What is the reason that the product of the electrophilic addition to asymmetnc a

Question

What is the reason that the product of the electrophilic addition to asymmetnc alkene occurs regioselectively, the electrophilic component of the reagent adds to the least substituted carbon of the double bond while the nucleophilic component of the reagent adds to the most substituted carbon of the double bond? Addition of 2 mol of HCI to 1-butyne would yield: The major product of the acid-catalyzed hydration of (Z)-3-methyl-3-hexene is Which of the following is the major product in the electrophilic addition of HCI to (3S)-2,3-dimethylpent-1-ene? Which of the following is the most resistant compound to the action of KMnO_4?

Explanation / Answer

1.Ans(c)
What is the reason that the product of the electrophilic addition to asymmetric alkene occurs regioselectively, the electrophilic componenet of the reagent
adds to the least substituted carbon of the double bond while the nucleophilic component of the reagent adds to the most substituted carbon of the double bond.
(c) It is formed through the most stable carbocation.
The more substituted carbon atom will have more adjacent C-H or C-C bonds attached to it, so the positive charge can be dispersed more easily by hyperconjugation of

these sigma bonds. As the charge is being dispersed, the
carbocation is stabilized and it forms quickly.

2.(e) According to Markownikoff's rule H adds to C atom having more H or least substituted.
Addition of 2 mol of HCl to 1-butyne would yield
C2H5CCH + HCl = C2H5CClCH2 + HCl = C2H5CCl2CH3

3(a) 3-methyl-3-hexanol racemic mixture of two enantiomers.
The major product of the acid-catalyzed hydration of (Z)-3-methyl-3-hexene is

C2H5-CH=C(CH3)-C2H5 + H2O = C3H7-C(CH3)OH-C2H5 + ENANTIOMER
C3 atom is chiral with 4 different substituents(C3H7,CH3,C2H5 and OH), hence it will have enantiomer
and product will be racemic mixture.

4 c.
Major product in the electrophilic addition of HCl to (3S)-2,3-dimethylpent-1-ene

C2H5-CH(CH3)-C(CH3)=CH2 + HCl = C2H5-CH(CH3)-CCl(CH3)-CH3
There are two stereocenters at C2 and C3.
So the major products will be racemic mixture of two enantiomers of
2-chloro-2,3-dimethylpentane

5.b.
Resistance to KMnO4
pentane having no unsaturation is most resistant compound to the action of KMnO4
Bayer test is done to detect the unsaturated compounds.

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