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# Br has a low reactivity in $CH_2 =CH-Br$ because (i) of -M effect of bromine (ii) of more electro negativity of bromine (iii) of partial double bond character between $C-Br$ bond. (iv) of the +M effect of bromine.

$(a)\;(i)\; and\;(ii) \\ (b) \; (ii)\;and \;(iii) \\ (c) \;(ii)\;and \;(iv) \\ (d)\;(iii)\;and (iv)$

$CH_2 = CH -Br :\leftrightarrow CH_2 -CH =Br^+$
$+M$ character because of resonance which also develop double bond character.
Hence d is the correct answer.