11, 2 and 3.
Which statement is/are correct about free radical halogenation of alkanes :-
1) The number of product molecules formed by one photon is very high
2) If O2 is added, initially the rate decreases then it increases.
3) Inhibitors combine with free radical and terminate the chain reaction
4) Presence of Ph-COO-COO-Ph inhibits the free radical mechanism.
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13 Answers
39Free radical reactions are greatly catalysed by light. Photons cause rapid collisions in the molecules(supplying the high energy required) and also cause them to rip homolytically for radical formation. One photon starts off a chain reaction(initiation) which in itself causes various other radicals to be formed in the propagation steps. So part 1 is right.
My guess for part 2 is that when oxygen is added initially, some amount of light is used for its own homolytic fission, which slows down the reaction. But once oxygen radicals participate in the propagation steps, the rate of reaction becomes high again.
3 mein kya hai? Free radical inhibitors do NOT terminate the reaction. They cause such a radical to be formed which is either too sterically hindered and thus neutral to further radical formation, or is very stable(aromaticity for example).
Dunno about 4th..
19there is no doubt that 2 and 3 are correct....i am a bit hesitant with the other two !
14 is incorrect, that toh i'm sure about.
I had doubt regarding 2 and 3.
Answer given by injun is correct...
1if inhibitors do not terminate the reaction then why is the 3rd option correct???
39http://usm.maine.edu/~newton/Chy251_253/Lectures/Free%20Radicals/FreeRadicalFS.html
Check the inhibitors section. It says unreactive radicals are formed, unreactive due to steric hindrance or extra stability. It doesn't mention termination..
11,2,3 Un doubtedly.Free radical inhibitors do not terminate the reaction. They cause such a radical is to be formed which is either too stearically hindered and thus neutral to further radical formation for stability