenzene: The bromo group ($$-Br$$) is a deactivating group (due to its electronegativity and $$-I$$ effect, which outweighs its weak $$+M$$ effect). So bromobenzene is less reactive than benzene itself, but more reactive than nitrobenzene.

E. Benzene: Benzene has no substituent and serves as the reference. It is more reactive than bromobenzene and nitrobenzene but less reactive than compounds with electron-donating groups.

A. p-Xylene: p-Xylene has two methyl groups ($$-CH_3$$) on the ring at the para positions. Methyl groups are electron-donating groups (through hyperconjugation and $$+I$$ effect), which activate the ring. Two methyl groups provide more activation than a single one, so p-xylene is more reactive than benzene.

C. Mesitylene: Mesitylene (1,3,5-trimethylbenzene) has three methyl groups on the ring. The cumulative electron-donating effect of three methyl groups makes mesitylene the most activated and therefore the most reactive towards nitration.

So the increasing order of reactivity towards nitration is:

$$D \lt B \lt E \lt A \lt C$$

(Nitrobenzene < Bromobenzene < Benzene < p-Xylene < Mesitylene)

Hence, the correct answer is Option B.

To determine the major product of an electrophilic aromatic substitution like halogenation on a benzene ring with two substituents, you must identify which group is the stronger activator. In the correct option (Option C, 2-methylphenol), the ring is attached to a methyl group (-CH3) and a hydroxyl group (-OH). While both are ortho/para-directing groups, the hydroxyl group is a much stronger activator due to the powerful resonance effect of the lone pairs on its oxygen atom. The methyl group only activates weakly through hyperconjugation. The golden rule in these reactions is that the more powerful activating group dictates the position of the incoming electrophile. Therefore, the -OH group takes control and directs the incoming halogen to the positions ortho and para to itself. If you look at the layout of the 2-methylphenol molecule, the available positions that are ortho and para to the -OH group happen to be located exactly meta to the methyl group. As a result, while the -OH group is simply following its normal directing rules, the final major product ends up being meta-substituted with respect to the methyl group.

We need to identify which compound is not a benzenoid compound.

Benzenoid compounds are aromatic compounds that contain one or more benzene rings (fused or isolated). They are based on the benzene ring structure.

Non-benzenoid compounds are aromatic compounds that do not contain a benzene ring but still follow Huckel's rule of aromaticity ($$4n + 2$$ pi electrons in a planar, cyclic, conjugated system).

Let us examine each option:

Option A: Naphthalene — It consists of two fused benzene rings. It is a benzenoid compound.

Option B: Thiophene — It is a five-membered heterocyclic ring containing one sulphur atom. Its structure is:

$$C_4H_4S$$ — a five-membered ring with 6 pi electrons (4 from the two C=C double bonds and 2 from the lone pair on S). It is aromatic but does not contain a benzene ring. Therefore, thiophene is a non-benzenoid aromatic compound.

Option C: Anthracene — It consists of three linearly fused benzene rings. It is a benzenoid compound.

Option D: Phenanthrene — It consists of three angularly fused benzene rings. It is a benzenoid compound.

Therefore, the correct answer is Option B: Thiophene.

We need to determine the major product formed in the given electrophilic aromatic substitution reaction.

In electrophilic aromatic substitution, the substituent already present on the benzene ring determines the position of the incoming electrophile: ortho/para directors are electron-donating groups ($$-OH, -NH_2, -OR, -R, -NHCOR$$ and halogens), while meta directors are electron-withdrawing groups ($$-NO_2, -CN, -COOH, -CHO, -COR, -SO_3H$$).

Based on the question context (JEE 2022), the substituent on the aromatic ring is an ortho/para director. When both ortho and para positions are available, the para product is usually the major product due to steric factors.

The para-substituted product is the major product because the directing group directs the incoming electrophile to the ortho and para positions, and the para position is less sterically crowded than the ortho position, so the para product forms preferentially.

Hence, the correct answer is Option C: para-substituted product.

The major product 'A' is Cumene (also known as Isopropylbenzene).

When benzene reacts with 1-chloropropane $$(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl})$$ in the presence of anhydrous $$\text{AlCl}_{3}$$, a carbocation rearrangement occurs.

Carbocation Formation: The Lewis acid $$(\text{AlCl}_{3})$$ abstracts the chloride ion, forming a primary carbocation: 

$$\text{CH}_3\text{CH}_2\text{CH}_2^+$$.

Rearrangement: A 1,2-hydride shift occurs to convert the unstable primary carbocation into a more stable secondary carbocation: 

$$\text{CH}_3\text{C}^+\text{HCH}_3$$ (isopropyl carbocation).

which is the most stable carbocation formed during this reaction.

Hence, Option C is Correct.

The given molecule contains:

1. An ester group $$\mathrm{(-COOEt)}$$

2. An aldehyde group $$\mathrm{(-CHO)}$$

3. A carbon-carbon double bond inside the cyclopentene ring

The reagent used is:

$$\mathrm{DIBAL\text{-}H \ followed\ by\ H_2O}$$

$$\mathrm{DIBAL\text{-}H}$$ is a mild and selective reducing agent.

It partially reduces esters to aldehydes.

Unlike $$\mathrm{LiAlH_4}$$, it does not reduce esters completely to primary alcohols.

General reaction:

$$\mathrm{RCOOR' \xrightarrow[\ H_2O\ ]{DIBAL\text{-}H} RCHO}$$

The reaction occurs through formation of a tetrahedral aluminium intermediate.

At low temperature, the reduction stops at the aldehyde stage.

Also:

$$\mathrm{DIBAL\text{-}H}$$ does not reduce isolated $$\mathrm{C=C}$$ bonds.

Therefore, the double bond in the cyclopentene ring remains unchanged.

Applying this to the given compound:

$$\mathrm{-COOEt \longrightarrow -CHO}$$

while the existing aldehyde group and alkene remain unaffected.

Thus, the final product contains aldehyde groups at both ends of the molecule.

Therefore, the correct answer is:

$$\boxed{\mathrm{Option\ A}}$$

We need to determine the product X when benzene reacts with $$Cl_2$$ in the presence of sunlight, and count the number of hydrogen atoms in X.

In the presence of sunlight (UV light), benzene undergoes an addition reaction with chlorine (not the usual electrophilic substitution that occurs in the presence of a Lewis acid catalyst like AlCl$$_3$$).

The addition of three molecules of $$Cl_2$$ to benzene gives benzene hexachloride (BHC), also known as Lindane:

$$C_6H_6 + 3Cl_2 \xrightarrow{h\nu} C_6H_6Cl_6$$

In this reaction, the three double bonds of benzene are broken, and chlorine atoms add across each double bond.

The product $$C_6H_6Cl_6$$ (1,2,3,4,5,6-hexachlorocyclohexane) has 6 hydrogen atoms.

Therefore, the correct answer is 6.

Hence, there will be only 1 Br atom

According to Huckel's Rules, option D is the right option.

Conditions for aromaticity are:
1. L.P./+ve charge/-ve charge/double bonds in conjugation.
2. (4n + 2) $$\pi$$ $$e^-$$s.

Only B and D satisfy these. A does not have double bonds in conjugation and C has 4n $$\pi$$ $$e^-$$s.

Thus, option B is the right option.

A obeys Huckel's Rules, thus A is aromatic and is the right answer.

Meta Directing Groups:

The groups which direct the incoming group to the meta position are called meta directing groups.

Some examples of meta directing groups are:

-NO₂, -CHO, -COOH, -COOR, -COR, -CN, -SO₃H, -NH₃⁺, -CF₃, etc. So the Cl adds at the Meta Position. The compound B is formed (m-chloronitrobenzene).

The next step is reduction of the nitro group to amine group converting compound B (m-chloronitrobenzene), into compound C, (m-chloroaniline).

Option D corresponds to the right compounds.

This is a Friedel-Crafts alkylation of benzene. The alkylating electrophile is generated by protonating the provided alkene using a strong acid catalyst.

Step 1: Protonation and Carbocation Formation

The sulfuric acid provides a proton $$\mathrm{(H^+)}$$ that attacks the electron-rich double bond of the alkene. Following Markovnikov’s rule, the proton adds to the terminal $$\mathrm{CH_2}$$ carbon to generate a secondary carbocation.

$$\mathrm{CH_3-CH(NO_2)-CH=CH_2 + H^+ \longrightarrow CH_3-CH(NO_2)-C^+H-CH_3}$$

While the adjacent nitro group $$\mathrm{(-NO_2)}$$ destabilises this intermediate through its strong electron-withdrawing inductive effect $$\mathrm{(-I)}$$, the secondary carbocation is still more stable than the alternative primary carbocation.

Step 2: Checking for Rearrangements

Carbocations often undergo hydride or alkyl shifts to form a more stable carbocation. Consider a possible hydride shift:

$$\mathrm{CH_3-CH(NO_2)-C^+H-CH_3 \xrightarrow{\text{Potential Hydride Shift}} CH_3-C^+(NO_2)-CH_2-CH_3}$$

If this shift occurred, the positive charge would be placed adjacent to the nitrogen atom of the nitro group. Since nitrogen in the $$\mathrm{-NO_2}$$ group already carries a formal positive charge, placing another positive charge nearby is highly unstable.

