$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{89}_{36}Kr+3 ^{1}_{0}n+X_{X}^{A}$
$92+0=36+Z$
$\Rightarrow Z=56$
$235+1=89+3+A$
$\Rightarrow A=144$
So, $^{144}_{56}Ba$ is generated
Predict the major product $ P $ in the following sequence of reactions:
(i) HBr, benzoyl peroxide
(ii) KCN
(iii) Na(Hg), $C_{2}H_{5}OH$