JEE Hydrocarbons - Alkanes Questions

We begin by examining statement (i). On combustion, dihydrogen reacts with dioxygen according to the equation $$2\,H_2(g)+O_2(g)\longrightarrow 2\,H_2O(l)+\text{energy}.$$ The only product is water, so no $$CO_2$$, no unburnt hydrocarbons and no soot are formed. Petrol, on the other hand, is a mixture of hydrocarbons and its combustion invariably gives $$CO_2$$, $$CO$$, $$SO_x$$, unburnt hydrocarbons and particulates. Therefore, dihydrogen indeed produces far fewer pollutants than petrol. So, statement (i) is true.

Now we come to statement (ii). The energy released per unit mass of dihydrogen is high, about $$120\;{\rm MJ\,kg^{-1}}$$, but the density of compressed gaseous hydrogen is extremely low. To store a useful amount of hydrogen gas, we need a very strong and therefore very heavy steel cylinder that can withstand high pressure (often >200 atm). NCERT explicitly mentions that “a cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol producing the same amount of energy.” Hence the overall weight (cylinder + hydrogen) becomes roughly thirty-fold compared with an equivalent petrol tank. So, statement (ii) is correct.

Next, let us verify statement (iii). One modern method of storing hydrogen is in the interstitial spaces of metal hydride-forming alloys. The typical alloy cited in the textbook is $$NaNi_5$$, which can absorb and release large amounts of hydrogen reversibly: $$NaNi_5 + \dfrac{x}{2}H_2 \rightleftharpoons NaNi_5H_x.$$ Thus, dihydrogen is indeed stored in tanks containing metal alloys such as $$NaNi_5$$. Consequently, statement (iii) is also correct.

Finally, we analyse statement (iv). According to data given in standard references and NCERT, the energy liberated per gram on complete combustion is about $$150\;{\rm kJ\,g^{-1}}$$ for liquid dihydrogen and about $$50\;{\rm kJ\,g^{-1}}$$ for LPG. The statement provided in the question interchanges these two numbers, saying $$50\;{\rm kJ\,g^{-1}}$$ for hydrogen and $$142\;{\rm kJ\,g^{-1}}$$ for LPG. This is the reverse of the correct values, so statement (iv) is false.

Collecting our results, statements (i), (ii) and (iii) are true, while statement (iv) is false. The option that lists exactly (i), (ii) and (iii) is Option D.

Hence, the correct answer is Option D.

First, let us write the general balanced combustion equation for an alkane. For an alkane of the form $$\text{C}_n\text{H}_{2n+2}$$, complete combustion in oxygen is given by

$$\text{C}_n\text{H}_{2n+2} \;+\;\frac{3n+1}{2}\,\text{O}_2 \;\longrightarrow\; n\,\text{CO}_2 \;+\;(n+1)\,\text{H}_2\text{O}.$$

This relation comes from balancing carbon atoms first ($$n\;\text{CO}_2$$), hydrogen atoms next ($$(n+1)\;\text{H}_2\text{O}$$ provides $$2(n+1)$$ hydrogens), and finally oxygen atoms, giving the factor $$\dfrac{3n+1}{2}$$ in front of $$\text{O}_2$$.

Now we apply this result to each hydrocarbon present in the mixture.

1. Propane has $$n = 3$$, i.e. its formula is $$\text{C}_3\text{H}_8$$. Substituting $$n = 3$$ into the oxygen coefficient $$\dfrac{3n+1}{2}$$ gives

$$\dfrac{3(3)+1}{2} \;=\; \dfrac{9+1}{2} \;=\; \dfrac{10}{2} \;=\; 5.$$

So the balanced combustion reaction for propane is

$$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \;\longrightarrow\; 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}.$$

Hence, 1 mole of propane requires $$5$$ moles of $$\text{O}_2$$.

2. Butane has $$n = 4$$, i.e. its formula is $$\text{C}_4\text{H}_{10}$$. Substituting $$n = 4$$ into the same expression $$\dfrac{3n+1}{2}$$ gives

$$\dfrac{3(4)+1}{2} \;=\; \dfrac{12+1}{2} \;=\; \dfrac{13}{2} \;=\; 6.5.$$

Thus the balanced combustion reaction for butane is

$$\text{C}_4\text{H}_{10} + \dfrac{13}{2}\,\text{O}_2 \;\longrightarrow\; 4\,\text{CO}_2 + 5\,\text{H}_2\text{O}.$$

Therefore, 1 mole of butane needs $$6.5$$ moles of $$\text{O}_2$$. Since the question contains 2 moles of butane, the oxygen required for butane is

$$2 \times 6.5 = 13 \text{ moles of } \text{O}_2.$$

We now add the oxygen required for propane and for butane:

$$\text{Total } \text{O}_2 = 5 \;+\; 13 = 18 \text{ moles}.$$

So, the answer is $$18$$.

Analyze the Starting Material

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

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

Step-by-Step Reaction Sequence

Step 1: Ozonolysis (O₃, Zn/H₂O)

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

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

Step 2: Oxidation (KMnO₄)

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

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

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

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

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

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

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

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

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

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

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

We need to identify the environmental consequence of gas released during anaerobic degradation of vegetation.

Anaerobic degradation means decomposition of organic matter (vegetation) in the absence of oxygen. This process is carried out by methanogenic bacteria. The primary gas released is methane ($$CH_4$$).

Methane is a greenhouse gas. Greenhouse gases trap infrared radiation emitted by the Earth's surface and prevent it from escaping into space. This leads to an increase in the average temperature of the Earth's atmosphere, a phenomenon known as global warming.

Methane is approximately 25 times more effective than carbon dioxide ($$CO_2$$) at trapping heat over a 100-year period. Therefore, even small amounts of methane can contribute significantly to global warming.

Additionally, methane in the atmosphere undergoes photochemical reactions that produce ground-level ozone and other secondary pollutants. Prolonged exposure to such pollutants has been associated with increased cancer risk in living organisms.

The other options do not correctly describe the primary effect of methane. Corrosion of metals is caused by electrochemical reactions, not methane. The ozone hole is caused by chlorofluorocarbons (CFCs), not methane. Acid rain is caused by $$SO_2$$ and $$NO_x$$ emissions, not methane.

Hence, the gas released during anaerobic degradation of vegetation may lead to global warming and cancer.

Hence, the correct answer is Option 3.

