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The question is asking for the major organic product. This seems like an $S_N1$ reaction to me as there is a weak base in $\ce{CH3OH}$ and heat. If it is, shouldn't there be 2 products. One with the $\ce{OCH3}$ in the equatorial position (shown in the answer key here) and another with the $\ce{OCH3}$ in the axial position. I think there are 2 products because the C attached to Cl is an asymmetric center, thus there will be an $\ce{OCH3}$ that is wedge and an $\ce{OCH3}$ that is dash.

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    $\begingroup$ The acetal has 2 conformations, one with the CH3O equatorial and the other with it axial. Adolph von Baeyer had a similar problem with cyclohexane carboxylic acid in a chair conformation. He assumed that there were 2 compounds, not realizing that they are in rapid equilibrium and not 2 different compounds. $\endgroup$ – user55119 May 31 '19 at 21:48
  • $\begingroup$ It is not very likely to have a weak base in the medium when there is sulphuric acid around $\endgroup$ – Raoul Kessels May 31 '19 at 22:12
  • $\begingroup$ Wait, is it because the major organic product would have OCH3 in its equatorial form as it is unstable to have OCH3 in its axial position. Even though there is some of OCH3 in its axial position conformation, more of the OCH3 in its equatorial position will be favoured? $\endgroup$ – Lat May 31 '19 at 23:26
  • $\begingroup$ The axial conformation is preferred known as the anomeric effect. pubs.acs.org/doi/pdf/10.1021/ja00437a007 $\endgroup$ – user55119 May 31 '19 at 23:38

A lone pair on the ring oxygen forces Chlorine leave. The anomeric carbon in the resulting oxocarbenium ion is sp 2 hybridized, so that part of the molecule is planar. (An oxocarbenium ion has a positive charge that is shared by a carbon and an oxygen.)

When the alcohol approaches from the top of the plane, equitorial isomer is forrmed. Similarly if alcohol apporoches from bottom of sp2 hybridised oxo carbenium ion , then ,axial isomer is formed .So the given chloro compound should give two isomers (as shown in figure below 2& 3).enter image description here.

Which is the major product ?

Substituents on a cyclohexane ring prefer the equatorial position to relieve steric strain. cyclohexane ring in chair form with substituent in equitorial position ring-flipps to chair in which the substituent is axial.

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When an endocyclic heteroatom ( oxygen, nitrogen, or sulfur) is at a position adjacent to the methoxyoxy group,the hetroatom substituent now prefers the axial orientation (despite of the associated steric strain).The equilibrium preference of most such compounds for the alpha-isomer is referred to as the anomeric effect.

enter image description here

Such behavior is explained as

  • In α-methyl glucoside, the non bonding HOMO with a pair of electrons on the ring oxygen is antiperiplanar to the antibonding LUMO of C-O bond in methoxy group. This allows hyperconjugation between them and thus by stabilizing the α-form.
  • In axial position,such interactions are possible.Whereas, in equitorial the methoxy group is at equatorial position and cannot involve in hyperconjugation since it is not antiperiplanar to the lone pair on ring oxygen. Therefore equitorial isomer is less stable than the axial isomer.

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Other Explanations:

  • The dipole interactions of the two ether functions are in the equatorial conformer are nearly parallel;while, in the axial conformer these dipoles are oriented in opposite directions.

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image source :https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd2.htm#anomr

  • Electrostatic Repulsion: Decreased electrostatic interactions are favored enter image description here

All the above combination of factors result in equitorial isomer being less stable than the axial isomer.



Organic Chemistry ,SEVENTH EDITION , Paula Yurkanis Bruice

| improve this answer | |
  • $\begingroup$ For the record, the equAtor circles the earth as do equAtorial groups cyclohexane. ;) $\endgroup$ – user55119 Jun 1 '19 at 22:35
  • $\begingroup$ "When the alcohol approaches from the top of the plane, equitorial isomer is forrmed." - Why? $\endgroup$ – Zachr Jun 2 '19 at 2:01

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