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Chapter 4: Q.14. (page 128)

Question: Draw the staggered and eclipsed conformations that result from rotation around the C-C bond in CH3-CH2Br.

Short Answer

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Answer

The staggered and eclipsed conformations are shown below:

Step by step solution

01

Newman projection

In the Newman projection, the dot represents the front carbon whereas the circle represents the back carbon.

02

Staggered conformation

In the staggered conformation, the dihedral angle between the atom of the front carbon and the atom of the back carbon is 600 . The staggered conformation is shown below:

Staggered conformation

03

Eclipsed conformation

In the eclipsed conformation, the dihedral angle between the atom of the front carbon and the atom of the back carbon is 00. The eclipsed conformation is shown below:

Eclipsed conformation

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Most popular questions from this chapter

Question: Classify each pair of compounds as constitutional isomers, stereoisomers, identical molecules, or not isomers of each other.

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b.

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Question: Although penicillin G has two amide functional groups, one is much more reactive than the other. Which amide is more reactive and why?

Question: Cyclopropane and cyclobutane have similar strain energy despite the fact that the C-C-C bond angles of cyclopropane are much smaller than those of cyclobutane. Suggest an explanation for this observation, considering all sources of strain discussed in Chapter 4.

Question: Give the structure corresponding to each IUPAC name.

  1. 3-methylhexane
  2. 3,3-dimethylpentane
  3. 3,5,5-trimethyloctane
  4. 3-ethyl-4-methylhexane
  5. 3-ethyl-5-isobutylnonane

Question: The melting points and boiling points of two isomeric alkanes are as follows: CH3(CH2)6CH3, mp=-570C and bp=1260C;(CH3)3CC(CH3)3 , mp=1020C and bp=1060C.

a. Explain why one isomer has a lower melting point but higher boiling point.

b. Explain why there is a small difference in the boiling points of the two compounds but a huge difference in their melting points.
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