of 21 /21
Al ka ne R–CH 2 CH 2 –R C n H 2n+2 This is the maximum H/C ratio for a given number of carbon atoms. Al ke ne R– CH=CH– R C n H 2n Each double bond reduces the number of hydrogen atoms by 2. Al ky ne R–C≡C– R C n H 2n-2 Each triple bond reduces the number of hydrogen atoms by 4. Functional Group Ending only C and H atoms e (alkanes, alkenes, and alkynes) alcohol ol aldehyde al ketone one carboxylic acid ic acid Compounds containing halogens Example 1: Write the structural formula for 1,1,1- trichloroethane. This is a two carbon chain (eth) with no double bonds (an). There are three chlorine atoms all on the first carbon atom. Example 2: Write the structural formula for 2-bromo- 2-methylpropane.

# Alkan r–ch2– e Ch2–r

Embed Size (px)

### Text of Alkan r–ch2– e Ch2–r

Alkan RCH2 e CH2R

CnH2n+ This is the maximum H/C ratio for a given number of carbon atoms. 2 Each double bond reduces the number of hydrogen atoms by 2.

Alken RCH=CH CnH2n e R Alkyn RCCR eFunctional Group

CnH2n- Each triple bond reduces the number of hydrogen atoms by 4. 2Ending

only C and H atoms e (alkanes, alkenes, and alkynes) alcohol aldehyde ketone carboxylic acid ol al one ic acid

Compounds containing halogens

Example 1: Write the structural formula for 1,1,1-trichloroethane. This is a two carbon chain (eth) with no double bonds (an). There are three chlorine atoms all on the first carbon atom.

Example 2: Write the structural formula for 2-bromo-2methylpropane. First sort out the carbon skeleton. It's a three carbon chain with no double bonds and a methyl group on the second carbon atom.

Draw the bromine atom which is also on the second carbon.

And finally put the hydrogen atoms in.

If you had to name this yourself: Notice that the whole of the hydrocarbon part of the name is written together - as methylpropane - before you start adding anything else on to the name. Example 2: Write the structural formula for 1-iodo-3-methylpent-2ene. This time the longest chain has 5 carbons (pent), but has a double bond starting on the number 2 carbon. There is also a methyl group on the number 3 carbon.

Now draw the iodine on the number 1 carbon.

Giving a final structure:

Note: You could equally well draw this molecule the other way round, but normally where you have, say, 1-bromo-something, you tend to write the bromine (or other halogen) on the right-hand end of the structure.

]

Alcohols

All alcohols contain an -OH group. This is shown in a name by the ending ol. Example 1: Write the structural formula for methanol. This is a one carbon chain with no carbon-carbon double bond (obviously!). The ol ending shows it's an alcohol and so contains an -OH group.

Example 2: Write the structural formula for 2-methylpropan-1-ol. The carbon skeleton is a 3 carbon chain with no carbon-carbon double bonds, but a methyl group on the number 2 carbon.

The -OH group is attached to the number 1 carbon.

The structure is therefore:

Example 3: Write the structural formula for ethane-1,2-diol. This is a two carbon chain with no double bond. The diol shows 2 -OH groups, one on each carbon atom.

Note: There's no particular significance in the fact that this formula has the

carbon chain drawn vertically. If you draw it horizontally, unless you stretch the carbon-carbon bond a lot, the -OH groups look very squashed together. Drawing it vertically makes it look tidier!

Aldehydes All aldehydes contain the group:

If you are going to write this in a condensed form, you write it as -CHO - never as -COH, because that looks like an alcohol. The names of aldehydes end in al. Example 1: Write the structural formula for propanal. This is a 3 carbon chain with no carbon-carbon double bonds. The al ending shows the presence of the -CHO group. The carbon in that group counts as one of the chain.

Example 2: Write the structural formula for 2-methylpentanal. This time there are 5 carbons in the longest chain, including the one in the -CHO group. There aren't any carbon-carbon double bonds. A methyl group is attached to the number 2 carbon. Notice that in aldehydes, the carbon in the -CHO group is always counted as the number 1 carbon.