Coal gasification is the reaction of pulverised coal with a controlled amount of $$\text{O}_2$$ and $$\text{H}_2\text{O}$$ at about 1270 - 1570 K.
The main reactions are $$$\text{C} + \text{H}_2\text{O} \rightarrow \text{CO} + \text{H}_2$$$ and $$$\text{C} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}$$$.
The gaseous mixture obtained, rich in $$\text{CO}$$ and $$\text{H}_2$$, is called synthesis gas (syn-gas). Producing this mixture from coal is, by definition, gasification of coal. Therefore Statement I is true.

Syn-gas contains mainly $$\text{CO}$$ and $$\text{H}_2$$ in approximately a 1 : 1 ratio. $$\text{CO}_2$$, if present, is only a minor impurity and never equals the other two components. Hence the composition $$\text{CO} + \text{CO}_2 + \text{H}_2\; (1\!:\!1\!:\!1)$$ given in Statement II is incorrect. Statement II is false.

Thus, Statement I is true but Statement II is false. Hence the correct choice is Option B.

First, we recall the definition of the water-gas shift reaction. In industrial chemistry this name is reserved for the reversible catalytic conversion in which carbon monoxide reacts with steam to give carbon dioxide and hydrogen. Symbolically we write

$$\text{CO}(g) + \text{H}_2\text{O}(g) \;\rightleftharpoons\; \text{CO}_2(g) + \text{H}_2(g).$$

The standard operating temperature for the high-temperature stage of this process is about $$673\;\text{K}$$, and it is always carried out in the presence of a suitable catalyst such as iron oxide promoted with chromium oxide. Thus, any correct option must explicitly contain (i) carbon monoxide on the reactant side, (ii) steam on the reactant side, (iii) carbon dioxide and hydrogen on the product side, and (iv) a temperature near $$673\;\text{K}$$ along with a catalyst mention.

Now we examine the four alternatives one by one.

Option A gives $$\text{CH}_4(g) + \text{H}_2\text{O}(g) \xrightarrow{1270\,\text{K}} \text{CO}(g) + 3\text{H}_2(g).$$ This is the steam-reforming reaction of methane, not the water-gas shift. The reactants and products do not match the required $$\text{CO}/\text{H}_2\text{O}$$ ⇌ $$\text{CO}_2/\text{H}_2$$ pattern.

Option B gives $$2\text{C}(s) + \text{O}_2(g) + 4\text{N}_2(g) \xrightarrow{1273\,\text{K}} 2\text{CO}(g) + 4\text{N}_2(g).$$ This merely represents the combustion of carbon in air to carbon monoxide; it again fails to involve steam and therefore cannot be the water-gas shift reaction.

Option C gives $$\text{C}(s) + \text{H}_2\text{O}(g) \xrightarrow{1270\,\text{K}} \text{CO}(g) + \text{H}_2(g).$$ Although steam is present, the solid carbon reactant shows that this is the water-gas (or carbon-steam) reaction, not the shift reaction. The essential carbon dioxide product is missing, so this option is also ruled out.

Option D gives $$\text{CO}(g) + \text{H}_2\text{O}(g) \xrightarrow{673\,\text{K}}_{\text{catalyst}} \text{CO}_2(g) + \text{H}_2(g).$$ Here every required feature is present: carbon monoxide and steam on the left, carbon dioxide and hydrogen on the right, a temperature around $$673\;\text{K}$$, and an explicit mention of a catalyst. Therefore this reaction exactly matches the definition of the water-gas shift reaction.

All other options fail to satisfy one or more of these essential criteria, so they must be rejected.

Hence, the correct answer is Option D.

We recall that any straight-chain or branched alkane has the general molecular formula $$C_nH_{2n+2}$$. For the complete combustion of such an alkane in oxygen, the balanced chemical equation (with water taken in the gaseous form so that every species can be treated as a gas at the given temperature and pressure) is first written as

$$C_nH_{2n+2}+O_2 \rightarrow n\,CO_2+\frac{n+1}{2}\,H_2O$$

Now we must balance the oxygen atoms. On the right side we have $$2n$$ oxygen atoms in $$n\,CO_2$$ and $$\frac{n+1}{2}\times 1$$ oxygen atoms in $$\frac{n+1}{2}\,H_2O$$. Adding them, the total number of oxygen atoms required is

$$2n+\frac{n+1}{2}=2n+\frac{n}{2}+\frac{1}{2}=\frac{5n+1}{2}.$$

Because one molecule of $$O_2$$ contributes two oxygen atoms, the number of $$O_2$$ molecules needed is half of this value. Therefore, the correctly balanced equation becomes

$$C_nH_{2n+2}+\frac{3n+1}{2}\,O_2\;\rightarrow\;n\,CO_2+\frac{n+1}{2}\,H_2O.$$

The coefficient in front of $$O_2$$ is clearly $$\dfrac{3n+1}{2}$$.

At constant temperature and pressure, equal volumes of gases contain equal numbers of moles (Gay-Lussac’s law of combining volumes). Hence the ratio of the coefficients in the balanced equation gives us directly the ratio of the volumes consumed. Thus, the volume ratio

$$\frac{\text{Volume of }O_2}{\text{Volume of alkane}}=\frac{3n+1}{2}.$$

According to the data, $$5\,\text{L}$$ of the alkane need $$25\,\text{L}$$ of oxygen. So the experimental ratio is

$$\frac{25}{5}=5.$$

Equating the theoretical and experimental ratios, we get

$$\frac{3n+1}{2}=5.$$

Multiplying both sides by $$2$$ to clear the denominator,

$$3n+1=10.$$

Subtracting $$1$$ from both sides,

$$3n=9.$$

Finally, dividing by $$3$$,

$$n=3.$$

Thus the alkane contains three carbon atoms, and its molecular formula is obtained by substituting $$n=3$$ into $$C_nH_{2n+2}$$:

$$C_3H_8.$$

$$C_3H_8$$ is propane. Checking the options, propane corresponds to Option D.

Hence, the correct answer is Option D.

First, we have to recognise what “the simplest optically active alkene” is. In hydrocarbons the first stereogenic (chiral) carbon atom can appear only when the molecule contains at least five carbon atoms, because with four carbon atoms or fewer some two of the four substituents on any sp3 carbon are necessarily identical. The smallest alkene fulfilling this requirement is $$CH_{2}=CH-CH(CH_{3})-CH_{2}-CH_{3}$$ whose IUPAC name is $$3\text{-methyl-1-pentene}.$$ The italicised carbon in the middle bears four different groups - $$CH_{3},\;H,\;CH_{2}=CH-$$ and $$-CH_{2}CH_{3}$$ - so the alkene is optically active.

