Chapter 26: Structures of Organic Compounds s1

Chapter 26: Structures of Organic Compounds

CHAPTER 26

STRUCTURES OF ORGANIC COMPOUNDS

PRACTICE EXAMPLES

1A (E) We have shown only the C atoms and the bonds between them. Remember that there are four bonds to each C atom; the remaining bonds not shown are to H atoms. First we realize there is only one isomer with all six C atoms in one line. Then we draw the isomers with one 1-C branch. The isomers with two 1-C branches can have them both on the same atom or on different atoms. This accounts for all five isomers.

1B (E) We have shown only the C atoms and the bonds between them. Remember that there are four bonds to each C atom; the remaining bonds not shown are to H atoms.

2A (M)

Numbering starts from the right and goes left so that the substituents appear with the lowest numbers possible. This is 3,6,6-trimethyloctane.

2B. (M)

This is a symmetrical molecule. Therefore it does not matter whether numbering starts from left or right. The compound’s name is 3,6-dimethyloctane.

3A. (M)

3B. (M)

4A. (M) The structural diagram for 2-methyl-pentane is given below:

When the molecule is viewed along C1-C2 bond there are several possible staggered conformations:

There are no gauche interactions in this molecule. Therefore, conformations (a), (b) and (c) all have the same energy.

4B. (D) When 1-chloropropane molecule is viewed along the C1-C2 bond, several eclipsed and staggered conformations can be identified:

5A. (M) We are dealing with a trans isomer, and so both methyl groups are adjacent to opposite face of the ring. The conformation of the lowest energy will be the one that has the methyl groups in the equatorial positions.

5B. (M) For this molecule, it is more favorable to place tert-butyl group in the equatorial position.

6A. (M)

(a) / / All three carbon atoms in this molecule are attached to at least two groups of the same type; thus, the molecule is achiral
(b) / / This molecule contains a carbon atom that is bonded to four different groups; consequently, the molecule is chiral.
(c) / / This molecule contains a carbon atom that is attached to four different groups; consequently, the molecule is chiral.
6B. (M)
(a) / / None of the three carbon atoms in this alcohol are bonded to four different groups; consequently the molecule is achiral
(b) / / This molecule contains a carbon atom that is bonded to four different groups; consequently, the molecule is chiral.
(c) / / This molecule contains a carbon atom that is bonded to four different groups; consequently, the molecule is chiral. The molecule is also unstable and eliminates HCl to form propanal CH3CH2CHO.

7A. (M) Chiral carbon atoms (*) are indicated below:

7B. (M) The molecular structure of 1,1,3-trimethylcyclohexane is shown below and chiral carbon atoms indicated by “*”:

8A. (D)

(a) (b) (c)

8B.

/ (D)(a) / / / Thus, the structures are
enantiomers.
(b) / / / Thus, the structures are enantiomers.

9A.

/ (M)(a) / This is the (E) stereoisomer / (b) / This is the (Z) stereoisomer
(c) / This is the (Z) stereoisomer

9B.

/ (M)(a) / (Z) stereoisomer / (E) stereoisomer
(b) / (Z) stereoisomer / (E) stereoisomer
(c) / (Z) stereoisomer / (E) stereoisomer

10A. (M) We need to establish the degree of unsaturation in the molecule and then construct one example of each type of molecule that can be formed. The maximum number of hydrogen atoms is 5x2+2=12. The molecular formula has only 11 hydrogen atoms, so we know the degree of unsaturation is one. Several possibilities are possible:

10B. (D) We need to establish the degree of unsaturation in the molecule and then construct one example of each type of molecule that can be formed. The maximum number of hydrogen atoms is 5x2+2=12. The molecular formula has only 10 hydrogen atoms, so we know the degree of unsaturation is one. Several possibilities are possible and some of them are shown below:

INTEGRATIVE EXAMPLE

11A. (M) Compound A has a formula C3H8O and it is coordinatively saturated. It reacts with sodium metal to produce gas and also is further oxidized by the treatment with chromic acid. Compound A must therefore be a primary alcohol, CH3-CH2-CH2-OH. On treatment with chromic acid, CH3-CH2-CH2-OH is completely oxidized to carboxylic acid, CH3-CH2-CH2-COOH (compound B). CH3-CH2-CH2-COOH further reacts with base Na2CO3 to yield CH3-CH2-CH2-COO-Na+. It also reacts with ethanol to produce ester CH3-CH2-CH2-COO-CH3-CH2 (compound C).

