Hydrocarbons for NEET

Hydrocarbons is the first chapter where the ideas from general organic chemistry turn into reactions you can predict. Why does HBr add to propene the way it does? Because a secondary carbocation is more stable than a primary one. Why is ethyne acidic? Because an sp carbon holds a negative charge well. If you read each reaction as a consequence of GOC, the chapter needs far less memorising.

NEET questions here are usually the product of an addition to an unsymmetrical alkene (with or without peroxide), the alkene that gives certain ozonolysis products, which compound is aromatic, the major product of substitution on a substituted benzene, or a preparation such as Wurtz or decarboxylation. The NMC syllabus also names the mechanisms of free-radical halogenation, electrophilic addition and electrophilic substitution, so learn the steps, not just the products.

Alkanes

Preparation:

• Hydrogenation of alkenes or alkynes with H₂ over finely divided Pt, Pd or Ni.
• Reduction of alkyl halides with Zn and dilute HCl.
• Wurtz reaction: 2RX + 2Na → R–R + 2NaX, in dry ether. It gives alkanes with an even number of carbons from a single halide, and it cannot give methane.
• Decarboxylation: heating a sodium carboxylate with soda lime (NaOH + CaO) removes CO₂: RCOONa + NaOH → R–H + Na₂CO₃. The alkane has one carbon fewer than the acid.
• Kolbe's electrolysis: electrolysis of an aqueous sodium or potassium carboxylate gives R–R at the anode. 2CH₃COONa + 2H₂O → CH₃–CH₃ + 2CO₂ + H₂ + 2NaOH. Methane cannot be made this way.

Free-radical halogenation (for example, methane and chlorine in light):

1. Initiation: Cl₂ → 2Cl• (homolysis by light or heat).
2. Propagation: CH₄ + Cl• → CH₃• + HCl, then CH₃• + Cl₂ → CH₃Cl + Cl•. The chlorine radical is regenerated, so the chain continues.
3. Termination: radicals combine, for example Cl• + Cl• → Cl₂, CH₃• + Cl• → CH₃Cl, or CH₃• + CH₃• → C₂H₆.

Reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂. Ease of replacing hydrogen: 3° > 2° > 1°, following free-radical stability.

Conformations of ethane. Rotation about the C–C single bond gives endless conformations; the two extremes are:

The energy difference is only about 12.5 kJ mol⁻¹, so rotation is effectively free at room temperature and the conformers cannot be isolated. Both are drawn as Sawhorse (a side view along the bond) and Newman projections (looking straight down the C–C bond, front carbon as a point, back carbon as a circle).

Alkenes

Preparation: partial reduction of alkynes (Lindlar's catalyst gives the cis alkene; sodium in liquid ammonia gives the trans alkene), dehydrohalogenation of alkyl halides with alcoholic KOH, dehalogenation of vicinal dihalides with zinc, and acid-catalysed dehydration of alcohols with concentrated H₂SO₄.

Geometrical isomerism arises because rotation about C=C is restricted. It needs two different groups on each doubly bonded carbon. cis-But-2-ene has the higher dipole moment and boiling point; trans-but-2-ene packs better and has the higher melting point.

Electrophilic addition is the characteristic reaction, because the π electrons are exposed to electrophiles.

Why Markovnikov? H⁺ adds first and forms a carbocation. Adding to the CH₂ end gives a secondary carbocation (CH₃–CH⁺–CH₃), which is more stable than the primary alternative. Br⁻ then attacks that carbon.

Why only HBr shows the peroxide effect: with peroxide, the reaction goes by a free-radical path, where Br• adds first to give the more stable secondary radical. For HCl, the H–Cl bond is too strong to break homolytically; for HI, the I• radicals combine to I₂ instead of adding.

Ozonolysis is used to locate the double bond: each C=C becomes two C=O. Replace the double bond with =O on each side and you have the products. Alkenes also polymerise: ethene gives polythene under high pressure with a trace of oxygen or peroxide.

Alkynes

Preparation: calcium carbide and water, CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂; or double dehydrohalogenation of vicinal dihalides (alcoholic KOH, then NaNH₂).

Acidic character. Hydrogen on a triple-bonded carbon is weakly acidic, because the sp carbon (50% s-character) holds the negative charge well:

• HC≡CH + Na → HC≡C⁻Na⁺ + ½H₂
• Acidity: HC≡CH > H₂C=CH₂ > H₃C–CH₃
• Only terminal alkynes react. But-2-yne has no acidic hydrogen.

Additions: H₂ (to alkene, then alkane), halogens (Br₂ is decolourised), and HX following Markovnikov's rule, twice, to give geminal dihalides.