Now we pass this alkene through step (i): catalytic hydrogenation with $$H_{2}/Ni$$ at high temperature. The general reaction is stated first:

$$R-CH=CH-R' + H_{2} \;\xrightarrow{Ni,\;\Delta\; R-CH2-CH2-R'}$$

Applying it to our substrate, the double bond between the first two carbons is reduced, giving

$$CH_{3}-CH_{2}-CH(CH_{3})-CH_{2}-CH_{3},$$

which is $$3\text{-methylpentane}.$$ After saturation, the central carbon is bonded to two identical ethyl groups, a methyl group and a hydrogen atom, so the molecule now has an internal plane of symmetry and becomes achiral.

Step (ii) is halogenation with a halogen $$X_{2}$$ in the presence of heat: $$R-H + X_{2} \;\xrightarrow{\Delta\; R-X + HX}.$$ Under these free-radical conditions any hydrogen atom of the alkane may be replaced by one halogen atom, and we have to count every distinct monohalogenated product, including enantiomers when they appear.

Let us label the carbon atoms of $$3\text{-methylpentane}$$ for clarity:

$$C_{1}H_{3} - C_{2}H_{2} - C_{3}H(CH_{3}) - C_{4}H_{2} - C_{5}H_{3}$$
and the branched $$CH_{3}$$ on C3 is $$C_{3}'.$$

Because the molecule is symmetric about C3, the pairs (C1, C5) and (C2, C4) are equivalent. Thus there are only four kinds of hydrogen positions:

Type I : the six terminal hydrogens on C1 or C5.
Type II : the four hydrogens on C2 or C4.
Type III : the single hydrogen on the central carbon C3.
Type IV : the three hydrogens on the side-chain methyl C3′.

We now replace one hydrogen of each type by X and inspect the stereochemistry.

Type I substitution gives $$XCH_{2}-CH_{2}-CH(CH_{3})-CH_{2}-CH_{3}$$ (1-halo-3-methylpentane). Introducing X on one end breaks the internal symmetry, so the formerly achiral carbon C3 now has four different groups:
$$-CH_{2}XCH_{2}, \; -CH_{2}CH_{3}, \; -CH_{3},\; H.$$
C3 therefore becomes chiral, and a pair of enantiomers is produced. Because the two ends are equivalent, the product from C1 and the product from C5 are the same pair. Number of distinct products here = 2.

Type II substitution gives $$CH_{3}-CHX-CH(CH_{3})-CH_{2}-CH_{3}$$ (2-halo-3-methylpentane). Now two carbons are stereogenic: C2 (bearing X) and C3. With two chiral centres and no internal symmetry four stereoisomers are possible: RR, SS, RS and SR. (RR and SS are enantiomeric; RS and SR are another enantiomeric pair.) Substituting at C4 generates exactly the same four structures, so the total number of new products here is 4.

Type III substitution replaces the lone hydrogen on C3 itself: $$CH_{3}-CH_{2}-C(X)(CH_{3})-CH_{2}-CH_{3}.$$
C3 now carries two identical ethyl groups, X and CH3, so it is not chiral; the molecule has no other stereogenic centre. Hence only 1 product arises from this site.

Type IV substitution gives $$CH_{3}-CH_{2}-CH(CH_{2}X)-CH_{2}-CH_{3}.$$ C3 again has two identical ethyl groups and is therefore achiral, so this also contributes just 1 product.

Adding all the distinct mono-halogenated structures together:

$$2\;(\text{from Type I}) + 4\;(\text{from Type II}) + 1\;(\text{from Type III}) + 1\;(\text{from Type IV}) = 8.$$

So, the answer is $$8$$.

We begin by recalling the general formula for an acyclic (open-chain) alkane. For any alkane containing $$n$$ carbon atoms, the number of hydrogen atoms is given by the relation $$H = 2n + 2$$. Putting $$n = 6$$ gives $$H = 2(6) + 2 = 14$$, so the molecular formula $$C_6H_{14}$$ indeed represents an alkane.

Next, we need to enumerate every distinct way of connecting six carbon atoms in a single, continuous skeleton without forming any rings, because each distinct connectivity corresponds to a different structural (constitutional) isomer. Whenever a carbon chain is shortened in the parent skeleton and one or more side chains (alkyl groups) are attached, a new structural isomer can arise, provided that its carbon-carbon connectivity is not identical (upon any possible rotation or flipping) to a structure already counted.

We systematically construct each possibility, beginning with the longest unbranched chain and then considering all permissible branchings.

1. The straight chain containing all six carbons end-to-end is called normal hexane, written as $$CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$$. This is the first isomer.

2. Now we shorten the parent chain by one carbon atom (making a five-carbon backbone) and attach the removed carbon as a single-carbon branch. Attaching that branch to the second carbon of the chain yields $$2$$-methylpentane. Because numbering from either end puts the branch on carbon $$2$$, there is only one distinct isomer of this kind for a single branch on a five-carbon backbone. This is the second isomer.

3. When the methyl group is attached to the middle carbon of the same five-carbon backbone, we obtain $$3$$-methylpentane. Rotation or reflection cannot convert this into $$2$$-methylpentane, so this is a distinct structure, giving the third isomer.

4. We now shorten the parent chain to four carbons and keep two carbons as branches. If both methyl branches are placed on the same second carbon of the butane backbone, the structure is $$2,2$$-dimethylbutane. This differs from every structure already listed, providing the fourth isomer.

5. Keeping the four-carbon backbone but placing the two methyl branches on adjacent carbons—one on carbon $$2$$ and the other on carbon $$3$$—gives $$2,3$$-dimethylbutane. There is no way to rotate or flip this skeleton so that it becomes identical to $$2,2$$-dimethylbutane$, to any of the earlier monomethyl isomers, or to $$n$$-hexane. Hence this is the fifth and final structural isomer.

At this point every conceivable carbon skeleton for six carbons has been exhausted. There are no further distinct arrangements: a parent chain of three carbons with more and longer branches would necessarily duplicate one of the skeletons already obtained, and any structure containing a ring would violate the requirement of an acyclic alkane.

So, the total count of unique carbon connectivities compatible with $$C_6H_{14}$$ is $$5$$.

Hence, the correct answer is Option B.

The problem asks for the alkane with exactly two secondary hydrogens. Secondary hydrogens are attached to secondary carbon atoms, which are bonded to exactly two other carbon atoms.

We will analyze each option by drawing the carbon skeleton and identifying the types of carbon atoms and their attached hydrogens.

Option A: 4-Ethyl-3,4-dimethyloctane

The parent chain is octane (8 carbons). Substituents: a methyl group at position 3, and at position 4, an ethyl group and a methyl group.

