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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Hydrocarbons
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When Grignard reagent is treated with water,alcohol,ammonia,$1^o$ amine,$2^o$ amine,alkyne-1 or carboxylic acid.We get alkane corresponding to alkyl part of Grignard reagent.In Grignard reagent C-atom is more electronegative than magnesium,hence its alkyl part acts are nucleophile and thus it will take an active hydrogen atom(H-atom that undergoes acylation) or acidic hydrogen atom to give its corresponding alkane.Also it is known that stronger acid displaces weak acid from weak acid salt.$C_6H_5SO_3-H+CH_3-COONa\rightarrow$ Product.The product in the given reaction is

$\begin{array}{1 1}(a)\;C_6H_6+CH_3-COOH\\(b)\;C_6H_5SO_3Na+CH_3-COOH\\(c)\;C_6H_5SO_3Na+CH_4\\(d)\;\text{None of these}\end{array}$
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$C_6H_5SO_3-H+CH_3-COONa\rightarrow C_6H_5SO_3Na+CH_3-COOH$
Stronger acid displaces weak acid from weak acid salt.
Hence (b) is the correct answer.
answered Mar 19, 2014 by sreemathi.v
 

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