UNIT 11: Organic Chemistry-Structure

First, we need to review a few structural characteristics of the carbon atom. These are ideas which were part of your general chemistry courses, but it will help if we briefly restate them.

  1. Carbon is tetracovalent. That means that a carbon atom typically makes four bonds to other atoms and that these bonds are covalent--formed by sharing an electron pair between the two atoms joined by the bond. Such arrangements provide eight valence electrons for a carbon atom, so that it's electronic configuration is like that of the very stable noble gas neon. Similarly, hydrogen forms one covalent bond, oxygen two, and nitrogen three.
  2. Carbon can form multiple covalent bonds. That is, a single carbon atom can form a double (to C, O or N) or triple (to C or N) bond to another atom. A double bond would involve two electron pairs between the bonded atoms and a triple bond would involve three electron pairs.
  3. Bonds between carbon and atoms other than carbon or hydrogen are polar. That is, in a bond between carbon and oxygen or nitrogen the electrons are closer to the more electronegative element (oxygen or nitrogen) than to the carbon, so the carbon has a slightly positive charge. (Fluorine is the most electronegative element, and the elements close to fluorine in the periodic table are also quite electronegative.)
  4. Bonds between one carbon atom and another and between a carbon and a hydrogen are non polar. That is, the electron pair forming the bond is quite evenly shared by the atoms.
  5. We can predict the geometry of the bonds around an atom by using the idea that electron pairs and groups of electron pairs (such as in double or triple bonds) repel each other (Valence Shell Electron Pair Repulsion--VSEPR--Theory).

 

 

 

1.      What is the difference between a structural formula and a molecular formula?

 

Answer:

Structural Formulas are diagrams that show which atoms are bonded to which other atoms in a molecule of a compound.  There are two different types of structural formulas:

(1)   Complete Lewis Structures, and

(2)   Condensed Structural Formulas

 

Lewis Structures symbolize a bonding pair of electrons as a pair of dots or as a dash between two bonding atoms. Lone pairs of nonbonding electrons are shown as pairs of dots. Condensed Structural Formulas are written without showing all of the individual bonds.  Each central atom is shown with the atoms bonded to it.

 

Molecular formulas give the number of atoms of each element in one molecule of a compound.

 

Example:    1-butanol            

                                                 Molecular Formula =  C4H10O

 

                               Condensed Structural Formula =  CH3(CH2)3OH

 

                                                                                           H     H    H     H

                                                                                            |       |      |       |

                                                     Lewis Structure =     H─C ─ C ─ C ─ C ─ O ─ H

                                                                                            |       |      |       |

                                                                                           H     H    H     H

 

 

 

2.      What is the difference between the condensed structural, the expanded structural and the line structure/ line-angle drawing of a compound?

 

Answer:

Expanded structural formulas show all the covalent bonds, while condensed structural formulas show no covalent bonds or only selected bonds.

Let's look at an example case, one with several carbon atoms. A molecular formula is not enough to specify the structure of an organic compound which includes more than three carbons. We need information about which atoms, particularly carbons, nitrogens and oxygens, are connected to each other. This is often represented in a condensed formula like the one for butanal, also called butyraldehyde:

CH3CH2CH2CHO

To take a more detailed look at the structure of this compound, we need to expand its representation. We begin by ignoring the hydrogens and focusing on the carbons and the oxygen to arrive at a skeleton. A trial skeleton might be (where the unused valences are again shown as lines that connect to only one atom):

If we compare this skeleton to the condensed formula above, the left end carbon seems to be associated with three hydrogens in the condensed formula and there are three unused valences on the left end carbon as shown in the skeleton, so that matches well. Similarly, the two middle carbons have two hydrogens each, and there are two unused valences on each of those carbons in the skeleton. If we match all these up we arrive at:

The right end carbon and the oxygen are possibly troublesome. We have only one hydrogen left and there are three unused valences to deal with. Since oxygen has two valences and can form a double bond, we can try a double bond between carbon and oxygen and see if it helps (it does). That gives us this skeleton:

Adding the final hydrogen, we arrive at this expanded structure:

 