Structure:

$$ \text{C1} - \text{C2} - \text{C3} - \text{C4} - \text{C5} - \text{C6} - \text{C7} - \text{C8} $$

At C3: methyl group → C3 bonded to C2, C4, and CH₃ → tertiary carbon → one hydrogen (tertiary).

At C4: ethyl group (-CH₂CH₃) and methyl group (-CH₃) → C4 bonded to C3, C5, CH₂ (of ethyl), and CH₃ → quaternary carbon → no hydrogen.

Ethyl group: -CH₂CH₃ → the CH₂ group is secondary carbon (bonded to C4 and CH₃) → two hydrogens (secondary).

Methyl groups: all primary carbons → primary hydrogens.

Chain analysis:

C1: primary carbon → three hydrogens (primary).

C2: bonded to C1 and C3 → secondary carbon → two hydrogens (secondary).

C5: bonded to C4 and C6 → secondary carbon → two hydrogens (secondary).

C6: bonded to C5 and C7 → secondary carbon → two hydrogens (secondary).

C7: bonded to C6 and C8 → secondary carbon → two hydrogens (secondary).

C8: primary carbon → three hydrogens (primary).

Secondary hydrogens: C2 (2), C5 (2), C6 (2), C7 (2), ethyl CH₂ (2) → total 10 secondary hydrogens.

Not the required alkane.

Option B: 2,2,3,3-Tetramethylpentane

The parent chain is pentane (5 carbons). Substituents: two methyl groups at C2 and two methyl groups at C3.

Structure:

$$ \text{C1} - \text{C2} - \text{C3} - \text{C4} - \text{C5} $$

At C2: two methyl groups → C2 bonded to C1, C3, CH₃, and CH₃ → quaternary carbon → no hydrogen.

At C3: two methyl groups → C3 bonded to C2, C4, CH₃, and CH₃ → quaternary carbon → no hydrogen.

Chain analysis:

C1: bonded to C2 → primary carbon → three hydrogens (primary).

C4: bonded to C3 and C5 → secondary carbon → two hydrogens (secondary).

C5: bonded to C4 → primary carbon → three hydrogens (primary).

Methyl groups: four methyl groups (all primary carbons) → each has three primary hydrogens.

Secondary hydrogens: only from C4 → two secondary hydrogens.

This matches the requirement.

Option C: 2,2,4,5-Tetramethylheptane

The parent chain is heptane (7 carbons). Substituents: two methyl groups at C2, one methyl at C4, one methyl at C5.

Structure:

$$ \text{C1} - \text{C2} - \text{C3} - \text{C4} - \text{C5} - \text{C6} - \text{C7} $$

At C2: two methyl groups → C2 bonded to C1, C3, CH₃, and CH₃ → quaternary carbon → no hydrogen.

At C4: methyl group → C4 bonded to C3, C5, CH₃, and H → tertiary carbon → one hydrogen (tertiary).

At C5: methyl group → C5 bonded to C4, C6, CH₃, and H → tertiary carbon → one hydrogen (tertiary).

Chain analysis:

C1: primary carbon → three hydrogens (primary).

C3: bonded to C2 and C4 → secondary carbon → two hydrogens (secondary).

C6: bonded to C5 and C7 → secondary carbon → two hydrogens (secondary).

C7: primary carbon → three hydrogens (primary).

Methyl groups: four methyl groups (all primary carbons) → primary hydrogens.

Secondary hydrogens: C3 (2) and C6 (2) → total four secondary hydrogens.

Not the required alkane.

Option D: 2,2,4,4-Tetramethylhexane

The parent chain is hexane (6 carbons). Substituents: two methyl groups at C2 and two methyl groups at C4.

Structure:

$$ \text{C1} - \text{C2} - \text{C3} - \text{C4} - \text{C5} - \text{C6} $$

At C2: two methyl groups → C2 bonded to C1, C3, CH₃, and CH₃ → quaternary carbon → no hydrogen.

At C4: two methyl groups → C4 bonded to C3, C5, CH₃, and CH₃ → quaternary carbon → no hydrogen.

Chain analysis:

C1: primary carbon → three hydrogens (primary).

C3: bonded to C2 and C4 → secondary carbon → two hydrogens (secondary).

C5: bonded to C4 and C6 → secondary carbon → two hydrogens (secondary).

C6: primary carbon → three hydrogens (primary).

Methyl groups: four methyl groups (all primary carbons) → primary hydrogens.

Secondary hydrogens: C3 (2) and C5 (2) → total four secondary hydrogens.

Not the required alkane.

Only option B has exactly two secondary hydrogens.

Final Answer: B

The Assertion states that hydrogen is an environment friendly fuel. This is true because when hydrogen burns, it produces only water as a byproduct:

$$2H_2 + O_2 \rightarrow 2H_2O$$

No harmful gases like CO$$_2$$, SO$$_2$$, or NO$$_x$$ are produced, making it environmentally friendly.

Now, the Reason states that the atomic number of hydrogen is 1 and it is a very light element. This is also true — hydrogen has atomic number 1 and is indeed the lightest element.

However, the reason is not the correct explanation of the assertion. Hydrogen being environment friendly is because its combustion product is water (non-polluting), not because of its atomic number or mass.

Hence, the correct answer is Option B: Both A and R are true but R is NOT the correct explanation of A.

Ethane (C₂H₆) when treated with excess Br₂ in diffused sunlight undergoes free radical halogenation. With excess bromine, all possible bromo derivatives can be formed.

The general formula for bromo derivatives is C₂H₆₋ₙBrₙ where n = 1, 2, 3, 4, 5, 6.

n = 1: CH₃CH₂Br (bromoethane) → 1 isomer

n = 2: CH₃CHBr₂ (1,1-dibromoethane), CH₂BrCH₂Br (1,2-dibromoethane) → 2 isomers

n = 3: CH₃CBr₃ (1,1,1-tribromoethane), CHBr₂CH₂Br (1,1,2-tribromoethane) → 2 isomers

n = 4: CBr₃CH₂Br (1,1,1,2-tetrabromoethane), CHBr₂CHBr₂ (1,1,2,2-tetrabromoethane) → 2 isomers

n = 5: CBr₃CHBr₂ (pentabromoethane) → 1 isomer

n = 6: C₂Br₆ (hexabromoethane) → 1 isomer

Total number of bromo derivatives = 1 + 2 + 2 + 2 + 1 + 1 = 9.

