A gas that decolourises KMnO_4 solution but g | vedclass

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(C) $1$. $KMnO_4$ solution (Baeyer's reagent) is decolourised by unsaturated hydrocarbons like alkenes and alkynes due to the presence of multiple bon

Vedclass · Accepted Answer

Ethene

Vedclass · Answer

(C) $1$. $KMnO_4$ solution (Baeyer's reagent) is decolourised by unsaturated hydrocarbons like alkenes and alkynes due to the presence of multiple bonds.$2$. Ammoniacal cuprous chloride $(Cu_2Cl_2 + NH_4OH)$ reacts specifically with terminal alkynes to form a red precipitate of copper acetylide.$3$. Ethane and methane are saturated hydrocarbons (alkanes) and do not decolourise $KMnO_4$.$4$. Acetylene (ethyne) is a terminal alkyne and gives a red precipitate with ammoniacal cuprous chloride.$5$. Ethene $(CH_2=CH_2)$ is an alkene; it decolourises $KMnO_4$ but does not react with ammoniacal cuprous chloride to form a precipitate.$6$. Therefore,the correct answer is Ethene.

$A$ gas that decolourises $KMnO_4$ solution but gives no precipitate with ammoniacal cuprous chloride is

Similar Questions

What is known as an asymmetrical alkene? Give examples.

$I$. $(CH_3)_2 C=CH_2 \xrightarrow{KMnO_4 / H^{+}} X + CO_2 + H_2 O$
$II$. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4 / H^{+}} Y$
The functional groups in $X$ and $Y$ are respectively

The major product obtained in the following reaction is

$CH_3-CH_2-CH=CH_2 \xrightarrow[CCl_4]{Br_2} \text{Product}$

Identify $(Z)$ in the above sequence of reactions:
$CH_3-CH_2-CH=CH_2$ $\xrightarrow{HCl/\text{peroxide}} (A)$ $\xrightarrow{EtONa/\Delta} (Z) \text{ (major)}$

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