Ph-CH_3 xrightarrow{CrO_2Cl_2} (A) xrightarro | vedclass

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(B) The reaction sequence is as follows:$1$. $Ph-CH_3$ reacts with $CrO_2Cl_2$ (Etard reaction) to form benzaldehyde $(A)$,which is $Ph-CHO$.$2$. $Ph-

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$Ph-CO_2^-$

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(B) The reaction sequence is as follows:$1$. $Ph-CH_3$ reacts with $CrO_2Cl_2$ (Etard reaction) to form benzaldehyde $(A)$,which is $Ph-CHO$.$2$. $Ph-CHO$ reacts with concentrated $KOH$ (Cannizzaro reaction). Since benzaldehyde has no $\alpha$-hydrogen,it undergoes disproportionation to form benzyl alcohol $(Ph-CH_2OH)$ and potassium benzoate $(Ph-COO^-K^+)$.$3$. Thus,product $(B)$ is the benzoate ion,$Ph-COO^-$.

$Ph-CH_3$ $\xrightarrow{CrO_2Cl_2} (A)$ $\xrightarrow{conc. KOH} Ph-CH_2OH + (B)$
Product $(B)$ of the above reaction is:

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$Ph-CH_3$ $\xrightarrow{CrO_2Cl_2} (A)$ $\xrightarrow{conc. KOH} Ph-CH_2OH + (B)$Product $(B)$ of the above reaction is: