Can some one please explain alkyl shifts for me

Question

Provide a mechanism for the following reaction. Include all intermediates, resonance structures, charges, and electron pushing arrows to obtain full credit.

I’m stuck with this problem. I got as far as step 2 of the mechanism. I know the double bond acts as a nucleophile and attacks one of the hydrogen on sulfuric acid thus creating its conjugated base $\ce{HSO4-}$ leaving a carbocation on carbon 1.

In previous problems I have done the carbocation will rearrange to form a more stable cation. I’m told it’s done by either a hydride shift (which I understand) or an alkyl shift. I’m having a hard time grasping the concept of a six-membered ring going into a five-membered ring. And essentially understanding why there is a shift.

Answer 1

The reaction of the ketal 1 with aqueous H₂SO₄ is more likely to follow the pathway 1 --> 2 --> 3 to form α,β-unsaturated ketone 3. This is a typical ketal hydrolysis in acid medium. For the proposed reaction pathway 1 --> 4 --> 5 --> 6, species 6 is unstable. Indeed, if it were formed as an intermediate, it would afford ester 7a and ultimately the carboxylic acid 7b. Is there a possibility that the actual reactant doesn't bear methoxy groups but methyl groups? If this is the case, then the reaction 8 --> 9 is more reasonable.

Answer 2

• Many organic reactions' steps may be described by equilibria. In other words, at least on microscopic scale, both «forward» and «backward reaction» is possible. So a homework for you is to become fluent in this reversibility. As an example relevant here:

• I'm not aware about an better approach than to learn a handfull of rearrangement reactions till you recognize «typical pattern», e.g. molecular substrates which are more likely than others to undergo this type of chemistry. Here, the formation of the product bears similarity to the Pinacol rearrarangement. As already identified by yourself, an intermediate carbocation is needed to trigger the cascade like,

  and to memorize larger alkyly and phenyl groups are more likely to move, than methyl groups (e.g., Beckmann rearrangement). In early exercises it helped me to recall that each single dash not only represents to carbon atoms connected with each other, but equally a pair of two electrons forming a bond between these two atoms in questions. Depending on teacher and school, a drawing like the following may be permissible until creating enough «muscle memory»:

  Probably soon you will learn that drawing a bent arrow anyway symbolizes the movement of two electrons (contrasting to the fish hooks to symbolize moves of single electrons) which renders drawing to dots for the two electrons in question frequently redundant.

• The formation of six- and five-membered carbocyles is thermodynamically favourable. If a six-membered ring interconverts into a five-membered carbocycle might be of lesser importance for your current level of study, than to recognize patterns typical for this kind of reaction. It depends both on thermodynamics as well as kinetics, and over time you develop a reasoning when one, or the other favours the starting materials, or the products. In some way, it is like learning grammar of a foreign language (e.g., I speak, you speak, s/he speaks, we speak, you speak, they speak). As such, keep speaking Chemistry.