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प्रश्न
Explain. Aryl halides are less reactive than alkyl halides towards nucleophilic substitution reactions.
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उत्तर
- The low reactivity of aryl halides is due to the resonance effect and sp2 hybrid state of carbon to which halogen atom is attached.
- In aryl halides, one of the lone pairs of electrons on the halogen atom is in conjugation with π-electrons of the ring. Due to resonance, the C–X bond acquires partial double bond character. Thus, the C–X bond in aryl halides is stronger and shorter than alkyl halides. Hence, it is difficult to break C–X bond in aryl halides.
- Further, the phenyl cation produced due to the self-ionization of aryl halide will not be stabilised by resonance. This rules out the possibility of SN1 mechanism. Also, the backside attack of nucleophiles is blocked by the aromatic ring. This rules out the possibility of SN2 mechanism. As a result, nucleophilic substitution reaction involving cleavage of C–X bond in haloarenes proceeds with difficulty.
- Therefore, aryl halides are less reactive than alkyl halides towards nucleophilic substitution reactions.
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\[\begin{array}{cc}\ce{CH3\phantom{................}}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3\phantom{................}}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{.......}\\
|\phantom{.........}\\
\ce{CH3 - c - CH2 - Cl ->[Na/dry ether]}\\
|\phantom{.........}\\
\ce{CH3}\phantom{.......}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{................}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{...............}\\
|\phantom{..................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether]A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
Complete the following reaction giving major product.
\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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