In Cannizzaro reaction given below \[ 2 \mathrm{Ph} \mathrm{CHO} \xrightarrow{: \stackrel{(-)}{\mathrm{O}} \mathrm{H}} \mathrm{Ph} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{(-)} \] $2 \mathrm{Ph} \mathrm{CHO} \xrightarrow{: \mathrm{OH}} \mathrm{Ph} \mathrm{CH} 2 \mathrm{OH}+\mathrm{PhCO}_{2}^{(-)}$the slowest step is :

(B) the transfer of hydride to the carbonyl group

(A) the attack of : $\stackrel{(-)}{\mathrm{OH}}$ at the carboxyl group

(C) the abstraction of proton from the carboxylic group

(D) the deprotonation of $\mathrm{Ph} \mathrm{CH}_{2} \mathrm{OH}$

JEE Main · 2009 · 12 Apr 2009 · Offline · Aldehydes Ketones and Carboxylic Acids

In Cannizzaro reaction given below

2 Ph CHO : O (-) H Ph CH_{2} OH + PhCO_{2}^{(-)}

2 Ph CHO : OH Ph CH 2 OH + PhCO_{2}^{(-)}

the slowest step is :

(A)the attack of : $OH (-)$ at the carboxyl group
(B)the transfer of hydride to the carbonyl group✓ Correct
(C)the abstraction of proton from the carboxylic group
(D)the deprotonation of $Ph CH_{2} OH$

Answer: (B)

Solution

In the Cannizzaro reaction hydroxide first adds to the carbonyl of one aldehyde to make a tetrahedral intermediate, and the slow, rate-determining step that follows is the transfer of a hydride ion from that intermediate to the carbonyl carbon of a second aldehyde molecule. It is this hydride shift to the carbonyl group that governs the overall rate.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Aldehydes Ketones and Carboxylic Acids

Solution

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