It is often inconvenient to show all the carbons and hydrogens in detail, especially since the parts of a molecule which are made up of only carbon and hydrogens joined by single bonds usually do not play a significant role in the reactions of that molecule. (These portions of the molecule are known as "R-groups.") We can represent these atoms and how they are connected by using an abbreviated structural type known as a bond-line structure (also called a line-angle formula,  skeletal structure, or a stick figure ).  Such a structure for the compound, butanal, that we just worked with is shown below:

In a bond-line structure (also called a line-angle formula,  skeletal structure, or a stick figure ), carbons are shown by the "empty" ends of lines and by junctions (corners) between lines. Letters at the end of a line represent atoms of the designated heteroatom (Heteroatoms are atoms other than carbon or hydrogen. Hydrogen is often shown where necessary for clarity.) Double or triple bonds are represented by two or three parallel lines joining the same two atoms. Hydrogens are added as needed to fill up the remaining unused valences.

We can "flesh out" the skeleton above by first filling in the carbons at the corners and "empty" ends:

Then we count how many bond each carbon has showing, and add enough hydrogens to each carbon to bring its number of bonds up to four. (For other atoms like oxygen or nitrogen which can make more than one bond, hydrogens are added as needed to arrive at an appropriate number of bonds.) For example, the left end carbon has one bond showing, so we need to add three hydrogens:

Continuing in the same way with the middle carbons, each of which has two bonds showing, we arrive at the final expanded structure:

Notice that we didn't need to do anything to the right end carbon (the "carbonyl" carbon) or the oxygen, since these atoms already have the appropriate number of bonds showing.

Sometimes we'll represent a molecule by showing its R-groups in a condensed fashion and its reactive parts (functional groups) in an expanded fashion:

 

3.      What are structural isomers?

 

Answer:

In general Isomers are different compounds with the same molecular formula.  But as we know from above, the molecular formula is not sufficient to specify the structure of the compound.  Different compounds with different chemical properties can have the same molecular formula. Structural isomers (also called constitutional isomers) are compounds with the same molecular formula that differ in their bonding sequence; that is, their atoms are connected differently. 

 

Example:        n-butane     and     isobutane     ( Molecular formula C4H10 ) are structural isomers.

 

n-butane (normal butane) has a condensed structural formula  CH3CH2CH2CH3

 

isobutane has a condensed structural formula    CH(CH3)3

 

The expanded structural formula for n-butane is  CH3─CH2─CH2─CH3

 

                                                                                       CH3

                                                                                        |

The expanded structural formula for isobutane is     CH3─CH─CH3 

 

 

4.      What are the four types of tetrahedral carbon?

            I have been unable to find a reference to answer this rather vague question.  Is he thinking of sp3 hybridization?  Diamond versus graphite? Who can tell? This is a very vague question.

            My guess is probably wrong in that it is likely too complex and not what the instructor intended.  See my wild guess below. I recommend you look to your textbook for the answer to this one. 

 

            1. acyclic hydrocarbons, saturated and unsaturated, branched and straight-chain

            2. monocyclic hydrocarbons,  substituted and unsubstituted and aromatics

            3. fused polycyclic hydrocarbons

            4. bridged hydrocarbons, bicyclic systems, polycyclic systems, and hydrocarbon bridges

 

5.      What are functional groups?

 

Answer:

Functional groups are the reactive nonalkane part of an organic molecule. Most nonalkane organic compounds are characterized and classified by the functional groups they contain.  A molecule that contains a functional group may be represented as the functional group with alkyl groups attached.  An alkyl group is the alkane portion of a molecule with one hydrogen atom removed to allow bonding to the functional group.  Often the alkyl groups bonded to a functional group are often designated using the symbol R to refer to the substituent.  The R group is merely an alkyl group.  Using this notation, we presume that the exact nature of the alkyl group is not important.

