For a Cannizzaro reaction, R-CHO + NaOH→ Rate law is derived as: Rate=k[RCHO]2[HO-]2 From the above rate law, it can be concluded that

Question

For a Cannizzaro reaction, R-CHO + NaOH→ Rate law is derived as: Rate=k[RCHO]2[HO-]2 From the above rate law, it can be concluded that (A) hydride donor is a dianion (B) reaction involves hydride ion transfer (C) reaction does not show kinetic isotopic effect (D) reaction is pseudo 2nd order

Tardigrade Admin · Accepted Answer

For formation of dianion hydride donor, an additional mole of NaOH is consumed <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/8c24632a9af6f3753249a47cc933648b-.png" /> Hence, rate becomes 2nd order with respect to HO- and fourth order overall.

Q. For a Cannizzaro reaction,
$R - C H O + N a O H \to$ Rate law is derived as: Rate $= k [RC H O]^{2} [H O^{-}]^{2}$ From the above rate law, it can be concluded that

hydride donor is a dianion

reaction involves hydride ion transfer

reaction does not show kinetic isotopic effect

reaction is pseudo $2^{n d}$ order

For formation of dianion hydride donor, an additional mole of $N a O H$ is consumed

Hence, rate becomes $2^{n d}$ order with respect to $H O^{-}$ and fourth order overall.

Q. For a Cannizzaro reaction, R−CHO+NaOH→ Rate law is derived as: Rate=k[RCHO]2[HO−]2 From the above rate law, it can be concluded that