NCERT Exercise Solution Of Chemistry (CH1-16) Class-12

(VI) Ethanol To Ethyl fluoride (VII) Bromomethane To Propanone (VIII) But-1-ene To But-2-ene (IX) 1- Chlorobutane To n-Octane (X)Benzene To Biphenyl Other Links :- You May Also Visit - 1)  https://chemiask.blogspot.com/2020/08/a-hydrocarbon-c5h10-does-not-react-with.html 2)  https://chemiask.blogspot.com/2020/08/what-are-isomers-of-compound-having.html 3)   https://chemiask.blogspot.com/2020/08/write-equations-for-prepration-of-1.html 4)  https://chemiask.blogspot.com/2020/08/what-are-ambidient-nucleophiles-explain.html 5)  https://chemiask.blogspot.com/2020/08/which-compound-in-each-of-following.html 6)  https://chemiask.blogspot.com/2020/08/predict-all-alkenes-that-would-be.html 7)  https://chemiask.blogspot.com/2020/08/how-will-you-bring-about-following.html

Here Is The Answer Of This Question - Alcoholic KOH Is Used As Dehydrogenating Agent. In Each Part Two Alkenes Are Formed , One Of Which One Is Major And Other One Is Minor. Major Alkene Is One In Which Double Bond Lies Between The Carbons Which Are Highly Substituted . Minor Alkene Is One In Which Double Bond Lies Between The Carbons Which Are Not Highly Substituted . (I) 1-Bromo-1-MethylCycloHexane (II) 2-Chloro-2-MethylButane (III) 2,2,3-Trimethyl-3-BromoPentane Other Links :- You May Also Visit - 1) https://chemiask.blogspot.com/2020/08/which-of-following-has-highest-dipole.html 2) https://chemiask.blogspot.com/2020/08/a-hydrocarbon-c5h10-does-not-react-with.html 3) https://chemiask.blogspot.com/2020/08/what-are-isomers-of-compound-having.html 4) https://chemiask.blogspot.com/2020/08/write-equations-for-prepration-of-1.html 5) https://chemiask.blogspot.com/2020/08/what-are-ambidient-nucleophiles-explain.html 6) https://chemiask.blogspot.com/202

(I)   CH3Br Or CH3I (II) (CH3)3CCl Or CH3Cl Answer(I) As   We Know , Rate Determining Factor Of Sterric Hindrances . The Less The Sterric Hindrances Are , More Faster Will Be The SN2 . In This Case Iodine Is Better Leaving Group Than Bromine Because If It's Size . So , CH3I Will React Faster In SN2 Reaction With OH . Answer(II)  As   We Know , Rate Determining Factor Of Sterric Hindrances . The Less The Sterric Hindrances Are , More Faster Will Be The SN2 . So , In Case Of CH3Cl Sterric Hindrances Are Less . So , CH3Cl Will React Faster In SN2 Reaction With OH. Other Links:- You Can Also Visit - 1)  https://chemiask.blogspot.com/2020/08/which-of-following-has-highest-dipole.html 2)  https://chemiask.blogspot.com/2020/08/a-hydrocarbon-c5h10-does-not-react-with.html 3)  https://chemiask.blogspot.com/2020/08/what-are-isomers-of-compound-having.html 4)  https://chemiask.blogspot.com/2020/08/write-equations-for-prepration-of-1.html 5)  https://chemiask.blogspot.com/2020/08/what-are-ambi

AMBIDIENT NUCLEOPHILES Are Nucleophiles Having Two Nucleophilic Sites . Thus , Ambidient Nucleophiles Have Two Sites Through Which They Can Attack . For Example , Nitrite Ion Is An Ambidient Nucleophile .                                    NITRITE ION Nitrite Ion Can Attack Through Oxygen Resulting In The Formation Of Alkyl Nitrites . Also , It Can Attack Through Nitrogen Resulting In The Formation Of Nitroalkanes.                                    ALKYLNITRITE                                  NITROALKANE OTHER LINKS :- You May Also Visit - 1) https://chemiask.blogspot.com/2020/08/which-of-following-has-highest-dipole.html 2) https://chemiask.blogspot.com/2020/08/a-hydrocarbon-c5h10-does-not-react-with.html 3) https://chemiask.blogspot.com/2020/08/what-are-isomers-of-compound-having.html 4) https://chemiask.blogspot.com/2020/08/write-equations-for-prepration-of-1.html AMAZON AD OTHER PRODUCTS YOU MAY LIKE

(I) 1-Butanol To 1-Iodobutane (II) 1-Chlorobutane To 1-Iodobutane (III) But-1-ene To 1-Iodobutane Hope It Helps!! Other Links :- You Can Also Visit - 1) https://chemiask.blogspot.com/2020/08/which-of-following-has-highest-dipole.html 2) https://chemiask.blogspot.com/2020/08/a-hydrocarbon-c5h10-does-not-react-with.html 3) https://chemiask.blogspot.com/2020/08/what-are-isomers-of-compound-having.html             AMAZON AD <

ISOMERS OF COMPOUND HAVING FORMULA C4H9Br                                1-Bromobutane                                2-Bromobutane                     1-Bromo-2-MethylPropane                     2-Bromo-2-MethylPropane Other Links :- You Can Also Visit- 1)  https://chemiask.blogspot.com/?m=1 2) https://chemiask.blogspot.com/?m=2 AMAZON ADVERTISEMENT

CCl4 is a Symmetrical Molecule . Therefore , The Dipole Moments Of All Four C-Cl Bonds Cancel Each Other . Hence , its resultant Dipole Moment Is Zero . As Shown In Figure , In CHCl3 , The resultant of Dipole Moments Of Two C-Cl Bonds Is Opposed By The Resultant Of Dipole Moments Of One C-H Bond and One C-Cl Bond . Since the resultant of one C-H Bond And One    C-Cl Bond Dipole moments is smaller than two C-Cl Bonds , the opposition is to a small extent . As a Result , CHCl3 Has A Small Dipole Moment Of 1.08 D. On The Other Hand , In Case Of CH2Cl2 , The Resultant Of The Dipole Moments Of Two  C-Cl Bonds Is Strengthened By The Resultant Of The Dipole Moments Of Two C-H Bonds . As a result , CH2Cl2 Has A Higher Dipole Moment Of 1.60 D Than ChCl3 i.e. , CH2Cl2  Has The Highest Dipole Moment . Hence , The Given Compound Can Be Arranged In The Increasing Order Of Dipole Moments As :-  CCl4<CHCl3<CH2Cl2 AMAZON ADVERTISEMENT