JEE Hydrocarbons - Alkanes Questions

The acidity of a particular hydrogen atom is judged by the stability of the conjugate base obtained after removing that hydrogen as a proton, $$H^+$$. The more stable the conjugate base (carbanion), the stronger (more acidic) the original C-H bond.

Four main factors control the stability of a carbanion:
  1. Hybridisation of the carbon bearing the negative charge: $$sp \; (50\%\,s) \gt sp^2 \; (33\%\,s) \gt sp^3 \; (25\%\,s)$$.
  2. Resonance delocalisation of the negative charge.
  3. Inductive $$(-I)$$ withdrawal by electronegative or electron-withdrawing groups.
  4. Destabilising $$+I$$ effects of alkyl groups (they increase electron density on the already negative carbon).

Analyse each option by applying these points:

Case A: The depicted hydrogen is attached to a simple primary $$sp^3$$ carbon of an alkane. Removing it gives a primary carbanion that is not resonance-stabilised and is slightly destabilised by the adjacent alkyl group’s $$+I$$ effect. Hence, very weakly acidic.

Case B: The depicted hydrogen is the terminal proton of an alkyne (≡C-H). That carbon is $$sp$$ hybridised, so it possesses 50 % $$s$$-character. Higher $$s$$-character means the electron pair in the conjugate base is held closer to the nucleus, making the carbon more electronegative and the negative charge more stabilised. No other option offers a carbanion on an $$sp$$ carbon; therefore this hydrogen is expected to be the most acidic.

Case C: The hydrogen is allylic/benzylic. After deprotonation the negative charge resides on an $$sp^2$$ carbon and is resonance-delocalised, giving moderate acidity. However, the $$sp^2$$ carbon (33 % $$s$$) is still less electronegative than an $$sp$$ carbon, so it is less stabilised than the acetylide anion from Case B.

Case D: The hydrogen is attached to a tertiary $$sp^3$$ carbon. The resulting tertiary carbanion is the least stable because three alkyl groups donate electron density by the $$+I$$ effect. Consequently, this hydrogen is the least acidic of all four.

Arranging in increasing order of acidity (weakest acid first):
Case D < Case A < Case C < Case B

Thus, the terminal alkyne hydrogen in Option B is the most acidic.

Option B which is: the terminal alkyne (≡C-H) hydrogen.

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 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

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.

We need to identify the yellow coloured precipitate formed in the qualitative test for phosphorus.

Describe the test procedure

In the qualitative detection of phosphorus in organic compounds:

1. The organic compound is first heated with an oxidising agent (such as sodium peroxide or concentrated nitric acid) to convert all the phosphorus into phosphate ions ($$PO_4^{3-}$$).

2. The resulting solution is treated with concentrated nitric acid ($$HNO_3$$) to ensure an acidic medium.

3. Ammonium molybdate $$(NH_4)_2MoO_4$$ solution is then added.

Identify the product

The phosphate ions react with ammonium molybdate in the presence of nitric acid to form a characteristic canary yellow precipitate of ammonium phosphomolybdate:

$$PO_4^{3-} + 12(NH_4)_2MoO_4 + 21HNO_3 \rightarrow (NH_4)_3PO_4 \cdot 12MoO_3 \downarrow + 21NH_4NO_3 + 12H_2O$$

The yellow precipitate has the formula $$(NH_4)_3PO_4 \cdot 12MoO_3$$ (ammonium phosphomolybdate). This is a standard confirmatory test for the presence of phosphate ions.

The correct answer is Option (2): $$(NH_4)_3PO_4 \cdot 12MoO_3$$.

We need to identify the acidic radical that gives a pale yellow precipitate soluble with difficulty in NH$$_4$$OH when tested with AgNO$$_3$$. When the sodium carbonate extract of a salt is acidified with dilute HNO$$_3$$ and treated with AgNO$$_3$$, the halide ions form silver halide precipitates with characteristic colours and solubilities: AgCl (from Cl$$^-$$) gives a white curdy precipitate easily soluble in dilute NH$$_4$$OH (forms soluble [Ag(NH$$_3$$)$$_2$$]Cl), AgBr (from Br$$^-$$) gives a pale yellow precipitate soluble with difficulty in NH$$_4$$OH (partially soluble, requires concentrated NH$$_4$$OH), and AgI (from I$$^-$$) gives a yellow precipitate insoluble in NH$$_4$$OH.

Since the observed precipitate is pale yellow, Cl$$^-$$ is ruled out and Br$$^-$$ is consistent; moreover, its solubility with difficulty in NH$$_4$$OH matches AgBr, whereas AgI would be completely insoluble and AgCl would dissolve easily.

Furthermore, CO$$_3^{2-}$$ would not give a precipitate with AgNO$$_3$$ in acidic medium, as it reacts with acid to form CO$$_2$$, eliminating Option (4).

The correct answer is Option (1): Br$$^-$$.

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.

Let us first draw the structure of 2-methylbutane. It can be written as:

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

Next, we identify all distinct types of hydrogen atoms that can be replaced by chlorine.

1. Primary hydrogens at C1 (three equivalent H)
2. Tertiary hydrogen at C2 (one H)
3. Secondary hydrogens at C3 (two equivalent H)
4. Primary hydrogens at C4 (three equivalent H)
5. Primary hydrogens on the methyl substituent at C2 (three equivalent H)

Each distinct type of H gives a different monochlorinated product. Thus we have five structural isomers:

(i) 1-chloro-2-methylbutane
(ii) 2-chloro-2-methylbutane
(iii) 3-chloro-2-methylbutane
(iv) 4-chloro-2-methylbutane
(v) 2-(chloromethyl)butane

Now we check for chirality. In 3-chloro-2-methylbutane the carbon bearing Cl (C3) has four different groups, so it is a chiral centre. Hence 3-chloro-2-methylbutane exists as two enantiomers (R and S).

Therefore, total isomers including stereoisomers =

$$1\;(\text{from C1}) \;+\;1\;(\text{from C2}) \;+\;2\;(\text{from C3, R \& S}) \;+\;1\;(\text{from C4}) \;+\;1\;(\text{from methyl substituent}) \;=\;6.$$

Final Answer: 6

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.

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.

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.

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.

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.

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 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 $$8$$.

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.

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.

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.

Molecule (III): Both methyl groups are in equatorial positions. This diequatorial arrangement minimizes steric repulsion, making it the most stable conformation in the set.

Molecules (I) and (II): In both structures, one methyl group is equatorial and the other is axial. Because each contains exactly one axial group causing 1,3-diaxial interaction, their energy levels, and thus their stabilities are approximately equal.

Molecule (IV): Both methyl groups are in axial positions. This diaxial arrangement creates significant steric strain (1,3-diaxial interactions) from both groups simultaneously, making it the least stable conformation.

Stability: $$(III) > (I) \approx (II) > (IV)$$