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Solutions to Problems Basic Principles

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1. Organic chemistry is the study of molecules containing a set of unique carbon atoms. The uniqueness of carbon is based on its ability to connect with other carbon atoms. This uniqueness is referred to as an expanded octet.

2. Friedrich Wӧhler, a Heidelberg physician, performed a set of interesting experiments around the middle of the nineteenth century. These experiments essentially demonstrated that organic compounds could be made from inorganic compounds.

By the end of the nineteenth century, the Vital Force Theory (the belief that all organic compounds must come from organic materials) died with the extraordinary synthesis of acetic acid from carbon disulfide. Adolph Wilhelm Hermann Kolbe accomplished this work. Kolbe prepared carbon tetrachloride

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by treating carbon disulfide with chlorine gas. The pyrolysis of carbon tetrachloride resulted in the formation of tetrachloroethylene, and aqueous chlorination of tetrachloroethylene led to the formation of trichloroacetic acid. The final step in Kolbe’s synthesis was the electrolysis of trichloroacetic which resulted in the formation of acetic acid. Frederick Wӧhler is considered to be the father of organic chemistry.

3. The Vital Force Theory is the belief that all organic compounds come from organic materials. The demise of the Vital Force Theory opened the field of organic chemistry to the syntheses of a vast number of compounds. As a consequence, more than 100,000 new organic compounds are synthesized annually. There are more than four million organic compounds today compared to 300,000 inorganic compounds.

4. Wӧhler’s work led to the demise of the Vital Force Theory.

5. Adolph Wilhelm Hermann Kolbe definitively demonstrated that an organic compound could be synthesized from an inorganic compound. He prepared acetic acid from carbon disulfide. First, he prepared carbon tetrachloride by treating carbon disulfide with chlorine gas. Secondly, the pyrolysis of carbon tetrachloride resulted in the formation of tetrachloroethylene, and, thirdly, aqueous chlorination of tetrachloroethylene led to the formation of trichloroacetic acid. The final step in Kolbe’s was the electrolysis of trichloroacetic resulting in the production of acetic acid. Friedrich Wohler is considered to be the father of organic chemistry.

6. (1) The demise of the Vital Force Theory.

(2) The birth of structural theory including the introduction of quantum mechanics to explain the nature of the chemical bond.

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(3) The use of modern instrumental techniques for separating, analyzing, and identifying organic compounds.

7.

8. The ionic bond can be described as the transfer of electrons where two electrons in the calcium atom are transferred to two chlorine atoms. The result is the formation of a calcium ion with +2 charge (referred as cation), and the formation two chloride ions each with a -1 charge (referred to as anions). In reality, calcium chloride does not exist as a single entity, but is more adequately described as the interconnection of ionic bonds in a complex matrix lattice structure.

9. The four C-I bonds in tetraiodomethane are sigma bonds with a I-C-I bond angle of approximately 109o.

10.Each C-I bond is formed from the linear combination of an electron in a 2 sp3 hybridized atomic orbital of C with an electron in a 5 p atomic orbital of iodine. The linear combination of an electron in a 2 sp3 hybridized atomic orbital of C with an electron in a 5 p atomic orbital of iodine results in the formation of two molecular orbitals, the bonding molecular orbital (where the two electrons reside) and the antibonding molecular orbtal (where there are no electrons).

11. “a”

C CH

H

H

H+ H H

H HH

H

H

H

C C120o

180o

π (2p + 2p)

σ(2sp2 + 2sp2)

(2sp2 + 1s)σ

(1s + 1s)σ

(2sp3 + 2sp3)σ

(2sp2 + 1s)σ

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12. “e”

Hund’s rule states that for a set of degenerate (equal in energy) orbitals, each orbital must have one electron in it before the second electron can be added. Pauli’s Exclusion Principle states that no two electrons in the same orbital can have the same set of four quantum numbers. For example, quantum numbers n, l, and ms can be identical for two electrons in the same orbital, but the spin quantum number must be different.

13. “c”

14.The electron pairs in the four C-H sigma bonds repel one another in such a manner that the bond angles are approximately 109.5o.

15.The electron pairs in the three N-H sigma bonds repel one another in such a manner that the bond angles are approximately 107o. The electron pair in the non-bonding ”fat” orbital causes the H-N-H bond angle can be 107o instead of 109o.

16.The electron pairs in the two O-H sigma bonds repel one another in such a manner that the bond angles are approximately 105o. The two electron pairs in the non-bonding ”fat” orbital causes the H-O-H bond angle can be 105o instead of 109o.

17.The double bond is made up of a pi , π, bond and a sigma, σ, bond. The geometry about the carbon atoms that are a part of the double bond is trigonal planar. The H-C=C bond is 120o. The geometry about the carbon atom of the methyl is tetrahedral with bond angles that approximate 109.5o. The three CH bonds of the methyl group are sigma bonds.