 

 

6.      What are the names and structures of the most common functional groups?

 

Answer:

Other than combustion, Alkanes undergo very few reactions and are therefore not included as functional groups by most authors. When a molecule contains an alkane portion and a nonalkane portion, we often ignore the alkane portion because it is relatively unreactive. On the other hand, Alkenes contain one or more carbon-carbon double bond(s). Because a carbon-carbon double bond is  relatively reactive, it is considered to be a functional group and thus Alkenes are often included in the list of functional groups. Some authors also include Alkynes and Aromatics as functional groups due to their reactivity.  The major functional groups are named and shown below:

 

 

Class                           Structure                                 Functional Group

 

Alkenes                        R─CH=CH─R*                      carbon-carbon double bond

 

Alkynes                        R─C≡C─R*                            carbon-carbon triple bond

 

Alkyl Halide                 R─X                                        X = F, Cl, Br, or I

 

 

Aromatics                             R                      benzene ring

 

 

 

Alcohols                       R─OH                                     hydroxyl group

 

 

Phenols                              OH                     hydroxyl group on Aromatic ring

 

 

 

Thiols                           R─SH                                      sulfhydril group

 

 

Ethers                           R─O─R*                                 oxygen between two alkyl groups

 

                                         O

                                         

Ketones                       R─C─R*                                 carbonyl group

 

 

 

                                         O

                                         

Aldehydes                    R─C─H                                   carbonyl group

 

 

 

                                         O

                                         

Carboxylic Acid           R─C─OH                                carboxyl group

 

 

 

                                         O

                                         

Esters                           R─C─O─R*                           carboalkoxy group

 

 

 

                                         O

                                         

Amides                         R─C─NH2                              carboxamide group

 

 

                                        

Amines                         R─NH2                                    amino group

 

 

Nitriles                         R─C≡N                                   cyano group

 

 

Nitroalkanes                 R─NO2                                    nitro group

 

 

Is there any difference between vanillin made synthetically and vanillin extracted from the vanilla bean?

Answer:   No, there is no difference.

Historically, when chemists discovered the Law of Definite Proportions at the beginning of the 19th century, it appeared that this law did not apply to the various compounds they had isolated from plant and animal sources.  Carbon compounds can be so complex that the ratios of elements in them did not appear to be simple numbers.  For example, ordinary table sugar has the molecular formula C12H22O11, not the kind of simple ratio seen with the oxides of copper, Cu2O or CuO, for example.  Chemists imagined that organic compounds were held together by a mysterious "vital force".

The beginning of the end of the "vital force" hypothesis is generally considered to be Friedrich Wφhler's synthesis of urea in 1828.  He started with lead cyanate, which is about as "dead" as any chemical can be, and ammonium hydroxide or chloride, also "dead", which generated ammonium cyanate, NH4+OCN- (empirical formula CH4N2O) (still "dead").  When he heated the ammonium cyanate, he got urea, H2NCONH2 (molecular formula also CH4N2O, but atoms arranged differently).  Urea is just what the name sounds like, a major ingredient in urine, and was thought at the time to be a purely "organic" chemical.  Other syntheses of "organic" compounds from "inorganic" materials soon convinced chemists that organic compounds obeyed the same laws of chemistry as other chemicals.

Although chemists gave up the "vital force" hypothesis at least 150 years ago, a shadow of it lingers on in the popular notion that "natural" organic materials are somehow safer or more healthful than synthetic chemicals.  This popular notion ignores the fact that we would not know, for example, what vitamin C is if we could not find out its molecular structure, synthesize it, and show that the synthetic material is in every way identical with the vitamin C that the famous Hungarian chemist Szent-Gyorgy (pronounced "Saint-George") first isolated from Hungarian peppers.  This bit of popular culture also ignores the toxicity of nicotine, strychnine, pufferfish toxin, and botulism toxin, the last of which, although quite natural and produced naturally by the bacteria Clostridium Botulinum, is the most poisonous chemical known. 

List three reasons why functional groups are important in organic chemistry

            This is another vague question that will depend on exactly what your textbook has to say about it.  Here are my ideas on the best way to answer the question.

  1. Functional groups are important as a means of organizing, characterizing, and classifying many thousands of organic compounds into different families of compounds according to the nature and the structure of their functional groups.
  2. When organized in this way, the features and reactions of the various families of organic molecules become readily comprehensible.
  3. The necessity of memorizing vast numbers of seemingly unrelated facts disappears, for we find that the reactions of the simplest member of a family of organic compounds are often the same for hundreds and sometimes thousands of compounds in the same family.

