Question

Convert:
PhCHO into PhCH = CHCOPh .

• A. CH${}_3$MgI, H${}_3$O${}^{+}$, H${}^{+}$/K${}_2$Cr${}_2$O${}_7$, PhCHO/$\stackrel{⊝}{O}$H, $\Delta$
• B. CH${}_3$MgI, H$O^{-}$, H${}^{+}$/K${}_2$Cr${}_3$O${}_6$, PhCHO/$\stackrel{⊝}{O}H,\Delta$
• C. CH${}_3$MgBr, H${}_3$O${}^{+}$, H${}^{+}$/K${}_2$Cr${}_4$O${}_7$, PhCHO/$\stackrel{⊝}{O}$H, $\Delta$
• D. None of these

Answer 1

The correct option is A CH${}_3$MgI, H${}_3$O${}^{+}$, H${}^{+}$/K${}_2$Cr${}_2$O${}_7$, PhCHO/$\stackrel{⊝}{O}$H, $\Delta$

Benzaldehyde reacts with $C{H}_{3}MgI,H_3{O}^{+}$ to form $PhCH\left(OH\right)\left(C{H}_3\right)$, then it reacts with potassium dichromate in acidic medium to form aceto phenone, then it reacts with benzaldehyde in presence of alkali to form aldol product.

$PhCHO+C{H}_{3}MgI\stackrel{H_3{O}^{+}}{\to }PhCH\left(OH\right)\left(C{H}_3\right)\stackrel{K_{2}C{r}_2{O}_7/H^{+}}{\to }PhCOC{H}_3$

$PhCOC{H}_3\stackrel{PhCHO,O{H}^{-}}{\to }PhCOCH=CHPh$

So the answer is A.