
A gas decolourised by KMnO4 solution but gives no precipitate with ammoniacal cuprous chloride is-
A. C2H6 (ethane)
B. CH4 (methane)
C. C2H4 (ethene)
D. C2H2 (acetylene)
Answer
163.8k+ views
Hint: KMnO4 and CuCl are strong oxidising agents and alkanes are saturated hydrocarbons, in a saturated hydrocarbon carbon’s valence is completely fulfilled by sigma bonds and the maximum number of Hydrogen is attached with it. Thus, saturated hydrocarbons are highly stable and to break the C-H bond large amounts of energy is required.
Complete Step by Step Answer:
As we can see that,
- C2H6 is the formula of ethane
- CH4 is the formula of methane
- C2H4 is the formula of ethene
- C2H2 is the formula of acetylene
Now since methane and ethane are saturated hydrocarbons so they will not react with either KMnO4 or CuCl
When ethene is mixed with KMnO4 diol is formed.
$CH_2=CH_2 + \ \ KMnO_4\ \ \rightarrow CH_2(OH)-CH_2(OH)$
As a result of the reaction pink colour of KMnO4 disappears
But, when ethene is mixed with CuCl no reaction takes place:
$CH_2=CH_2 + \ \ CuCl\ \ \rightarrow N o\ reaction$
Now acetylene when mixed with KMnO4 and CuCl. It gives a reaction with both when it reacts with KMnO4 the product will be acid and the pink colour of KMnO4 disappears:
$CH \equiv CH + \ \ KMnO_{4\ } H_2O\ \ \ \rightarrow O=CH-CH=O \left [ O \right ] → HOOC-COOH$
At first Ethane 1,2-dial is formed which on further oxidation gives oxalic acid. When acetylene reacts with ammoniacal CuCl the product formed will be:
$CH \equiv CH + \ \ CuCl(NH_3)\ \ \ \rightarrow CuC \equiv CCu+2NH_4Cl+2NH_3$
The product formed in this reaction is cuprous acetylide which is red in colour. So, we can see that alkanes that is methane and ethane does not react with either KMnO4 or CuCl and ethene and acetylene both reacts with KMnO4 and CuCl but acetylene also reacts with CuCl for a product which is red in colour but ethene does not react with CuCl So, the correct answer is ethene.
Thus, Option (C) is correct
Note: Cu2C2 when produced in dry form can be used as an explosive. The reaction
$ \left [ CH \equiv CH + \ \ CuCl(NH_3) \ \ \ \rightarrow CuC \equiv CCu + 2NH_4Cl + 2NH_3 \right ]$ when further proceeds in the presence of antioxidant, dimerization of acetylene takes place and but-1,3-yne is formed.
Complete Step by Step Answer:
As we can see that,
- C2H6 is the formula of ethane
- CH4 is the formula of methane
- C2H4 is the formula of ethene
- C2H2 is the formula of acetylene
Now since methane and ethane are saturated hydrocarbons so they will not react with either KMnO4 or CuCl
When ethene is mixed with KMnO4 diol is formed.
$CH_2=CH_2 + \ \ KMnO_4\ \ \rightarrow CH_2(OH)-CH_2(OH)$
As a result of the reaction pink colour of KMnO4 disappears
But, when ethene is mixed with CuCl no reaction takes place:
$CH_2=CH_2 + \ \ CuCl\ \ \rightarrow N o\ reaction$
Now acetylene when mixed with KMnO4 and CuCl. It gives a reaction with both when it reacts with KMnO4 the product will be acid and the pink colour of KMnO4 disappears:
$CH \equiv CH + \ \ KMnO_{4\ } H_2O\ \ \ \rightarrow O=CH-CH=O \left [ O \right ] → HOOC-COOH$
At first Ethane 1,2-dial is formed which on further oxidation gives oxalic acid. When acetylene reacts with ammoniacal CuCl the product formed will be:
$CH \equiv CH + \ \ CuCl(NH_3)\ \ \ \rightarrow CuC \equiv CCu+2NH_4Cl+2NH_3$
The product formed in this reaction is cuprous acetylide which is red in colour. So, we can see that alkanes that is methane and ethane does not react with either KMnO4 or CuCl and ethene and acetylene both reacts with KMnO4 and CuCl but acetylene also reacts with CuCl for a product which is red in colour but ethene does not react with CuCl So, the correct answer is ethene.
Thus, Option (C) is correct
Note: Cu2C2 when produced in dry form can be used as an explosive. The reaction
$ \left [ CH \equiv CH + \ \ CuCl(NH_3) \ \ \ \rightarrow CuC \equiv CCu + 2NH_4Cl + 2NH_3 \right ]$ when further proceeds in the presence of antioxidant, dimerization of acetylene takes place and but-1,3-yne is formed.
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