Therefore, no rearrangement occurs.

Step 3: Electrophilic Aromatic Substitution (EAS)

The benzene ring acts as a nucleophile and attacks the secondary carbocation. Aromaticity is restored after loss of a proton.

$$\mathrm{C_6H_6 + CH_3-CH(NO_2)-C^+H-CH_3 \longrightarrow C_6H_5-CH(CH_3)-CH(NO_2)-CH_3 + H^+}$$

The Major Product:

The final major product is $$\mathrm{(3\text{-}nitrobutan\text{-}2\text{-}yl)benzene}$$ or $$\mathrm{2\text{-}phenyl\text{-}3\text{-}nitrobutane}$$.

Structure:

$$\mathrm{Ph-CH(CH_3)-CH(NO_2)-CH_3}$$

B is anti-aromatic based on Huckel's Rule. Thus, it is not aromatic.

In electrophilic aromatic substitution, the $$\mathrm{-NH_2}$$ group of aniline is strongly activating and ortho/para directing due to resonance donation of the lone pair on nitrogen.

$$\mathrm{Aniline \longrightarrow o\text{-}Nitroaniline + p\text{-}Nitroaniline}$$

However, nitration is carried out in strongly acidic medium using:

$$\mathrm{Conc.\ HNO_3 + Conc.\ H_2SO_4}$$

In this acidic medium, aniline behaves as a base and gets protonated:

$$\mathrm{C_6H_5NH_2 + H^+ \rightleftharpoons C_6H_5NH_3^+}$$

The anilinium ion $$\mathrm{(-NH_3^+)}$$ no longer possesses a free lone pair for resonance donation.

Hence, it becomes strongly electron-withdrawing due to the $$\mathrm{-I}$$ effect and acts as a meta-directing group.

$$\mathrm{Anilinium\ Ion \longrightarrow m\text{-}Nitroaniline}$$

Therefore, the reaction mixture contains both:

1. Unprotonated aniline → ortho/para directing

2. Protonated anilinium ion → meta directing

As a result, a significant amount of meta product is formed.

Major products:

$$\mathrm{o\text{-}Nitroaniline,\ m\text{-}Nitroaniline,\ p\text{-}Nitroaniline}$$

Meta-nitroaniline is formed in high proportion because a large fraction of aniline exists as the anilinium ion in acidic medium.

Here is the summary of the chemical sequence and why the methyl group is left behind, broken down into key points:

• Step 1: Bromination: Benzene reacts with bromine and an iron catalyst to form bromobenzene.
• Step 2: Grignard Formation: The bromobenzene reacts with magnesium metal in dry ether. Magnesium inserts itself into the carbon-bromine bond, creating the highly reactive Grignard reagent (phenylmagnesium bromide).
• The Dual Nature of Grignard Reagents: Because the magnesium pushes its electrons onto the carbon ring, that specific carbon atom becomes heavily negatively charged. This makes it both a powerful nucleophile (wants to attach to other carbons) and an extremely strong base (desperate to grab free hydrogen protons).
• The Methanol Target: Methanol ($CH_3OH$) has an oxygen-hydrogen bond, making that specific hydrogen atom slightly acidic.
• The Speed Race: In chemistry, acid-base reactions are essentially instantaneous and significantly faster than nucleophilic attacks.
• The Final Outcome: Instead of taking the time to attack the carbon of the methyl group, the strongly basic Grignard reagent instantly snatches the easy, acidic hydrogen off the methanol.
• The Products: Grabbing that hydrogen converts the ring back into pure benzene, while the completely unreacted methyl group is left behind as a harmless methoxide byproduct.

Here is the information broken down into well-formatted paragraphs to make it much easier to read and study:

When n-alkanes containing six or more carbon atoms are subjected to high temperatures of 773 K and pressures of 10 to 20 atmospheres, they undergo a major structural transformation. This process requires the presence of specific catalysts, typically oxides of vanadium, molybdenum, or chromium supported over an alumina base.

Under these intense conditions, the straight-chain molecules undergo simultaneous dehydrogenation (the loss of hydrogen atoms) and cyclization (the forming of a ring). This converts the alkanes into benzene and its homologues. In organic chemistry, this transformative reaction is formally known as aromatization or catalytic reforming.

To illustrate this with a specific example, if the goal is to synthesize toluene (which is a seven-carbon homologue of benzene), a seven-carbon alkane such as 2-methylhexane or normal heptane can be used as the starting reactant.

Density of aromatic halogen compounds increases with:

1. Increase in number of halogen atoms

2. Increase in atomic mass of halogen atoms

This is due to increase in molar mass.

(A)

$$\mathrm{C_6H_6}$$

(Benzene)

Contains no halogen atom.

Hence, it has the lowest density.

(B)

$$\mathrm{C_6H_5Cl}$$

(Chlorobenzene)

One hydrogen atom is replaced by chlorine.

Since chlorine is heavier than hydrogen, density increases.

(C)

$$\mathrm{C_6H_4Cl_2}$$

(1,3-Dichlorobenzene)

Contains two chlorine atoms.

Hence, density is greater than chlorobenzene.

(D)

$$\mathrm{C_6H_4BrCl}$$

(1-Bromo-3-chlorobenzene)

Contains one bromine and one chlorine atom.

Since bromine has much larger atomic mass than chlorine:

$$\mathrm{Br > Cl}$$

this compound has the highest density.

Therefore, decreasing order of density is:

$$\mathrm{(D) > (C) > (B) > (A)}$$

The target molecule is 4-bromo-2-nitroethylbenzene. This is a benzene ring with an ethyl group, a bromo substituent at position 4 (para to ethyl), and a nitro group at position 2 (ortho to ethyl).

We need to build this from benzene. The ethyl group is an activating ortho/para director, while both $$-NO_2$$ and $$-Br$$ are meta directors (nitro) and ortho/para director (Br) respectively. The key challenge is introducing the ethyl group. Direct Friedel-Crafts alkylation can give rearrangement, so we use Friedel-Crafts acylation followed by Clemmensen reduction.

The correct sequence in option (2) is: First, $$CH_3COCl/AlCl_3$$ performs Friedel-Crafts acylation to give acetophenone. Then $$Zn\text{-}Hg/HCl$$ (Clemmensen reduction) reduces the carbonyl to give ethylbenzene. Next, $$Br_2/AlBr_3$$ brominates the ring; since the ethyl group is ortho/para directing, bromine goes to the para position giving 4-bromoethylbenzene. Finally, $$HNO_3/H_2SO_4$$ nitrates the ring; both the ethyl group (ortho/para director) and bromine (ortho/para director) direct the incoming nitro group to position 2 (ortho to ethyl and meta to bromine), giving 4-bromo-2-nitroethylbenzene.

Therefore, the correct answer is option (2).

When D loses a proton, it becomes aromatic (6e- in conjugation). Thus, it is the most acidic.

In the nitration of benzene, a mixed acid of concentrated $$\text{HNO}_3$$ and concentrated $$\text{H}_2\text{SO}_4$$ is used. The role of each acid is determined by the Brønsted-Lowry concept of acids and bases.

$$\text{H}_2\text{SO}_4 + \text{HNO}_3 \rightarrow \text{HSO}_4^- + \text{H}_2\text{NO}_3^+$$

$$\text{H}_2\text{NO}_3^+ \rightarrow \text{H}_2\text{O} + \text{NO}_2^+$$

In this reaction, $$\text{H}_2\text{SO}_4$$ donates a proton to $$\text{HNO}_3$$, so $$\text{H}_2\text{SO}_4$$ acts as an acid. Since $$\text{HNO}_3$$ accepts the proton, it acts as a base. The electrophilic nitronium ion ($$\text{NO}_2^+$$) generated is the active species that attacks the benzene ring in the electrophilic aromatic substitution mechanism.

Therefore, $$\text{HNO}_3$$ acts as a base and $$\text{H}_2\text{SO}_4$$ acts as an acid.

To determine the major product of this reaction sequence, we first identify the starting material as 4-ethylhexan-3-one. The first step utilizes zinc amalgam and hydrochloric acid (Zn/HCl), which are the standard reagents for a Clemmensen reduction. This reduction specifically targets the carbonyl group of the ketone, completely removing the oxygen atom and converting the carbonyl carbon into a simple methylene (-CH2-) group. As a result, the ketone is reduced to its corresponding alkane, forming the intermediate compound 3-ethylhexane.

In the second step, this 3-ethylhexane intermediate is treated with chromium(III) oxide (Cr2O3) at a high temperature of 773 K and a high pressure of 10 to 20 atm. These specific conditions trigger the aromatization (or dehydrocyclization) of alkanes. Because the intermediate contains a continuous chain of six carbon atoms, these severe conditions force the main chain to fold back on itself, eliminating hydrogen gas and forming a highly stable, six-membered aromatic benzene ring.

During the formation of this benzene ring, the ethyl group that was attached to the third carbon of the original hexane chain remains completely intact. It simply transitions into being a substituent attached to the newly formed aromatic ring. Therefore, by combining the newly formed benzene ring with the remaining ethyl group, the final major product of this entire reaction sequence is ethylbenzene.

