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Chapter 22: Problem 43

Draw the following. a. cis \(-2\) -hexene b. trans-2-butene c. cis-2,3-dichloro-2-pentene

Short Answer

Expert verified
a. cis-2-hexene: \[ H_3C-CH=CH-CH_2-CH_2-CH_3 \] b. trans-2-butene: \[ H_3C-CH=CH-CH_3 \] (substituents on opposite sides of the double bond) c. cis-2,3-dichloro-2-pentene: \[ Cl-CH_2-CH=CH-CH_2-CH_3 \] (chlorine atoms on the same side of the double bond)

Step by step solution

01

Draw the base structure

Draw the unbranched carbon backbone for each compound: a. For cis-2-hexene, draw 6 carbons connected in a straight chain. b. For trans-2-butene, draw 4 carbons connected in a straight chain. c. For cis-2,3-dichloro-2-pentene, draw 5 carbons connected in a straight chain.
02

Add the double bond

Place a double bond between the specified carbon atoms: a. For cis-2-hexene, add a double bond between the 2nd and 3rd carbons. b. For trans-2-butene, add a double bond between the 2nd and 3rd carbons. c. For cis-2,3-dichloro-2-pentene, add a double bond between the 2nd and 3rd carbons.
03

Add the substituents

Add substituents based on the cis/trans notation: a. For cis-2-hexene, no additional substituents are needed, so just add the hydrogen atoms to complete the structure. b. For trans-2-butene, we have a trans configuration, which means that the substituents are on opposite sides of the double bond. Add the hydrogen atoms to complete the structure. c. For cis-2,3-dichloro-2-pentene, we have a cis configuration, which means that the substituents are on the same side of the double bond. Add two chlorine atoms on carbons 2 and 3 and complete the structure with hydrogen atoms.
04

Review the final structures

Verify that the structures are correct for the given compounds: a. cis-2-hexene: \[ H_3C-CH=CH-CH_2-CH_2-CH_3 \] b. trans-2-butene: \[ H_3C-CH=CH-CH_3 \mathrm{with\;substituents\;on\;opposite\;sides\;of\;the\;double\;bond} \] c. cis-2,3-dichloro-2-pentene: \[ Cl-CH_2-CH=CH-CH_2-CH_3 \mathrm{with\;chlorine\;atoms\;on\;the\;same\;side\;of\;the\;double\;bond} \] The structural formulas drawn are correct representations of the given compounds.

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