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Chapter 4: Q.5. (page 128)

Question: a. Which compounds (B鈥揊) are identical to A? b. Which compounds (B鈥揊) represent an isomer of A?

Short Answer

Expert verified

Answer

a. B, C, and E are identical to A.

b. D and F are isomers of A.

Step by step solution

01

Isomer

An isomer is represented by changing the location of groups or atoms of the main compound. It has the same molecular composition as the main compound.

An identical structure is obtained when the groups are at the same positions, and the number of carbon and hydrogen atoms is also the same.

02

Explanation

a. Compound A has a chain of six carbon atoms and a methyl group at the third position.

Compounds B, C, and E are identical to A because they all have a chain of six carbon atoms and a methyl group substituted at the third position.

b. Compound A has a total of seven carbon and sixteen hydrogen atoms. Therefore, the isomers of A are D and F because they both have the same number of carbon and hydrogen atoms.

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Most popular questions from this chapter

Question: The eclipsed conformation of CH3CH2Cl is role="math" localid="1648620042367" 15kJ/mol less stable than the staggered conformation. How much is the H, Cl eclipsing interaction worth in destabilization?

Question: Calculate the barrier to rotation for each bond highlighted in pink.

a.

b.

Question: Consider the substituted cyclohexane shown in the ball-and-stick model.

  1. Label the substituents on C1, C2, and C4 as axial or equatorial.
  2. Are the substituents on C1 and C2 cis or trans to each other?
  3. Are the substituents on C2 and C4 cis or trans to each other?
  4. Draw the second possible conformation in the chair form, and classify it as more stable or less stable than the conformation shown in the three-dimensional model.

Question: Draw the products of combustion of each alkane.

a.

b.

Question: a. Using Newman projections, draw all staggered and eclipsed conformations that result from rotation around the bond highlighted in pink in each molecule;

b. draw a graph of energy versus dihedral angle for rotation around this bond.
See all solutions

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