Q.Will CCl4 give a white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Concept: Nucleophilic Substitution Reactivity — specifically, whether the carbon–halogen bond can be broken by a nucleophile like $\text{Ag}^+$ under the given conditions.

Reasoning:

1. $\text{CCl}_4$ has four chlorine atoms bonded to a single carbon. The carbon is already in its highest oxidation state and is fully substituted — there are no hydrogen atoms on the carbon.
2. For $\text{AgNO}_3$ to give a white precipitate of $\text{AgCl}$, the $\text{Cl}$ must leave as $\text{Cl}^-$. This requires a nucleophilic attack by $\text{Ag}^+$ (or $\text{H}_2\text{O}$) on the carbon, but the carbon in $\text{CCl}_4$ is sterically hindered and lacks a $\text{C–H}$ bond to help stabilise any partial positive charge. …

The question tests your understanding of nucleophilic substitution reactivity in organic halides. Silver nitrate in ethanol (or aqueous) is a classic test for halide ions — but only if those halides are free or easily ionisable. CCl₄ (carbon tetrachloride) is a tetrahedral molecule with four polar C–Cl bonds, yet it is famously unreactive toward nucleophiles like Ag⁺.

Why? Because the reaction Ag⁺ + Cl⁻ → AgCl (white precipitate) requires the chloride to leave as an anion. In CCl₄, the carbon is fully substituted with four chlorines — there is no hydrogen to help stabilise a leaving group, and the molecule has no dipole moment that would allow ionisation. More importantly, the C–Cl bond in CCl₄ is strengthened by the electron-withdrawing effect of the other three chlorines (through inductive effect), making it even harder to break.

Let’s walk through the reasoning step by step.

1. Nature of CCl₄

   CCl₄ is a symmetrical, non-polar molecule (net dipole = 0). Each C–Cl bond is polar, but the vector sum cancels out. This means CCl₄ does not dissolve in polar solvents like water or ethanol — it is immiscible. For AgNO₃ to react, the halide must be accessible in solution.

2. Mechanism of AgCl precipitation

   The test works when an organic halide can undergo Sₙ1 or Sₙ2 substitution with Ag⁺ (which acts as a Lewis acid, coordinating to the halogen). For example, alkyl halides like CH₃CH₂Cl give AgCl slowly; allylic or benzylic halides give it rapidly. But CCl₄ has no hydrogen on the carbon — it cannot form a carbocation (Sₙ1 impossible) and is too sterically hindered for Sₙ2 (four large chlorines block backside attack).

3. No ionisable chloride

   The decisive point is not bond strength but the mechanism. For a white AgCl

   precipitate the C–Cl bond would have to break heterolytically to release a free

   $\ce{Cl-}$ ion that $\ce{Ag+}$ can capture. $\ce{CCl4}$ is a symmetrical, non-polar,

   covalent molecule that does not ionise: it undergoes neither $S_N1$ (it cannot form a

   carbocation) nor $S_N2$ (the four bulky chlorines block backside attack), so no free

   $\ce{Cl-}$ is produced and no $\ce{AgCl}$ forms.

4. Experimental observation …