 

 

 

UNIT 10: Acids and Bases

 

Workshop Questions

 

1. Write the formula of the conjugate base of the ammonium ion

Answer:    NH3

Explanation:  According to the Bronsted-Lowry theory of acids and bases, an acid is a proton donor and a base is a proton acceptor. Once, an acid has given up a proton, the remaining part can be a proton acceptor, and thus a base. In this regard, an acid and a base are closely related to one another.

H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of Acid-

For example:

NH3 + H2O = NH4+ + OH-
HAc = H+ + Ac-

 

Thus, NH4+ and NH3 are a pair of conjugate acids and bases, as are HAc and Ac-.

 

2. Write the formula of the conjugate acid of the carbonate ion

Answer:  HCO3–

Explanation: Carbonate ion is  CO3–2

            H+  +  CO3–2  =  HCO3–

 

3. List the following acids in order of decreasing strength: H2SO4, HF, H2S

Answer:                        H2SO4     HF     H2S

Explanation:  Acid strength depends only on the value of Ka and not on the concentration.

                       In general, the larger the value of Ka, the stronger the acid, and vice versa

                        Strong acids have large values of Ka, weak acids have small values of Ka and

                        very weak acids have very small values of Ka.  See the Table of Ka values.

 

 

4. List the following bases in order of decreasing strength: HSO4-, F-, HS-

Answer:                       HS–           F–       HSO4–

Explanation:  In general, the larger the value of Ka, the weaker the base and vice versa.

                        Strong bases have very small values of Ka, weak bases have small values of Ka and

                        very weak bases have large values of Ka.  See the Table of Ka values.

 

 

4.      Explain the relationship between your answers to 3 and 4

Answer:          A strong acid has a very weak conjugate base,

                        a weak acid has a weak conjugate base, and

                        a very weak acid has a strong conjugate base.

 

 

5.      Write the net ionic equation for the following reactions. For each reaction, predict if the formation of products will be favored at equilibrium

 

i.      HI and HC2O4-

 

Answer:          HI (hydroiodic acid) is a very strong acid ( Ka = 3.2 x 109 ) and so it will dissociate 100% in aqueous solution.           

 

HI  =  H+ +  I–      Formation of the dissociated ionic products is favored at equilibrium

 

HC2O4–  (hydrogen oxalate ion) is a weak base (Ka = 5.4 * 10-2) and therefore oxalic acid (H2C2O4) is a weak acid.

A weak acid is only sparingly dissociated in water and therefore HC2O4–  will readily accept and hold free H+ ions forming H2C2O4  

 

H+ +  HC2O4–  =  H2C2O4       Formation of the oxalic acid product is favored at equilibrium

 

The net ionic equation is  HI  +  HC2O4–  =  H2C2O4  +  I–     Formation of the products is favored at equilibrium.

 

 

ii.      H3BO3 and HC2H3O2 2-

 

Answer:           H3BO3  (Boric acid) is a weak acid (Ka = 5.8 * 10-10)and will dissociate sparingly.

                        I regret that I am unable to identify the compound having the molecular formula HC2H3O2 2-

                                Therefore, Solution still pending upon further thought

 

 

 

 

 

*Will carbon dioxide be evolved as a gas when sodium bicarbonate us added to an aqueous solution of each compound? Explain

 

sulfuric acid

Ethanol C2H5OH

Ammonium chloride NH4CL

 

Solution still pending upon further thought

 

 

6.      What is neutralization reaction in terms of: Bronsted-Lowry concept and Arrhenius concept? Give two examples of each.

 

Answer:   In general, neutralization consists of   Acid + Base =  Salt + Water

 

Solution still pending upon further thought

 

 

 

7.      Without using a calculator, fill in the blanks in the following table:

 

pH

pOH

H30+

OH-

Acidic, Basic or Neutral

3

11

1*10-3

1*10-11

Acidic

6

8

1*10-6

1*10-8

Acidic

10

4

1*10-10

1*10-4

Basic

9

5

1*10-9

10-5

Basic

 

Explanation:   

 

pH = - log [H3O+]        pOH = - log [OH-]       pH  +  pOH   =  14      [H3O+] =10–pH              [OH-] = 10–pOH

Detailed Discussion of pH and pOH

Adding an acid to water increases the H3O+ ion concentration and decreases the OH- ion concentration. Adding a base does the opposite. Regardless of what is added to water, however, the product of the concentrations of these ions at equilibrium is always 1.0 x 10-14 at 25oC.

