IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

danny_007

a hydrocarbon A of molecular weight 54 reacts with excess of BBr2 in CCl4 to give the compound B whose molecular weight 493% more than that of A . how ever on hydrogenation with excess of Hydrogen , A form C whose molecular weight is just 7.4% more than A . A reacts with CH3CH2Br in presence of NaNH2 to give another Hydrocarbon D which on ozonolysis yields diketonbe E , E on oxidation gives propanoic acid . give all reaction and structures of ABCDE and reasons too

tarinbansal

Ur ques seems to be a bit faulty but still probable answers cud be-

For molecular mass=54, see the nearest no of C possible, i.e. 4, as 12x4=48 and rest can be 6H atoms to make the formula C4H6, that is an alkyne.

Now to find whether it is a terminal or internal alkyne, start backwards.

D on ozonolysis gives E which is a diketone and which on oxidation gives only ONE product, propanoic acid. As it is a diketone and still giving only one product, it means it must be a symmetric diketone otherwise it wud hav given 2 diff. acids.

For it to be symmetric, A has to be terminal.

SO, A----> CH3-CH2-CCH.

C---> CH3-CH2-CH2-CH3

For D,

A+CH3CH3Br + NaNH2------> CH3-CH2-CC-CH2-CH3

O O

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Most probably, E is----> CH3-CH2-C-C-CH2-CH3

Acc to the data given by U, there is some doubt in B. It shud be 593% heavier not 493% taking mass of Br=80.

But still B-------> CH3-CH2-C(Br)₂-C(Br)₂

It shud be a double halogenation reaction.

Hope U understand.

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