Question

In the given reaction
$C_6{H}_5-\stackrel{O}{\stackrel{||}{C}}-\stackrel{O}{\stackrel{||}{C}}-C_6{H}_5\frac{\left(i\right)C{H}_{3}MgBr\left(excess\right)}{\left(iii\right)HOH/H^{@}}Product$

Product will be:

$\left(a\right)C_6{H}_{5}C-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C_6{H}_5$

$\left(b\right)C_6{H}_{5}C-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-\stackrel{O}{\stackrel{|}{C}}-C_6{H}_5$
$\left(c\right)C_6{H}_5-COOH$
$\left(d\right)C_6{H}_5-CHOH-CHOH-C_6{H}_5$

• A. a
• B. b
• C. c
• D. d

Answer 1

The correct option is A a

$C_6{H}_5-\stackrel{O}{\stackrel{||}{C}}-\stackrel{O}{\stackrel{||}{C}}-C_6{H}_5\stackrel{C{H}_3{H}_5\left(excess\right)}{\to }{C}_6{H}_5-\stackrel{O^{-}}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-\stackrel{O^{-}}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C_6{H}_5\stackrel{H_3{O}^{-}}{\to }{C}_6{H}_5-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C_6{H}_5$