Question

$Ca{C}_2\stackrel{H_{2}O}{\to }A\underset{N{H}_{4}Cl}{\stackrel{C{u}_{2}C{l}_2}{\to }}B\underset{\underset{\left(3\right)C{O}_2/H_{2}O/H^{+}}{\left(2\right)Mg/ether}}{\stackrel{\left(1\right)1eqv.HCl}{}}C$

In above reaction product $\left(C\right)$ is?

• A. $C{H}_2=CH-COOH$
• B. $HC\equiv C-COOH$
• C. $HOOC-C\equiv C-COOH$
• D. $C{H}_2=CH-\underset{\underset{COOH}{|}}{C}=C{H}_2$

Answer 1

The correct option is A $C{H}_2=CH-COOH$

$Ca{C}_2\stackrel{H_{2}O}{\to }\underset{A}{CH\equiv CH}\underset{N{H}_{4}Cl}{\stackrel{C{u}_{2}C{l}_2}{\to }}\underset{B}{CH\equiv C^{-}C{u}^{+}}\stackrel{1equivHCl}{\to }CH\equiv C-Cl\stackrel{Mg/ether}{\to }CH\equiv CMgCl\stackrel{C{O}_2/H_{2}O/H^{+}}{\to }C{H}_2=CH-COOH$

$Ca{C}_2$ on hydrolysis gives $C_2{H}_2$ which on reacting with $C{u}_{2}C{l}_2$ in $N{H}_{4}Cl$ forms vinyl acetylene. Vinyl acetylene with 1 eqv. $HCl$ undergoes electrophilic addition reaction giving chloroprene which reacts with $Mg/ether$ gives a grignard reagent $\left(RMgX\right)$ followed by carboxylation giving carboxylic acid.
Option A is correct.