11B. (M) Compounds (a) and (b) are both ethers. Ethers are relatively unreactive and the ether linkage is stable in the presence of most oxidizing and reducing agents, as well as dilute acids and alkalis. Compound (c) is unsaturated secondary alcohol. It will decolorize a Br2/CCl4 solution. Compound (d), on the other hand is an aldehyde. On treatment with chromic acid it will be oxidized to carboxylic acid. The same reaction would not happen with compound (c). Although the principal oxidation product of the unsaturated secondary alcohols with chromic acid will be a ketone, the reaction of chromic acid with the aldehyde will appear identical to that of the secondary alcohol (i.e., the chromic acid will change from a red solution to a green solution). A subsequent NaHCO3 treatment would be needed to test for the presence of carboxylic acid (evolution of CO2).

EXERCISES

______

Organic Structures

1. (E) In the following structural formulas, the hydrogen atoms are omitted for simplicity. Remember that there are four bonds to each carbon atom. The missing bonds are C— H bonds.

(a)
(b)

(c)

2. (E) In the structural formulas drawn below, we omit the hydrogen atoms. Remember that there are four bonds to each C atom. The bonds that are not shown are C— H bonds.

(a) 3-isopropyloctane (b) 2-chloro-3-methylpentane

(c) 2-pentene (d) dipropyl ether

C–C=C–C–C C–C–C–O–C–C–C

3. (E) (a) (b) (c)

4. (E) In the following structural formulas, the hydrogen atoms are omitted for simplicity. Remember that there are four bonds to each carbon atom. The missing bonds are C—H bonds.

(a)

(CH3)3CCH2CH(CH3)CH2CH2CH3 (b) (CH3)2CHCH2C(CH3)2CH2Br

5. (M) (a) (b) (c)

6. (M)

7. (M) (a) Each carbon atom is hybridized. All of the bonds in the structure (drawn on the next page) are sigma bonds, between the 1s orbital of H and the orbital of C. The bond is between orbitals on each C atom.

(b) Both carbon atoms are hybridized. All of the bonds in the structure (see next page) are sigma bonds between the 1s orbital of H and the orbital of C. The bond is between the orbital on C and the 3p orbital on Cl. The double bond is composed of a sigma bond between the orbitals on each C atom and a pi bond between the orbitals on the two C atoms.

(c) The left-most C atom (in the structure drawn on the next page) is hybridized, and the bonds to that C atom are between the orbitals on C and the 1s orbital on H. The other two C atoms are sp hybridized. The right-hand bond is between the sp orbital on C and the 1s orbital on H. The C º C triple bond is composed of one sigma bond formed by overlap of sp orbitals, one from each C atom, and two pi bonds, each formed by the overlap of two 2p orbitals, one from each C atom (that is a overlap and a overlap).

(a) (b) (c)

8. (M) (a) The left- and right-most C atoms in the structure (drawn below) are hybridized. All bonds are sigma bonds formed by the overlap of an orbital on C with a 1s orbital on H. The central C atom is hybridized; both bonds are sigma bonds, formed by the overlap between the orbital on the terminal C atom and an orbital on the central C atom. The double bond is composed of a bond between the orbital on the central C atom and a orbital on the O atom, and a bond between the orbital on the central C atom and the orbital on the O atom.

(b) The left C atom in the structure (drawn below) is hybridized. All bonds are sigma bonds formed by the overlap of an orbital on C with a 1s orbital on H. The central C atom is hybridized; the bond is a sigma bond, formed by the overlap between the orbital on the terminal C atom and an orbital on the central C atom. The double bond is composed of a bond between the orbital on the central C atom and an sp2 orbital on the O atom, and a bond between the orbital on the central C atom and the orbital on the O atom. The right O atom is hybridized; the sigma bond forms by the overlap of with O(sp3) and the OH with the H(1s).

(c) The two end C's are hybridized; all bonds form by the overlap of C(sp2) with H(1s). The central C is sp hybridized. Both bonds consist of a bond formed by Ccentral(sp) with Cend(sp2) overlap and a bond formed by with Cend(2p) overlap. Note: The 2p orbitals that make up each pi bond are mutually perpendicular relative the left or right side of the molecule.

