A trisubstituted cyclohexane compound is given below in its chair conformation. Draw the corresponding planar (overhead) represe
ntation, using wedge-and-dash bonds to indicate the substituent positions. To be graded properly, include the hydrogen atoms on the chirality centers (asymmetric carbons). Be sure that both wedge/dash bonds are drawn on the outside of the ring, or else the stereochemistry may be interpreted as square planar.
2 answers:
Answer:
Find attached
Explanation:
The planar structure shown represents the stereochemical position of the substituted atoms in the cyclohexane chair-shaped ring.
First, the chiral centres (asymmetric carbons) are located at positions 1,2 and 4 on the ring, each having a substituted atom Br, F and Cl respectively.
A Chiral centre is usually recognized when a carbon atom is substituted by four different atoms or groups of atoms.
In organic chemistry, especially for chair-ringed structures, the chiral centre is usually bonded to at least two other carbon atoms, and the remaining two atoms bonded to it assume two positions, namely; axial and equatorial positions.
An atom attached by a bond that is perpendicular to the axis of the ring is said to be in an equatorial position.
An atom attached by a bond that is parallel to the axis of the ring is said to be in an axial position.
Hence, the chiral Hydrogen atom and the other substituted atoms assume either an axial or an equatorial position.
According to the chair-ringed structure in the question, the Br and Cl atoms are at an axial position, while the F atom is at the equatorial position. Consequently, the wedge bonds represent the position of the chiral hydrogen atom and the substituted atoms on the planar structure with respect to their axial position on the chair-ringed structure. The dash bonds represent the position of the atoms on the planar structure with respect to their equatorial position on the chair-ringed structure shown in the question.
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Answer:
a) But-1-ene
b) E-But-2-ene
c) Z-But-2-ene
d) 2-Methylpropene
Explanation:
In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u> in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>
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For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>
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Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>
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See figure 1 to further explanations.
I hope it helps!
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