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प्रश्न
Propanoic acid to ethylamine.
उत्तर
\[\ce{\underset{\text{Propanoic acid}}{CH_3CH_2COOH} + NH_3 -> CH_3 CH_2COONH_4 ->[{Δ}][{-H_2O}] CH_3 CH_2CONH_2 ->[{Br_2+4NaOH}][{Δ}]\underset{\text{Ethyl amine}}{CH_3CH_2NH_2}+ {Na_2 CO_3} + {2NaBr +2H_2 O}}\]
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\[\begin{array}{cc}
\ce{CH3 - CH2 - O - CH - CH2 - CH3}\\
\phantom{...}|\\
\phantom{.....}\ce{CH3}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3-CH-OCH3}\\
|\phantom{....}\\
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\end{array}\]
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|\phantom{....................}\\
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\[\begin{array}{cc}
\phantom{...............}\ce{CH3}\\
\phantom{............}|\\
\ce{CH3 - CH - CH - C - CH3}\\
|\phantom{......}|\phantom{......}|\\
\phantom{...}\ce{CH3}\phantom{...}\ce{OH}\phantom{...}\ce{CH3}
\end{array}\]
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\[\begin{array}{cc}
\phantom{...............}\ce{CH3}\\
\phantom{.............}|\\
\ce{CH3 - CH - CH - C - CH3}\\
|\phantom{......}|\phantom{......}|\\
\phantom{...}\ce{CH3}\phantom{..}\ce{OH}\phantom{...}\ce{CH3}
\end{array}\]