Hydration: with dilute H₂SO₄ and HgSO₄ at about 333 K, ethyne gives ethanal (acetaldehyde) through an unstable enol; propyne gives propanone (acetone).

Polymerisation: ethyne passed through a red-hot iron tube at 873 K trimerises to benzene.

Aromatic hydrocarbons

Benzene is a planar hexagon. All six C–C bonds are the same length (139 pm), between a single and a double bond, because the six π electrons are delocalised over the ring. This resonance makes benzene unusually stable, so it prefers substitution, which keeps the ring intact, over addition.

Hückel's rule: a compound is aromatic if it is cyclic, planar, fully conjugated and has (4n + 2) π electrons (2, 6, 10 …).

Electrophilic substitution happens in three steps: the electrophile is generated, it attacks the π cloud to give a carbocation (the arenium ion, or σ-complex, in which one ring carbon is sp³), and a proton is lost to restore aromaticity.

The nitronium ion forms as HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Under UV light with excess chlorine, benzene does add: it gives hexachlorocyclohexane (C₆H₆Cl₆).

Directive influence of substituents

A group already on the ring decides where the next group goes and how fast.

Worked NEET-style questions

Example 1: ozonolysis in reverse

An alkene C₄H₈ gives only ethanal on ozonolysis. Another C₄H₈ gives methanal and propanone. Identify both.

• Two ethanal (CH₃CHO) pieces joined at the carbonyl carbons: CH₃CH=CHCH₃, but-2-ene.
• Methanal (HCHO) + propanone ((CH₃)₂C=O) joined: (CH₃)₂C=CH₂, 2-methylpropene.

Example 2: with and without peroxide

Give the products of 2-methylpropene with HBr, without and with peroxide.

• Without peroxide, H⁺ adds to CH₂ to give the tertiary cation (CH₃)₃C⁺: 2-bromo-2-methylpropane.
• With peroxide, Br• adds to CH₂ to give the tertiary radical, then H is picked up: 1-bromo-2-methylpropane.

Example 3: a mixed Wurtz reaction

A mixture of CH₃Br and C₂H₅Br is treated with sodium in dry ether. What alkanes form?

• CH₃ + CH₃ → ethane; CH₃ + C₂H₅ → propane; C₂H₅ + C₂H₅ → butane.
• Three alkanes, which is why a mixed Wurtz reaction is a poor way to make an odd-carbon alkane.

Example 4: directing effects

What is the major product of nitrating chlorobenzene? And of nitrating nitrobenzene?

• Cl is an ortho/para director: 1-chloro-4-nitrobenzene (major) with some 1-chloro-2-nitrobenzene.
• NO₂ is a meta director: 1,3-dinitrobenzene, and the reaction is slower than for benzene.

Practice MCQs

1. Sodium propanoate heated with soda lime gives: (a) methane (b) ethane (c) propane (d) butane
2. The more stable conformation of ethane is: (a) staggered (b) eclipsed (c) skew (d) both equally stable
3. The peroxide effect is shown by: (a) HCl (b) HBr (c) HI (d) all three
4. Propyne with dilute H₂SO₄ and HgSO₄ gives: (a) propanal (b) propan-1-ol (c) propanone (d) propanoic acid
5. Which is aromatic? (a) cyclopentadienyl anion (b) cyclooctatetraene (c) cyclopentadienyl cation (d) cyclobutadiene
6. The electrophile in the nitration of benzene is: (a) NO₂⁻ (b) HNO₃ (c) NO⁺ (d) NO₂⁺
7. Which is a meta-directing group? (a) –OH (b) –CH₃ (c) –CHO (d) –Cl
8. Ethyne reacts with sodium but ethene does not, because: (a) the sp carbon of ethyne is more electronegative (b) ethyne is a gas (c) ethene has more hydrogens (d) ethyne has a σ bond

Answers

1. (b) Decarboxylation removes one carbon: C₂H₅COONa gives C₂H₆.
2. (a) Hydrogens on the two carbons are farthest apart.
3. (b) Only HBr adds by the free-radical path.
4. (c) Markovnikov hydration gives an enol that becomes the ketone.
5. (a) 6 π electrons in a planar, cyclic, conjugated ring.
6. (d) The nitronium ion, NO₂⁺.
7. (c) –CHO withdraws electrons by −R and −I.
8. (a) 50% s-character holds the acetylide ion's negative charge well.

What to do next

• Draw the mechanisms of free-radical chlorination, Markovnikov addition of HBr and nitration of benzene, step by step, from memory.
• Solve ten ozonolysis questions in both directions: alkene to products, and products to alkene.
• Learn the directing-group table and the halogen exception.
• Revise carbocation and radical stability in GOC, then carry the substitution ideas into haloalkanes and haloarenes.