NEET Hydrocarbons Previous Year Questions with Solutions

The carbon-magnesium bond in a Grignard reagent is highly polarized: C^{δ-} - Mg^{δ+}Br. The alkyl group behaves as a extremely strong Brønsted-Lowry base (conjugate base of an alkane, pKa ≈ 50).

• • Whenever an active hydrogen donor is present, acid-base proton transfer dominates.

Primary amines possess two acidic protons bonded to the electronegative nitrogen atom (-NH₂). The cyclohexyl group of the Grignard reagent abstracts a proton from the amino group of cyclohexylamine.

• • The carbanion equivalent C₆H₁₁⁻ receives a proton H⁺ to become neutral cyclohexane (C₆H₁₂).
• • The byproduct is cyclohexylaminomagnesium bromide.

Structure (1) is dicyclohexylamine; Structure (2) is 2-cyclohexylcyclohexanamine; Structure (3) is bicyclohexyl; Structure (4) is a plain cyclohexyl ring (cyclohexane).

• • Matches Option (4).

Addition of Br₂ across an alkene in non-polar CCl₄ proceeds through a cyclic bromonium ion intermediate, resulting in stereospecific ANTI addition. According to the stereochemical addition rule (TAM: Trans + Anti = Meso), bromination of trans-but-2-ene MUST produce MESO-2,3-dibromobutane.

• • In meso-2,3-dibromobutane in its anti conformation, there is an inversion center: if front carbon has Br UP and Me DOWN-RIGHT, the back carbon must have Br DOWN and Me UP-LEFT.
• • In the Newman projection given in Statement I, both Me groups are on the right and both H are on the left. Inverting through the center fails. This depicts the (2R, 3R) or (2S, 3S) chiral enantiomer (threo-pair), NOT the meso isomer!
• • Statement I is incorrect.

Reaction of an alkene with cold, alkaline KMnO₄ (Baeyer's reagent) proceeds via a cyclic permanganate ester intermediate, delivering both hydroxyl groups to the SAME face of the double bond (SYN addition). According to the stereochemical addition rule (CSM: Cis + Syn = Meso), syn-dihydroxylation of cis-but-2-ene produces MESO-butane-2,3-diol.

• • In the Newman projection in Statement II, the groups are in a near-eclipsed conformation where front OH aligns with back OH, front H aligns with back H, and front Me aligns with back Me.
• • This eclipsed conformation contains an internal plane of symmetry (σ), which is the definitive visual hallmark of the meso isomer.
• • Statement II is correct.

Statement I is incorrect, but Statement II is correct.

• • Matches Option (4).

Reaction of 1-phenylprop-1-yne with molecular hydrogen over deactivated palladium (Lindlar's catalyst, Pd/CaCO₃ or Pd/C poisoned with lead salts or quinoline) results in stereoselective syn-addition of hydrogen across the triple bond to form the (Z) or cis-alkene.

• • K is cis-1-phenylpropene (Z-isomer).

Reduction of an alkyne with sodium in liquid ammonia proceeds via sequential single-electron transfer and protonation through a radical anion. The radical anion adopts the more thermodynamically stable trans-configuration to minimize steric repulsion between the phenyl and methyl groups, yielding the (E) or trans-alkene.

• • L is trans-1-phenylpropene (E-isomer).

Both K and L share identical molecular formulas and identical atom connectivity, but differ in the spatial orientation of groups around the carbon-carbon double bond (cis vs trans). Therefore, K and L are geometrical isomers (cis-trans diastereomers).

• • Matches Option (1).

In the presence of benzoyl peroxide, HBr undergoes homolytic cleavage to produce bromine free radicals (Br•). The Br• radical adds regioselectively to the less substituted double-bond carbon (C2) to generate the more stable 3° tertiary radical at C1:

• • Bromine is positioned on C2, and the methyl group remains on C1 (1,2-regiochemistry).

Potassium cyanide (KCN) is an ionic cyanide source. The cyanide ion (CN⁻) acts as an ambident nucleophile and attacks through the carbon atom, displacing bromide in an S_N2 reaction:

• • The nitrile group (-C≡N) is now located at C2.

Sodium amalgam in ethanol (Na(Hg) / C₂H₅OH) provides nascent hydrogen to reduce the nitrile group completely to a primary aminomethyl group (-CH₂NH₂):

• • The product is (2-methylcyclopentyl)methanamine, represented by Option (1) with adjacent -CH₃ and -CH₂NH₂ groups.

Cis-trans isomerism arises when rotation is restricted (either by a double bond or a ring structure) and each of the two bridgehead/olefinic carbons carries two distinct groups:

• • If either carbon bears two identical groups, interchange of groups produces the identical molecule, precluding isomerism.

