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More substituted alkene should be major product while the less substituted as minor product as it is E1 mechanism.

But in my answer sheet it is given reverse, i.e. major as less substituted and minor as more substituted by giving the reason that it is happening because of steric hindrance of tert-butyl group. But I am confused now that it should not be the case because there is no role of steric hindrance in E1 mechanism, and answer should be as usual.

Please help me in this question with other reason. Reaction image

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    $\begingroup$ Perhaps, you could provide the name of the base that is used for the elimination reaction. The size of the base is also crucial in allowing us to determine whether the Zaitsev product or Hofmann product is formed. For example, a large, bulky base (e.g. DBU) would offer the Zaitsez product. $\endgroup$ – Tan Yong Boon Jun 24 '20 at 10:40
  • $\begingroup$ Actually base was not present in reaction..rather AcOH/∆ was written above arrow and below arrow it was written E1..can some reaction happen with this? $\endgroup$ – Romir Mehta Jun 24 '20 at 11:34
  • $\begingroup$ @TanYongBoon i added a screenshot of ques..please check it $\endgroup$ – Romir Mehta Jun 24 '20 at 11:42
  • $\begingroup$ @AniruddhaDeb is it totally experimental or any reason for this..? Other than steric effect as it does not affects E1... $\endgroup$ – Romir Mehta Jun 24 '20 at 12:08
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    $\begingroup$ @TanYongBoon This is basically solvolysis, so the solvent is the base. Furthermore, as an E1 reaction, the size of the base is significantly less important. $\endgroup$ – Zhe Jun 24 '20 at 18:40
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The major product in this case is favoured due to Steric hindrance in the product. Drawing the full structure of the molecule in the reaction makes this clearer

enter image description here

We can compare the degree of steric hindrance in the major product and the minor product by using A-values. The Wikipedia article gives the following A-values for common substituents:

$$\begin{array}{|c|c|} \hline \text{Substituent} & \text{A-value} \\ \hline \ce{-CH3} & 1.7 \\ \ce{-CH2C(CH3)3} & 2 \\ \ce{-C(CH3)3} & >4 \\ \hline \end{array}$$

Clearly, the major product has a lower set of A-values for it's substituents (2 & 1.7) compared to the minor product (4 & 1.7). Also, the minor product does not have any scope for forming stereoisomers, as both the groups on one end of the double bond are methyl. If one of them was a hydrogen group, then the E isomer would be the major product.

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  • $\begingroup$ Even clearer is if one were to draw out the hydrogen atoms on all of the methyl groups. Not suggesting that you do that in the answer... $\endgroup$ – Zhe Jun 24 '20 at 18:41
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    $\begingroup$ Even clearer if you show that there can be rotation about the C-C single bond such that the bulky t-butyl group can be positioned further away from the methyl group. $\endgroup$ – Tan Yong Boon Jun 24 '20 at 23:16

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