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In the following compounds , we have to find order of compounds of their reactivity towards SN1 reaction. enter image description here

According to me , we should compare carbo cation stability which is formed as an intermediate.

So the 4th compound (tertiary butyl chloride) should be in top in order of reactivity.

But how to compare other three compounds.

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  • $\begingroup$ @Freddy the 1,2,3 compounds are 1°, now how to compare. $\endgroup$ – Koolman May 12 '16 at 0:12
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First of all, why the carbocation of fourth one is more stable. The major factor is hyperconjugation. Greater the number of hydrogen atoms on beta-carbons, greater is the stability achieved, as there are more resonance structures possible.

Therefore, 3rd one is least stable among. Likewise, you can easily point out that 1st one is more stable than the 2nd one.

Addition: After reading comments and other posts, I would like to share an image. enter image description here

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  • $\begingroup$ But in third after carbocation rearrangement (methyl shift) it will become mote stable. $\endgroup$ – Koolman May 12 '16 at 0:10
  • $\begingroup$ yes. however that requires bond breaking. $\endgroup$ – adianadiadi May 12 '16 at 6:12
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The (4th) would be the first. Then the third (explanation ) since methide shift is a preliminary property of a carbocation(C+) therefore the carbocation would do the same and become a 3 degree hense would be second.

Simillar is for the secon one here hydride shift comes into effect and thus carbocation becomes 2 degree

FINAL ANS.4>3>2>1.

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    $\begingroup$ Bond breaking is involved in alkyl or hydride shift, that means one more extra step. Thus reactivity is also affected. $\endgroup$ – adianadiadi May 12 '16 at 6:13
  • $\begingroup$ THE shift is a step definately but it is a fast step because carbocation approach to a more stable state the RDS(rate determining step) in all cases would be the formation of carbocation initially . $\endgroup$ – Mukhar Jain May 12 '16 at 17:04
  • $\begingroup$ The shift requires breaking of C-H bond and it requires more energy and hence requires more time.... $\endgroup$ – adianadiadi May 12 '16 at 18:06
  • $\begingroup$ We ahould not see the thermodynamics when there is a posibility of 1->2 or 2->3 degree carbocation formation i have been taught this in my coaching $\endgroup$ – Mukhar Jain May 12 '16 at 18:30
  • $\begingroup$ so you think there is no relation between thermodynamics and kinetics? Then why more stable tertiary carbocation is formed easily? I suggest you to think rather following what is taught in coaching. $\endgroup$ – adianadiadi May 13 '16 at 12:36

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