Question

$HC\equiv CH\underset{Pd-{BaSO}_4}{\overset{Quinoline}{\to }}A\underset{HCl}{\overset{{AlCl}_3}{\to }}B\underset{Na}{\overset{Dryether}{\to }}C$. Here, ‘C’ is :

• A. $C_2{H}_6$
• B. $C_4{H}_{10}$
• C. $C_2{H}_{5}Cl$
• D. $C_3{H}_{7}Cl$

Answer 1

Answer: (B)

$C_4{H}_{10}$

$HC\equiv CH\underset{Pd-{BaSO}_4}{\overset{Quinoline}{\to }}C{H}_2=C{H}_2\underset{HCl}{\overset{{AlCl}_3}{\to }}C{H}_{3}C{H}_{2}Cl$

Pd/BaSO4 along with quinoline is termed as Lindlar Catalyst and used to reduce Alkynes to alkenes. $2^{nd}$ step is electrophilic addition reaction and then by Wurtz reaction (in presence of dry ether/Na), it forms higher alkane.

$C{H}_{3}C{H}_{2}Cl+2Na+C{H}_{3}C{H}_{2}Cl\stackrel{dryether}{\to }{C}_4{H}_{10}+2NaCl$

Hence, option $B$ is correct.