C. Correct! The abbreviation Ar stands for an aromatic ring, sometimes called an aryl ring.

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1 Organic Chemistry - Problem Drill 05: Drawing Organic Structures No. 1 of What does the abbreviation Ar stand for? (A) Acetyl group (B) Benzyl group (C) Aromatic or Aryl group (D) Benzoyl group (E) Ethyl group Acetyl groups are abbreviated as Ac. Go back and review the most common abbreviations used in Benzyl groups are abbreviated as Bn. Go back and review the most common abbreviations used in C. Correct! The abbreviation Ar stands for an aromatic ring, sometimes called an aryl ring. Benzoyl groups are abbreviated as Bz. Go back and review the most common abbreviations used in Ethyl groups are abbreviated as Et. Go back and review the most common abbreviations used in (1) Recall the common organic abbreviations. You will need to know the abbreviations and the structures they represent. Ar is the abbreviation for an aromatic ring (also called an aryl ring.) Therefore, the correct answer is (C).

2 No. 2 of What abbreviation would be used for the following structure? O (A) Bn (B) Ac (C) Py (D) Bz (E) Ph The abbreviation Bn, which stands for a benzyl group, describes a structure where a benzene ring is bonded to a C 2 (methylene) group. Go back and review the common abbreviations used in organic chemistry. The abbreviation Ac, which stands for an acetyl group, describes a structure where a methyl group is bonded to a carbonyl. Go back and review the common abbreviations used in The abbreviation Py, which stands for a pyridine ring, describes a six-membered aromatic ring where one of the atoms in the ring is nitrogen. Go back and review the common abbreviations used in organic chemistry. D. Correct! The structure pictured above is a benzoyl group which is abbreviated as Bz. The abbreviation Ph stands for a phenyl group. Go back and review the common abbreviations used in (1) Recall the common organic abbreviations. You will need to know the abbreviations and the structures they represent. This structure containing a carbonyl bonded to an aromatic ring is known as a benzoyl group. Its abbreviation is Bz. Therefore, the correct answer is (D).

3 No. 3 of What kind of structure appears below? (A) Lewis structure (B) Skeletal structure (C) Sawhorse structure (D) Condensed structure (E) None of the above. C 3 C 2 C 2 C 2 O In a Lewis structure, all individual bonds and nonbonding electron pairs are shown. In the above structure, no bonds or lone pairs are explicitly drawn. Go back and review the different types of structures that can be used to illustrate organic compounds. In a skeletal structure, lines represent bonds and carbon atoms are assumed to be where two lines meet or at the end of a line. ydrogens are not shown. Go back and review the different types of structures that can be used to illustrate organic compounds. In a sawhorse structure, the molecule is drawn so that one is looking down a carbon-carbon bond. Go back and review the different types of structures that can be used to illustrate organic compounds. D. Correct! In a condensed structure, atoms bonded to a single atom are listed after the atom. No bonds are explicitly drawn. The correct answer is in the above choices. Go back and review the different types of structures that can be used to illustrate organic compounds. (1) Recall the different types of structures that may be drawn to represent an organic compound. Lewis structures show all individual atoms and nonbonding pairs of electrons. They are drawn using the valence electrons of the compound and show most, if not all, bonds explicitly. Condensed structures are more like a shorthand version of Lewis structures with atoms bonded to one atom being shown after that atom. Bonds are not typically shown in a condensed structure. Skeletal structures are an even quicker way of drawing a structure. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. Carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. eteroatoms (atoms that are not carbon or hydrogen) are shown as are any nonbonding pairs of electrons. ydrogens, however, are not shown. If a carbon is shown with less than 4 bonds, it is assumed that the remaining atoms bonded to it are hydrogens. This is the most common structure used when drawing organic compounds. Sawhorse structures are drawn as to give a more three-dimensional look to the structure. They are drawn looking down a carbon-carbon bond in the molecule. Which carbon-carbon bond used may vary. (2) Look at the structure above and determine what kind of structure it is. This structure is drawn by showing atoms bonded to one atom being shown after that atom. No bonds are explicitly drawn. It must be a condensed structure. Therefore, the correct answer is (D).