First, let us recall why the photochemical iodination of methane is reversible. After the forward photochemical step

$$CH_4 + I_2 \xrightarrow{h\nu} CH_3I + HI,$$

the hydrogen iodide that is formed is a very strong reducing agent. It immediately reacts back with the product methyl iodide and reproduces the starting materials:

$$CH_3I + HI \longrightarrow CH_4 + I_2.$$

This back-reaction is fast, so the overall process settles at an equilibrium and the arrow becomes $$\rightleftharpoons.$$ To convert the system into an irreversible one we must remove, destroy or otherwise make $$HI$$ unavailable as soon as it appears; then the reverse step cannot occur and the reaction will keep moving to the right until one of the reactants is exhausted.

Now we test each reagent given in the options to see whether it can do this job.

Option A : $$HOCl$$ (hypochlorous acid)
$$HOCl$$ can indeed oxidise iodide ions, but in doing so it itself gets reduced to $$HCl$$ and water, and it works only under very dilute, neutral media. In the strongly acidic medium in which $$HI$$ exists, $$HOCl$$ disproportionates rapidly and does not survive long enough to consume all the $$HI$$ formed continuously in the photochemical reaction. Hence the equilibrium is not driven completely to the right.

Option B : dilute $$HNO_2$$ (nitrous acid)
$$HNO_2$$ is a mild oxidising agent. The reaction $$2\,HI + HNO_2 \;\longrightarrow\; I_2 + NO + H_2O$$ is slow and incomplete; the small amount of $$HI$$ removed is quickly replenished by the back-reaction. Thus the system still remains reversible.

Option C : liquid $$NH_3$$
Being a base, ammonia neutralises hydrogen iodide : $$NH_3 + HI \;\longrightarrow\; NH_4I.$$ Although free $$HI$$ disappears, the iodide ion $$(I^-)$$ that stays in the medium keeps reducing $$CH_3I$$ by $$S_N2$$ attack to reproduce methane: $$CH_3I + I^- \;\longrightarrow\; CH_3^- + I_2 \;\longrightarrow\; CH_4 + I_2.$$ Therefore the reverse path is still available and the reaction is not rendered irreversible.

Option D : concentrated $$HIO_3$$ (iodic acid)
Concentrated iodic acid is a much stronger oxidising agent for hydrogen iodide. The well-known redox reaction is

$$5\,HI + HIO_3 \;\longrightarrow\; 3\,I_2 + 3\,H_2O.$$

Observe each consequence carefully:

• The freshly produced $$HI$$ is immediately destroyed, so its concentration in the reaction vessel remains practically zero.
• Because $$HI$$ never accumulates, the reverse step $$CH_3I + HI \rightarrow CH_4 + I_2$$ becomes impossible.
• The extra $$I_2$$ formed in the redox process simply joins the starting $$I_2$$ pool and continues to react with more $$CH_4$$ under the influence of light, producing still more $$CH_3I$$. In other words, the presence of $$HIO_3$$ continuously pulls the overall reaction to the right.

Therefore, concentrated $$HIO_3$$ removes the key species $$HI$$ as soon as it forms, blocking the backward path and converting the equilibrium into a one-way (irreversible) reaction.

Hence, the correct answer is Option D.

We consider 2-methylbutane of formula $$C_5H_{12}$$ and number its longest chain as C1-C4 so that the methyl substituent is at C2. The structure is:

$$ \text{C1: }CH_3\;-\;\text{C2: }CH(CH_3)\;-\;\text{C3: }CH_2\;-\;\text{C4: }CH_3 $$

Next we identify the distinct types of hydrogen atoms and count their number:

Type 1 (C1-H): three equivalent primary hydrogens at C1.

Type 2 (C4-H): three equivalent primary hydrogens at C4, which is not equivalent to C1.

Type 3 (C2-H): one secondary hydrogen at C2.

Type 4 (C3-H): two equivalent secondary hydrogens at C3.

On monochlorination each hydrogen type gives a distinct chlorinated product. We must also check whether the carbon bearing chlorine becomes a chiral center.

Case 1: Substitution at C1 or at C4 converts a $$CH_3$$ into $$CH_2Cl$$. In each case the carbon has two identical hydrogens left, so the product is achiral and gives only one isomer for C1 and one for C4.

Case 2: Substitution of the single hydrogen at C2 gives $$CHCl$$ at C2 bonded to -CH_3, -C_3H_7 (from C1), and -C_2H_5 (from C3-C4). These three substituents are all different, so C2 becomes a chiral center. Therefore this case yields two enantiomers (2 isomers).

Case 3: Substitution of one of the two equivalent hydrogens at C3 also creates $$CHCl$$ at C3 with four different groups, making C3 a chiral center. Replacement of either prochiral hydrogen gives an enantiomeric pair, so this case also yields two isomers.

Summing the isomers from all cases:

1 (from C1) + 1 (from C4) + 2 (from C2) + 2 (from C3) = 6.

Therefore, the total number of isomeric monochlorinated products of 2-methylbutane is 6.

We need to identify the products A and B in the atmospheric reaction: $$\dot{C}l + CH_4 \rightarrow A + B$$

The reaction involves a chlorine free radical ($$\dot{C}l$$) reacting with methane ($$CH_4$$). This is the propagation step of the free radical substitution (halogenation) of methane.

In the propagation step of free radical halogenation, the chlorine radical abstracts a hydrogen atom from methane:

The chlorine radical ($$\dot{C}l$$) has an unpaired electron and is highly reactive. It attacks the C-H bond in methane, removing a hydrogen atom to form HCl (a stable molecule), while leaving behind a methyl radical ($$\dot{C}H_3$$).

From the reaction, $$A = \dot{C}H_3$$ (methyl radical) and $$B = HCl$$ (hydrogen chloride).

Option A: $$C_2H_6$$ and $$Cl_2$$ — Incorrect. This would require recombination, not a propagation step.

Option B: $$\dot{C}HCl_2$$ and $$H_2$$ — Incorrect. The chlorine radical abstracts H, it does not insert into the C-H bond.

Option C: $$\dot{C}H_3$$ and $$HCl$$ — Correct. This matches the propagation step.

Option D: $$C_2H_6$$ and $$HCl$$ — Incorrect. $$C_2H_6$$ cannot form from a single methane molecule in one step.

Hence, the correct answer is Option C: $$\dot{C}H_3$$ and $$HCl$$.

We need to find the total number of monobromo derivatives formed by all isomers of the alkane with molecular formula $$C_5H_{12}$$, excluding stereoisomers.