Examine each candidate compound against the distinct-substituent criterion:

• • 2-Methylhex-2-ene: (CH₃)₂C=CH-Pr ⟹ Two identical methyl groups on C2.
• • 1,1-Dimethylcyclopropane: C1 is bonded to two identical methyl groups, while C2 and C3 are identical -CH₂- groups. Rotation restriction does not produce diastereomers.

In 1,2-dimethylcyclohexane, the closed ring restricts free rotation:

• • When both methyl groups lie on the same side of the ring plane (e.g. 1-axial, 2-equatorial or both up), it forms the cis-isomer.
• • When the methyl groups lie on opposite sides (one up, one down), it forms the trans-isomer (e.g. diequatorial 1e, 2e).
• • These are distinct, non-superimposable geometrical diastereomers.
• • Matches Option (4) perfectly.

Verify the products of the first three reactions from standard NCERT organic chemistry:

• • (2) n-Hexane (CH₃(CH₂)₄CH₃) undergoes aromatization over Mo₂O₃/Al₂O₃ at 773 K and 10–20 atm to yield Benzene + 4 H₂.
• • (3) 3 HC≡CH undergoes cyclic trimerization in a red hot iron tube at 873 K to yield Benzene.

When benzenediazonium chloride solution is warmed with water, nucleophilic substitution by H₂O takes place with evolution of nitrogen gas:

• • The organic product is PHENOL, not benzene.
• • To reduce benzenediazonium chloride to benzene, hypophosphorous acid (H₃PO₂ + H₂O) or ethanol (CH₃CH₂OH) must be used.

Since Reaction (4) produces phenol and does NOT give benzene, it is the correct answer to the question.

• • Matches Option (4) with 100% precision.

Bond dissociation energy (BDE) of a C-H bond is defined by homolytic cleavage: R-H ⟶ R• + H•. Thermodynamically, BDE reflects both orbital hybridization (s-character) and the stability of the resulting carbon radical:

• • Greater s-character holds the bonding pair closer to carbon, increasing bond strength.
• • Forming an unstable radical requires significantly higher energy input.

Compare the terminal alkyne and aromatic C-H bonds:

• • An sp-hybridized C-H bond is the strongest among all hydrocarbon C-H bonds.
• • I (Benzene): sp²-hybridized C-H (33.3% s-character) has BDE ≈ 470 kJ/mol. The phenyl radical resides in an sp² orbital orthogonal to the aromatic π-electron cloud and cannot be resonance-stabilized.
• • Thus BDE(II) > BDE(I).

Homolysis of the saturated C-H bond in cyclopropene yields an allylic free radical:

• • The unpaired electron is resonance-delocalized over the adjacent double bond, substantially lowering the energy needed to break the bond.
• • Therefore, Compound III has the lowest BDE.
• • Final decreasing order of BDE: II > I > III, matching Option (1).

Cyclohexane (C₆H₁₂) is a fully saturated cycloalkane without π-bonds or activating heteroatoms. Under normal laboratory conditions without high-intensity ultraviolet irradiation:

• • Cyclohexane cannot decolourize aqueous bromine solution.
• • Therefore, Option (1) is the correct answer.

Styrene (phenylethene) possesses an aliphatic carbon-carbon double bond which acts as an electron-rich nucleophile:

• • Rapid electrophilic addition across the alkene double bond discharges the reddish-brown color.

In aqueous medium, the -OH and -NH₂ groups are exceptionally powerful activating groups through +M resonance:

• • Aniline similarly reacts with bromine water to form 2,4,6-tribromoaniline precipitate.
• • Both reactions consume bromine rapidly, completely decolourizing bromine water.
• • Conclusion: Only cyclohexane does not decolourize bromine water, matching Option (1).

For constitutional isomers of formula C5H12, the experimental boiling points decrease monotonically with increasing branching: n-pentane (309.1 K) > isopentane / 2-methylbutane (301.0 K) > neopentane / 2,2-dimethylpropane (282.5 K). Hence, Statement I is correct.

Van der Waals dispersion forces act across the contact surface between neighboring molecules. As branching increases, molecular geometry becomes compact and approaches a sphere. Because a sphere possesses the minimum surface area for a given volume, intermolecular contact area diminishes, weakening London dispersion forces and lowering the boiling point. Hence, Statement II is correct.

