JEE MAIN - Chemistry (2009 - No. 9)
In Cannizzaro reaction given below
2PhCHO $$\buildrel {\mathop {:OH}\limits^{\left( - \right)} } \over \longrightarrow $$ PhCH2OH + $$PhC\mathop O\limits^{..} $$2(-)
the slowest step is :
2PhCHO $$\buildrel {\mathop {:OH}\limits^{\left( - \right)} } \over \longrightarrow $$ PhCH2OH + $$PhC\mathop O\limits^{..} $$2(-)
the slowest step is :
the transfer of hydride to the carbonyl group
the abstraction of proton from the carboxylic group
the attack of $$:\mathop {OH}\limits^{( - )} $$ at the carboxyl group
the deprotonation of Ph CH2OH
Обяснение
