Question

$HC\equiv CH\stackrel{2C{H}_{3}MgBr}{⟶}\left(A\right)\underset{\left(ii\right)HOH/H^{\oplus }}{\overset{\left(2\right)2C{O}_2}{\to }}\left(B\right)\underset{HgS{O}_4}{\overset{H_{2}S{O}_4}{\to }}\left(C\right)\stackrel{\Delta }{\to }\left(D\right)$
If the Molecular weight of D is $xgm$, then write your answer as $\frac{x}{2}$

Answer 1

$HC\equiv CH\underset{-2C{H}_4}{\overset{2C{H}_{3}MgBr}{\to }}BrMg-C\equiv C-MgBr\underset{\left(ii\right)H_{2}O/H^{\oplus }}{\overset{\left(i\right)2C{O}_2}{\to }}HOOC-C\equiv C-COOH{H}_{2}S{O}_4\downarrow HgS{O}_{4}HOOC-\stackrel{\stackrel{O}{||}}{C}-C{H}_3\underset{\Delta }{\overset{-C{O}_2}{\leftarrow }}HOOC-\stackrel{\beta }{\underset{\underset{O}{||}}{C}}-\stackrel{\alpha }{C{H}_2}-COOH$

Molecular weight of the product (x) = 88
$\therefore \frac{x}{2}=44$