(a) (b) (c)

Isomers

9. (E) Structural (skeletal) isomers differ from each other in the length of their carbon atom chains and in the length of the side chains. The carbon skeleton differs between these isomers. Positional isomers differ in the location or position where functional groups are located attached to the carbon skeleton. Geometric isomers differ in whether two substituents are on the same side of the molecule or on opposite sides of the molecule from each other; usually they are on opposite sides or the same side of a double bond.

(a) The structures are identical.

(b) The two compounds are constitutional isomers.

(d) The two compounds are constitutional isomers.

(e) The two compounds are stereoisomers.

10. (M) (a) The two compounds are identical.

(b) The two compounds are tautomers.

(d) The two compounds are constitutional isomers.

(e) The two compounds are constitutional isomers. They differ in the mode of attachment of bromine atoms. In the first one, bromine atoms are attached to the same carbon, whereas in the second one they are attached to two adjacent carbon atoms.

11. (D) We show only the carbon atom skeleton in each case. Remember that there are four bonds to carbon. The bonds that are not indicated in these structures are bonds.

12. (D) In each case, we draw only the carbon skeleton. It is understood that there are four bonds to each carbon atom. The bonds that are not shown are bonds.

(a)

(b)

(c)

Functional Groups

17. (E) (a)

alkyl halide (bromide)

(b)

aldehyde

(c)

ketone

(d) phenol, hydroxyl group, phenyl group

18. (E) (a)

carboxylic acid

(b) ether

isobutylmethyl ether

2-aminobutane

(d)

ester

19. (M) (a) A carbonyl group is >C=O , whereas a carboxyl group is –C(O)-OH

The essential difference between them is the hydroxyl group, — OH.

(b) An aldehyde has a carbonyl group with a hydrogen and a carbon group attached.

In a ketone, the carbonyl group is attached to two carbons.

The essential difference is the presence of the — OH group in acetic acid.

20. (M) (a) aromatic nitro compound, , nitrobenzene

(b) aliphatic amine, , ethylamine

(d) aliphatic diol, , 1,2-ethanediol

(e) unsaturated aliphatic alcohol, , 3-butene-1-ol

(f) alicyclic ketone, , cyclohexanone (drawn below)

(g) halogenated alkane, , 2-iodobutane

(h) aromatic dicarboxylic acid, o-C6H4(COOH)2, phthalic acid (drawn below)

25. (M)

26. (M)

27. (M)

28. (M)

29. (M)

30. (D) There are numerous isomers; here are a few of them

Nomenclature and Formulas

31. (E) (a) The longest chain is eight carbons long, the two substituent groups are methyl

groups, and they are attached to the number 3 and number 5 carbon atom. This is 3,5-dimethyloctane.

(b) The longest carbon chain is three carbons long, the two substituent groups are methyl groups, and they are both attached to the number 2 carbon atom. This is 2,2-dimethylpropane.

(c) The longest carbon chain is 7 carbon atoms long, there are two chloro groups attached to carbon atom 3, and an ethyl group attached to carbon 5. This is 3,3-dichloro-5-ethylheptane.

32. (E) (a) There are 2 chloro groups at the 1 and 3 positions on a benzene ring. This is 1,3-

dichlorobenzene or, more appropriately, meta-dichlorobenzene.

(b) There is a methyl group at position 1 on a benzene ring, and a nitro group at position 3. This is 3-nitrotoluene or meta-nitrotoluene.

(c) There is a — COOH group at position 1 on the benzene ring, and a group at position 4, or para to the — COOH group. This is 4-aminobenzoic acid or p-aminobenzoic acid.

33. (M) (a) The longest carbon chain has four carbon atoms and there are 2 methyl groups attached to carbon 2. This is 2,2-dimethylbutane.

(b) The longest chain is three carbons long, there is a double bond between carbons 1 and 2, and a methyl group attached to carbon 2. This is 2-methylpropene.

(d) The longest chain is 5 carbons long, there is a triple bond between carbons 2 and 3 and a methyl group attached to carbon 4. This is 4-methyl-2-pentyne.

(e) The longest chain is 6 carbons long. This compound is 3,4-dimethylhexane

(f) The longest carbon chain containing the double bond is 5 carbons long. The double bond is between carbons 1 and 2. There is a propyl group on carbon 2, and carbons 3 and 4 each have one methyl group. This is 3,4-dimethyl-2-propyl-1-pentene.