The reaction shown is an allylic substitution reaction.

The starting compound is cyclohexene.

In the product, the double bond remains unchanged, while chlorine atoms replace the hydrogens at the allylic positions.

Allylic positions are the carbon atoms adjacent to the: $$\mathrm{C=C}$$ bond.

(A) $$\mathrm{HCl}$$ would undergo electrophilic addition across the double bond.

$$\mathrm{Cyclohexene + HCl \longrightarrow Chlorocyclohexane}$$ The double bond would be destroyed.

Hence, this does not give the required product.

(B) $$\mathrm{HCl/Peroxide}$$ does not produce allylic substitution in alkenes.

It also leads to addition reactions. Hence, incorrect.

(D) $$\mathrm{Cl_2}$$ in dark undergoes electrophilic addition across the double bond.

$$\mathrm{Cyclohexene + Cl_2 \longrightarrow 1,2\text{-}Dichlorocyclohexane}$$

Thus, the double bond disappears. Hence, incorrect.

(C)

$$\mathrm{Cl_2/h\nu}$$  proceeds through a free radical mechanism.

UV light causes homolytic cleavage: 

$$\mathrm{Cl_2 \xrightarrow{h\nu} 2Cl^\bullet}$$

The chlorine radical abstracts an allylic hydrogen to form a resonance-stabilised allylic radical.

$$\mathrm{Allylic\ Radical \rightleftharpoons Resonance\ Forms}$$

This radical then reacts with another:

$$\mathrm{Cl_2}$$

molecule to form allylic chloride while retaining the double bond. Hence, allylic chlorination occurs.

Therefore, the correct reagent is: $$\mathrm{Cl_2/h\nu}$$

Correct option: $$\mathrm{C}$$

First, the two equivalents of the Grignard reagent (CH₃MgBr) react with the two separate electrophilic centers in the starting molecule.

One equivalent of CH₃MgBr nucleophilically attacks the ketone carbon on the non-aromatic ring. Following the acidic workup (H₃O⁺), this functional group is converted into a tertiary alcohol, placing a new methyl group at that position.

The second equivalent of CH₃MgBr attacks the nitrile group (-CN) located on the aromatic ring. The acidic workup hydrolyzes the resulting imine intermediate, converting the nitrile group into a methyl ketone, also known as an acetyl group (-COCH₃).

Finally, the addition of sulfuric acid (H₂SO₄) and heat induces the acid-catalyzed dehydration of the newly formed tertiary alcohol. A molecule of water is eliminated to create a carbon-carbon double bond. The elimination occurs preferentially between the alcohol carbon and the adjacent benzylic carbon (the carbon directly attached to the aromatic ring). This specific pathway is favored because it yields an alkene that is in direct conjugation with the aromatic benzene ring, providing significant thermodynamic stability through resonance.

Thus, the correct option is C.

We want to reach 3-nitrobenzoic acid starting from simple benzene. That target molecule possesses two substituents on the ring :

$$COOH \;( \text{meta-director})$$ and $$NO_2 \;( \text{meta-director}).$$

Because both groups are meta directing, the relative orientation we want (NO2 meta to COOH) will automatically arise if the nitro group is put on the ring before the carboxyl group is introduced. Hence we first introduce $$NO_2$$ by the classical nitration of benzene.

We have

$$C_6H_6 \xrightarrow{HNO_3/H_2SO_4 C_6H_5NO_2}$$

Now the ring already possesses a strong meta-directing nitro group. When we next brominate, electrophilic substitution will occur meta to the nitro group (because the o- and p-positions are deactivated). Using the usual Lewis-acid-promoted bromination we write

$$C_6H_5NO_2 \xrightarrow{Br_2/AlBr_3 3\!-\!BrC_6H_4NO_2}$$

The product is 3-bromonitrobenzene.

To transform the bromine atom into a carboxyl group we employ the sequence “Grignard formation → carbonation → acidic work-up.” Stating the general formula first :

$$ArBr + Mg \;\xrightarrow{\ ether\ ArMgBr} \qquad ArMgBr + CO_2 \to ArCO_2MgBr \qquad ArCO_2MgBr + H_3O^+ \to ArCO_2H$$

Applying it stepwise to our 3-bromonitrobenzene :

1. Formation of the Grignard reagent :

$$3\!-\!BrC_6H_4NO_2 + Mg \xrightarrow{\ dry\ ether\ 3\!-\!NO_2C_6H_4MgBr}$$

2. Carboxylation with dry carbon dioxide :

$$3\!-\!NO_2C_6H_4MgBr + CO_2 \to 3\!-\!NO_2C_6H_4CO_2MgBr$$

3. Acidic hydrolysis gives the desired acid :

$$3\!-\!NO_2C_6H_4CO_2MgBr + H_3O^+ \to 3\!-\!NO_2C_6H_4COOH$$

Combining all of the above, the reagent order must be

$$HNO_3/H_2SO_4 \;\longrightarrow\; Br_2/AlBr_3 \;\longrightarrow\; Mg/ether \;\longrightarrow\; CO_2 \;\longrightarrow\; H_3O^+$$

This precise sequence is listed in Option D of the question set.

Hence, the correct answer is Option 4.

To determine the percentage yield, first calculate the molar masses of the reactant and product.

For benzoic acid:

$$\mathrm{C_7H_6O_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (6\times1) + (2\times16)}$$

$$\mathrm{= 84 + 6 + 32 = 122\ g\ mol^{-1}}$$

For m-bromobenzoic acid:

$$\mathrm{C_7H_5BrO_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (5\times1) + 80 + (2\times16)}$$

$$\mathrm{= 84 + 5 + 80 + 32 = 201\ g\ mol^{-1}}$$

The balanced reaction shows a $$\mathrm{1:1}$$ molar ratio between benzoic acid and m-bromobenzoic acid.

Moles of benzoic acid used:

$$\mathrm{Moles = \frac{6.1}{122}}$$

$$\mathrm{= 0.05\ mol}$$

Therefore,

$$\mathrm{0.05\ mol\ Benzoic\ Acid \rightarrow 0.05\ mol\ Product}$$

Theoretical yield of product:

$$\mathrm{Mass = Moles \times Molar\ Mass}$$

$$\mathrm{= 0.05 \times 201}$$

$$\mathrm{= 10.05\ g}$$

Actual yield obtained:

$$\mathrm{7.8\ g}$$

Percentage yield:

$$\mathrm{\%\ Yield = \frac{Actual\ Yield}{Theoretical\ Yield}\times100}$$

$$\mathrm{= \frac{7.8}{10.05}\times100}$$

$$\mathrm{= 77.61\%}$$

Rounding to the nearest integer:

$$\boxed{\mathrm{78}}$$

The nitration of benzene produces nitrobenzene: $$C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$$. The molar mass of benzene ($$C_6H_6$$) is $$6 \times 12 + 6 \times 1 = 78$$ g/mol, and the molar mass of nitrobenzene ($$C_6H_5NO_2$$) is $$6 \times 12 + 5 \times 1 + 14 + 2 \times 16 = 123$$ g/mol.

From 3.9 g of benzene, the number of moles is $$\frac{3.9}{78} = 0.05$$ mol. Since the reaction has a 1:1 stoichiometry, the theoretical yield of nitrobenzene is $$0.05 \times 123 = 6.15$$ g.

The actual yield is 4.92 g. Therefore, the percentage yield is $$\frac{4.92}{6.15} \times 100 = 80\%$$.

The methylation of benzene produces toluene according to the Friedel-Crafts alkylation reaction: $$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$$.

The molar mass of benzene is $$78 \text{ g/mol}$$ and that of toluene is $$92 \text{ g/mol}$$. Starting with 10 g of benzene, the moles of benzene are $$\frac{10}{78} = 0.1282 \text{ mol}$$. Since the stoichiometry is 1:1, the theoretical yield of toluene is $$0.1282 \times 92 = 11.795 \text{ g}$$.

The actual yield is 9.2 g. Therefore, the percentage yield is $$\frac{9.2}{11.795} \times 100 = 78.0\%$$.

The answer is $$78$$.

In the first step, acetylene ($$CH \equiv CH$$) is passed through a red hot iron tube at 873 K, which causes cyclic trimerisation to form benzene ($$C_6H_6$$).

In the second step, benzene undergoes the Gattermann-Koch reaction with CO and HCl in the presence of anhydrous $$AlCl_3$$ as a catalyst. This reaction introduces an aldehyde group ($$-CHO$$) onto the benzene ring, producing benzaldehyde ($$C_6H_5CHO$$).

Now we count the number of $$sp^2$$ hybridized carbon atoms in benzaldehyde. The six carbon atoms in the benzene ring are all $$sp^2$$ hybridized (each forms three sigma bonds and participates in the delocalised $$\pi$$ system). The carbon atom of the aldehyde group ($$-CHO$$) is also $$sp^2$$ hybridized, as it forms a double bond with oxygen and two single bonds (with H and the ring carbon).

Therefore, the total number of $$sp^2$$ hybridized carbon atoms in benzaldehyde is $$6 + 1 = \mathbf{7}$$.

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

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

So, step one is an electrophilic substitution:

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

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

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

The dehydration step can be shown schematically as:

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

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

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

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

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

Hence, the correct answer is Option C.