4 No. 4 of What kind of structure is shown below? C C C C (A) Lewis structure (B) Condensed structure (C) Sawhorse structure (D) Skeletal structure (E) Ionic structure A. Correct! This structure is drawn using the valence electrons of the compound. Bonds are explicitly shown so it must be a Lewis structure. Condensed structures are a shortened version of Lewis structures with atoms bonded to one atom being shown after that atom. Bonds are not typically shown in a condensed structure. ere, the bonds are drawn out so it can not be a condensed structure. Go back and review the different types of structures used to represent organic compounds. Sawhorse structures are drawn to give the structure a more three-dimensional look by drawing it as though one was looking down a carbon-carbon bond in the molecule. This is a very flat representation of a molecule so it can not be a sawhorse structure. Go back and review the different types of structures used to represent organic compounds. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. In the above structure, all carbons and hydrogens are shown. Go back and review the different types of structures used to represent organic compounds. Ionic structures refer to compounds composed of a metal and nonmetal (salt) that bond through the complete transfer of electrons between atoms. Go back and review the different types of structures used to represent organic compounds. (1) Recall the different types of structures that may be drawn to represent an organic compound. Lewis structures show all individual atoms and nonbonding pairs of electrons. They are drawn using the valence electrons of the compound and show most, if not all, bonds explicitly. Condensed structures are more like a shorthand version of Lewis structures with atoms bonded to one atom being shown after that atom. Bonds are not typically shown in a condensed structure. Skeletal structures are an even quicker way of drawing a structure. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. Carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. eteroatoms (atoms that are not carbon or hydrogen) are shown as are any nonbonding pairs of electrons. ydrogens, however, are not shown. If a carbon is shown with less than 4 bonds, it is assumed that the remaining atoms bonded to it are hydrogens. This is the most common structure used when drawing organic compounds. Sawhorse structures are drawn as to give a more three-dimensional look to the structure. They are drawn looking down a carbon-carbon bond in the molecule. Which carbon-carbon bond used may vary. (2) Look at the structure above and determine what kind of structure it is. This structure is drawn using the valence electrons of the compound. The bonds are explicitly drawn. It must be a Lewis structure. Therefore, the correct answer is (A).

5 No. 5 of ow many carbons are in the skeletal structure below? Cl (A) 6 (B) 8 (C) 10 (D) 12 (E) 14 O There are more than 6 carbons in the skeletal structure above. Remember, carbons are assumed to be anywhere a line ends or where two lines meet (at the vertices). There are more than 8 carbons in the skeletal structure above. Remember, carbons are assumed to be anywhere a line ends or where two lines meet (at the vertices). C. Correct! Carbons are assumed to be anywhere a line ends or where two lines meet. There are 10 carbons in this structure. There are less than 12 carbons in the skeletal structure above. Remember, carbons are assumed to be anywhere a line ends or where two lines meet (at the vertices). There are less than 14 carbons in the skeletal structure above. Remember, carbons are assumed to be anywhere a line ends or where two lines meet (at the vertices). (1) Recall what the tutorial said about skeletal structures. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. Carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. eteroatoms (atoms that are not carbon or hydrogen) are shown as are any nonbonding pairs of electrons. ydrogens, however, are not shown. If a carbon is shown with less than 4 bonds, it is assumed that the remaining atoms bonded to it are hydrogens. (2) Study the structure and count the number of carbons it contains. Number all the carbons in the structure below. Remember, carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. Cl O 10 Therefore, the correct answer is (C).

6 No. 6 of What is the relationship between the two molecules below? C 2 C 3 C 2 C 3 C 3 C 3 (A) Constitutional isomers (B) Configurational isomers (C) Conformational isomers (D) Trans/Cis isomers (E) Stereoisomers Constitutional isomers are compounds that share the same molecular formula but have different points of attachment. The points of attachment, or how the atoms are bonded together, in the above molecules are the same. Go back and review the different types of isomers. Configurational isomers have the same molecular formula and same points of attachment, but different orientations in space. Alkene cis/trans isomers would be an example of this type of isomer. Go back and review the different types of isomers. C. Correct! Conformational isomers have the same formula and points of attachment but differ by a rotation about a single bond. One can see the structures above differ by a 60 0 rotation about the C2-C3 bond. Trans/cis isomers pertain to alkene configurations. There are no double bonds in the structure above. Go back and review the different types of isomers. Stereoisomers are a type of isomer where the atoms in a molecule have different three-dimensional spatial orientations. Go back and review the different types of isomers. (1) Recall the different type of isomers mentioned in the tutorial. Constitutional isomers are compounds that share the same molecular formula but have different points of attachment. An example would be an alcohol and an ether that shared the same molecular formula. Configurational isomers have the same molecular formula and same points of attachment, but different orientations in space. Alkene cis/trans isomers would be an example of this type of isomer. Conformational isomers have the same formula and points of attachment but differ by a rotation about a single bond. Examples include the eclipsed and staggered conformers of ethane. (2) Look at the structures above and determine what kind of isomer they are. Looking at these structures, one can determine they share the same molecular formula. Next, look at how the atoms are bonded together. Above, the atom connections are the same in both molecules. Lastly, determine if there is a different orientation in space or if a rotation about a single bond has occurred. In the above structures, one can see the only difference between the two is that the front carbon in the C2-C3 bond of the sawhorse structure has been rotated The structures must be conformational isomers. Therefore, the correct answer is (C).

