Doubt from hydrocarbons

Question no 6 second part

6 B

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Is Kolbes Electrolysis Method and Decarboxylation method is same

No , decarboxylation happen with CaO+NaOH and there carbanion is formed and here carbon free radical is formed

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Ok but they form same product

@abhijit_2020 in post #4 i wrote about one specific decarboxylation reaction i.e soda lime decarboxylation .

Decarboxylation is not any method or reaction , basically any reaction in which Co2 is removed is known as decarboxylation reaction .
So we can say that in kolbe’s electrolysis , decarboxylation happens . That’s it , there’s nothing like decarboxylation method

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Question no 5

It should be B because only 2 opp Br will come out as free radical and thus form a bond along the diagonal

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Question no 9

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@abhijit_2020 is the answer (c)?

No answer is option A

Oh right... frequency of collision will be lowest since the molecules (here) are reacting with the correct orientation and hence their collision will be successful i.e reaction will take place..Therefore each time the no. Of colliding species will be decreasing and hence it's frequency will decrease... @Shwetanshu_2018 pls check..

@pushkar_2020 it is sort of correct , there can be other explanations also , that this product can be formed by one more alternate mechanism which has more tendency .

This question is pretty much of no use , so better not waste time on this .
I guess @abhijit_2020 there would be one more similar question in that book , simply ignore it

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Ok

Can anyone tell me how this has been calculated relative amount

Please help i forgot how it is calculated

1st you have to check that how many 1,2,3degree H are present then check the selectivity of halogen for eg
Cl- 1: 3.8 : 5. So for 1 degree it will be 1 x no of 1 degree H
For 2 degree it will be 3.8 X no of 2 degree H
And so on

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So how to do question no 49

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Thanks