Carbon atoms in an alkane framework are classified based on the number of other carbon atoms attached to them: - Primary (1°): Attached to 1 carbon - Secondary (2°): Attached to 2 carbons - Tertiary (3°): Attached to 3 carbons (possessing a -CH< group) - Quaternary (4°): Attached to 4 carbons

Drawing the expanded structural formula of 2,3-dimethylbutane: CH3 - C²H(CH3) - C³H(CH3) - CH3 - C² is directly bonded to: C¹ (methyl), C³ (methine), and methyl branch -> 3 carbons -> 3° carbon. - C³ is directly bonded to: C² (methine), C⁴ (methyl), and methyl branch -> 3 carbons -> 3° carbon. - Total 3° carbons = 2. - Total 1° carbons = 4. This satisfies the given condition of formula C6H14 with two tertiary carbons.

Evaluating the remaining isomers: - 2,2-Dimethylbutane: (CH3)3C-CH2-CH3 contains one 4° carbon (C2), one 2° carbon (C3), and four 1° carbons. Zero 3° carbons. - n-Hexane: CH3-CH2-CH2-CH2-CH2-CH3 contains four 2° carbons and two 1° carbons. Zero 3° carbons. - 2-Methylpentane: (CH3)2CH-CH2-CH2-CH3 contains one 3° carbon (C2), two 2° carbons (C3, C4), and three 1° carbons. Only ONE 3° carbon.

When sodium ethanoate is heated with soda-lime (a dry mixture of sodium hydroxide and calcium oxide in an approximate 3:1 mass ratio), it undergoes decarboxylation to produce an alkane with one less carbon atom than the parent carboxylate salt:

Of the two products formed, Na₂CO₃ is an inorganic salt, so the organic compound obtained is methane (CH₄).

Calculate the molecular weight of methane from standard atomic masses (C = 12 u, H = 1 u):

The problem explicitly asks for the weight in grams of TWO MOLES of this organic compound:

For a compound to obey Hückel’s rule of aromaticity, it must satisfy three structural criteria: (a) cyclic and planar ring system, (b) complete delocalization of π electrons across the ring, and (c) contain (4n + 2) π electrons where n = 0, 1, 2, 3...

Naphthalene (i) has 5 conjugated double bonds = 10 π electrons (4n+2 with n=2). Anthracene (vii) has 7 conjugated double bonds = 14 π electrons (4n+2 with n=3). Both are planar, fully conjugated, and aromatic:

Cyclopentadienyl anion (ii) has two double bonds (4 e⁻) and one localized lone pair in a p-orbital (2 e⁻) = 6 π electrons (n=1, aromatic). Cyclopropenyl cation (v) has one double bond (2 e⁻) and a vacant p-orbital = 2 π electrons (n=0, aromatic):

Cyclobutadiene (iii) has 4 π electrons (4n, anti-aromatic). Cyclopropenyl anion (iv) has 4 π electrons (4n, anti-aromatic). Cyclooctatetraene (vi) has 8 π electrons and adopts a non-planar tub conformation (non-aromatic). None of these three obey Hückel’s rule:

The dihedral angle (also called torsional angle $\theta$) in ethane ($CH_3-CH_3$) is the angle between the two intersecting planes formed by the $H-C-C$ bonds of the adjacent carbon atoms when viewed along the carbon-carbon single bond axis.

Ethane exhibits infinite conformations resulting from free rotation about the $C-C$ $\sigma$-bond. The two extreme conformations are: - Staggered Conformation: Hydrogen atoms are as far apart as possible (dihedral angle $\theta = 60^\circ$). Torsional strain is minimized, rendering it the most stable conformer. - Eclipsed Conformation: Hydrogen atoms on adjacent carbons are closest to each other, resulting in maximum repulsive interaction between adjacent bonding electron pairs (torsional strain of $\approx 12.5\text{ kJ/mol}$). Consequently, it is the least stable conformer.

The dihedral angle of the least stable (eclipsed) conformer is identically $0^\circ$, corresponding to Option (4).

Benzoyl peroxide undergoes homolytic cleavage upon heating or exposure to light to yield benzoyloxy radicals, which decarboxylate into phenyl radicals and abstract hydrogen from $HBr$ to generate bromine radicals:

The electrophilic bromine radical attacks the unsymmetrical double bond of 3-methylbut-1-ene. Addition at the less hindered terminal $CH_2$ carbon generates a stable secondary ($2^\circ$) free radical rather than an unstable primary ($1^\circ$) radical:

The secondary radical abstracts a hydrogen atom from another $HBr$ molecule to produce 1-bromo-3-methylbutane and regenerates a bromine radical:

Break down the IUPAC components: - Root word: dec $\implies$ longest continuous carbon chain containing the double bond has 10 carbon atoms. - Suffix: -4-ene $\implies$ double bond is between carbon-4 and carbon-5. - Substituents: 2,6-dimethyl $\implies$ methyl groups ($-CH_3$) are located at carbon-2 and carbon-6.

