Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Please explain the addition of carbon atom as per markovnikov’s rule to primary, secondary or tertiary positions.

Please explain the addition of carbon atom as per markovnikov’s rule to primary, secondary or tertiary positions.

Grade:12

1 Answers

Vikas TU
14149 Points
4 years ago

The rule states that with the addition of a protic acid HX to an unsymmetrical alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and thehalide (X) group becomes attached to the carbon with more alkyl substituents. Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greater number of hydrogen atoms while the X component is added to the carbon with the fewer number of hydrogen atoms.

The same is true when an alkene reacts with water in an addition reaction to form an alcohol which involve formation of carbocations. The hydroxyl group (OH) bonds to the carbon that has the greater number of carbon–carbon bonds, while the hydrogen bonds to the carbon on the other end of the double bond, that has more carbon–hydrogen bonds.

The chemical basis for Markovnikov's Rule is the formation of the most stable carbocation during the addition process. The addition of the hydrogen ion to one carbon atom in the alkene creates a positive charge on the other carbon, forming a carbocation intermediate. The more substituted the carbocation, the more stable it is, due to induction andhyperconjugation. The major product of the addition reaction will be the one formed from the more stable intermediate. Therefore, the major product of the addition of HX (where X is some atom more electronegative than H) to an alkene has the hydrogen atom in the less substituted position and X in the more substituted position. But the other less substituted, less stable carbocation will still be formed at some concentration, and will proceed to be the minor product with the opposite, conjugate attachment of X.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free