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Dehydrohalogenation involves removal of ...

Dehydrohalogenation involves removal of the halogenn atom together with a hydrogen atom from carbon adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product.
Q. The ease of dehydrohalogenation for different halogens is in the order

A

dehydrogenation

B

dehalogenation

C

dehydration

D

dehydrohalogenation

Text Solution

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The correct Answer is:
D
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Dehydrohalogenation involves removal of the halogenn atom together with a hydrogen atom from carbon adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product. Q. What are the products of dehydrohalogenation of 2-iodopentane?

Dehydrohalogenation involves removal of the halogenn atom together with a hydrogen atom from carbon adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product. Q. Arrange the following alkyl halides in decreasing order of the rate of beta -elimination reaction with alcoholik KOH. underset((i))(CH_(3)CH_(2)CH_(2)Br) underset((ii))(CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)Br) underset((iii))(CH_(3)-CH_(2)-underset(Br)underset(|)(C)H-CH_(3))

J.C. Slater proposed an empirical constant that represents the cumulative extent to which the other electrons of an atom shield (or screen) any particular electron from the nuclear charge. Thus, slater's screening contant sigma is as : Z^(**)=Z-sigma Here, Z is the atomic number of the atom, and hence is equal to the actual number of protons in the atom. the parameter Z^(**) is the effective nuclear charge, which according to is smaller than Z, since the electron in question is screened (shielded) from Z by an amount sigma . Conversely, an electron that is well shielded from the nuclear charge Z experiences a small effective nuclear charge Z^(**) . The value of sigma for any one electron in a given electron configuration (i.e., in the presence of the other electrons of the atom in question) is calculated using a set of empirical rules developed by slater. according to these rules, the value of sigma for the electron in question is the cumulative total provided by the various other electrons of the atom. Q. According to Slater's rule, order of effective nuclear charge (Z^(**)) for last electron in case of Li, Na and K.

It we see the reaction of methane with halogen, the rate determining step for chlorination is, endothermic reaction of the chlorine atom with methane to form methyl radical and a molecule of HCl. So free radical is the intermediate of the reaction. Formation of free radical depends upon the energy required to break a bond between a hydrogen atom and a carbon atom. Chlorination of propane and Bromination of propane. when compared it is found that bromination is more selective than chlorination. The probability factor for 3^(@),2^(@),1^(@)H atom is 5.0:3.8:1.0 at 25^(@)C for chlorination. Isobutane when reacts with chlorine in presence of ultra violet radiations yield 2 products primary hydrogen substituted and 3^(@) hydrogen substituted Find their % in product mixture

Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with non metals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different. The oxidation state of the most electronegative element in the products of the reaction between BaO_(2) and H_(2)SO_(4) are

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarrZ . Z is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. Y is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarrZ . Z is :

Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)) . The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product.

Comprehension given below is followed by some multiple choice question, Each question has one correct options. Choose the correct option. Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and hte energy of the bonding orbital is lowered than the parent atomic orbitals. energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order sigma1s lt sigma^(star)1s lt sigma^(star)2s lt ((pi2p_(x))=(pi2p_(y))) lt sigma2p_(z) lt (pi^(star)2p_(x) = pi^(star)2p_(y)) lt sigma^(star)2p_(z) and For oxygen and fluorine order of enregy of molecules orbitals is given below. sigma1s lt sigma^(star)1s lt sigma2s lt sigma^(star)2s lt sigmap_(z) lt (pi2p_(x) ~~ pi2p_(y)) lt (pi^(star)2p_(x)~~ pi^(star)2py) lt sigma^(star)2p_(z) Different atomic orbitalsof one atom combine with those atoms orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called sigma, sigma andif the overlap is lateral, the molecular orbital is called pi, pi . The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds. 67) Which of the following pair is expected to have the same bonod order?

ICSE-HYDROCARBONS -OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS
  1. Liquid hydrocarbon is converted to a mixture of gaseous hydrocarbons b...

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  2. Bromoethane on treatment with alcoholic KOH gives

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  3. Dehydrohalogenation involves removal of the halogenn atom together wit...

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  4. The compound that decolourises alk. KMnO4 is

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  5. The product formed when but-1-ene is subjected to HBr in the presence ...

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  6. CH2 = CHCl reacts with HCl to form

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  7. The final products formed by ozonolysis of compound R - CH = CR2 is

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  8. What is Baeyer's reagent ?

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  9. Dilute aqueous KMnO4 at room temperature reacts with R-CH=CH-R to giv...

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  10. Acetylene can be prepared from

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  11. Number of acidic hydrogen atoms in but-1-yne is

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  12. Which of the following reacts with metal by displacing the hydrogen at...

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  13. In the above reaction X is

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  14. Which one of the following reagents distinguish ethylene from acetylen...

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  15. A hydrocarbon that reacts with sodium in liquid NH3 is

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  16. A metallic carbide on treatment with water gives a colourless gas whic...

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  17. In alkenes, pi -electrons forming carbon-carbon pi-bond are

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  18. The process C7H16 underset(670K) overset(Pt)(to) C6H5CH3 + 4H2 is a

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  19. Which of the following when used as a fuel have the maximum tendency t...

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  20. The quality of diesel is expressed in terms of

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