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CBSE Class 12 Chemistry 2017 Delhi Set 2 Solved Paper

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Question : 7 of 10
Marks: +1, -0
Write the reactions involved in the following:
(i) Hell-Volhard-Zelinsky reaction
(ii) Decarboxylation reaction
Solution:  
(i) In Hell-Volhard-Zelinsky (HVZ) reaction, carboxylic acid having an α\alpha-hydrogen is halogenated at α\alpha-position on treatment with chlorine or bromine in the presence of red phosphorus to give α\alpha-halogenated carboxylic acid.
RCH2COOH→(ii) H2O(i) X2/Red PRCHCOOH∣X\mathrm{RCH}_2\mathrm{COOH} \xrightarrow[(ii)\,\mathrm{H}_2\mathrm{O}]{\text{(i)}\,\mathrm{X}_2/\mathrm{Red\ P}} \underset{\mathrm{X}}{\underset{\vert}{\mathrm{RCHCOOH}}}
(ii) In decarboxylation reaction, carboxylic acid loses CO2\mathrm{CO}_2 to form hydrocarbons when their sodium salts are heated with sodalime (NaOH(\mathrm{NaOH} and CaO\mathrm{CaO} ) in the ratio 3:13:1.
RCOONa→ΔNaOH and CaORH+Na2CO3\mathrm{RCOONa} \xrightarrow[\Delta]{\mathrm{NaOH}\text{ and }\mathrm{CaO}} \mathrm{RH} + \mathrm{Na}_2\mathrm{CO}_3
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