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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.In the below equation product is

$\begin{array}{1 1}(a)\;CH_4&(b)\;CH_3CH_3\\(c)\;CH_3CH_2-CH_3&(d)\;CH_3CH_2-CH_2-CH_3\end{array}$

1 Answer

As in the given passage we get alkane.Corresponding to alkyl part of Grignard reagent.
Hence (c) $CH_3CH_2-CH_3$ is the correct answer.
answered Mar 19, 2014 by sreemathi.v
 

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