{"id":6345,"date":"2012-10-18T07:00:28","date_gmt":"2012-10-18T11:00:28","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=6345"},"modified":"2026-04-18T06:22:06","modified_gmt":"2026-04-18T11:22:06","slug":"the-e2-reaction-and-cyclohexane-rings","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2012\/10\/18\/the-e2-reaction-and-cyclohexane-rings\/","title":{"rendered":"Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings"},"content":{"rendered":"

Antiperiplanar Relationships Between C-H And The Leaving Group: The E2 Reaction and Cyclohexane Rings<\/strong><\/p>\n

Here we come to a very\u00a0testable<\/em> application of the E2 reaction – how to draw the products of E2 reactions in cyclohexane rings!<\/p>\n

\"in<\/a><\/p>\n

Table of Contents<\/strong><\/p>\n

    \n
  1. In The E2 Reaction, The Leaving Group Is Always “Anti-Periplanar” To The Hydrogen That Is Removed On The Adjacent Carbon (“Beta-Carbon”)<\/a><\/li>\n
  2. In Cyclohexane Rings, E2 Reactions Can Only Occur When The Leaving Group Is Axial<\/a><\/li>\n
  3. In E2 Reactions Of Cyclohexane Rings, The Only Way The Leaving Group And C-H Bond Can Be Anti-Periplanar Is If They Are On Opposite Faces Of The Ring<\/a><\/li>\n
  4. Some Examples: What Would Be The Major E2 Product In Each Case?<\/a><\/li>\n
  5. Cyclohexane Substituents Can Affect The Rate Of E2 Elimination Reactions In Cyclohexane Rings<\/a><\/li>\n
  6. The Higher The Concentration Of The Conformer With An Axial Leaving Group, The Faster The Elimination Will Be<\/a><\/li>\n
  7. Notes<\/a><\/li>\n
  8. Quiz Yourself!<\/a><\/li>\n
  9. (Advanced) References and Further Reading<\/a><\/li>\n<\/ol>\n
    \n

    <\/a>1. In The E2 Reaction, The Leaving Group Is Always “Anti-Periplanar” To The Hydrogen That Is Removed On The Adjacent Carbon (i.e. the “Beta-Carbon”)<\/strong><\/h2>\n

    Last time we compared the E1 and E2 reactions<\/a> and mentioned one of the key differences was the stereochemistry of the E2 reaction. Remember that in the E2, the leaving group is always “anti” to the hydrogen that is removed on the adjacent carbon. [That means that they’re directly opposed to each other, or 180\u00b0; kind of like the minute hand and the hour hand when a clock reads 6:00].<\/em><\/span><\/p>\n

    This is an extremely important detail to be able to apply in reactions.<\/p>\n

    <\/a>2. In Cyclohexane Rings, E2 Reactions Only Occur When The Leaving Group Is Axial<\/h2>\n

    One way this often comes up is in discussions of cyclohexane rings. If you’ll recall, in the cyclohexane chair conformation, groups can either be axial (pointing straight up or down) or equatorial (pointing “somewhat up” or “somewhat down”).<\/p>\n

    In order for a hydrogen to be “anti” to a leaving group, it’s required that\u00a0both groups be axial.\u00a0<\/strong>Look closely at the cyclohexane ring on the left, where the leaving group is equatorial – see how the group that is “anti” is the C-C bond [highlighed in red]?<\/p>\n

    That E2 is never gonna work!<\/p>\n

    \"in<\/p>\n

    So if you draw the leaving group equatorial in a cyclohexane chair, you’ll have to do a chair flip so that the leaving group is axial.<\/strong>\u00a0That’s shown in the right hand example, where an E2 can actually happen.<\/p>\n

    <\/a>3. In E2 Reactions Of Cyclohexane Rings, The Only Way The Leaving Group And C-H Bond Can Be Anti-Periplanar Is If They Are On Opposite Faces Of The Ring<\/strong><\/h2>\n

    This brings us to the second point. If the leaving group is, let’s say, on the “top” face of the cyclohexane, you can\u00a0only\u00a0<\/strong>form an alkene to adjacent carbons where the hydrogen is on the\u00a0opposite face.\u00a0<\/strong><\/p>\n

    You might remember \u00a0the example from last time where we couldn’t form the “Zaitsev” alkene because the Br was a wedge and there was an alkyl group on the carbon next door that was on the opposite face. In this case we can only form the less substituted alkene. If the methyl group is switched, however, then the E2 to give the Zaitsev product becomes possible:<\/p>\n

    \"e2<\/p>\n

    The bottom line here [and trust me, this comes up in tests, a lot!<\/strong><\/span>]<\/em> is that you always want to pay attention to what side of the ring your leaving group is on, and make sure that the E2 you draw is indeed possible.<\/p>\n

    <\/a>4. Some Examples: What Would Be The Major E2 Product In Each Case?<\/h2>\n

    Here are some more examples to think about. What would be the major E2 product in each case?<\/p>\n