[H3O+][OH-] = 1.0 x 10-14

The table below lists pairs of H3O+ and OH- ion concentrations that can coexist at equilibrium in water at 25oC.

Pairs of Equilibrium Concentrations of H3O+ and OH- Ions That Can Coexist in Water

Concentration (mol/L)

[H3O+]

 

[OH-]

 

 

1

 

1 x 10-14

---->

 

1 x 10-1

 

1 x 10-13

 

1 x 10-2

 

1 x 10-12

 

1 x 10-3

 

1 x 10-11

Acidic Solution

1 x 10-4

 

1 x 10-10

 

1 x 10-5

 

1 x 10-9

 

1 x 10-6

 

1 x 10-8

 

1 x 10-7

 

1 x 10-7

 

Neutral Solution

1 x 10-8

 

1 x 10-6

---->

 

1 x 10-9

 

1 x 10-5

 

1 x 10-10

 

1 x 10-4

 

1 x 10-11

 

1 x 10-3

Basic Solution

1 x 10-12

 

1 x 10-2

 

1 x 10-13

 

1 x 10-1

 

1 x 10-14

 

1

 

Data from this table are plotted in the figure below over a narrow range of concentrations between 1 x 10-7 M and 1 x 10-6 M. The point at which the concentrations of the H3O+ and OH- ions are equal is called the neutral point. Solutions in which the concentration of the H3O+ ion is larger than 1 x 10-7 M are described as acidic. Those in which the concentration of the H3O+ ion is smaller than 1 x 10-7 M are basic.

graph

It is impossible to construct a graph that includes all the data from the table given above. In 1909, the Danish biochemist S. P. L. Sorenson proposed using logarithmic mathematics to condense the range of H3O+ and OH- concentrations to a more convenient scale. By definition, the logarithm of a number is the power to which a base must be raised to obtain that number. The logarithm to the base 10 of 10-7 for example, is -7.

log (10-7) = -7

Since the concentrations of the H3O+ and OH- ions in aqueous solutions are usually smaller than 1 M, the logarithms of these concentrations are negative numbers. Because he considered positive numbers more convenient, Sorenson suggested that the sign of the logarithm should be changed after it had been calculated. He therefore introduced the symbol "p" to indicate the negative of the logarithm of a number. Thus, pH is the negative of the logarithm of the H3O+ ion concentration.

pH = - log [H3O+]

Similarly, pOH is the negative of the logarithm of the OH- ion concentration.

pOH = - log [OH-]

pH + pOH = 14

The equation above can be used to convert from pH to pOH, or vice versa, for any aqueous solution at 25?C, regardless of how much acid or base has been added to the solution. By converting the H3O+ and OH- ion concentrations in the table above into pH and pOH data, we can fit the entire range of concentrations onto a single graph, as shown in the figure below.

graph

 

8.      Use a calculator to fill in the blanks in the following table:

 

pH

pOH

H30+

OH-

Acidic, Basic or Neutral

8.12

5.88

7.6 x10-9

1.3 x10-6

Basic

10.56

3.44

2.8 x10-11

3.6 x10-4

Basic

8.47

5.53

3.4 x 10-9

3.0 x10-6

Basic

7.08

6.92

8.3 x 10-8

1.2 x10-7

Almost Neutral

 

 

10. What are acid-base indicators? Why are they used? Give two examples of acid-base

     indicators.

 

Answer:

 

What are acid-base indicators?

An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.

 

Why are they used?

Acid-base indicators are useful in titrations and for determination of the pH of a solution. Weak acids should be titrated in the presence of indicators which change color under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change color under slightly acidic conditions.

 

Some common acid-base indicators

Tried-and-true indicators include: thymol blue, tropeolin OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, phenolphthalein, thymolphthalein, alizarin yellow, tropeolin O, nitramine, and trinitrobenzoic acid. Data in this table are for sodium salts of thymol blue, bromphenol blue, tetrabromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, and cresol red.