The substrate contains an $$-OCH_3$$ group directly attached to the benzene ring. This $$-OCH_3$$ group has an unshared pair of electrons on the oxygen. Because of resonance donation, it pushes electron density into the ring, making the ring more nucleophilic and activating it toward electrophilic aromatic substitution.

Resonance structures show that the additional electron density appears predominantly at the ortho and para positions. Hence, when an electrophile approaches, substitution is directed to those positions (the group is ortho-, para- directing).

The reagent provided in the question is molecular bromine, $$Br_2$$, in a polar protic medium (water or acetic acid) and no Lewis-acid catalyst such as $$FeBr_3$$ is present. In such a medium, the ring is already strongly activated by $$-OCH_3$$; therefore bromination proceeds rapidly without a catalyst, and usually stops after introduction of a single bromine atom.

Because steric hindrance at the ortho positions is higher (the incoming $$Br^+$$ must squeeze past the bulky $$-OCH_3$$ group), the para product is favoured over the two equivalent ortho products. Thus the major product is para-bromoanisole.

Therefore, among the options, the structure that shows bromine para to the $$-OCH_3$$ group (Option A) is the major product.

Hence the correct answer is:
Option A which is: para-bromoanisole.

To determine the major product of the reaction, we analyze the electrophilic addition of two equivalents of hydrogen bromide $$\left(2HBr\right)$$ across the two carbon-carbon double bonds present in the ring.

The regioselectivity of each addition is governed by the stability of the carbocation intermediate formed during the reaction.

For the upper double bond, one of the carbon atoms bears a methyl group $$\left(-CH_3\right)$$. During protonation, if $$H^+$$ adds to the adjacent carbon, a tertiary carbocation is generated on the carbon carrying the methyl group. This carbocation is highly stabilized by the electron-donating inductive effect and hyperconjugation provided by the methyl substituent. On the other hand, protonation at the methyl-bearing carbon would produce a less stable secondary carbocation.

Therefore, the preferred pathway involves formation of the tertiary carbocation, followed by nucleophilic attack by $$Br^-$$. As a result, the methyl group and bromine atom become attached to the same carbon atom.

For the lower double bond, one of the carbon atoms is attached to a nitro group $$\left(-NO_2\right)$$. The nitro group is a strong electron-withdrawing substituent due to both its $$-I$$ and $$-R$$ effects. If protonation occurs at the adjacent carbon, the resulting carbocation would be located directly on the carbon bonded to the nitro group, making it highly unstable. In contrast, protonation at the nitro-substituted carbon places the positive charge on the neighbouring carbon, avoiding direct destabilization by the nitro group.

Hence, the preferred pathway is for $$H^+$$ to add to the carbon bearing the nitro group, after which $$Br^-$$ attacks the adjacent carbon.

Combining the outcomes of both additions:

• The carbon bearing the methyl group acquires a bromine atom, so it contains both $$-CH_3$$ and $$-Br$$ substituents.
• The nitro-substituted carbon receives the proton.
• The carbon adjacent to the nitro group receives the second bromine atom.

Therefore, the major product is the structure in which:

• one bromine atom is attached to the same carbon as the methyl group, and
• the other bromine atom is attached to the carbon adjacent to the nitro-substituted carbon.

This corresponds to the option D.

In a Friedel-Crafts (alkylation or acylation) reaction the aromatic ring behaves as a nucleophile while the electrophile is generated with the help of a Lewis‐acid catalyst such as $$AlCl_3$$. Hence, the rate and overall yield of the reaction depend mainly on two factors:

1. The electron density of the aromatic ring (controlled by the substituents already present).
2. Possible inhibition caused by the substituent forming a tight complex with the Lewis acid, thereby tying up the catalyst.

Step 1 - Classify the substituents in the four compounds
• Electron-withdrawing, $$-I$$ and $$-M$$ (deactivating) groups such as $$-NO_2$$ decrease the ring’s nucleophilicity and give very poor, often zero, Friedel-Crafts yield.
• Electron-withdrawing but weakly deactivating halogens ($$-I$$, weak $$+M$$) also lower the yield.
• Alkyl groups ($$+I$$, hyperconjugation) activate the ring moderately and usually provide good yields.
• Strongly electron-donating groups having lone pairs in conjugation with the ring (e.g. $$-OCH_3$$, $$-NH_2$$) possess a powerful $$+M$$ effect and give the fastest reactions and the best isolated yields.

Step 2 - Examine each option
Option A and Option B contain deactivating substituents (for example $$NO_2$$ or a halogen), so they will suffer from very low conversion.
Option C contains only an alkyl group. The ring is activated, but not as strongly as it could be.
Option D is anisole (methoxybenzene). The $$-OCH_3$$ group donates electron density through resonance ($$+M$$ effect), creating a highly nucleophilic ortho-/para-rich ring. Although oxygen can coordinate weakly with $$AlCl_3$$, this complex is reversible, and the strong activation dominates, giving the highest overall yield among the four compounds.

Therefore, the substrate that delivers the highest yield in a Friedel-Crafts reaction is the one bearing the methoxy group.

Option D which is: anisole (methoxybenzene)

The molecular formula of the monobrominated product is $$C_8H_7O_2Br$$.
During bromination one hydrogen atom is replaced by bromine, so the parent compound [P] must be $$C_8H_8O_2$$.

Heating [P] with sodalime (NaOH + CaO) carries out decarboxylation, i.e. $$R-COOH \rightarrow R-H + CO_2$$.
Because the decarboxylation product is toluene ($$C_7H_8$$), [P] must be an aromatic carboxylic acid containing one extra carbon compared to toluene. Therefore [P] is some isomer of $$CH_3-C_6H_4-COOH$$ (methyl-benzoic acid, also called toluic acid).

The three possible isomers are:
  • o-methylbenzoic acid (2-toluic acid)
  • m-methylbenzoic acid (3-toluic acid)
  • p-methylbenzoic acid (4-toluic acid)

We now use the bromination data. Electrophilic bromination in $$CCl_4/FeBr_3$$ is governed by the directing effects of the substituents present:

• $$-COOH$$ is a strong deactivating, meta-directing group.
• $$-CH_3$$ is an activating, ortho/para-directing group.

Case 1: o-methylbenzoic acid
  • The ring positions 3, 4, 5, 6 are all chemically nonequivalent and at least two of them are activated for substitution, so a mixture of brominated isomers would be obtained.

Case 2: m-methylbenzoic acid
  • Again several positions (2, 4, 5, 6) differ in reactivity, giving more than one monobromo product.

Case 3: p-methylbenzoic acid
  • Place $$-COOH$$ at position 1 and $$-CH_3$$ at position 4.
  • Both groups favour substitution at positions 3 and 5:
    - For $$-COOH$$ these are its meta positions.
    - For $$-CH_3$$ these are its ortho positions.
  • Positions 3 and 5 are equivalent by symmetry, so only one distinct monobromo isomer can form.

Thus only p-methylbenzoic acid satisfies the statement “produced a single isomer $$C_8H_7O_2Br$$”.

Therefore, the compound [P] is p-methylbenzoic acid (4-methylbenzoic acid).

Option A which is: p-methylbenzoic acid.

Let us begin with compound A. The statement “A is a lowest molecular weight alkyne” means the alkyne with the smallest possible number of carbon atoms, i.e. ethyne (acetylene), whose formula is $$HC\equiv CH$$.

Now this ethyne is passed over a red-hot copper tube. The textbook reaction that occurs under these conditions is the trimerisation of ethyne to benzene: three molecules of ethyne cyclise and lose hydrogen as shown below

$$3\,HC\equiv CH \;\xrightarrow[\;873\;{\rm K}\;]{\text{Red-hot Cu}}\; C_6H_6 + 3\,H_2$$

Thus, compound B is benzene, $$C_6H_6$$.

Next, benzene is treated with one equivalent of methyl chloride in the presence of anhydrous $$AlCl_3$$. This is a classical Friedel-Crafts alkylation. First we state the general formula:

Friedel-Crafts alkylation: $$\quad C_6H_6 + RCl \;\xrightarrow{AlCl_3}\; C_6H_5R + HCl$$

Substituting $$R = CH_3$$ and taking only one equivalent (so that only a single substitution occurs), we obtain

$$C_6H_6 + CH_3Cl \;\xrightarrow{AlCl_3}\; C_6H_5CH_3 + HCl$$

Therefore compound C is methylbenzene, commonly called toluene, whose condensed formula is $$C_6H_5CH_3$$.

We are now asked: “what is the maximum number of atoms present in molecule C in one plane?” To answer this we must look at the geometry of every atom in toluene.

• The six ring carbons in benzene are $$sp^2$$ hybridised; all $$sp^2$$ atoms and the σ-bonds between them lie in a single plane.
• The five hydrogens that remain on the ring (one hydrogen is replaced by the methyl group) are each bonded through an $$sp^2$$ carbon; the C-H σ-bond in an $$sp^2$$ centre is also contained in the same trigonal plane. Hence each of these five hydrogens can lie in that plane.
• The ring carbon that bears the side chain is itself $$sp^2$$ and already counted among the six ring carbons.
• The first carbon of the side chain (the methyl carbon, $$CH_3$$) is $$sp^3$$. In a tetrahedral $$sp^3$$ centre the central carbon and any one of its three substituents can be chosen to lie in the same plane as the benzene ring; the other two substituents must protrude above and below the plane. Hence the methyl carbon itself and exactly one of its three hydrogens can be co-planar with the ring.

