Hydrocarbons Test - 17

The two hydrocarbons can be easily distinguished by simple chemical tests, as ethyne molecule is supposed to have acidic hydrogen.

Ethyne forms a red precipitate of copper acetylide (Cu2C2) when it is passed through ammoniacal cuprous chloride solution.

Ethene does not react with Cu2Cl2 solution.

Hence C is the correct answer.

Photochemical chlorination of alkane take place by free radical mechanism which are possible by Homolysis of C - C bond 

CH₃ – CH = CH₂ → CH₃ – CH⁺ – CH₂
The reason for this is only that carbocation is formed which has maximum stability. In this case, we have 6 α-H while for option a, b and d; we have 0, 2 and 2 α-H respectively. So only carbocation in option c forms.

CH₃ - CH= CH₂ + HOSO₂OH→CH₃—CH(OSO₂OH)—CH₃

Ethylene reacts with HBr to form Ethyl bromide. The reaction propagates as follow:-
H₂C=CH₂  +  HBr → H₂C+-CH₃ →H₂BrC-CH₃ 
Since π cloud is electron rich, so HBr dissociates into H⁺ and Br^-. H⁺ attacks on alkene to give a carbocation and then Br^- attacks to get ethyl bromide.

When propene on ozonolysis it yields a new structure called ozonide 
and there cleavage takes place and it yields two products namely 
1.acetaldehyde
2.formaldehyde

The reductive ozaonalysis of 2,3 -Dimethyl-2-butene yields acetone.The reaction is as follows: 2,3 -Dimethyl-2-butene acetone

They form vicinal glycols as shown below:

In organic chemistry, an electrophilic addition reaction is an addition reaction where, in a chemical compound, a π bond is broken and two new σ bonds are formed. The substrate of an electrophilic addition reaction must have a double bond or triple bond.

Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment . The dipole moment enables electronic interactions which hold molecules together . Trans 2-Butene has more symmetry thanits cis isomer which results in better packing and hence higher melting point .