Question

The product $A,BandC$ in the reaction sequence,

$(H-\underset{\underset{O}{\parallel }}{C}-O{)}_{2}Ca\stackrel{\Delta }{⟶}A\stackrel{NaOH}{⟶}B\underset{{360}^{\circ }}{\stackrel{Heat}{⟶}}C$ are

• A. $HCHO,HCOONa,C{H}_{3}OH$
• B. $HCHO,N{a}_{2}C{O}_3,NaHC{O}_3$
• C. $HCHO,HCOONa,(COONa{)}_2$
• D. $HCHO,HCOONa,N{a}_{2}C{O}_3$

Answer 1

The correct option is C $HCHO,HCOONa,(COONa{)}_2$

The reaction sequence is
$(H-\underset{\underset{O}{\parallel }}{C}-O{)}_{2}Ca\stackrel{\Delta }{⟶}\underset{A}{HCHO}\stackrel{NaOH}{⟶}\underset{B}{HCOONa}\underset{{360}^{\circ }}{\stackrel{Heat}{⟶}}\underset{C}{(COONa{)}_2}$.
Thus, the product $A,BandC$ are $HCHO,HCOONa,(COONa{)}_2$.
Dry distillation of calcium formate gives formaldehyde (compound A).
Formaldehyde reacts with NaOH to form sodium formate (compound B).
Sodium formate on heating at ${360}^0$ forms disodium oxalate (compound C).