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प्रश्न
Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.
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उत्तर
Decreasing order of reactivity of sodium metal is:
1° > 2° > 3°
Alcohols react with sodium metal to form alkoxides and hydrogen is liberated:
\[\ce{R - O - H + Na -> RO - Na+ {+} 1/2 H2}\]
The order of reactivity of alcohols is primary > secondary > tertiary. This can be explained on the basis of cleavage of O – H bond. The alkyl groups are electron releasing groups (+I effect) and they increase the electron density around the oxygen. As a result, the electrons of O – H bond cannot be withdrawn strongly towards oxygen and O – H remains strong. Therefore, greater is the number, of alkyl groups present, smaller will be reactivity of alcohol.
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संबंधित प्रश्न
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\[\begin{array}{cc}
\phantom{.......................}\ce{Br}\\
\phantom{......................}|\\
\ce{CH3 - CH - CH - CH3 ->[HBr] CH3 - C - CH2 - CH3}\\
\phantom{.}|\phantom{......}|\phantom{......................}|\phantom{........}\\
\phantom{}\ce{CH3}\phantom{...}\ce{OH}\phantom{...................}\ce{CH3}\phantom{.....}
\end{array}\]
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