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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

The pKa of ascorbic acid (vitamin C, page 9) is 4.17 , showing that it is lightly more acidic than acetic acid (CH3COOH,pKa =4.74) .

(a)Show four different conjugate bases that would be formed by deprotonation of the four different OH groups in ascorbic acid.

(b) Compare the stabilities of these four conjugate bases, and predict which OH group of ascorbic acid is the most acidic.

(c) Compare the most stable conjugate base of ascorbic acid with the conjugate base of acetic acid, and suggest why these two compounds have similar acidities, even though ascorbic acid lacks the carboxylic acid (COOH ) group

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)

Each of these compounds can react as an electrophile. In each case, use curved arrows to show how the electrophile would react with strong nucleophile sodium ethoxide, Na+ - OCH2CH3.

(a)

(b)

NH+4

(c)

CH3CH2Br

(d)

BH3

(e)

CH3COOH

(f)

Write equations for the following acid- base reactions. Use the information in Table 2.2 or Appendix 4 to predict whether the equilibrium will favor the reactants or the products.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

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).
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