<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n How do I draw a cha<\/strong>ir conformation correctly?<\/strong><\/p>\n First, draw the structure<\/span> VERY BIG<\/strong>.\u00a0 This will make it much more easy to draw in the substituents.\u00a0 To draw this projection, 4 steps are needed:<\/p>\n Step 1: Draw a very flat and wide \u201cV\u201d.<\/p>\n Step 2: Draw two lines at an angle away from the \u201cV\u201d.<\/p>\n Step 3: Draw one line that comes up at a steeper angle than the \u201cV\u201d and then shorter and off of the center of the \u201cV\u201d (i.e. sorter than where the dotted line is).<\/p>\n Step 4: Draw the last line that closes the ring.<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n![\"\"](\"https:\/\/www.studyorgo.com\/blog\/wp-content\/uploads\/2018\/08\/fig-1-1.gif\")
<\/a>Because there are six sp3 hybridized carbons in a cyclohexane ring, the chair conformation becomes the most stable conformation. The ring can be twisted in several orientations that are all higher in energy and thus less stable, but because the ring can be twisted it can potentially become inverted, which would flip all of the substituents of the ring in the opposite order.\u00a0 In the example below, if we twisted the red and green carbon in the direction of the arrows, the ring would become \u201cflipped\u201d have an equally stable conformation.\u00a0 In solution, both are possible unless there is steric strain that prevents one conformation over the other.<\/p>\n<\/p>\n