 

 

11.  Will the solution formed by dissolving each of the following salts in water be basic,

       neutral or acidic? Write the equation to show how each ion of each salt affects the  

       pH of the solution.

 

i.Na3PO4

Solution still pending upon further thought

 

 

 

ii.NH4Cl

Solution still pending upon further thought

 

 

        iii.CH3COONa

Solution still pending upon further thought

 

 

 

12.  Write the equation for the reaction of hydrochloric acid with the following:

 

i. Sodium carbonate

 

            HCl + Na2CO3  =  H+  +  Cl–  +  2Na+    +   CO3–2    

 

           All ions remain solvated and dissociated in water and no precipitate forms

 

ii. Ammonia

 

            HCl   +   NH3   =   NH4+  +  Cl–

 

 

iii. Sodium hydroxide

 

            HCl   +   NaOH  =  H+  +  Cl–  +  Na+   +   OH–

 

 

13.  A buffer was prepared from equal molar quantities of acetic acid, HC2H3O2 and sodium acetate, NaC2H3O2.  Explain how the buffer keeps the pH of the solution at about the same value (a) after the addition of a few drops of a dilute HCl solution and (b) after the addition of a few drops of a dilute NaOH solution. Use reaction equations to describe the chemical changes that occur.

 

Solution still pending upon further thought

 

 

 

 

 

 

 

 

 

14.  How would your answer to question 12 be different if the buffer was prepared from

      equal molar quantities of sodium bicarbonate and sodium carbonate?

 

 

Solution still pending upon further thought

 

 

 

 

 

 

15.  Determine the equivalent mass of each of the following:

 

Answer:

For acid-base reactions, equivalent weights or equivalent masses are based on the fact that

one H+(aq) ion reacts with one OH-(aq) ion. One equivalent of an acid is the mass of the acid

that supplies one mole of H+(aq) ions, and one equivalent of a base is the amount of the base

which supplies one mole of OH-(aq) ions.

 

The equivalent weight or equivalent mass is therefore given by the general equation,

 

equivalent mass = molar mass/a

 

where "a" for an acid is the number of moles of H+(aq) supplied by one mole of acid in the reaction

taking place, and for a base "a" is the number of moles of OH-(aq) supplied by one mole of base in the

reaction taking place.

 

 

 

i.                     HCl

 

        molar mass = 36.4 grams per mole

                       a =  1

equivalent mass  = 36.4 grams per mole

 

 

ii.                    H2SO4

 

        molar mass =  98 grams per mole

                       a =  2

equivalent mass  = 49 grams per mole

 

 

iii.      KOH

        molar mass = 56 grams per mole

                       a =  1

equivalent mass  = 56 grams per mole

 

iv.      Ba(OH)2

        molar mass = 171 grams per mole

                       a =  2

equivalent mass  = 85.5 grams per mole

     

 

v.   Al(OH)3

        molar mass = 78 grams per mole

                       a =  3

equivalent mass  = 26 grams per mole

 

 

 

 

16.  How is the formula VacidNacid = VbaseNbase affected when the acid used in the

titration is HCl instead of H2SO4?  How is the formula affected when the base is Mg(OH)2 instead of NaOH?

 

This equation is derived from the idea that equivalents of acid will equal equivalents of base when a system is at the endpoint of a titration.

If the units of the terms on the left side of the equal sign and the units of the terms on the right side of the equal sign are analyzed,

the resulting units will be (equivalents acid) = (equivalents base).

 

In the case of using HCl instead of H2SO4, HCl provides only half the protons per mole compared to H2SO4

So for the same volume and normality of base, one would need HCL in twice as much volume as H2SO4

 

In the case of using Mg(OH)2 instead of NaOH,  Mg(OH)2 provides twice as many OH–  compared to NaOH

So for the same volume and normality of acid, one would need Mg(OH)2 in half as much volume as NaOH

 

 

17.  Your group is to prepare an HCO3- solution that approximates the normal

      concentration in blood, which is 24meq/L. Give a complete description of how you

      would make 10.0L of this solution.

 

 

Solution still pending upon further thought