Tag: acetylene

Questions Related to acetylene

Sodium fumarate on Kolbes electrolysis gives :

  1. Ethene

  2. Ethyne

  3. Ethane

  4. Methane


Correct Option: B

The compound used in Hawker's lamp and in light houses for illumination purposes is :

  1. methane

  2. ethane

  3. ethylene

  4. acetylene


Correct Option: D
Explanation:

Acetylene is used in-
Illumination of cycle lamps, hawker's lamps and buoys in light houses.
Producing oxyacetylene flame for welding and cutting.

A hydrocarbon of molecular formula , ${C _6}{H _{10}}$ reacts with sodamide and the same on ozonolysis followed by hydrogen peroxide gives two molecules of carboxylic acids, one being optically active. Then the hydrocarbon may be

  1. $1$ hexyne

  2. $2- hexyne$

  3. $3$ hexyne

  4. $3$ methyl $1$ pentyne


Correct Option: A

Product formed on electrolysis of potassium salt of fumaric and maleic acid is

  1. Ethene

  2. Ethane

  3. Butane

  4. Ethyne

  5. None of the above


Correct Option: D

A mixture of oxygen and ethyne is burnt for welding. Why a mixture of ethyne and air is not used?

  1. ethene can not burn in air

  2. incomplete combustion takes place

  3. ethene burn in air with explosion 

  4. none of these


Correct Option: B
Explanation:

When ethyne is burnt in air incomplete burning takes place and produce sooty flame due to limited supply oxygen present in the air.


 However, if ethyne is burnt in oxygen, it gives a clean flame with high temperature (3000°C ) due to complete combustion of ethyne.

 Hence, This oxy-acetylene flame is used for welding and it is not possible to attain such a high temperature with air. Due this reason a mixture of ethyne and air is not used for welding.

Acetylene may be prepared using Kolbes electrolytic method employing:

  1. Pot. acetate

  2. Pot. succinate

  3. Pot. fumarate

  4. None of these


Correct Option: C
Explanation:

Acetylene may be prepared using Kolbes electrolytic method employing potassium fumerate. Acetylene is obtained at anode. The solution is basic due to formation of KOH at cathode.
$\displaystyle  (CHCOONa) _2 \xrightarrow {Electrolysis} C _2H _2+2CO _2+2KOH + H _2$

$C _{2}H _{2}$ on reaction with Na followed by reaction with $C _{2}H _{5}I$ gives 'X'. The molecular formula of 'X' is:

  1. $C _{2}H _{6}$

  2. $C _{4}H _{8}$

  3. $C _{4}H _{6}$

  4. $C _{3}H _{6}$


Correct Option: C
Explanation:

Reaction: $CH \equiv CH \xrightarrow{Na} Na - C \equiv  CH \xrightarrow{C _2H _5I} C _2H _5 - C \equiv  CH + NaI$


Hence, the correct option is $C$

1, 1, 2, 2-Tetra bromoethane $\xrightarrow {Zn}A$ $\xrightarrow{red hot \ tube}B$,


 Then B is:

  1. Acetylene

  2. Ethylene

  3. Ethane

  4. Benzene


Correct Option: D
Explanation:

$H-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{C}}}}-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{C}}}}-H\xrightarrow[\triangle]{2n}H-C\underset{(A)}{\equiv} C-H\xrightarrow{ red hot \ tube} C _6H _6$


Option D is correct.

$H-C\equiv C-H+NaNH _{2}\longrightarrow A\overset{2 \ mole \ CH _3I}{\longrightarrow}B$


Then B is :

  1. 1-Butyne

  2. 2-Butyne

  3. 2-Pentyne

  4. Propyne


Correct Option: B
Explanation:

Reaction of process:
$CH\equiv CH+2NaNH _2\rightarrow NaC\equiv CNa\xrightarrow {2 mole CH _3I} CH _3CH\equiv CHCH _3$


Option B is correct.

$X+2Zn\overset{Alcohol}{\rightarrow}H-C\equiv C-H$ 


Here X is :

  1. 1, 1-Dibromoethane

  2. 1, 2-Dibromoethane

  3. Dibromo ethane

  4. 1, 1, 2, 2-Tetrachloroethane


Correct Option: D
Explanation:

This is dehalogenation process:


$CHCl _2-CHCl _2+2Zn\overset{Alcohol}{\longrightarrow}H-C\equiv C-H$

Option D is correct.