Counting these atoms explicitly:

$$\begin{aligned} \text{Ring carbons} &= 6\\ \text{Hydrogens on the ring} &= 5\\ \text{Methyl (side-chain) carbon} &= 1\\ \text{One hydrogen of the CH}_3 &= 1\\ \hline \text{Total atoms that can be in one plane} &= 6 + 5 + 1 + 1 = 13 \end{aligned}$$

No other atom of toluene can be added to this plane without forcing a tetrahedral angle to straighten, so 13 is the absolute maximum.

So, the answer is $$13$$.

(4n+2) pi electrons = Aromatic

4n pi electrons = Anti- Aromatic

Negative Charge = Lone Pair = 2 electrons

A = 2+2+2 = 6 pi electrons

B = 2+2+2 = 6 pi electrons, nitrogen's lone pair don't undergo resonance

C = 2+2+2 = 6 pi electrons including the electrons from lone pair of Nitrogen.

D = 2+2 = 4 pi electrons anti-aromatic

Nitration is favourable by +M or +I effect. Hence Para would be major due to steric factors

The ease of Electrophilic substitution is directly proportional to the strength of +M or +I group on the aromatic ring. Hence, B>C>A>D would be the right answer.

Here is the clean explanation, completely free of math formatting or special text tags, so you can copy and paste it directly.

Breakdown of Each Compound

(A) Cyclopropenyl cation

• Structure: A 3-membered ring with one double bond and a carbocation.
• Conjugation: It is cyclic, planar, and completely conjugated because the positive charge allows the electrons to delocalize around the entire ring.
• Pi-electrons: It has 2 pi-electrons from the single double bond.
• Classification: It follows Huckel's rule (4n + 2 pi-electrons where n = 0). Therefore, it is Aromatic.

(B) Cyclopentadienyl cation

• Structure: A 5-membered ring with two double bonds and a carbocation.
• Conjugation: It is cyclic, planar, and completely conjugated.
• Pi-electrons: It has 4 pi-electrons from the two double bonds.
• Classification: It follows the 4n pi-electron rule (where n = 1). Therefore, it is Antiaromatic (Not aromatic).

(C) Cycloheptatrienyl cation (Tropylium ion)

• Structure: A 7-membered ring with three double bonds and a carbocation.
• Conjugation: It is cyclic, planar, and completely conjugated. The positive charge allows the pi-electrons to delocalize freely across all seven carbons.
• Pi-electrons: It has 6 pi-electrons from the three double bonds.
• Classification: It perfectly follows Huckel's rule (4n + 2 pi-electrons where n = 1). Therefore, it is Aromatic.

(D) Cycloheptatriene

• Structure: A 7-membered ring with three double bonds and one CH2 carbon.
• Conjugation: The CH2 carbon at the bottom is sp3 hybridized. This breaks the continuous cyclic loop of conjugation.
• Classification: Because it lacks complete cyclic conjugation, it is Non-aromatic (Not aromatic).

Final Summary

• Aromatic: (A) and (C)
• Not Aromatic: (B) and (D)

We begin by recalling the meaning of the term $$\text{polysubstitution}$$. In aromatic electrophilic substitution, polysubstitution means that after the first substituent is introduced, further substitutions keep taking place on the same benzene ring, often yielding di- or tri-alkylated products rather than the desired mono-alkylated compound. This situation is undesirable because it lowers the yield of the required mono-substituted product and complicates purification.

Now we examine each of the given reactions one by one and analyse whether polysubstitution is an inherent drawback.

First, consider the Reimer-Tiemann reaction. In this reaction, phenol is treated with chloroform and a strong base to give salicylaldehyde. The reaction proceeds through a dichlorocarbene intermediate that inserts at the ortho and para positions. Once the formyl group $$\left( -\!\!CHO \right)$$ is installed, the ring becomes deactivated because the $$-CHO$$ group is $$-M$$ and $$-I$$ withdrawing, so further electrophilic attack is suppressed. Therefore polysubstitution is not a major issue in the Reimer-Tiemann reaction.

Next, we analyse Friedel-Crafts alkylation, in which an alkyl halide $$R-X$$ reacts with benzene in the presence of a Lewis acid such as $$AlCl_3$$ to give an alkylbenzene. The key mechanistic step is the generation of a carbocation $$R^+$$ (or a Lewis-acid-complexed equivalent), which is a very strong electrophile. After the first alkyl group attaches to the benzene ring, it exerts a $$+I$$ inductive effect and a $$+H$$ hyperconjugative effect, both of which activate the ring and make it still more electron-rich. Because of this increased activation, the freshly formed alkylbenzene reacts faster than the original benzene, so more alkyl cations attack it in subsequent steps. This chain of events produces di- and tri-alkylated products very easily. Thus polysubstitution is indeed a major drawback in Friedel-Crafts alkylation.

Now we look at the acetylation of aniline, typically performed with acetic anhydride $$\left( (CH_3CO)_2O \right)$$. In this reaction the amino group $$-NH_2$$ is converted into an amide $$-NHCOCH_3$$. The product contains the $$-NHCOCH_3$$ group, which is less activating than the parent $$-NH_2$$ group because the lone pair on nitrogen becomes partially delocalised into the carbonyl. Therefore the ring is not strongly activated toward further electrophilic attack, and polysubstitution is not a significant problem.

Finally, Friedel-Crafts acylation introduces an acyl group $$RCO-$$ using an acyl chloride $$RCOCl$$ and a Lewis acid such as $$AlCl_3$$. The acylium ion $$RCO^+$$ is generated as the electrophile. After the first acyl group enters the ring, the newly formed ketone group $$-COR$$ is strongly $$-M$$ and $$-I$$ withdrawing, which deactivates the ring. That deactivation greatly diminishes the likelihood of further substitutions, so polysubstitution is almost never observed in Friedel-Crafts acylation.

Comparing all four possibilities, we see that only Friedel-Crafts alkylation suffers from the intrinsic problem that the first substitution step makes the aromatic ring more reactive instead of less. This leads to the formation of multiple alkylated products unless stringent reaction control (large excess of benzene, low temperature, etc.) is employed.

Hence, the correct answer is Option 2.

Electrophilic aromatic substitution (EAS) proceeds faster when the benzene ring possesses higher electron density, because the aromatic π-cloud then stabilises the positively-charged σ-complex (arenium ion) that forms in the rate-determining step. In general, an electron-donating group (EDG) activates the ring, while an electron-withdrawing group (EWG) deactivates it. The magnitude of activation or deactivation depends on the balance of two electronic effects:

Inductive effect $$\; (\pm I)$$ and $$\text{Mesomeric / Resonance effect} \; (\pm M)$$

We now examine the substituent already attached to the ring in each of the three given compounds.

For toluene (compound II) the substituent is the methyl group, $$-CH_3$$. A methyl group exhibits a $$+I$$ (electron-releasing inductive) effect and can also donate electron density through hyperconjugation, which behaves like weak resonance donation. Both mechanisms increase electron density on the ring. Therefore toluene is activated toward EAS and reacts faster than benzene.

For chlorobenzene (compound I) the substituent is chlorine. Chlorine is more electronegative than carbon, so it withdraws electron density through the $$-I$$ inductive effect. At the same time, its lone pairs can participate in $$+M$$ resonance donation into the ring. Quantitatively, the $$-I$$ effect of chlorine is stronger than its $$+M$$ effect, giving a net deactivation. However, because some resonance donation is present, the deactivation is only moderate; chlorobenzene reacts a little more slowly than benzene but much faster than rings bearing strongly withdrawing carbonyl groups.

For acetophenone (compound III) the substituent is the acyl group, $$-COCH_3$$. The carbonyl oxygen pulls electron density strongly toward itself by both a powerful $$-M$$ (resonance withdrawal) and a $$-I$$ inductive effect. The combined withdrawal greatly lowers the π-electron density of the ring, making acetophenone strongly deactivated. Its rate toward EAS is far below that of benzene.

Summarising the relative strength of these effects:

$$\text{Rate (EAS)} \;\propto\; \text{Electron density on ring}$$

$$\begin{aligned} -COCH_3 &: \; -M \; \text{(very strong)} + -I \; \Longrightarrow \text{strongly deactivated} \\ Cl &: \; -I \; \text{(strong)} + +M \; \text{(moderate)} \Longrightarrow \text{moderately deactivated} \\ CH_3 &: \; +I \; \text{(moderate)} + \text{hyperconjugation} \Longrightarrow \text{activated} \end{aligned}$$

Hence the rate order is

Acetophenone (III)} \;\lt \; \text{Chlorobenzene (I)} \;\lt \; \text{Toluene (II)

So, the increasing order of reactivity towards electrophilic aromatic substitution is $$\mathrm{III \; \lt \; I \; \lt \; II}$$.

Hence, the correct answer is Option C.

