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Hydrocarbons: Alkanes revision notes

Every idea in the Hydrocarbons: Alkanes chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 7 and download the PDF.

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Preparation of alkanes

Methods of preparation
1× asked, last 2025
2 RX+2 Na→dry etherR−R+2 NaX2 RCOONa+2 HX2O→electrolysisR−R+2 COX2+HX2+2 NaOHRCOONa+NaOH→ΔCaORH+NaX2COX3,RX→Zn, dil. HClRHRX→Mg, etherRMgX→HX2ORH,RC≡CH→HX2/Pt, Pd or NiRCHX2CHX3\begin{gathered}\ce{2RX + 2Na ->[\text{dry ether}] R-R + 2NaX}\\ \ce{2RCOONa + 2H2O ->[\text{electrolysis}] R-R + 2CO2 + H2 + 2NaOH}\\ \ce{RCOONa + NaOH ->[\text{CaO}][\Delta] RH + Na2CO3},\qquad \ce{RX ->[\text{Zn, dil. HCl}] RH}\\ \ce{RX ->[\text{Mg, ether}] RMgX ->[H2O] RH},\qquad \ce{RC#CH ->[\text{H}_2/\text{Pt, Pd or Ni}] RCH2CH3}\end{gathered}
Wurtz, Kolbe: R–R, never CH₄; one RX (or RCOONa) gives even-C; two different R give 3 alkanes, R–R′ may be odd-C.
Br
CH₃Br
+
Br
C₂H₅Br
Na, dry ether
 
ethane
+
propane
(odd C)
+
butane
MethodC inC outProduct
Wurtz: 2 C₂H₅Br + 2 Na24
Kolbe: 2 CH₃COONa22
soda lime: C₂H₅COONa32
Zn/HCl: C₂H₅Br22
Wurtz and Kolbe join two R (C doubles); soda lime removes the COO carbon; reduction keeps the count.
  • Soda lime: one C fewer. Reduction keeps the C count; Zn/dil. HCl reduces RCl, RBr, RI but not RF.
  • Quick method. Halides that give one alkane by reduction (Zn/HCl, or RMgX then water) = its monohalo structural isomers: 2-methylbutane has four distinct C–H sets, so 4 RBr.

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6 more concepts in 2 topics

Free-radical halogenation and Conformations and properties. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Free-radical halogenation

  • Mechanism: initiation to termination2×
  • Selectivity: 3° > 2° > 1° H3×
  • Counting halogenation products4×

Conformations and properties

  • Staggered vs eclipsed ethane
  • Boiling and melting points1×
  • Combustion and controlled oxidation2×

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Practise Hydrocarbons: Alkanes

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