Question

$P{h}_{2}CH-\stackrel{O}{\stackrel{||}{C}}-H\stackrel{aqueousacid}{\to }\underset{81\%}{\left(A\right)}+\underset{2\%}{enol}+\underset{17\%}{aldehyde}$
Product (A) of given reaction will be.

• A. $Ph-\underset{Ph}{\underset{|}{C}}=CH-O$
• B. $P{h}_{2}CH-\stackrel{OH}{\stackrel{|}{C}}{H}_2$
• C. $P{h}_{2}CH-\stackrel{OH}{\stackrel{|}{C}}H-OH$
• D. $P{h}_{2}CH-\stackrel{O}{\stackrel{||}{C}}-C{H}_3$

Answer 1

Answer: (C)

$P{h}_{2}CH-\stackrel{OH}{\stackrel{|}{C}}H-OH$

In aqueous solutions at aldehydes hydrates are formed which are more stable.

$O$ $OH$ $OH$ $O$

$p{h}_{2}CH-\stackrel{||}{C}-H⟶p{h}_{2}CH-\underset{|}{\stackrel{|}{C}}-H+p{h}_2-C=\stackrel{|}{C}-H+p{h}_{2}C-\stackrel{||}{C}-H$

$OH$ $17$%

$OH$ $81$%

$\Rightarrow p{h}_{2}CH-\stackrel{|}{CH}-OH$ ; Hydrates are formed if the $\alpha -$ carbocation is stable.