Question

Explain the effect of the inductive effect on the reactivity of alkyl halides.

Answer 1

Hint: Halogen group is an electron-withdrawing group because it requires only one electron to become stable. So it acquires a negative charge and the alkyl group will acquire a positive charge.

Complete step by step solution:
The displacement of $\sigma -electrons$ along the saturated carbon chain whenever an electron-withdrawing or electron-donating group is present at the end of the carbon chain is called inductive effect or I-effect.
–I-Effect is when the substituent attached to the end of the carbon chain is electron-withdrawing.
+I-Effect is when the substituent attached to the end of the carbon chain is electron-donating.
So, when the atom or group like halogen which is an electron-withdrawing group attached to the alkyl carbon chain, the $\sigma -electrons$ of $C-X$ are attracted by or displaced to the more electronegative atom i.e., halogen atom. Due to which the halogen atom will acquire a small negative charge and the carbon atom of the alkyl group will acquire a small positive charge.

Now in this chain, there is a positive charge on ${{C}_{1}}$ atom, this, in turn, attracts the $\sigma -electrons$ of ${{C}_{1}}-{{C}_{2}}$ bond towards it. Due to this ${{C}_{2}}$ will also have some positive charge and this, in turn, will attract the electrons of ${{C}_{2}}-{{C}_{3}}$ bond towards it. Due to this ${{C}_{3}}$ will also acquire a very small positive charge.
So, the reactivity of alkyl halide due to the inductive effect is that the alkyl part acts as an electron-donating group and the halide acts as an electron-withdrawing group.

Note: The same process occurs when an electron-donating group is attached to the alkyl group, but the difference is that the alkyl group will acquire the negative charge and the electron-donating group will acquire the positive charge.