Question

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $1^{\circ },2^{\circ },3^{\circ }$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
$\underset{\left(1600\right)}{3^{\circ }}>\underset{\left(82\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
The relative rate of abstraction of hydrogen by Cl is
$\underset{\left(5\right)}{3^{\circ }}>\underset{\left(3.8\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
G. How many dichloro products (including stereoisomers) will be formed when R - 2 - chloropentane reacts with $C{l}_2$ in presence of UV radiation ?

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is C 7

Given compound is $R-2-Chloropentane$ which is shown in $C-C-C-\stackrel{Cl}{\stackrel{|}{C}}-C$

To find dichlolo stereoisomers.

Products on reaction of $C{l}_2$ in $h\nu$ are shown in

$C-C-C-\stackrel{Cl}{\stackrel{|}{\underset{\ast }{C}}}-\stackrel{Cl}{\stackrel{|}{C}}$ $C-C-\stackrel{Cl}{\stackrel{|}{\underset{\ast }{C}}}-\stackrel{Cl}{\stackrel{|}{C}}-C$ $C-\stackrel{Cl}{\stackrel{|}{C}}-C-\stackrel{Cl}{\stackrel{|}{C}}-C$ $\underset{Cl}{\underset{|}{C}}-C-C-\underset{Cl}{\underset{|}{\stackrel{\ast }{C}}}-C$

$C-C-C-\underset{Cl}{\underset{|}{\stackrel{Cl}{\stackrel{|}{C}}}}-C$

Hence, $8$ stereoisomers are formed.