In the given Cannizzaro reaction,the slowest step is:
$2PhCHO \xrightarrow{KOH} PhCH_2OH + PhCOO^-$

  • A
    The attack of $OH^-$ at the carbonyl group
  • B
    The transfer of hydride to the carbonyl group
  • C
    The abstraction of proton from the carboxylic acid
  • D
    The deprotonation of $PhCH_2OH$

Explore More

Similar Questions

$C_6H_5-CH=CHCHO \xrightarrow{X} C_6H_5-CH=CHCH_2OH$. In the above sequence,$X$ can be:

$Ph_2CH-C(=O)H \xrightarrow{\text{aqueous acid}} (A) (81\%) + \text{enol} (2\%) + \text{aldehyde} (17\%)$. Product $(A)$ of the above reaction will be:

Which of the following is not an $o, p-$ directing group?

Schiff's reagent is

What is the product obtained in the reaction?
$CH_3-CH=CH-CH_2-CHO \xrightarrow[(ii) H_3O^{+}]{(i) LiAlH_4} \text{product}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo