[saira16]

What product(s) would you expect to obtain if propene labeled with 14C at C1 were subjected to allylic chlorination or bromination?                                   
                                              High Temperature
¹⁴CH_{2double bond}CHCH₃+X₂  ?
                                   Low concentration of X₂
My question is that if i attack the labeled carbon-14, what would be my product. My answer is

                                 
¹⁴CH₂Xsingle bondCHCH₃, but the correct answer is CH₂double bond¹⁴CH single bond CH₂X
                                   
can someone please explain the right answer! Thanks

[Quaff]

I do not know but my guess is that the protons on the methyl are slightly acidic due to the electron withdrawing effect of the double bond, analogous to enolates.   If in some tiny proportion of the propene a methyl proton is liberated, a stable radical CH2-CH-CH2 results in which bond order is 1.5 between each of the carbons, delocalized.  Then statistically there are two CH2's for every molecule and on the ends which are more kinetically accessible than on the middle C.   At high temps, X2 dissociation equilibrium is shifted so more X* radicals are present, and the tiny number of protons that dissociate off the methyl are immediately sucked up by an X* to form HX and the halogenated propene, which by LeChatlier then forces the equilibrium to generate more of the CH2-CH-CH2 delocalized radical, continuing until all the X2 is gone.

But, I could be wrong.

[savy2020]

@quaff
I completely agree with you.   
@saira16
Well your answer corresponds to the electrophilic addition to alkene but the product asked for in the question is that due to allylic halogenation