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Predict all order of reactivity of the following compounds in dehydrohalogenation.
a,. I. `CH_(3)CH_(2)CH_(2)CH_(2)Cl`
II. `(CH_(3))_(2)CHCH_(2)Cl`
III. `(CH_(3))_(2)CH - CH_(2)Br`
IV. `CH_(3)CH(Br)CH_(2)CH_(3)`
V. `(CH_(3))_(3) C - Br`
b. I. `CH_(3)CH(Br)CH_(3)`
II. `CH_(3)CH_(2)CH_(2)Br`
III. `(CH_(3))_(2) CH - CH_(2) Br`
IV. `(CH_(3))_(3)C - CH_(2) Br`

Text Solution

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Ease fo dehdrohalogenation is due to the formation of carbocation and the stability of carbocation is `3^(@) gt 2^(@) gt 1^(@)`, and case of fornation of carbocation is `(R-1) gt (R-Br) gt (R-Cl)`
So the order of dehydrogenation is `(V) gt (III) gt (II) gt (IV) gt (I)`.
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Predict the order of reactivity of the following compounds in dehydrohalogenation: (a) CH_(3)CH_(2)CH_(2)CH_(2)Cl, (CH_(3))_(2)CHCH_(2)Cl, (CH_(3))_(2)CH-CH_(2)Br, CH_(3)CH(Br)CH_(2)CH_(3), (CH_(3))_(3)C-Br (b) CH_(3)CH(Br)CH_(3), CH_(3)CH_(2)CH_(2)Br, (CH_(3))_(2)CHCH_(2)Br, (CH_(3))_(3)C CH_(2)Br

Arrange the following compounds in increasing order of S_(N^(1)) reactivity. (a). (I). ClCH_(2)CH=CHCH_(2)CH_(3) , (II). CH_(3)C(Cl)=CHCH_(2)CH_(3) , (III). CH_(3)CH=CHCH_(2)CH_(2)Cl (b). (I). CH_(3)CH_(2)Br , (II). CH_(2)=CHCH(Br)CH_(3) , (III). CH_(2)=CHBr, (IV). CH_(3)CH(Br)CH_(3) (c). (I). (CH_(3))_(3)CBr , (II). (CH_(3))_(2)CHBr , (III). CH_(3)CH_(2)CH_(2)Br ,

Give the decreasing order of reactivity of the following alkyl halides in the Williamson's reaction. i. (CH_(3))_(3)C-CH_(2)Br ii. CICH_(2)CH=CH_(2) iii. (CICH_(2)CH_(2)CH_(3) iv. BrCH_(2)CH_(2)CH_(3)

Arrange the folliwing compounds in increasing order of SN^(-2) reactivity. a. I.m ClCH_(2)CH = CHCH_(2)CH_(3) II. CH_(3)C(Cl) = CHCH_(2) CH_(3) III. CH_(3)CH = CHCH_(2)CH_(2)Cl IV. CH_(3)CH = CHCH_(2)(Cl) CH_(3) b. I. CH_(3)CH_(2)Br II. CH_(2) = CHCH(Br) CH_(3) III. CH_(2) = CHBr IV. CH_(3)CH (Br) CH_(3) C. I. (CH_(3))_(3)CCl , II. C_(6)H_(5)C(CH_(3))_(2)Cl III. (CH_(3))_(2)CHCl , IV. CH_(3)CH_(2)CH_(2)Cl

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