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The figure illustrates three ways to represent molecules of n dash butane and 2 dash methlylpropane. In the first row of the figure, Lewis structural formulas show element symbols and bonds between atoms. The n dash butane molecule shows 4 carbon atoms represented by the letter C bonded in a straight horizontal chain with hydrogen atoms represented by the letter H bonded above and below all carbon atoms. H atoms are bonded at the ends to the left and right of the left-most and right-most C atoms. In the second row, ball-and-stick models are shown. In these representations, bonds are represented with sticks, and elements are represented with balls. Carbon atoms are black and hydrogen atoms are white in this image. In the third row, space-filling models are shown. In these models, atoms are enlarged and pushed together, without sticks to represent bonds. The molecule names are provided in the fourth row.

The compounds n -butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The n -butane molecule contains an unbranched chain , meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal , or the prefix n , to refer to a chain of carbon atoms without branching. The compound 2–methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms)

Identifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in [link] all represent the same molecule, n -butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.

The figure illustrates three ways to represent molecules of n dash butane. In the first row of the figure, Lewis structural formulas show carbon and hydrogen element symbols and bonds between the atoms. The first structure in this row shows three of the linked C atoms in a horizontal row with a single C atom bonded above the left-most carbon. The left-most C atom has two H atoms bonded to it. The C atom bonded above the left-most C atom has three H atoms bonded to it. The C atom bonded to the right of the left-most C atom has two H atoms bonded to it. The right-most C atom has three H atoms bonded to it. The C atoms and the bonds connecting all the C atoms are red. The second structure in the row similarly shows the row of three linked C atoms with a single C atom bonded below the C atom to the left. The left-most C atom has two H atoms bonded to it. The C atom bonded below the left-most C atom has three H atoms bonded to it. The C atom bonded to the right of the left-most C atom has two H atoms bonded to it. The right-most atom has three H atoms bonded to it. All the C atoms and the bonds between them are red. The third structure has two C atoms bonded in a row with a third C atom bonded above the left C atom and the fourth C atom bonded below the right C atom. The C atom bonded above the left C atom has three H atoms bonded to it. The left C atom has two H atoms bonded to it. The right C atom has two H atoms bonded to it. The C atom bonded below the right C atom has three H atoms bonded to it. All the C atoms and the bonds between them are red. In the second row, ball-and-stick models for the structures are shown. In these representations, bonds are represented with sticks, and elements are represented with balls. Carbon atoms are black and hydrogen atoms are white in this image. In the third row, space-filling models are shown. In these models, atoms are enlarged and pushed together, without sticks to represent bonds.
These three representations of the structure of n-butane are not isomers because they all contain the same arrangement of atoms and bonds.

The basics of organic nomenclature: naming alkanes

The International Union of Pure and Applied Chemistry ( IUPAC ) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules:

  1. To name an alkane, first identify the longest chain of carbon atoms in its structure. A two-carbon chain is called ethane; a three-carbon chain, propane; and a four-carbon chain, butane. Longer chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10). These prefixes can be seen in the names of the alkanes described in [link] .
  2. Add prefixes to the name of the longest chain to indicate the positions and names of substituents . Substituents are branches or functional groups that replace hydrogen atoms on a chain. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.
This figure shows structural formulas for propane, 2 dash chloropropane, 2 dash methylpropane, 2 comma 4 dash difluorohexane, and 1 dash bromo dash 3 dash chlorohexane. In each of the structures, the carbon atoms are in a row with bonded halogen atoms and a methyl group bonded below the figures. Propane is listed as simply C H subscript 3 C H subscript 2 C H subscript 3, with the numbers 1, 2, and 3 appearing above the carbon atoms from left to right. 2 dash chloropropane similarly shows C H subscript 3 C H C H subscript 3, with the numbers 1, 2, and 3 appearing above the carbon atoms from left to right. A C l atom is bonded below carbon 2. The C l atom is red. 2 dash methylpropane similarly shows C H subscript 3 C H C H subscript 3, with the numbers 3, 2, and 1 appearing above the carbon atoms from left to right. A C H subscript 3 group is bonded beneath carbon 2 and is red. 2 comma 4 dash difluorohexane similarly shows C H subscript 3 C H subscript 2 C H C H subscript 2 C H C H subscript 3, with the numbers 6, 5, 4, 3, 2, and 1 appearing above the carbon atoms from left to right. F atoms are bonded to carbons 4 and 2 at the bottom of the structure and are red. 1 dash bromo dash 3 dash chlorohexane similarly shows C H subscript 2 C H subscript 2 C H C H subscript 2 C H subscript 2 C H subscript 3, with numbers 1, 2, 3, 4, 5, and 6 appearing above the carbon atoms from left to right. B r is bonded below carbon 1 and C l is bonded below carbon 3. Both B r and C l are red.

When more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ide at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with -ide like chloride; in organic compounds, such atoms are treated as substituents and the -o ending is used). The number of substituents of the same type is indicated by the prefixes di- (two), tri- (three), tetra- (four), and so on (for example, difluoro- indicates two fluoride substituents).

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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