Identify all structural isomers of $$C_5H_{12}$$

There are three structural isomers:

(i) n-Pentane: $$CH_3-CH_2-CH_2-CH_2-CH_3$$

(ii) Isopentane (2-methylbutane): $$(CH_3)_2CH-CH_2-CH_3$$

(iii) Neopentane (2,2-dimethylpropane): $$(CH_3)_4C$$

n-Pentane — types of hydrogen

$$\underset{a}{CH_3}-\underset{b}{CH_2}-\underset{c}{CH_2}-\underset{b}{CH_2}-\underset{a}{CH_3}$$

Due to the plane of symmetry at C-3, there are 3 types of hydrogens (a, b, c), giving 3 monobromo products:

1-bromopentane, 2-bromopentane, and 3-bromopentane.

Isopentane — types of hydrogen

The structure $$(CH_3)_2CH-CH_2-CH_3$$ has 4 types of hydrogens:

(a) H on the two equivalent methyl groups at C-1 and C-1' (6H)

(b) H on the methine (CH) at C-2 (1H)

(c) H on the methylene ($$CH_2$$) at C-3 (2H)

(d) H on the terminal methyl ($$CH_3$$) at C-4 (3H)

This gives 4 monobromo products: 1-bromo-2-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane (by substitution at C-3), and 1-bromo-3-methylbutane.

Neopentane — types of hydrogen

All four $$CH_3$$ groups are equivalent, so there is only 1 type of hydrogen (12H), giving 1 monobromo product: 1-bromo-2,2-dimethylpropane.

Total count

$$\text{Total monobromo derivatives} = 3 + 4 + 1 = 8$$

The answer is 8.

We have isobutane, whose condensed formula is $$(CH_3)_3CH$$ and whose structural skeleton looks like a central carbon attached to three methyl groups and one hydrogen.

The reagent is molecular bromine, $$Br_2$$, and the reaction conditions are light and a temperature of about $$125^\circ{\rm C}$$. Under these conditions bromine undergoes homolytic fission giving two bromine radicals: $$Br_2 \;\xrightarrow{h\nu,\,125^\circ{\rm C}}\; 2\,Br^{\bullet}$$. Thus the reaction proceeds by the free-radical halogenation mechanism.

Radical bromination shows a very strong preference for abstracting a hydrogen from the carbon that forms the most stable radical; the order of stability (and hence of ease of abstraction) is

$$3^{\circ}\,H \;>\; 2^{\circ}\,H \;>\; 1^{\circ}\,H$$

In isobutane there are:

• one tertiary hydrogen on the central carbon, and
• nine primary hydrogens on the three methyl groups.

Although there are many more primary hydrogens, the rate of abstraction from the tertiary position is so much higher that the tertiary radical is formed predominantly. We write the propagation step producing that radical:

$$(CH_3)_3CH \;+\; Br^{\bullet}\;\longrightarrow\;(CH_3)_3C^{\bullet}\;+\;HBr$$

The tertiary carbon radical then rapidly combines with another bromine molecule:

$$(CH_3)_3C^{\bullet}\;+\;Br_2\;\longrightarrow\;(CH_3)_3CBr\;+\;Br^{\bullet}$$

Because an excess of isobutane is present, the concentration of $$Br_2$$ is kept low, so further substitution (di- or tri-bromination) is statistically suppressed. Therefore the chief product is tert-butyl bromide, whose condensed structural formula is:

$$CH_3-C(Br)(CH_3)_2$$

This structure corresponds exactly to Option D.

Hence, the correct answer is Option D.

Statement I says that 2-methylbutane on oxidation with $$\text{KMnO}_4$$ gives 2-methylbutan-2-ol. In 2-methylbutane (isopentane), the carbon at position 2 is a tertiary carbon bonded to three other carbons. Oxidation with $$\text{KMnO}_4$$ preferentially attacks tertiary C$$-$$H bonds because they have the lowest bond dissociation energy. The tertiary hydrogen at C-2 is abstracted, and the resulting radical is hydroxylated, forming 2-methylbutan-2-ol (a tertiary alcohol). This is a known reaction specific to branched alkanes with tertiary carbons. So Statement I is correct.

Statement II says that n-alkanes can be easily oxidised to corresponding alcohols with $$\text{KMnO}_4$$. Straight-chain alkanes have only primary and secondary C$$-$$H bonds, which are much more resistant to oxidation. Under mild conditions, $$\text{KMnO}_4$$ cannot oxidise n-alkanes to alcohols. Under harsh conditions, n-alkanes undergo uncontrolled oxidation leading to a mixture of carboxylic acids and other products rather than cleanly forming alcohols. So Statement II is incorrect.

Therefore, Statement I is correct but Statement II is incorrect.

This problem involves Kolbe's electrolysis of a mixture of two sodium carboxylates.

Key Concept: Kolbe's Electrolysis

In Kolbe's electrolysis, sodium salts of carboxylic acids are electrolysed. At the anode, the carboxylate ions lose electrons and undergo decarboxylation to form free radicals:

$$RCOO^- \rightarrow RCOO \cdot + e^- \rightarrow R \cdot + CO_2$$

These free radicals then combine (couple) to form alkanes.

From $$CH_3COONa$$: the methyl radical $$CH_3 \cdot$$ is produced.

From $$C_2H_5COONa$$: the ethyl radical $$C_2H_5 \cdot$$ is produced.

When two types of radicals are present in the reaction mixture, they can combine in all possible pairwise ways:

1. $$CH_3 \cdot + CH_3 \cdot \rightarrow CH_3CH_3$$ (ethane, $$C_2H_6$$)

2. $$CH_3 \cdot + C_2H_5 \cdot \rightarrow CH_3C_2H_5$$ (propane, $$C_3H_8$$)

3. $$C_2H_5 \cdot + C_2H_5 \cdot \rightarrow C_2H_5C_2H_5$$ (butane, $$C_4H_{10}$$)

We obtain three distinct alkanes: ethane, propane, and butane.

The answer is 3.

When an alkene reacts with $$Br_2$$ in the presence of light ($$h\nu$$) or heat, the reaction proceeds via a free-radical mechanism that favors substitution at the allylic position.

The major product is formed by replacing the hydrogen atom at the most stable allylic position with a bromine atom. In this case, the carbon at the 3-position is both allylic and tertiary, making it the most reactive site.

Therefore, the hydrogen at the carbon at the 3-position is replaced by a Bromine atom.

In the free radical bromination of sec-butylcyclohexane, bromine selectively attacks the most stable radical intermediate, which is the tertiary ($$3^\circ$$) carbon at the ring junction (Carbon I). This specific position is prioritized over the side-chain tertiary carbon because it is stabilized by a higher number of $$\alpha$$-hydrogen atoms through hyperconjugation.

