Question

$C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}-}CH=C{H}_2\stackrel{dil.H_{2}S{O}_4}{\to }\underset{\Delta }{\overset{conc.H_{3}P{O}_4}{\to }}\underset{\left(ii\right)Zn/H_{2}O}{\overset{\left(i\right)O_3}{\to }}$

Final product is / are:

• A. $\underset{C{H}_3}{\underset{|}{C}=}O$
  
  
• B. $HCHO+C{H}_3-\stackrel{O}{\stackrel{||}{C}}-\underset{C{H}_3}{\underset{|}{C}H}-C{H}_3$
  
• C. $C{H}_{3}CHO+$ $C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3$
• D. $C{H}_{3}CHO+HCHO$

Answer 1

Answer: (A)

$\underset{C{H}_3}{\underset{|}{C}=}O$

$C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}-}CH=C{H}_2\stackrel{dil.H_{2}S{O}_4}{\to }C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}-}\underset{OH}{\underset{|}{C}H}-C{H}_3\underset{\Delta }{\overset{conc.H_{3}P{O}_4}{\to }}C{H}_3-\underset{C{H}_3}{\underset{|}{C}=}\underset{C{H}_3}{\underset{|}{C}-}C{H}_3$

$\underset{\left(ii\right)Zn/H_{2}O}{\overset{\left(i\right)O_3}{\to }}C{H}_3-\underset{C{H}_3}{\underset{|}{C}=}O$

Here first $2^0$ alcohol forms as a product which forms symmetrical alkene as secondary product.
Then this alkene in presence of ozone forms ketone as a product.