Take this example:

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Multiple hydride shifts are possible only if in each step, the formed intermediate is more stable than the previous one.

(1) to (2) converts $2^\circ$ carbocation to $3^\circ$ carbocation, thus increases stability.

Now, for (2), there are 5 hyperconjugating structures and for (3), there are 4 hyperconjugating structures. Therefore, (2) to (3) should not be favourable. There is a distant $\pu{+I}$ effect of $\pu{-Me}$ group in (3) but hyperconjugation dominates over inductive effect.

(3) to (4) hydride shift converts to the most stable carbocation possible in this scheme.

Then the final products (5) and (6) are obtained.

Is this the major product obtained from intermediate (2)?

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    $\begingroup$ Most probably yes because it results in the alkene with highest hyperconjugating structures and the cation won't rearrange into (3) because as you said, "...formed intermediate should be more stable than previous one." $\endgroup$ – Robin Singh Aug 12 '20 at 11:31
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    $\begingroup$ Given a lack of stereochemistry, it is difficult to assess the number of hyperconjugating C-H bonds. A difference of 4 versus 5 such bonds may imply an equilibrium. In your scheme, why is 5 the major product and not 6? I'd opt for your last structure as the product from cation 2. Please avoid light green. Tough to read. $\endgroup$ – user55119 Aug 12 '20 at 13:05
  • $\begingroup$ @user55119 (a) What can be done (stereochemistry) to make it a simple problem? (b) Are you suggesting that 4 vs. 5 hyperconjugating structure is not much to decide the stability of intermediate? (c) In 6, one bridged carbon is sp2 and the other is sp3 hybridised carbon, thus, more strain than 5? Though, 6 is most substituted product. $\endgroup$ – Apurvium Aug 12 '20 at 16:52
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    $\begingroup$ (a) Stereochemistry may have an effect on rates of rearrangement. (b) You may be overcounting C-H hyperconjugation of methylenes with vicinal cations. Hard to have both C-H bonds overlapping the cation at the same time. (c) Whether 5 or 6 is more stable may depend on whether or not 5 has a cis or trans ring fusion. $\endgroup$ – user55119 Aug 12 '20 at 20:17

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