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The compounds $X(C_5H_8)$ reacts with ammoniacal $AgNO_3$ to give a white precipitate and on oxidation with hot alkaline $KMnO_4$ gives the acid,$(CH_3)_2CHCOOH$.Therefore X is

$\begin{array}{1 1}(a)\;CH_2=CHCH=CHCH_3\\(b)\; CH_3(CH_2)_2C\equiv CH\\(c)\;(CH_3)_2CH-C\equiv CH\\(d)\;(CH_3)_2C=C=CH_2\end{array}$

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1 Answer

Since compound $X(C_5H_8)$ reacts with ammoniacal $AgNO_3$ solution to give a white precipitate,therefore $X$ must be a terminal alkyne.Further since X gives $(CH_3)_2CHCOOH$ on oxidation with hot $KMnO_4$ solution.
$X$ must be $(CH_3)_2CHC\equiv CH$
$(CH_3)_2CHC\equiv\quad \underrightarrow {HOT\; KMnO_4} \quad (CH_3)_2CHCOOH+[HCOOH]\quad\underrightarrow {[O]}\quad CO_2+H_2O$
Hence (c) is the correct answer.
answered Mar 18, 2014 by sreemathi.v
 

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