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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.Structure of hydrocarbon 0.34gm of which when treated with $CH_3MgI$ liberates 112ml of $CH_4$ at STP,will be

$\begin{array}{1 1} a \\ b \\ c \\ d\end{array}$

1 Answer

$R-C\equiv C-H+CH_3-MgI\rightarrow CH_4+R-C\equiv C-Mg-I$
Number of moles of $CH_4$=Number of moles of hydrocarbon
$\large\frac{112}{22400}=\frac{0.34}{MM}$
$\therefore MM=68$
$C_nH_{2n-2}=68$
$\Rightarrow n=\large\frac{68+2}{14}$$=5$
Molecular formula of hydrocarbon $R-C\equiv C-H=C_5H_8$
Hence $R=C_3H_7$ which may be propyl and isopropyl
Hence (a) is the correct answer.
answered Mar 19, 2014 by sreemathi.v
 

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