In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _...

In the given reaction

{CH}_{3} - COOH\to

\begin{array}{ll} \left(ii\right)NaCN \\ ( iii ) {H}_{2}O / H^{⊕}\ \end{array}

\left(i\right)

Red - P + ${Br}_{2}\left(X\right)
'X' will be $In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _ 2 O / H^ ⊕ \ array (i) Red - P + Br _ 2 (X) 'X' will$ $In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _ 2 O / H^ ⊕ \ array (i) Red - P + Br _ 2 (X) 'X' will$ $In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _ 2 O / H^ ⊕ \ array (i) Red - P + Br _ 2 (X) 'X' will$ $In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _ 2 O / H^ ⊕ \ array (i) Red - P + Br _ 2 (X) 'X' will$

Options

A
B $COOH - C{H}_{2} - C{H}_{2} - COOH$
C
D

Correct answer

A. Match List - I with List - II. List - I (Chemical Reaction) List - II (Reagent used) a CH _ 3 COOCH _ 2 CH _ 3 → CH _ 3

Step-by-step solution

Here the final product is malonic acid
$In the given reaction CH _ 3 - COOH→ array ll (ii)NaCN ( iii ) H _ 2 O / H^ ⊕ \ array (i) Red - P + Br _ 2 (X) 'X' will$

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