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Hydrocarbons for NEET UG

Hydrocarbons are the simplest organic compounds: just carbon and hydrogen. Their reactions (substitution, addition, and aromatic substitution) are the foundation of organic chemistry.

ChemistryClass 1168 MCQs3 lessons14 flashcards4 memory tricks

Practise Hydrocarbons free

What you'll learn

  1. Alkanes: conformations and free-radical substitution
    Draw conformations of ethane and explain the halogenation of alkanes.
  2. Alkenes and alkynes: addition reactions
    Apply Markovnikov's rule and the peroxide effect to addition reactions.
  3. Benzene and aromaticity
    Apply Hückel's rule and explain electrophilic substitution in benzene, including directing effects.

Key formulas

Hückel's rulePlanar, cyclic, fully conjugated with (4n + 2) π electrons → aromatic

Memory tricks

First four alkane prefixesMera Ek Pyaara Bhai.

Meth (1 C), Eth (2), Prop (3), But (4). After that the Greek numbers take over: pent, hex, hept, oct.

Markovnikov's ruleThe rich get richer: H goes to the carbon that already has more hydrogens.

In adding HX to an unsymmetrical alkene, X goes to the more substituted carbon. With HBr and peroxide, the anti-Markovnikov (Kharasch) product forms.

Why benzene prefers substitutionBenzene is a happily settled family: it won't break up the ring (addition); it only swaps one member (substitution).

Addition would destroy aromatic stability, so benzene undergoes electrophilic substitution: nitration, halogenation, sulphonation and Friedel–Crafts reactions.

Acidic terminal alkynesEthyne is the only one of the three that reacts with sodium: it's 50% s, so it holds its electrons close.

The sp carbon of a terminal alkyne is more electronegative, so its H is weakly acidic. Ethyne gives sodium acetylide; ethene and ethane do not react.

Try 3 NEET-style questions

Q1. In ethane, the energy difference between the staggered and eclipsed conformations is about

  1. 125 kJ mol⁻¹
  2. zero
  3. 12.5 kJ mol⁻¹, small enough that the conformations cannot be separated at room temperature
  4. 1250 kJ mol⁻¹
Show answer

Answer: (3) 12.5 kJ mol⁻¹, small enough that the conformations cannot be separated at room temperature

The energy difference is only about 12.5 kJ mol⁻¹. Thermal energy at room temperature is enough for free rotation, so the conformers cannot be isolated.

Q2. All carbon–carbon bonds in benzene are equal, with a length of about 139 pm. This lies

  1. below the C≡C bond length
  2. above the C–C single bond length
  3. between the C–C single bond length (154 pm) and the C=C double bond length (134 pm)
  4. exactly equal to the C=C bond length
Show answer

Answer: (3) between the C–C single bond length (154 pm) and the C=C double bond length (134 pm)

All six C–C bonds in benzene are 139 pm, between a single (154 pm) and a double bond (134 pm). This supports the resonance hybrid structure.

Q3. The major product of the addition of HBr to propene (no peroxide) is

  1. 1-Bromopropane
  2. 2-Bromopropane
  3. 1,2-Dibromopropane
  4. Propane
Show answer

Answer: (2) 2-Bromopropane

By Markovnikov's rule, H adds to the carbon with more hydrogens, giving 2-bromopropane.

Practise all 68 questions Free · explanations for every option · see where you go wrong

More Chemistry chapters (Class 11)

Some Basic Concepts of Chemistry · Structure of Atom · Classification of Elements and Periodicity · Chemical Bonding and Molecular Structure · Chemical Thermodynamics · Equilibrium · Redox Reactions · The s-Block Elements · Organic Chemistry: Basic Principles and Techniques