Can some one provide theory related to this. @Supreeta_Sen
For 17 and 18
I can be clearly seen to be A from +ve Baeyer's Test and NaHCO3 test (It is given by acids stronger than H2CO3)
J will be Cinnamic Acid.
For K see the image below.
Okk my mistake I was also reducing -COOH group which is not reduced by H2 Pd
Bro, first one is free radical stability basically. I don't have theory related to it. But by comparing stabilities of free radical and testing for possibility of forming carbonyl compound, we can get answer.
Br2 + CH3COOH
3 moles of bromine react with 1 mole of acetone.
1 mole bromine react with 1/3 mole of acetone.
Hence 1/3 mole of acetone undergoes haloform and gives haloform product.
2/3 mole acetone remains as it is in the solution.
Okk this was really silly. Not thought about bromine being the lr. Anyway thanks bro.
For the free radical one don't type the solution , I should better look for it in solution videos.