explain the law of marconix
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Macronix International is an integrated device manufacturer in the non-volatile memory (NVM) market. The company manufactures NOR Flash, NAND Flash, and ROM products for the consumer, communication, computing, automotive, and networking markets. Its headquarters are located in Taiwan,
Macronix International Co., Ltd.
旺宏電子股份有限公司
Type
PublicTraded asTWSE: 2337IndustrySemiconductorsFounded1989HeadquartersTaiwanWebsitewww.macronix.com 
.
hope this may help u✌️✌️✌️✌️
Macronix International Co., Ltd.
旺宏電子股份有限公司
Type
PublicTraded asTWSE: 2337IndustrySemiconductorsFounded1989HeadquartersTaiwanWebsitewww.macronix.com 
.
hope this may help u✌️✌️✌️✌️
pavanjadhao347:
sorry but this is not right answer
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hey mate here is your answer
The rule states that with the addition of a protic acid HX to an asymmetricalkene, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets 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 greatest number of hydrogen atoms while the X component is added to the carbon with the least number of hydrogen atoms.[3]
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 stablecarbocation 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 and hyperconjugation. 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.
Hope this will help you
The rule states that with the addition of a protic acid HX to an asymmetricalkene, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets 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 greatest number of hydrogen atoms while the X component is added to the carbon with the least number of hydrogen atoms.[3]
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 stablecarbocation 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 and hyperconjugation. 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.
Hope this will help you
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