Question

Markovnikov’s rule is applicable to:
a.) - C$_3$H$_6$ + HBr
b.) - C$_2$H$_4$ + HCl
c.) - C$_3$H$_6$ + Cl$_2$
d.) - C$_2$H$_4$ + HBr

Answer 1

Hint: Markovnikov’ rule is applicable to predict the regioselectivity of electrophilic addition reactions of alkenes, and alkynes. In the given options, look at the unsymmetrical alkene, or alkyne. The rule might be applied.

Complete step by step solution:
First, let us discuss the Markovnikov’s rule, it states that in the hydro-halogenation of unsymmetrical alkene, the hydrogen atom in the hydrogen halide like HBr, HF forms a bond with the doubly bonded carbon atom in the alkene having the greater number of hydrogen atoms.

Now, look at the given options, first is C$_3$H$_6$ + HBr, propene reacting with the hydrogen bromide, propene is an unsymmetrical alkene, and HBr is an unsymmetrical reagent.So, the reaction occurred by Markovnikov’s rule is

C$_3$H$_6$ + HBr $\rightarrow$ CH$_3$-CH(Br)-CH$_3$.

Here, C$_3H_6$ is CH$_3$-CH=CH$_2$.

Now, if we talk about C$_2$H$_4$ + HCl, and C$_2$H$_4$ + HBr there is an ethene molecule, it is a symmetrical alkene. So, Markovnikov’s rule cannot be applied.
Talking about the C$_3$H$_6$ + Cl$_2$, the reaction will not happen by Markovnikov’s rule as there is absence of hydrogen halide.
In the last, we can conclude that Markovnikov’s rule is applicable to C$_3$H$_6$ + HBr. As this satisfies both the condition. The correct option is (A).


Note: Don’t get confused between the symmetrical, and unsymmetrical alkenes. The symmetrical alkene is an alkene in whose molecule the doubly bonded carbons bear the same ligands, whereas the unsymmetrical alkene is an alkene in whose molecule the pair of ligands on one doubly bonded carbon is different from that on the other.