Trace the carbon skeleton of Structure (1) starting from the right-hand isopropyl terminus: - $C_1$: terminal methyl group - $C_2$: carbon with vertical methyl branch - $C_3$: $-CH_2-$ methylene carbon - $C_4=C_5$: internal double bond (dec-4-ene) - $C_6$: carbon with methyl branch - $C_7-C_8-C_9-C_{10}$: linear butyl tail ($-CH_2-CH_2-CH_2-CH_3$) Total carbon chain length $= 10$. The double bond is at position 4, and substituents are at 2 and 6. Thus, Structure (1) is identically 2,6-dimethyldec-4-ene.

- Structure (2): Double bond is located towards the bottom at position 5 or 6. - Structure (3): The double bond connects directly to the carbon adjacent to C2, placing the double bond at position 3 (dec-3-ene). - Structure (4): The double bond is located between the branch carbon and the next carbon, giving an incorrect substituent position.

Ozonolysis of alkenes involves the addition of an ozone molecule to the carbon-carbon double bond to form an ozonide intermediate. Subsequent cleavage of the ozonide with zinc dust and water ($$\text{Zn} + \text{H}_2\text{O}$$) cleaves the double bond completely, converting each doubly bonded carbon into a carbonyl ($$\text{C}=\text{O}$$) group: $$\text{R}_1\text{R}_2\text{C}=\text{CR}_3\text{R}_4 \xrightarrow{\text{1. O}_3, \text{ 2. Zn/H}_2\text{O}} \text{R}_1\text{R}_2\text{C}=\text{O} + \text{O}=\text{CR}_3\text{R}_4$$

Methanal (formaldehyde, $$\text{HCHO}$$) has the structure $$\text{H}_2\text{C}=\text{O}$$. Therefore, for methanal to be produced, one end of the alkene double bond must have been an unsubstituted methylene carbon ($$=\text{CH}_2$$), which only occurs in terminal alkenes.

- Structure (1): 1-methylcyclohexene. The double bond is inside the ring (endocyclic). Ozonolysis breaks the ring to yield 6-oxoheptanal (a single dicarbonyl compound), with no formaldehyde. - Structure (2): Cyclohexene with internal methyl substituent, gives a dicarbonyl upon ring opening. - Structure (3): Methylenecyclohexane (cyclohexane ring with an exocyclic $$=\text{CH}_2$$ group). Ozonolysis cleaves the exocyclic double bond, yielding Cyclohexanone and Methanal ($$\text{HCHO}$$). - Structure (4): Internal endocyclic double bond.

Option (3) is uniquely capable of generating methanal upon ozonolysis.

Alkyl halides on treatment with sodium metal in dry ethereal solution give higher alkanes containing twice the number of carbon atoms: $$2\text{R}-\text{X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R}-\text{R} + 2\text{NaX}$$ Because two identical alkyl groups are coupled, the product is always a symmetrical alkane with an even number of carbon atoms.

- $$n$$-Hexane ($$6\text{ C}$$): Formed by coupling two 1-propyl groups ($$2\text{ C}_3\text{H}_7\text{Br} + 2\text{Na} \to \text{C}_6\text{H}_{14}$$) in good yield. - 2,3-Dimethylbutane ($$6\text{ C}$$): Symmetrical alkane formed by coupling two isopropyl groups ($$2 (\text{CH}_3)_2\text{CH-Br} + 2\text{Na} \to (\text{CH}_3)_2\text{CH}-\text{CH}(\text{CH}_3)_2$$) in good yield. - $$n$$-Butane ($$4\text{ C}$$): Formed by coupling two ethyl groups ($$2\text{ C}_2\text{H}_5\text{Br} + 2\text{Na} \to \text{C}_4\text{H}_{10}$$) in good yield.

$$n$$-Heptane contains 7 carbon atoms (an odd number). To synthesize it, two different alkyl halides must be used: 1-halopropane ($$3\text{ C}$$) and 1-halobutane ($$4\text{ C}$$). Three different radical coupling reactions occur simultaneously: 1. $$\text{C}_3\text{H}_7\cdot + \text{C}_3\text{H}_7\cdot \to \text{Hexane (6 C)}$$ 2. $$\text{C}_3\text{H}_7\cdot + \text{C}_4\text{H}_9\cdot \to \text{Heptane (7 C)}$$ 3. $$\text{C}_4\text{H}_9\cdot + \text{C}_4\text{H}_9\cdot \to \text{Octane (8 C)}$$ This generates a complex mixture of three alkanes whose boiling points are very close, making separation extremely difficult and the yield of heptane very poor.

NCERT explicitly states that the Wurtz reaction cannot be used for the preparation of unsymmetrical alkanes having an odd number of carbon atoms. Thus, $$n$$-heptane is the correct answer.