Ketones Ketones contain a carbon-oxygen double bond just like aldehydes, but this time it's in the middle of a carbon chain. There isn't a hydrogen atom attached to the group as there is in aldehydes. Ketones are shown by the ending one.

Example 1: Write the structural formula for propanone. This is a 3 carbon chain with no carbon-carbon double bond. The carbon-oxygen double bond has to be in the middle of the chain and so must be on the number 2 carbon.

Ketones are often written in this way to emphasise the carbon-oxygen double bond. Example 2: Write the structural formula for pentan-3-one. This time the position of the carbon-oxygen double bond has to be stated because there is more than one possibility. It's on the third carbon of a 5 carbon chain with no carbon-carbon double bonds. If it was on the second carbon, it would be pentan-2-one.

This could equally well be written:

Where Write the structural formula for 3-methylbutanoic acid. This is a four carbon acid with no carbon-carbon double bonds. There is a methyl group on the third carbon (counting the -COOH carbon as number 1).

Example 2: Write the structural formula for 2-hydroxypropanoic acid.

The hydroxy part of the name shows the presence of an -OH group. Normally, you would show that by the ending ol, but this time you can't because you've already got another ending. You are forced into this alternative way of describing it.

The old name for 2-hydroxypropanoic acid is lactic acid. That name sounds more friendly, but is utterly useless when it comes to writing a formula for it. In the old days, you would have had to learn the formula rather than just working it out should you need it. Example 3: Write the structural formula for 2-chlorobut-3-enoic acid. This time, not only is there a chlorine attached to the chain, but the chain also contains a carbon-carbon double bond (en) starting on the number 3 carbon (counting the -COOH carbon as number 1).

Salts of carboxylic acids Example: Write the structural formula for sodium propanoate. This is the sodium salt of propanoic acid - so start from that. Propanoic acid is a three carbon acid with no carbon-carbon double bonds.

When the carboxylic acids form salts, the hydrogen in the -COOH group is replaced by a metal. Sodium propanoate is therefore:

Notice that there is an ionic bond between the sodium and the propanoate group. Whatever you do, don't draw a line between the sodium and the oxygen. That would represent a covalent bond. It's wrong, and makes you look very incompetent in an exam! In a shortened version, sodium propanoate would be written CH3CH2COONa or, if you wanted to emphasise the ionic nature, as CH3CH2COO- Na+.Note: The confusing thing about these salts (and even more so for the esters that are coming up next) is that they are named the wrong way round. In the formula, the sodium is at the end, but appears first in the name. Why? Salts are always named with the metal first - think of sodium chloride or potassium iodide. So for consistency you would need to reverse the formula of sodium propanoate NaOOCCH2CH3. But if you reverse the formula, you can't see immediately that it is related to propanoic acid. So you learn to live with the inconsistency.

Esters Esters are one of a number of compounds known collectively as acid derivatives. In these the acid group is modified in some way. In an ester, the hydrogen in the -COOH group is replaced by an alkyl group (or possibly some more complex hydrocarbon group). Example 1: Write the structural formula for methyl propanoate. An ester name has two parts - the part that comes from the acid (propanoate) and the part that shows the alkyl group (methyl). Start by thinking about propanoic acid - a 3 carbon acid with no carbon-carbon double bonds.

The hydrogen in the -COOH group is replaced by an alkyl group in this case, a methyl group.

Ester names are confusing because the name is written backwards from the way the structure is drawn. There's no way round this you just have to get used to it! In the shortened version, this formula would be written CH3CH2COOCH3. Example 2: Write the structural formula for ethyl ethanoate. This is probably the most commonly used example of an ester. It is based on ethanoic acid ( hence, ethanoate) - a 2 carbon acid. The hydrogen in the -COOH group is replaced by an ethyl group.

