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During the preparation of alkanes by Kolbe's electrolytic decarboxylatio of CH_3CH_2COONa , which of the following products are formed |
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Answer» `CH_3CH_2CH_2CH_3` `CH_3CH_2COONa to CH_3 CH_2COO^(-) + Na^(+)` `H_2O to H^(+) + OH^(-)` At anode as discharge POTENTIAL of `CH_3CH_2COO^-` is less than discharge potential of `OH^-`, HENCE discharging of `CH_3CH_2COO^-` TAKES place. `CH_2CH_2-undersetunderset(O)(||)C-underset(ddot)overset(theta)O: to CH_3CH_2-undersetunderset(O)(||)C-overset(dot)O+e^-` `CH_3CH_2-undersetunderset(O)(||)C-overset(dot)O to CH_3overset(dot)CH_2+CO_2` `CH_3overset(dot)CH_2+overset(dot)CH_3 to CH_3CH_2CH_2CH_3` `CH_3 overset(dot)CH_2 to CH_2=CH_2+overset(dot)H` `CH_3overset(dot)CH_2+overset(dot)H to CH_3CH_3` `CH_3CH_2-undersetunderset(O)(||)C-overset(dot)O -overset(dot)O + .^(dot)CH_2CH_3 to CH_3CH_2-undersetunderset(O)(||)C=O-CH_2-CH_3` `CH_3CH_2-undersetunderset(O)(||)C-OCH_2CH_3 overset(OH^-)to CH_3CH_2COO^(-) + CH_3CH_2OH` Hence , the various products formed at anode are `CH_3CH_2CH_2CH_3,CH_2=CH_2, CH_3CH_3 & CH_3CH_2OH`. Hence, (a),(b), ( d) are correct while ( c) is INCORRECT. |
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