Question

(i) Explain the nature of the C-X bond. Give an example of a compound having sp³ C-X bonding. (ii) Write chemical equations of the following: (a) Carbylamine reaction (b) Hinsberg reaction.

Hint:

The C-X bond is polar covalent (C is delta plus, X delta minus); use an sp³ haloalkane such as CH₃Cl, and recall carbylamine uses CHCl₃ with alcoholic KOH while Hinsberg uses C₆H₅SO₂Cl.

Answer 1

Step 1: Why the C-X bond is polar.
Picture the electron cloud between carbon and the halogen. Because the halogen holds electrons more tightly, the cloud is lop-sided towards X. The result is a permanent dipole: a fractional negative charge on X and a matching fractional positive charge on C. This is a covalent bond with ionic character, that is a polar covalent bond, and the electron-poor carbon is where reactions begin.
Step 2: How the bond changes down the halogens.
Two opposing trends matter. Polarity tracks electronegativity, largest for C-F. Bond strength tracks orbital overlap, which worsens as the halogen grows, so C-I is the longest and weakest link while C-F is the shortest and strongest.
Step 3: A concrete sp³ case.
Take chloromethane, CH₃Cl. Its carbon uses four equivalent sp³ orbitals arranged tetrahedrally, one of which overlaps with chlorine. That carbon-chlorine link is therefore an sp³ C-X bond.
Step 4: Carbylamine equation.
Aniline illustrates the test just as well as a simple alkyl amine. Warmed with trichloromethane and alcoholic potash it yields phenyl isocyanide.
C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O
Step 5: Hinsberg equation.
The Hinsberg reagent is benzenesulphonyl chloride. Reacting it with methylamine (a primary amine) swaps the chlorine for the amine and releases HCl.
C₆H₅SO₂Cl + CH₃NH₂ → C₆H₅SO₂NHCH₃ + HCl
The sulphonamide formed keeps one N-H, so it is soluble in KOH.