1. Carbocation Generation:

   In the presence of a Lewis acid or under ionizing conditions, the chloride leaving group ($$\text{Cl}^\ominus$$) departs from the saturated ring, generating a reactive carbocation intermediate.




2. Intramolecular Electrophilic Attack:

   The electron-rich aromatic ring, heavily activated by the strongly electron-donating methoxy group ($$\text{--OCH}_3$$) via resonance ($$+R$$ effect), acts as an internal nucleophile. The aromatic system attacks the newly formed carbocation center to close a new ring, generating the localized arenium ion (Wheland intermediate) shown on the right side of the scheme.




3. Deprotonation (Aromatization):

   As explicitly highlighted by the step labeled $$-\text{H}^\oplus$$, the arenium intermediate undergoes rapid loss of a proton from the site of substitution. This deprotonation removes the localized positive charge and restores the fully conjugated, thermodynamically stable aromatic system.

Conclusion:

The loss of the proton ($$\text{H}^\oplus$$) from the arenium intermediate regenerates the aromaticity of the benzene ring containing the methoxy substituent, completing the intramolecular cyclization framework.

Answer: Option B

The reaction begins with a Friedel-Crafts Acylation where the Lewis acid catalyst, AlCl3, reacts with phthalic anhydride to open the anhydride ring and generate a powerful electrophilic acylium ion. This ion attacks the chlorobenzene ring, primarily at the para-position due to the directing influence of the chlorine atom and steric hindrance at the ortho-position. This first step produces a complex that, upon aqueous workup H2O, yields o-(4-chlorobenzoyl)benzoic acid.

The reaction stops here because the newly formed carbonyl group strongly deactivates the aromatic rings, making them resistant to further nucleophilic attack under these mild conditions. To achieve the second ring closure the dehydration step you would need a much stronger dehydrating agent like concentrated H2SO4 or Polyphosphoric Acid (PPA) and significant heat to force the carboxylic acid group to cyclize onto the remaining ortho-position of the chlorobenzene ring

The carbocation is formed first then ring formation takes place by replacing on -H from Benzene. 

Simple question from Friedel Craft's Alkylation. Carbonyl attaches to the para position of Benzene ring then due to heating dewatering takes place and you are left with a 6 membered ring with Carbonyl attached on para position with respect to the -OH. 

1. Step 1: Radical Oxidation to Cumene Hydroperoxide

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

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



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

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

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

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

Conclusion:

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

Answer: Option C — Phenol and Acetone

1. Generation of the Carbocation

The primary alkyl chloride reacts with the Lewis acid, anhydrous aluminium chloride (AlCl3), to form a complex. Because it is a primary carbon, it can undergo a synchronized hydride shift as the leaving group departs to form a more stable carbocation directly:

• A hydride (H-) shifts from the adjacent secondary carbon to the terminal carbon.
• This expels the [AlCl4]- leaving group and creates a more stable secondary carbocation.

2. Further Carbocation Rearrangement (Hydride Shift)

The newly formed secondary carbocation is separated from the electron-rich aromatic ring by a chain of CH2 groups. To achieve the most stable intermediate before ring closure, another 1,2-hydride shift occurs along the alkyl chain.

• This moves the positive charge closer to the benzene ring, generating an even more stable secondary carbocation position that sets up a highly favorable 6-membered ring closure.

3. Intramolecular Electrophilic Aromatic Substitution (Ring Closure)

The methoxy group (-OCH3) attached to the benzene ring is a strong ortho/para-directing group due to resonance activation.

• The position para to the methoxy group is stericly less hindered and highly nucleophilic.
• The benzene ring attacks the rearranged carbocation intermediate intramolecularly.
• This ring-closure forms a fused 6-membered ring (a tetralin derivative).

4. Aromatization and Hydrolysis

Loss of a proton (H+) restores the aromaticity of the benzene ring. The subsequent addition of water (H2O) during the workup step breaks down the aluminum complexes.

Final Major Product Structure

The final product is a 6-membered saturated ring fused to the methoxybenzene ring system, specifically forming at the position para to the methoxy group.

In acidic medium KMnO4 is a strong oxidising agent in acidic medium and it will oxidise both the carbon substituted on benzene to carboxylic acid.

Furan (A), Pyridine (B), and Benzene (D) are aromatic.

4,4-dimethylcyclohexa-2,5-dien-1-one (C) is non-aromatic. Hence, it is the least stable by resonance.

Aniline has a basic amino group ($$-NH_2$$). In the presence of strong acids, the lone pair on the nitrogen atom accepts a proton ($$H^+$$) to form the anilinium ion ($$-NH_3^+$$).

Aniline ($$-NH_2$$): Under normal conditions, the amino group is a powerful activating group and is ortho/para directing.

Anilinium Ion ($$-NH_3^+$$): Once protonated, the nitrogen carries a full positive charge. This makes it a strongly deactivating group that pulls electron density away from the ring via the inductive effect. Consequently, the anilinium ion is meta-directing.

Because a significant equilibrium concentration of the anilinium ion exists in the acidic nitrating mixture, the reaction yields a surprisingly high amount of the meta-isomer. A typical product distribution for the direct nitration of aniline is:

Para-product: ~51%, Meta-product: ~47% (Significant amount), Ortho-product: ~2%

While the other functional groups ($$-OH, -OCOCH_3, -NHCOCH_3$$) also have lone pairs, they are much less basic than the amino group in aniline.

First we remind ourselves of a very common principle of acid-base chemistry: a base removes the most acidic hydrogen that is available. A hydrogen atom is called “acidic” when, after its removal as $$H^+$$, the conjugate base that remains is stabilised (generally by resonance or by an electron-withdrawing group). The better the stabilisation, the smaller the $$\mathrm{p}K_a$$ of the hydrogen-bearing compound and the more easily even a weak base can abstract that hydrogen.

In aqueous solution sodium carbonate provides the weak base $$CO_3^{2-}$$:

$$Na_2CO_3 \;\;\xrightarrow{H_2O}\;\; 2\,Na^+ + CO_3^{2-}$$

The carbonate ion itself is the species that can abstract an acidic hydrogen. Because $$CO_3^{2-}$$ is only a mild base, it can deprotonate a substrate only if the resulting conjugate base is appreciably stabilised. With this idea in mind we examine each of the four hydrocarbons offered in the question.

(i) Cyclopropane, $$C_3H_6$$
All six hydrogens are attached to ordinary sp3 carbon atoms. When one such proton is removed we would have a simple alkyl carbanion

$$C_3H_5^-$$

This carbanion has no resonance stabilisation and experiences strong angle strain; it is therefore highly unstable. The corresponding $$\mathrm{p}K_a$$ of cyclopropane is close to that of a normal alkane (well above 45). A base as weak as $$CO_3^{2-}$$ cannot remove such a proton.

(ii) Benzene, $$C_6H_6$$
All hydrogens in benzene are bound to sp2 ring carbons. Removal of one proton produces the phenyl carbanion

$$C_6H_5^-$$

Although this anion is conjugated with the ring, it is antiaromatic and hence severely destabilised. Consequently the $$\mathrm{p}K_a$$ of benzene is still extremely high (≈43). Again, $$CO_3^{2-}$$ is far too weak to effect deprotonation.

(iii) Toluene, $$C_6H_5CH_3$$
Toluene contains three types of hydrogen, but the ones of interest are the benzylic hydrogens on the $$CH_3$$ group. If one benzylic proton is abstracted we obtain the benzyl carbanion

$$C_6H_5CH_2^-$$

This anion is resonance-stabilised; the negative charge can be delocalised into the aromatic ring:

$$C_6H_5CH_2^- \;\;\rightleftharpoons\;\; C_6H_4CH_2 \; \text{(multiple resonance forms)}$$

Because of this stabilisation the $$\mathrm{p}K_a$$ of a benzylic hydrogen is appreciably lower (≈41) than that of a normal alkane and, more importantly, distinctly lower than the hydrogens in cyclopropane, benzene or naphthalene. Although $$CO_3^{2-}$$ is not strong enough to drive the equilibrium completely to the right, it can abstract a benzylic proton to a noticeable extent. Symbolically we can write

$$C_6H_5CH_3 + Na_2CO_3 \;\;\rightleftharpoons\;\; C_6H_5CH_2Na + NaHCO_3$$

The equilibrium lies far to the left, but the fact that it is accessible means that toluene is the most reactive of the four compounds toward aqueous sodium carbonate.

(iv) Naphthalene, $$C_{10}H_8$$
Like benzene, every hydrogen is attached directly to an sp2 carbon of the aromatic framework. Removal of a proton would break aromaticity in both fused rings, giving an extremely unstable anion. Its $$\mathrm{p}K_a$$ is even higher than that of benzene, so no reaction with $$CO_3^{2-}$$ occurs.

Putting all these points together, only toluene possesses hydrogens that are sufficiently acidic for even the mild base $$CO_3^{2-}$$ to abstract. Therefore toluene exhibits the greatest reactivity toward an aqueous solution of sodium carbonate, while the other hydrocarbons remain virtually inert.

Hence, the correct answer is Option C.

Friedel Crafts alkylation/acylation involves the formation of a stable electrophile (a carbocation).

The electrophiles formed from the compounds in options B, C, and D are stabilized by resonance. 

