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Chapter 2: Question 2.24 (page 129)

Classify the following hydrocarbons, and draw a lewis structure for each one. A compound may fit into more than one of the following classifications:

Alkane, alkene, alkyne, cycloalkane, cycloalkene, cycloalkyne, aromatic hydrocarbon.

  1. (CH3CH2)2CHCH(CH3)2
  2. CH3CHCHCH2CH3
  3. CH3CCCH2CH2CH3

e.

f.

g.

h.

i.

Short Answer

Expert verified

Answer:


a.)

(b.)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

Step by step solution

01

Explanation for (a)

(a) The molecule has bonds, therefore, it is alkane. In the case of alkane, alkene and alkyne molecule has single, double, and triple bond present respectively in their structural unit.

02

Explanation for (b)

(b)Molecule has a double bond, therefore, it is alkene.

03

Explanation for (c)

c)Molecule has a triple bond, therefore, alkyne.

04

Explanation for (d)

(d)Molecule has a double bond, triple bond and ring, therefore, cycloalkene and cycloalkyne.

05

Explanation for (e)

(e)Molecule has a double bond and saturated ring, therefore, cycloalkane and alkene.

06

Explanation for (f)

(f)Molecule has a benzene ring and triple bond, therefore, it is an aromatic hydrocarbon and alkyne.

07

Explanation for (g)

(g)Molecule has a benzene ring and double bond, therefore, it is an aromatic hydrocarbon and alkene

08

Explanation for (h)

(h)Molecule has single bonds and saturated ring, therefore, cycloalkane and alkane.

09

Explanation for (i)

(i)Molecule has a benzene ring and unsaturated ring (ring having double bond) therefore, aromatic hydrocarbon and cycloalkene.

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Most popular questions from this chapter

Consider the following proposed Bronsted-Lowry acid-base reactions. In each case, draw the products of a transfer of the most acidic proton on the acid to the most basic site on the base. Use Appendix 4 to find the pKa values for the acids and the pKb values for the bases. Then determine which side of the reaction is favored, either reactants of products.

(a)

(b)

(c)

(d)

(e)

(f)

For each of the following compounds,

  1. Draw the Lewis structure.
  2. Show how the bond dipole moments (and those of any nonbonding pairs of electrons) contribute to the molecular dipole moment.
  3. Estimate whether the compound will have a large, small, or zero dipole moment.

(a) CH3-CH=N-CH3

(b) CH3-CH2OH

(c) CBr4

(d)

Structure (d)

(e)

Structure (e)

(f)

Structure (f)

(g)

Structure (g)

(h)

Structure (h)

The following compounds can all react as bases.

  1. For each compound, show its conjugate acid. Show any resonance forms if applicable.
  2. Rank the conjugate acids in the order you would predict, from most stable to least stable.
  3. Rank the original compounds in order from strongest base to weakest base.

N-Methylpyrrolidine has a boiling point of81慰C, and piperidine has a boiling point of106慰C.

  1. Explain the large difference (25慰C) in boiling point for these two isomers.
  2. Tetrahydropyran has a boiling point of 88慰C, and cyclopentanol has a boiling point of 141慰C. These two isomers have a boiling point difference of 53慰C.Explain why the two oxygen-containing isomers have a much larger boiling point difference than two amine isomers.
  3. N,N-Dimethylformamide has a boiling point of 150慰C, and N-methylacetamide has a boiling point of 206慰C, for a difference of 56慰C.Explain why these two nitrogen-containing isomers have a much larger boiling point difference than the two amine isomers. Also explain why these two amides have higher boiling points than any of the other four compounds shown (two amines, an ether, and an alcohol).

Ozone has a dipole moment of 0.53 D. Carbon dioxide has a dipole moment of zero, even though C-O bonds are more polar than O-O bonds. Explain this apparent contradiction.
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