C₇H₁₆ is heptane. The chain isomers are:

1. n-heptane

2. 2-methylhexane

3. 3-methylhexane

4. 2,2-dimethylpentane

5. 2,3-dimethylpentane

6. 2,4-dimethylpentane

7. 3,3-dimethylpentane

8. 3-ethylpentane

9. 2,2,3-trimethylbutane

Total = 9 isomers.

The answer is 9.

The reactant is $$n$$-hexane ($$\text{CH}_3(\text{CH}_2)_4\text{CH}_3$$), a straight-chain alkane.

The reagent used is anhydrous $$\text{AlCl}_3 / \text{HCl}$$ with heating ($$\Delta$$).

For $$n$$-hexane, the rearrangement yields a mixture of branched isomers, primarily:

• 2-methylpentane (Major product)
• 3-methylpentane

$$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{Anhy. AlCl}_3/\text{HCl}, \Delta} \text{CH}_3-\underset{\substack{| \\ \text{CH}_3}}{\text{CH}}-\text{CH}_2-\text{CH}_2-\text{CH}_3$$

For an alkane having the general formula $$C_nH_{2n+2}$$, the balanced combustion reaction is

$$\text{}$$ 

$$C_nH_{2n+2} + \left(n + \frac{2n+2}{4}\right)O_2 \rightarrow nCO_2 + (n+1)H_2O$$

The number of $$O_2$$ molecules needed equals the sum of:

• $$n$$ (for converting $$n$$ carbon atoms to $$nCO_2$$)
$$\text{}$$
• $$\dfrac{n+1}{2}$$ (because each mole of $$H_2O$$ needs one atom of oxygen, and each mole of $$O_2$$ supplies two atoms)

Hence, moles of $$O_2$$ required per mole of alkane $$=$$

$$n + \frac{n+1}{2} = \frac{2n + n + 1}{2} = \frac{3n+1}{2}$$

The problem states that 1 mole of the alkane requires 8 moles of $$O_2$$:

$$\frac{3n+1}{2} = 8 \quad \Longrightarrow \quad 3n+1 = 16 \quad \Longrightarrow \quad 3n = 15 \quad \Longrightarrow \quad n = 5$$

So the alkane is $$C_5H_{2(5)+2} = C_5H_{12}$$.

The sum of the number of carbon and hydrogen atoms in one molecule is

$$5 + 12 = 17$$

Therefore, the required sum is 17.

To determine the major product P, we analyze each step of the reaction sequence systematically.

The first two steps involve the conversion of 1,2-dibromocyclooctane into an alkyne. Treatment with aqueous $$KOH$$ initially causes dehydrohalogenation to form a vinyl bromide, and the subsequent treatment with the strong base $$NaNH_2$$ removes another molecule of $$HBr$$ to produce cyclooctyne through double dehydrohalogenation.

In the third step, cyclooctyne is treated with $$Hg^{2+}/H^+$$. Under these conditions, the alkyne undergoes acid-catalyzed hydration to form an enol, which rapidly tautomerizes to the corresponding ketone. Thus, cyclooctyne is converted into cyclooctanone.

In the final step, cyclooctanone is subjected to Clemmensen reduction using $$Zn\text{-}Hg/H^+$$. The carbonyl group is completely reduced to a methylene group $$(-CH_2-)$$, converting the ketone into the corresponding saturated hydrocarbon.

Therefore, the overall transformation is

$$\text{1,2-dibromocyclooctane} ;\longrightarrow; \text{cyclooctyne} ;\longrightarrow; \text{cyclooctanone} ;\longrightarrow; \text{cyclooctane}.$$

Hence, the final major product P is cyclooctane, corresponding to the eight-membered saturated ring shown in option B

Structure of neopentane (also called $$2,2$$-dimethylpropane): it has a central $$C$$ atom attached to four $$CH_3$$ groups giving the formula $$C\,(CH_3)_4$$.

Case 1: Verifying Statement I - “Neopentane forms only one monosubstituted derivative.”
Every hydrogen in neopentane belongs to a $$CH_3$$ group that is equivalent by symmetry; each $$CH_3$$ is attached to the same central carbon. Therefore, replacing any one hydrogen (by $$Cl$$, $$Br$$, $$OH$$, etc.) always gives the same product. Hence only one type of monosubstituted derivative is possible.
Thus, Statement I is correct.

Case 2: Verifying Statement II - “Melting point of neopentane is higher than n-pentane.”
Melting point depends on how well molecules fit into a crystal lattice. Neopentane has an almost spherical shape, so its molecules pack efficiently, giving stronger intermolecular van der Waals attractions in the solid state. In contrast, n-pentane is elongated, packs less efficiently, and therefore melts at a lower temperature.
Experimental data confirm: mp (neopentane) ≈ $$-16^\circ\mathrm C$$, mp (n-pentane) ≈ $$-130^\circ\mathrm C$$. Hence, Statement II is also correct.

Since both statements are correct, the appropriate choice is Option B.

The degree of unsaturation (or DBE) for $$C_5H_{10}$$ is
$$\frac{2(5)+2-10}{2}=1$$.

A single degree of unsaturation can arise from either one C=C double bond or one ring. The question states that the compounds do not decolourise alkaline $$KMnO_4$$, so they contain no double bond. Hence every isomer must be a cycloalkane with exactly one ring and no multiple bonds.

List all unique cycloalkane skeletons that give the formula $$C_5H_{10}$$ (only constitutional isomers are needed, stereochemistry is ignored).

Case 1: Five-membered ring - cyclopentane  Skeleton: $$\text{cyclopentane}$$

Case 2: Four-membered ring with one carbon outside the ring - methylcyclobutane  Skeleton: $$\text{methylcyclobutane}$$

Case 3: Three-membered ring with two carbons outside the ring - three possibilities  
(a) 1,1-dimethylcyclopropane (gem-dimethyl)  
(b) 1,2-dimethylcyclopropane (two methyls on adjacent ring carbons)  
(c) 1-ethylcyclopropane (an ethyl group on one ring carbon)

Thus there are in total five structural isomers that match the conditions:

1. cyclopentane
2. methylcyclobutane
3. 1,1-dimethylcyclopropane
4. 1,2-dimethylcyclopropane
5. 1-ethylcyclopropane

Next, count the number of distinct monochloro products each isomer can give on free-radical chlorination (light). Only connectivity is considered; stereochemical differences are ignored.

