Question

The compound A in the following reaction is :

$A \xrightarrow[(ii)Conc.H_2SO_4/\Delta]{(i) CH_3MgBr/H_2O}$

$B \xrightarrow[(ii)Zn/H_2O]{(i)O_3} C + D$



$D \xrightarrow[\Delta]{Ba(OH)_2}H_3C-\overset{\underset{|}{C}H_3}{C} = CH - \overset{\underset{||}{O}}{C}-CH_3$

• A.
• B.
• C. $C _6 H _5- CH _2- \overset{\underset{||}{O}}C - CH _3$
• D.

Answer 1

Answer: (C)

$Ph - \overset{\underset{||}{O}}{C} \xrightarrow[NaOH (\text{cannizaro})]{\text{Conc.}}Ph - COOK + Ph-CH_2OH$