Formation of a vinylic carbocation is very difficult as the $$C-Cl$$ bond is on a sp^2 hybridized carbon, and the resulting vinylic carbocation is highly unstable.

Thus, the compound in option A does not undergo Friedal Crafts reaction with benzene.

Photochemical smog is a type of air pollution formed when sunlight reacts with pollutants like nitrogen oxides (NOx) and volatile organic compounds (VOCs) released from vehicles and industries. The chemical reactions under sunlight produce secondary pollutants.

The question states that photochemical smog contains an excessive amount of X, along with aldehydes, ketones, and peroxy acetyl nitrile (PAN). We need to identify X from the options.

Recall the key components of photochemical smog. The primary pollutants are nitrogen oxides (NO and NO2) and VOCs. Sunlight triggers reactions that form secondary pollutants, including:

• Ozone (O3)
• Aldehydes (e.g., formaldehyde)
• Ketones
• Peroxyacyl nitrates (PAN)

Ozone is a major product. It forms when nitrogen dioxide (NO2) absorbs sunlight and breaks down:

Then, the oxygen atom reacts with oxygen molecules:

This ozone accumulates and is a defining component of photochemical smog, causing respiratory issues and environmental harm.

Now, evaluate the options:

• Option A (CH4): Methane is a greenhouse gas but is not reactive enough to form photochemical smog. It does not produce aldehydes, ketones, or PAN in significant amounts under typical smog conditions.
• Option B (CO2): Carbon dioxide is a greenhouse gas from combustion but is not a key reactant or product in photochemical smog reactions. It does not contribute to smog formation.
• Option C (O3): Ozone is a central component of photochemical smog, produced in high concentrations from NOx and VOCs. It is present excessively and aligns with the given pollutants (aldehydes, ketones, PAN).
• Option D (CO): Carbon monoxide is emitted from incomplete combustion but is primarily a primary pollutant. It does not lead to the formation of aldehydes, ketones, or PAN in photochemical smog.

Thus, X must be ozone (O3), as it is excessively produced and characteristic of photochemical smog.

Hence, the correct answer is Option C.

To determine which reaction does not result in the formation of a carbon-carbon bond, we need to analyze each option step by step. Let's recall the key features of each reaction.

Starting with option A: Friedel Craft's acylation. This reaction involves an aromatic compound reacting with an acyl chloride (R-COCl) in the presence of a Lewis acid catalyst like AlCl₃. The acyl group attaches to the aromatic ring, forming a new bond between the carbonyl carbon of the acyl group and a carbon atom of the aromatic ring. For example, benzene reacts with acetyl chloride to form acetophenone: $$C_{6}H_{6} + CH_{3}COCl \xrightarrow{\text{AlCl}_3} C_6H_5COCH_3$$ Here, a new carbon-carbon bond is formed between the benzene ring carbon and the carbonyl carbon of the acetyl group. Thus, this reaction does form a carbon-carbon bond.

Next, option B: Reimer-Tieman reaction. This reaction is used for ortho-formylation of phenols. Phenol reacts with chloroform (CHCl₃) in the presence of a strong base like NaOH to produce salicylaldehyde (2-hydroxybenzaldehyde). The mechanism involves the formation of dichlorocarbene (:CCl₂), which attacks the ortho position of the phenoxide ion. This results in a new bond between the carbon of the benzene ring and the carbon of the formyl group. The reaction is: $$C_{6}H_{5}OH + CHCl_{3} + NaOH -\gt o-HOC_{6}H_{4}CHO$$ A carbon-carbon bond is formed between the aromatic carbon and the carbon of the -CHO group. Therefore, this reaction also forms a carbon-carbon bond.

Now, option C: Cannizaro reaction. This is a disproportionation reaction that occurs with aldehydes lacking an alpha hydrogen, such as formaldehyde or benzaldehyde. In this reaction, one molecule of aldehyde is oxidized to a carboxylic acid, while another molecule is reduced to an alcohol. For example, formaldehyde undergoes the Cannizaro reaction: $$2HCHO -\gt [\text{NaOH] HCOONa + CH3OH}$$ The mechanism involves a hydride ion transfer from one aldehyde molecule to another. There is no formation of a new carbon-carbon bond; instead, bonds are rearranged to form C-O bonds in the carboxylate and alcohol. The carbon atoms remain separate—no coupling occurs. Hence, this reaction does not form a carbon-carbon bond.

Finally, option D: Wurtz reaction. This reaction couples two alkyl halides using sodium metal in dry ether solvent. For instance, two molecules of bromoethane react to form butane: $$2CH_{3}CH_{2}Br + 2Na -\gt CH_{3}CH_{2}CH_{2}CH_{3} + 2NaBr$$ Here, the alkyl groups from the halides combine, forming a new carbon-carbon bond between the two carbon atoms. Thus, this reaction clearly forms a carbon-carbon bond.

In summary:

• Friedel Craft's acylation (A) forms a carbon-carbon bond.
• Reimer-Tieman reaction (B) forms a carbon-carbon bond.
• Cannizaro reaction (C) does not form a carbon-carbon bond.
• Wurtz reaction (D) forms a carbon-carbon bond.

Hence, the correct answer is Option C.

Option A is a tertiary ($$3^\circ$$) allylic carbocation. The positive charge sits on a carbon attached to three other carbons, and it is directly adjacent to a conjugated double-bond system. Furthermore, the double bond is in conjugation with a phenyl ($$\text{Ph}$$) ring, which allows extensive delocalization of the positive charge throughout the aromatic ring via resonance. Hence, it is the most stable.

The major component of Borsch reagent is 2,4- dinitrophenyl hydrazine which can be obtained by reaction of 2,4-dinitrochloro benzene and hydrazine.

To solve this question, we need to determine which option does not produce 2-phenylbutane as the major product in a Friedel-Crafts alkylation reaction. Friedel-Crafts alkylation involves attaching an alkyl group to a benzene ring, forming a new carbon-carbon bond. The electrophile is typically a carbocation, which can rearrange to form a more stable carbocation. The target molecule, 2-phenylbutane, has the structure $$Ph-CH(CH_{3})CH_{2}CH_{3}$$, where the carbon attached to the benzene ring is secondary.

Now, let's evaluate each option step by step.

Option A: 1-butene + HF

1-butene is $$CH_{2}=CH-CH_{2}-CH_{3}$$. With HF, protonation occurs at the terminal carbon, forming a secondary carbocation: $$CH_{2}=CH-CH_{2}-CH_{3} + H+ \rightarrow CH_{3}-CH+-CH_{2}-CH_{3}$$. This carbocation attacks benzene in a Friedel-Crafts alkylation, yielding $$Ph-CH(CH_{3})CH_{2}CH_{3}$$, which is 2-phenylbutane. Thus, this option gives the desired product as the major product.

Option B: 2-butanol + H$$_2$$SO$$_4$$

2-butanol is $$CH_{3}-CH(OH)-CH_{2}-CH_{3}$$. With H$$_2$$SO$$_4$$, dehydration occurs, protonating the hydroxyl group and losing water to form a secondary carbocation: $$CH_{3}-CH(OH)-CH_{2}-CH_{3} + H+ \rightarrow CH_{3}-CH(OH_{2}+)-CH_{2}-CH_{3} \rightarrow CH_{3}-CH+-CH_{2}-CH_{3} + H_{2}O$$. This carbocation attacks benzene, forming $$Ph-CH(CH_{3})CH_{2}CH_{3}$$ (2-phenylbutane). No rearrangement occurs, as the secondary carbocation is stable. Hence, this option gives 2-phenylbutane as the major product.

Option C: Butanoyl chloride + AlCl$$_3$$ then Zn, HCl

Butanoyl chloride is $$CH_{3}CH_{2}CH_{2}C(=O)Cl$$. With AlCl$$_3$$, Friedel-Crafts acylation occurs, forming an acylium ion: $$CH_{3}CH_{2}CH_{2}C(=O)Cl + AlCl_{3} \rightarrow CH_{3}CH_{2}CH_{2}C{\equiv}O^+ + AlCl_{4}-$$. This electrophile attacks benzene, yielding the ketone $$Ph-C(=O)-CH_{2}CH_{2}CH_{3}$$ (phenyl propyl ketone). The subsequent step, "then Zn, HCl", refers to Clemmensen reduction, which reduces the carbonyl group to a methylene group: $$Ph-C(=O)-CH_{2}CH_{2}CH_{3} \rightarrow Ph-CH_2-CH_2CH_2CH_3$$ (1-phenylbutane). This is a straight-chain alkylbenzene, whereas 2-phenylbutane is branched. Therefore, this option does not give 2-phenylbutane as the major product.

Option D: Butyl chloride + AlCl$$_3$$

Butyl chloride typically refers to n-butyl chloride, $$CH_{3}CH_{2}CH_{2}CH_{2}Cl$$. With AlCl$$_3$$, it forms a primary carbocation: $$CH_{3}CH_{2}CH_{2}CH_{2}Cl + AlCl_{3} \rightarrow CH_{3}CH_{2}CH_{2}CH_{2}+ + AlCl_{4}-$$. Primary carbocations are unstable and rearrange via a hydride shift: $$CH_{3}CH_{2}CH_{2}CH_{2}+ \rightarrow CH_{3}CH_{2}CH+CH_{3}$$ (a more stable secondary carbocation). This carbocation attacks benzene, forming $$Ph-CH(CH_{3})CH_{2}CH_{3}$$ (2-phenylbutane). Due to rearrangement, this option gives 2-phenylbutane as the major product.

In summary, Options A, B, and D all yield 2-phenylbutane as the major product, while Option C yields 1-phenylbutane. Therefore, Option C does not give 2-phenylbutane as the major product.

Hence, the correct answer is Option C.

Electrophilic aromatic substitution reactivity depends on electron density, where electron-donating groups (EDGs) activate the ring and electron-withdrawing groups (EWGs) deactivate it.

$$-CH_3$$ (B) activates via hyperconjugation, while $$-Cl$$ (C) and $$-NO_2$$ (D) deactivate the ring through $$-I$$ and $$-M$$ effects respectively.

Consequently, toluene (B) is the most reactive followed by benzene (A), with nitrobenzene (D) being the least reactive due to strong deactivation.

The correct order of reactivity is B > A > C > D

For a compound to be antiaromatic, it must be cyclic, planar, completely conjugated, and contain $$4n$$ $$\pi$$ electrons.

In furan, one lone pair of oxygen participates in conjugation. Thus, the ring contains $$6\pi$$ electrons and satisfies Hückel's $$4n+2$$ rule. Hence, furan is aromatic.

Cycloheptatriene contains an $$sp^3$$ hybridized carbon atom, which interrupts conjugation. Therefore, it is non-aromatic.

Cyclooctatetraene contains $$8\pi$$ electrons, but it adopts a non-planar tub-shaped structure to avoid antiaromaticity. Hence, it is non-aromatic.

In the cyclopentadienyl anion, the lone pair on the negatively charged carbon participates in conjugation, giving a total of $$6\pi$$ electrons. Therefore, it is aromatic.

The cyclopentadienyl cation contains $$4\pi$$ electrons and is cyclic, planar, and fully conjugated. Hence, it is antiaromatic.

Cyclobutadiene is cyclic, planar, fully conjugated, and contains $$4\pi$$ electrons. Hence, it is antiaromatic.

Therefore, the antiaromatic compounds are (V) and (VI) only.

The four compounds are assumed to be the common monosubstituted benzenes used in standard JEE problems:

(i) Aniline  $$C_6H_5NH_2$$
(ii) Phenol   $$C_6H_5OH$$
(iii) Toluene  $$C_6H_5CH_3$$
(iv) Nitrobenzene $$C_6H_5NO_2$$

In electrophilic aromatic substitution (EAS) the rate depends on how strongly the substituent donates or withdraws electrons from the ring.

Guiding principles:
1. An electron-donating group (EDG) increases electron density and therefore activates the ring.
2. An electron-withdrawing group (EWG) decreases electron density and therefore deactivates the ring.
3. The order of activation depends on the magnitude of the +M (mesomeric), +I (inductive) or +H (hyperconjugative) effects shown below.

Substituent analysis

(i) $$-NH_2$$ : Strong +M (lone pair conjugation) and weak +I → very strong activator.
(ii) $$-OH$$ : Strong +M (lone-pair donation) but a −I effect; overall donation slightly weaker than $$-NH_2$$ → strong activator.
(iii) $$-CH_3$$ : No +M, only +H hyperconjugation and a feeble +I → weak activator.
(iv) $$-NO_2$$ : Strong −M (draws π electrons into the nitro group) and strong −I → strong deactivator.

Resultant reactivity order
Highest activation gives the fastest EAS; strongest deactivation gives the slowest. Therefore

$$C_6H_5NH_2 \; \gt \; C_6H_5OH \; \gt \; C_6H_5CH_3 \; \gt \; C_6H_5NO_2$$

Translating this back to the given numbering, the decreasing order of reactivity is
$$(i) \; \gt \; (ii) \; \gt \; (iii) \; \gt \; (iv).$$

Hence, the correct option is
Option D which is: (i) > (ii) > (iii) > (iv).

N-Bromosuccinimide (NBS) in the presence of light or a peroxide is a reagent for selective radical bromination at an allylic or benzylic position. It maintains a very low concentration of molecular $$Br_2$$ so that substitution (not addition) occurs.

Ethyl benzene is $$C_6H_5CH_2CH_3$$. The carbon that is directly attached to the ring (the $$CH_2$$ group) is a benzylic carbon. Benzylic radicals are strongly resonance-stabilised, so the hydrogen atoms on this carbon are the ones that react fastest with NBS.

Step-wise outline:
1. Abstraction of a benzylic H gives the benzylic radical $$C_6H_5\dot{C}HCH_3$$.
2. The radical is trapped by $$Br_2$$ (generated in situ from NBS) to give the benzylic bromide and a new $$Br\cdot$$ radical, propagating the chain.

The overall transformation is therefore

$$C_6H_5CH_2CH_3 \;\xrightarrow[\;h\nu\;]{\;NBS\;} C_6H_5CH(Br)CH_3 + \text{succinimide}$$

The product $$C_6H_5CH(Br)CH_3$$ is named 1-bromo-1-phenylethane (also called α-bromoethylbenzene).

Hence the correct choice is
Option D which is: 1-bromo-1-phenylethane.

Chlorination of an aromatic compound can follow two different pathways, depending on the reaction conditions:

1. Radical (side-chain) chlorination: $$Cl_2$$ in the presence of heat, light, or peroxides produces benzyl chloride, benzal chloride, etc.
2. Electrophilic (ring) chlorination: $$Cl_2$$ in the presence of a Lewis-acid catalyst such as $$FeCl_3$$ produces chlorinated aromatic rings.

In this question the reagent system is $$Cl_2/FeCl_3$$, so the reaction proceeds through the electrophilic aromatic substitution (EAS) mechanism.
The methyl group in toluene is:

  • Electron-donating through hyperconjugation and +I effect.
  • Activating and ortho/para-directing in EAS.

Therefore the incoming electrophile $$Cl^+$$ attacks predominantly at the ortho and para positions of the benzene ring relative to the $$CH_3$$ group.

Hence the major products are $$o$$-chlorotoluene and $$p$$-chlorotoluene.

Option C which is: $$o-$$ and $$p-$$chlorotoluene

The nitro group is represented as $$-NO_2$$. Its electronic behaviour can be understood through two effects:

1. Strong $$-I$$ effect (inductive withdrawal): The electronegative oxygen atoms pull electron density away from the ring through the σ-bond framework.
2. Strong $$-R$$ effect (resonance withdrawal): One of the oxygen atoms can withdraw the π-electron pair from the ring, forming a resonance structure like

$$\text{Benzene-NO}_2 \longrightarrow \text{O=N}^+\text{-O}^- \xrightarrow{\text{Resonance}} \text{O}^-\text{-N}^+=\text{O (with positive charge in the ring)}$$

The resonance structures place a partial positive charge on the ortho and para carbon atoms of the ring. Thus electron density throughout the ring decreases. Because electrophilic substitution reactions require electron-rich aromatic rings to attract an incoming electrophile $$E^{+}$$, any factor that removes electron density deactivates the ring toward electrophilic attack.

Let us check each option:

Option A Activates the ring toward electrophilic substitution - Incorrect; the nitro group withdraws electrons, so it has the opposite effect.

Option B Renders the ring basic - Incorrect; lower electron density actually makes the ring less basic.

Option C Deactivates the ring toward nucleophilic substitution - Incorrect; an electron-withdrawing group like $$-NO_2$$ actually activates the ring toward nucleophilic aromatic substitution (by stabilising the Meisenheimer complex).

Option D Deactivates the ring toward electrophilic substitution - Correct; explained above.

Hence, the right choice is:
Option D which is: deactivates the ring towards electrophilic substitution.

Alkyl side-chains attached to an aromatic ring that contain at least one benzylic hydrogen are oxidised by strong oxidising agents such as alkaline or acidic $$KMnO_4$$ to a single product: the carboxylic acid $$\left(-COOH\right)$$ derivative of the ring, irrespective of the length of the side chain.

Ethyl benzene is $$C_6H_5CH_2CH_3$$. The benzylic carbon (the $$CH_2$$ directly bonded to the ring) possesses benzylic hydrogens, so it is susceptible to oxidation.

On treatment with hot, acidic or basic $$KMnO_4$$:

$$C_6H_5CH_2CH_3 \;\xrightarrow[\text{heat}]{KMnO_4} \; C_6H_5COOH + CO_2 + H_2O$$

The entire side-chain—here $$CH_2CH_3$$—is ultimately cleaved down to one carbon attached to the ring, which is converted into the carboxyl group $$-COOH$$. Hence the aromatic product is benzoic acid $$C_6H_5COOH$$.

Checking the options:
Option A: benzophenone - requires oxidation of two phenyl rings attached to a carbonyl; not formed here.
Option B: acetophenone - has a carbonyl group but no oxidising pathway from ethyl benzene under these conditions.
Option C: benzoic acid - matches the reaction product.
Option D: benzyl alcohol - is formed by mild reduction, not oxidation.

Therefore, the compound obtained is benzoic acid.

Option C which is: benzoic acid