Isomer 1: cyclopentane
All five ring carbons are equivalent, so only one monochloro product is possible.
Number of monochloro isomers = $$1$$

Isomer 2: methylcyclobutane
Identify symmetry-independent positions:
- C1 = ring carbon bearing CH3
- C2 (and C4) = ring carbons adjacent to C1 (equivalent to each other)
- C3 = ring carbon opposite C1
- CH3 group itself
Therefore four different sites give four constitutional monochloro products.
Number of monochloro isomers = $$4$$

Isomer 3: 1,1-dimethylcyclopropane
C1 (bearing the two methyls) has no H, so it cannot be substituted.
- C2 and C3 (ring CH2 groups) are equivalent → one product.
- The two methyl groups are equivalent → substitution on either gives the same product.
Number of monochloro isomers = $$2$$

Isomer 4: 1,2-dimethylcyclopropane
The molecule has a mirror plane that exchanges C1 and C2; C3 is unique.
Independent sites:
- C1 (≡ C2) → one product
- C3 → second product
- Either methyl group (equivalent) → third product
Number of monochloro isomers = $$3$$

Isomer 5: 1-ethylcyclopropane
Symmetry exchanges C2 and C3; C1 is unique.
Independent sites:
- C1 (ring carbon bearing ethyl)
- C2 (≡ C3)
- α-carbon of ethyl group (CH2 attached to ring)
- β-carbon of ethyl group (terminal CH3)
Number of monochloro isomers = $$4$$

Finally, add the monochloro products from all isomers:

$$1 + 4 + 2 + 3 + 4 = 14$$

Hence, the total number of distinct monochloro structural isomers that can be formed is 14.

Let the molecular formula of the alkane be (C_nH_{2n+2}).

The general combustion equation for an alkane is

$$C_nH_{2n+2}+\left(\frac{3n+1}{2}\right)O_2\rightarrow nCO_2+(n+1)H_2O.$$

Since 8 moles of (O_2) are required for complete combustion,

$$\frac{3n+1}{2}=8,$$

$$3n+1=16,$$

$$3n=15,$$

$$n=5.$$

Hence, the molecular formula of the alkane is

$$C_5H_{12}.$$

Among the three isomers of (C_5H_{12}):

• (n)-Pentane gives three monochlorinated products.
• 2-Methylbutane gives four monochlorinated products.
• 2,2-Dimethylpropane (neopentane) gives only one monochlorinated product because all twelve hydrogen atoms are equivalent.

Therefore, compound (Y) is neopentane with the structure

$$C(CH_3)_4.$$

A primary carbon is bonded to only one other carbon atom. In neopentane, each of the four methyl carbons is attached only to the central carbon and is therefore a primary carbon, while the central carbon is quaternary.

Hence, the number of primary carbon atoms in (Y) is 4.

Therefore, the correct answer is 4.

First, note that all three reactions mentioned in the statement — isomerisation with $$\text{anh.}\,AlCl_3/HCl$$, aromatisation and cyclisation — are characteristic transformations shown by straight-chain alkanes when treated with suitable catalysts at elevated temperature.

Let the unknown hydrocarbon be $$\mathrm{C_7H_{14}}$$ (an alkene or triene) or $$\mathrm{C_7H_{16}}$$ (an alkane). 

We now match the products obtained in each step with the known behaviour of C-7 hydrocarbons.

Step (1) Isomerisation:

Straight-chain alkanes in the presence of $$\text{anh.}\,AlCl_3$$ and dry $$HCl$$ undergo skeletal rearrangement to give more highly branched isomers having the same molecular formula. For the product to be $$2,4\text{-dimethylpentane}$$ (molecular formula $$\mathrm{C_7H_{16}}$$), the starting compound must also possess the same formula $$\mathrm{C_7H_{16}}$$. The only option given that matches this requirement is heptane (the straight-chain isomer).

Step (2) Aromatisation:

A normal C-7 alkane on dehydrogenation and ring-closure over a Pt/Cr2O3 catalyst forms a C-6 aromatic ring plus one carbon as a side chain, i.e. $$\mathrm{C_6H_5CH_3}$$ (toluene) $$+$$ $$4H_2$$. This is exactly the product stated, confirming that the substrate is the normal (unbranched) C-7 alkane.

Step (3) Cyclisation:

Normal alkanes containing 6-10 carbons cyclise over acidic catalysts (e.g.\ alumina) to give n-1 membered cycloalkanes carrying one alkyl substituent. Thus an n-heptane chain folds to give $$\mathrm{CH_3-(CH_2)_5-CH_3} \longrightarrow \mathrm{CH_3-C_6H_{11}}$$ which is methyl cyclohexane, the product mentioned.

All three observations are therefore satisfied only by the straight-chain alkane $$\mathbf{n\text{-}heptane}$$.

Hence, the correct name of compound (X) is Heptane.

Option C $$\longrightarrow\ $$Heptane

Step 1:

For a saturated hydrocarbon (alkane), the general formula is $$C_nH_{2n+2}$$.

Molar mass is $$72g/mol^{-1}$$

Therefore,

$$12n+(2n+2)=72$$
$$14n+2=72$$
$$14n=70$$
$$n=5$$

the molecular formula is $$C_5H_{12}$$

Step 2:

Now check its isomers:

1. $$n-pentane$$ → 2 primary carbons
2. $$2-methylbutane$$ → 3 primary carbons (given in question)
3. $$2,2-dimethylpropane$$ → 4 primary carbons

Therefore, the IUPAC name of the compound with 3 Primary Carbons and $$72 g/mol^{-1}$$ mass is 

$$2-methylbutane$$ 

$$i.e$$ Option C. 

In Statement I, methane can be prepared by decarboxylation of sodium ethanoate with soda lime:

$$CH_3COONa+NaOH\xrightarrow{\Delta,\ CaO}CH_4+Na_2CO_3.$$

Methane is also obtained by the hydrolysis of methyl magnesium bromide:

$$CH_3MgBr+H_2O\rightarrow CH_4+Mg(OH)Br.$$

However, Kolbe's electrolysis of sodium acetate does not produce methane. Instead, it gives ethane by coupling of methyl radicals:

$$2CH_3COO^-\xrightarrow{\text{electrolysis}}C_2H_6+2CO_2.$$

Therefore, Statement I is false.

In Statement II, methane cannot be prepared by hydrogenation of unsaturated hydrocarbons because hydrogenation only converts alkenes or alkynes into higher alkanes and cannot reduce the carbon count to one. Methane also cannot be prepared by the Wurtz reaction since Wurtz coupling always combines two alkyl halides to form a higher alkane containing at least two carbon atoms.

Therefore, Statement II is true.

Hence, Statement I is false, Statement II is true, and the correct answer is Option D.