7 No. 7 of Which term below correctly describes this configurational isomer? (A) Trans (B) Cis (C) Anti (D) Gauche (E) Staggered Trans isomers are those where similar groups are on opposite sides of the double bond. In this structure, where are the similar groups located? Go back and review the different types of isomers. B. Correct! The similar groups in the above compound are on the same side of the double bond. Therefore, it must a cis alkene. Anti refers to a type of conformational isomer where the two largest groups on adjacent carbons are 180 degrees apart. Go back and review the different types of isomers. Gauche refers to a type of conformational isomer where two substituents on adjacent carbons are 60 degrees apart. Go back and review the different types of isomers. Staggered refers to a type of conformational isomer where the groups on adjacent carbons are as far as possible from each other. Go back and review the different types of isomers. (1) Recall the different types of configurational isomers. The two types of configurational isomers covered in the tutorial are trans and cis isomers of alkenes. In a trans alkene, similar groups are on opposite sides of the double bond. In a cis alkene, similar groups are on the same side of the double bond. (2) Determine the position of similar groups on the alkene above. Look at the structure. It is a skeletal structure. There are two hydrogens, one bonded to each end of the double bond, that are not explicitly drawn in the structure. Go ahead and add them: Now, circle the similar groups on each end of the double bond: Determine if the similar are on the same side or opposite of each other. In the structure above, the groups are on the same side of the double bond so it is a cis alkene. Therefore, the correct answer is (B).

8 No. 8 of What is the molecular formula of the skeletal structure below? O (A) C 7 12 O (B) C 7 12 O (C) C 8 12 O (D) C 6 12 O (E) C 8 14 O Go back and review skeletal structures. Then determine the number of each type of atom present in the molecule. Go back and review skeletal structures. Then determine the number of each type of atom present in the molecule. Go back and review skeletal structures. Then determine the number of each type of atom present in the molecule. Go back and review skeletal structures. Then determine the number of each type of atom present in the molecule. E. Correct! There are 8 carbons, 14 hydrogens, and 1 oxygen in this molecule. (1) Recall how to interpret skeletal structures. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. Carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. eteroatoms (atoms that are not carbon or hydrogen) are shown as are any nonbonding pairs of electrons. ydrogens, however, are not shown. If a carbon is shown with less than 4 bonds, it is assumed that the remaining atoms bonded to it are hydrogens. (2) Determine what elements are present in the molecule. Looking at the structure, oxygen is the only heteroatom shown. The rest of the molecule is made up of carbon and hydrogen. (3) Count each type of atom present. Number each carbon in the molecule: O There are 8 carbons in the molecule. Now, count the hydrogens in the molecule. It may help to draw them in until you are comfortable with skeletal structures. O C 3 C 3 There are 14 hydrogens in the molecule. Lastly, how many oxygens are present? Just one. Now put all this information together in the form of a molecular formula: C 8 14 O. Therefore, the correct answer is (E).

9 No. 9 of What abbreviation should be used for the following structure? O (A) Me (B) Et (C) Py (D) Ac (E) Ph Me is the abbreviation for a methyl group (-C 3 ). Go back and review the common abbreviations used in The abbreviation Et stands for an ethyl group. Go back and review the common abbreviations used in The abbreviation Py stands for a six-membered aromatic ring where one atom of the ring is a nitrogen. Go back and review the common abbreviations used in D. Correct! This structure is an acetyl group. The abbreviation Ph stands for a phenyl group or benzene ring. Go back and review the common abbreviations used in (1) Recall the common organic abbreviations. You will need to know the abbreviations and the structures they represent. This structure contains a carbonyl bonded to a methyl group. It is called an acetyl group. Its abbreviation is Ac. Therefore, the correct answer is (D).

10 No. 10 of What is the relationship between these two molecules? Et C 3 C C C C 3 C 2 C 3 (A) Constitutional isomers (B) Configurational isomers (C) Conformational isomers (D) Stereoisomers (E) Identical Constitutional isomers have the same molecular formula but different points of attachment. In these structures, the atom connections are the same. Go back and review the different types of isomers. Configurational isomers differ in their orientation around a double bond. ere, these structures are the same. Go back and review the different types of isomers. Conformational isomers different by a rotation about a single bond. These structures do not. Go back and review the different types of isomers. Stereoisomers are a type of isomer where the atoms in a molecule have different three-dimensional spatial orientations. These molecules have the same three-dimensional structure. Go back and review the different types of isomers. E. Correct! These structures are the same molecule, just drawn in a different style of structure. (1) Recall the types of structures covered in the tutorial. Lewis structures show all individual atoms and nonbonding pairs of electrons. They are drawn using the valence electrons of the compound and show most, if not all, bonds explicitly. In a skeletal structure, carbon and hydrogen atoms are not explicitly drawn. Carbons are assumed to be at the point where two lines, representing bonds, meet or at the end of such a bond. The structure on the left you should recognize as a skeletal structure. The one on the right is a Lewis structure. (2) Study each structure and determine where they differ. If needed, convert the skeletal structure into a Lewis structure and then compare the two structures. These structures have the same molecular formula and the same points of attachment. They share the same configuration around the double bond. There seems to be no differences due rotation about a single bond. They are in fact identical structures. Therefore, the correct answer is (E).