Question

In the given reaction two products are expected.
$C{{H}_{3}}CH=C{{H}_{2}}+HBr\to C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br(A)+C{{H}_{3}}-CH(Br)-C{{H}_{3}}$
A) The carbocation \[C{{H}_{3}}-CH\left( - \right)-C{{H}_{3}}\] is formed which is more stable.
B) The carbocation \[C{{H}_{3}}-C{{H}_{2}}-\underset{{}}{\overset{+}{\mathop{C{{H}_{3}}}}}\,\] is formed which is more stable.
C) Both carbons are equally stable but the nucleophile attacks on the central atom.
D) \[C{{H}_{3}}-CH\left( - \right)-C{{H}_{3}}\] can easily gives proton to $B{{r}^{-}}.$

Answer 1

Hint: We know that in the reaction in which you have given any cyanide and water keep in mind that these conditions are counted for the increase of one carbon atom. So, the number of carbons increases in the product. Try to solve firstly by applying the Markovnikov rule as the addition of hydrogen bromide

Complete step-by-step answer:
Reaction given in the question is generally based on Markovnikov's rule which is also known by the name Markownikoff’s rule. This rule is used in organic chemistry to know the product formed during the addition reaction of alkenes.

We have a double-bonded compound as a substrate in which we have to add hydrogen bromide in such a way that Markovnikov addition takes place. In the Markovnikov addition rule, it says that in a doubly bonded compound the negative part of the attacking reagent gets attached to the carbon atom having a lesser number of hydrogen.

Here, addition of hydrogen bromide is in such a way that the negative part which is bromide here gets attached to the carbon center having lesser hydrogen atoms. This step is proved when we see its mechanism where an attack of electrophile takes place. The hydrogen ion is an electrophile by which the carbocation forms is stable. The reaction is given as:
Step I: \[HBr\xrightarrow{{}}\underset{(Electrophile)}{\mathop{{{H}^{+}}}}\,+B{{r}^{-}}\]
Step II: $C{{H}_{3}}-CH=C{{H}_{2}}+{{H}^{+}}\xrightarrow{{}}\underset{(MoreStable)}{\mathop{C{{H}_{3}}\overset{+}{\mathop{C}}\,HC{{H}_{3}}}}\,+\underset{(LessStable)}{\mathop{C{{H}_{3}}C{{H}_{2}}\overset{+}{\mathop{C}}\,{{H}_{2}}}}\,$
Step II: $C{{H}_{3}}\overset{+}{\mathop{C}}\,HC{{H}_{3}}+B{{r}^{-}}\xrightarrow{{}}\underset{(Major)}{\mathop{C{{H}_{3}}-CH(Br)-C{{H}_{3}}}}\,$
As Secondary carbocation is more stable than that of primary carbocation the carbocation \[C{{H}_{3}}-CH\left( - \right)-C{{H}_{3}}\]​ is formed which is more stable .

Hence, the correct option is A.

Note: Remember that here, the addition of hydrogen bromide is in such a way that the negative part which is bromide here gets attached to the carbon center having lesser hydrogen atoms. This step is proved when we see its mechanism where an attack of electrophile takes place. Hydrogen ion is an electrophile by which the carbocation forms is stable.