Question

The major product of the following addition reaction is
\[C{{H}_{3}}-CH=C{{H}_{2}}\xrightarrow{C{{l}_{2}}/{{H}_{2}}O}\]
A.

B.

C.

D.

Answer 1

Hint:. Whenever addition reaction is going to happen on alkenes then always the stable form of carbocation formed during the reaction should be either tertiary or secondary carbocation. Formation of primary carbocation during the reaction does not favor the formation of the addition product.

Complete step by step answer:
- In the given question chlorine and water are participating in addition reaction with propene.
- Propene is an unsaturated compound.
- The product formed during the reaction of chlorine and water with propene is as follows.

- The proper mechanism of the above reaction is as follows.

- The above reaction contains three steps.
- In the first step the double bond present in propene reacts with chlorine and forms a three membered ring as the product.
- In the second step the lone pair of electrons present on the oxygen atom of the water molecule reacts with secondary carbon which is present in the compound and cleaves the three membered and forms a stable product after losing a hydrogen atom in third step.
- Therefore the product forms after reaction chlorine and water with propene is 1-chloro-2-propanol.
So, the correct answer is “Option B”.

Note: The unsaturated compound (compound containing double or triple bond in its structure) is converted into a saturated compound at the end of the reaction after reacting with chlorine and water. Then this reaction is called addition reaction.
Conversion of saturated compounds to unsaturated compounds is called an elimination reaction.