Question

Among the isomeric alkanes of molecular formula $C_5{H}_{12}$, identify the one that on photochemical chlorination yields:
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.

Answer 1

$i)$ To have a single mono-chloride , there should be only one type of $H-atom$. This is because, replacement of any $H-atom$ leads to the formation of same product.

$\therefore$ The isomer $C_5{H}_{12}$ that leads to a single mono-chloride is neo-pentane.

$C{H}_3-\stackrel{\underset{|}{C{H}_3}}{\underset{\stackrel{|}{C{H}_3}}{C}}-C{H}_3$

$ii)$ To have $3$ isomeric mono-chlorides, the isomer should contain $3$ different types of $H-atoms.$

$\therefore$ The isomer should be $n-pentane.$

$\stackrel{c}{C{H}_3}-\stackrel{b}{C{H}_2}-\stackrel{a}{C{H}_2}-\stackrel{b}{C{H}_2}-\stackrel{c}{C{H}_3}$

$iii)$ To have $4$ isomeric mono-chlorides, the isomer should contain $4$ different types of $H-atoms.$

$\therefore$ The isomer is $2-methylbutane.$