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`,

Arrange the folliwing compounds in increasing order of SN^(-1) 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

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

Arrange the compounds in the order of increasing S_(N)1 reactivity: ClCH_(2)CH=CHCH_(2)CH_(3) (I), CH_(3)C Cl=CHCH_(2)CH_(3) (II) and CH_(3)CH = CHCH_(2)CH_(2)Cl (III)

Arrange the following halides in the decreasing order of S_(N^(1)) reactivity : underset((I))(CH_(3)CH_(2)CH_(2)Cl), underset((II)) (CH_(2)=CHCH(Cl)CH_(3)), underset((III))(CH_(3)CH_(2)CH(Cl)CH_(3))

Arrange the following compounds in the order of increasing S_(N)2 reactivity and explain the order: CH_(3)CH_(2)CH_(2)Br, CH_(3)CH_(3)Br, (CH_3)_(3)C CH_(2)Br, (CH_3)_(2)CHCH_(2)Br.