Rank the transition states that occur during the following reaction steps in order of increasing stability $(l e a s t \to m o s t s t a b l e)$
1. $H_{3} C - + O H_{2} \to C H_{3}^{+} + H_{2} O$
2. $(C H_{3})_{3} C - + O H_{2} \to - (C H_{3})_{3} C + H_{2} O$
3. $(C H_{3})_{2} C H - + O H_{2} \to (C H_{3})_{2} C H^{+} + H_{2} O$
1 < 2 < 3
2 < 3 < 1
1 < 3 < 2
2 < 1 < 3

Question

Rank the transition states that occur during the following reaction steps in order of increasing stability $(l e a s t \to m o s t s t a b l e)$
1. $H_{3} C - + O H_{2} \to C H_{3}^{+} + H_{2} O$
2. $(C H_{3})_{3} C - + O H_{2} \to - (C H_{3})_{3} C + H_{2} O$
3. $(C H_{3})_{2} C H - + O H_{2} \to (C H_{3})_{2} C H^{+} + H_{2} O$
1 < 2 < 3
2 < 3 < 1
1 < 3 < 2
2 < 1 < 3

Toppr Admin · Accepted Answer

-1- -Refer to Image 1-2- -Refer to Image 2-3- -Refer to Image 3-Now stability of carbocations- tertiary -gt- secondary -gt- primary- Therefore transition state corresponding to tertiary carbocation will be most stable- So stability order will be 1-lt-3-lt-2-

Rank the transition states that occur during the following reaction steps in order of increasing stability (least→moststable)1.H3C−+OH2→CH+3+H2O2.(CH3)3C−+OH2→−(CH3)3C+H2O3.(CH3)2CH−+OH2→(CH3)2CH++H2O1 < 2 < 32 < 3 < 11 < 3 < 22 < 1 < 3

(1) (Refer to Image 1)(2) (Refer to Image 2)(3) (Refer to Image 3)Now stability of carbocations: tertiary > secondary > primary. Therefore transition state corresponding to tertiary carbocation will be most stable. So stability order will be 1<3<2.

(1) (Refer to Image 1)
(2) (Refer to Image 2)
(3) (Refer to Image 3)
Now stability of carbocations: tertiary $>$ secondary $>$ primary. Therefore transition state corresponding to tertiary carbocation will be most stable. So stability order will be $1 < 3 < 2$ .