18.The triple bond is made up of two pi, π, bonds and a sigma, σ, bond. The geometry about the carbon atoms that are a part of

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the triple bond is linear. The H-C≡C bond is 180o. The geometry about the carbon atom of the methyl is tetrahedral with bond angles that approximate 109.5o. The three CH bonds of the methyl group are sigma bonds.

19. “a”

20. “a”

21. sp

22. sp2

23. sp3

24.Yes, the more s-character in the bond, the greater the acidity

25.Yes, the greater the electronegativity, the higher the bond dissociation energy.

26.The molecule exhibits a dipole moment when the geometry of the molecule is considered and the electronegativity differences of the atoms attached to the central atom create a vector sum.

27. CHBr3 exhibits a dipole moment; however, CBr4 has a dipole moment equal to zero.

BrBr

BrBr

C

µ = 0µ does notequal zero

C

BrBr

BrH

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28.

29.

30.NCl3 has a dipole moment. NCl3 has a higher dipole moment than NF3.

31. The sulfate radical

χBr − χH = 0.102 mol12

kJ12

ΔHHBr - ΔHBr2( ) ΔHH2( )

χBr − 2.2 = 0.102 mol12

kJ12

362 kJ/mole - 190 kJ/mol( ) 432 kJ/mole( )

χBr = 2.2 + 0.88 = 3.1

µHBr = 1.84 x 10−10 esu x 0.44 Ao

µHBr = 0.81 D

..

_

_

:

::....

....:

O

OOS

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Resonance of the sulfate anion:

32. Yes

33.

(a)

CH3COOH

ClCH2COOH

Cl2CHCOOH

H3O+⎡⎣ ⎤⎦ = 1.77 x 10−5 x 0.100 = 1.33 x 10−3 M

pH = - log H3O+⎡⎣ ⎤⎦ = 2.88

H3O+⎡⎣ ⎤⎦ = 1.36 x 10−3 x 0.100 = 1.17 x 10−2 M

pH = - log H3O+⎡⎣ ⎤⎦ = 1.93

H3O+⎡⎣ ⎤⎦ = 5.50 x 10−2 x 0.100 = 7.42 x 10−2 M

pH = - log H3O+⎡⎣ ⎤⎦ = 1.13

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Cl3 CCOOH

(b)

In order of acid strength,

Cl3 CCOOH > Cl2CHCOOH > ClCH2COOH> CH3COOH

34.

(a) Aluminum chloride is a Lewis acid

(b)

(c) The conjugate base is the tetrachloroaluminate anion, AlCl4-

(d) The conjugate acid is Cl+

35.

(a) The targeted carbon atom has two electron withdrawing groups (O and Cl) attached.

(b) CH3OH is the Lewis base, because it donates an electron pair to the carbon atom.

(c) CH3XCH2CO2Cl, the acyl chloride, is the acid, because it is accepting the electron pair of the methanol, CH3OH.

(d) The conjugate base is the ester, CH3XCH2CO2CH3.

(e) Hydrochloric acid, HCl, is the conjugate acid.

H3O+⎡⎣ ⎤⎦ = 0.23 x 0.100 = 0.15 M

pH = -log H3O+⎡⎣ ⎤⎦ = 0.82

AlCl3 + Cl2 → AlCl4− + Cl+

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36.

(a) If

is represented as C6H5COOH, then the equilibrium expression can be represented by the following equation:

(b) If

is represented as O2NC6H4COOH, then the equilibrium expression can be represented by the following equation:

Ka = C6H5COO−⎡⎣ ⎤⎦ H3O

+⎡⎣ ⎤⎦C6H5COOH[ ]

Ka = O2NC6H4COO−⎡⎣ ⎤⎦ H3O

+⎡⎣ ⎤⎦O2NC6H4COOH[ ]

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37.The stronger acid is m-nitrobenzoic acid:

The nitro group, an electron withdrawing group, facilitates the release of the hydrogen ion to form the hydronium ion in aqueous solution (the Arrhenius definition of an acid).

38.

(a)

(b)

Kb = CH3CH2CH2NH3 (aq)

+⎡⎣ ⎤⎦−OH⎡⎣ ⎤⎦

CH3CH2CH2NH2[ ]

Kb = 10−3.3 = 5.0 x 10−4

−OH⎡⎣ ⎤⎦ = 5.0 x 10−4 x 0.100 = 7.1 x 10−3

pOH = 2.2pH = 11.8

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39.

CH3CH2O

C

O

C

C

CH3CH2O

O

HH

+ Na+ -OCH2CH3

O

CH3CH2O

C

C

O

C

CH3CH2OHOCH2CH3+

.._ Na+

H