Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Hydrocarbons
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App