Structure in Chemistry

Concepts of Acidity

AB19.  Solutions to Selected Problems

With contributions from Nicholas Jones and Kate Graham, College of Saint Benedict / Saint John's University

Problem AB2.1.

A Lewis base must have lone pairs or non-bonding electron pairs so that it can donate them to a Lewis acid. 

Alternatively, in some cases the electrons in a π-bond can be donated instead, so sometimes compounds with π-bonds can be Lewis basic.

 

Problem AB2.2.

a. not a Lewis base
b. not a Lewis base
c. Lewis base
d. Lewis base
e. Lewis base

 

Problem AB3.1.

A Lewis acidis atom attracts electrons from a Lewis base.

The most common feature of a Lewis acid is an atom that is not "electronically saturated" or has not filled its octet.  For example, an aluminum with only six electrons rather than eight is Lewis acidic.

Other atoms, like transition metals, have "octets" of  eighteen electrons, so having fewer than eighteen electrons in their valence shell can make these atoms Lewis acidic.

 

Problem AB3.2.

a. Lewis acid
b. Lewis acid
c. not a Lewis acid
d. Lewis acid
e. not a Lewis acid

 

Problem AB3.3.

a)

 

b) cations

c)  possibly ion-dipole forces; alternatively, the oxygen atoms could act as Lewis bases, donating lone pairs to a cation.

d)

 

e)  anions

f)

 

g)

 

h) When the chloride ion binds to the sensor molecule it gives the overall complex a negative charge and addition of the potassium ion cancels out this charge.

 

 

 

Problem AB4.1.

 

Problem AB4.2.

Problem AB4.3.

 

Problem AB4.4.

 

 

Problem AB4.5.

 

Problem AB4.6.

a)

 

b) 

Also, the aromatic rings may be able to form a conjugated system with the empty p orbital; if that p orbital is partially filled, the boron atom becomes less Lewis acidic.

 

Problem AB5.1.

 

Problem AB5.2.

 

Problem AB5.3.

 

Problem AB6.1.

 

Problem AB6.2.

 

Problem AB6.3.

 

Problem AB7.1.

 

Problem AB7.2.

 

 

Problem AB7.3.

a)

b)

c)

d)

e) aromatic

f)

g)

h)  This nitrogen atom can donate a lone pair of electrons without disturbing the aromatic character of the molecule.

i) 

j) 

k) Each side chain has a Lewis base with a lone pair of electrons to donate the copper ion.

l)

m) CN = 4

n) tetrahedral

o)     Cu:  11 e-

Cu(II):  9 e-

4 x 2e- = 8 e-

8 e- + 9 e- = 17 e-

p) If the imidazole becomes occupied by a proton, then the imidazole no longer has the lone pair to donate to the copper ion.

q) trigonal

 

 

Problem AB8.1.

 

Problem AB8.2.

Problem AB8.3.

Problem AB8.4.

a)  The O-H bond is polar covalent.  The oxygen is much more electronegative than the hydrogen, so the bond is easily ionized to give H+.

b)  In a case like NaOH, the electronegativity difference between the sodium and oxygen is much greater than the electronegativity between the oxygen and the hydrogen.  The Na-O bond is ionic.  The removal of a proton from hydroxide ion is harder than the removal from a proton from hydroxide.  It would make an oxide anion, O2-; that buildup of negative charge is more difficult than the formation of -OH.

 

Problem AB8.5.

 

Problem AB9.1.

a) HNO3 (pKa = -1.3); HNO2 (pKa = 3.3)

HNO3 is a stronger Bronsted acid compared to HNO2

b) H2Se (pKa = 3.9); H2O (pKa = 14)

H2Se is a stronger Bronsted acid compared to H2O

c)  HCl (pKa = -8); H2SO4 (pKa = -3)

HCl is a stronger acid compared to H2SO4

d)  Ba(OH)2 (pKa > 50); HSeO3 (pKa = 6.6)

HSeO3- is a stronger Bronsted acid compared to Ba(OH)2

 

Problem AB 9.2.

a)  NH4+ (pKa = 9.2); HN3 (pKa = 4.7)

NH3 is a stronger Bronsted base compared to N3-

b)  HCN (pKa = 9.4); HSCN (pKa = 4)

-CN is a stronger Bronsted base compared to -SCN

c) NH3 (pKa = 35); H2O (pKa = 14)

-NH2 is a stronger Bronsted base compared to HO-

 

Problem AB10.1.

 

 Problem AB11.1.

a) H2S or SiH4                b)  GaH3 or AsH3

c) PH3 or AlH3                d) H2Se or HBr

 

Problem AB11.2.

a)  +NH4 or NH3            b) -PH2 or PH3       

c) H2O or +NH4            d) +H3O or CH4

 

Problem AB11.3.

a)  H2S or H2Te            b)  GeH4 or SnH4       

c)  HCl or HBr            d)  NH3 or AsH3

 

Problem AB11.4.

 

Problem AB11.5.

a. Polarizability would not be useful as the atom that will be anionic in the conjugate base is the same (oxygen).

b. Se is larger and therefore more polarizable leading to H2Se being more acidic compared to H2S.

c. Ge is larger and therefore more polarizable leading to GeH4 being more acidic compared to SiH4.

d. Polarizability would not be useful as the atoms are next to one another in the same row and therefore likely to have the same polarizability.

 

Problem AB11.3.

 

Problem AB12.1.

 

Problem AB12.2.

a. Vinyl alcohol would be more acidic as its anion is stabilized by resonance while the anionic charge on ethoxide would be localized on the oxygen atom.

b. Nitromethane would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine anion would be localized on the carbon atom.

c. Acetonitrile would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine would be localized on the carbon atom.

 

Problem AB12.3.

 

Problem AB12.4.

 

Problem AB13.1.

 

Problem AB14.1.

 

Problem AB14.2.

 

Problem AB15.1.

 

Problem AB15.2.

 

Problem AB15.3.

 

Problem AB15.4.

The thiol proton on cystine is more acidic due to the polarizability of the sulfur anion stabilizing the conjugate base's anionic charge. The oxygen anion is less polarizable and therefore less stable rendering the alcohol proton less acidic.

 

Problem AB15.5.

 

Problem AB15.6.

 

Problem AB15.7.

Problem AB16.1.

Ammonium bromide is more soluble because in water it can participate in both ion-dipole and hydrogen-bonding, while sodium bromide only benefits from ion-dipole interactions.

Problem AB16.2.

Hydrogen cyanide will be have a lower pKa in water as the resulting cyanide anion will be stabilized by both ion-dipole and hydrogen-bonding interactions. Hydrogen cyanide would have a higher pKa in pentanes as the resulting cyanide anion would only experience ion-induced dipole interactions which are relatively weak.

 

Problem AB17.1.

 

Problem AB17.2.

a. pKa = -6

b. pKa = 9

c. pKa = 24

d. pKa = 16

 

Problem AB17.3.

a. Ka = 103.5

b. Ka = 10-4.3

c. Ka = 10-25

d. Ka = 10-9

 

Problem AB17.4.

At equilibrium formation of the Lewis acid-base complex is slightly favored.

 

Problem AB17.5.

At equilibrium formation of the Lewis acid-base complex of dimethylether and BF3 is more favored than the corresponding diethylether BF3 complex. This is likely due to the decreased steric bulk of the methy groups compared to the ethyl groups.

 

Problem AB17.6.

 

Problem AB18.1.

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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