Make sure that you draw the ethyl group the right way round. A fairly common mistake is to try to join the CH3 group to the oxygen. If you count the bonds if you do that, you will find that both the CH3 carbon and the CH2 carbon have the wrong number of bonds.

Acyl chlorides (acid chlorides) An acyl chloride is another acid derivative. In this case, the -OH group of the acid is replaced by -Cl. All acyl chlorides contain the -COCl group:

Example: Write the structural formula for ethanoyl chloride. Acyl chlorides are shown by the ending oyl chloride. So ethanoyl chloride is based on a 2 carbon chain with no carbon-carbon double bonds and a -COCl group. The carbon in that group counts as part of the chain. In a longer chain, with side groups attached,

the -COCl carbon is given the number 1 position.

Acid anhydrides Another acid derivative! An acid anhydride is what you get if you dehydrate an acid - that is, remove water from it. Example: Write the structural formula for propanoic anhydride. These are most easily worked out by writing it down on a scrap of paper in the following way:

Draw two molecules of acid arranged so that the -OH groups are next to each other. Tweak out a molecule of water - and then join up what's left. In this case, because you want propanoic anhydride, you draw two molecules of propanoic acid.

Amides Yet another acid derivative! Amides contain the group -CONH2 where the -OH of an acid is replaced by -NH2. Example: Write the structural formula for propanamide. This is based on a 3 carbon chain with no carbon-carbon double bonds. At the end of the chain is a -CONH2 group. The carbon in that group counts as part of the chain.

Nitriles Nitriles contain a -CN group, and used to be called cyanides. Example 1: Write the structural formula for ethanenitrile. The name shows a 2 carbon chain with no carbon-carbon double bond. nitrile shows a -CN group at the end of the chain. As with the previous examples involving acids and acid derivatives, don't forget that the carbon in the -CN group counts as part of the chain.

The old name for this would have been methyl cyanide. You might think that that's easier, but as soon as the chain gets more complicated, it doesn't work - as the next example shows. Example 2: Write the structural formula for 2hydroxypropanenitrile. Here we've got a 3 carbon chain, no carbon-carbon double bonds, and a -CN group on the end of the chain. The carbon in the -CN group counts as the number 1 carbon. On the number 2 carbon there is an -OH group (hydroxy). Notice that you can't use the ol ending because you've already got a nitrile ending.

Primary amines A primary amine contains the group -NH2 attached to a hydrocarbon chain or ring. You can think of amines in general as being derived from ammonia, NH3. In a primary amine, one of the hydrogens has been replaced by a hydrocarbon group. Example 1: Write the structural formula for ethylamine. In this case, an ethyl group is attached to the -NH2 group.

This name (ethylamine) is fine as long as you've only got a short

chain where there isn't any ambiguity about where the -NH2 group is found. But suppose you had a 3 carbon chain - in this case, the -NH2 group could be on an end carbon or on the middle carbon. How you get around that problem is illustrated in the next example. Example 2: Write the structural formula for 2-aminopropane. The name shows a 3 carbon chain with an amino group attached to the second carbon. amino shows the -NH2 group.

Ethylamine (example 1 above) could equally well have been called aminoethane.

Secondary and tertiary amines You are only likely to come across simple examples of these. In a secondary amine, two of the hydrogen atoms in an ammonia molecule have been replaced by hydrocarbon groups. In a tertiary amine, all three hydrogens have been replaced. Example 1: Write the structural formula for dimethylamine. In this case, two of the hydrogens in ammonia have been replaced by methyl groups.

Example 2: Write the structural formula for trimethylamine. Here, all three hydrogens in ammonia have been replaced by methyl groups.

Amino acids An amino acid contains both an amino group, -NH2, and a carboxylic acid group, -COOH, in the same molecule. As with all acids the carbon chain is numbered so that the carbon in the -COOH group is counted as number 1. Example: Write the structural formula for 2-aminopropanoic acid. This has a 3 carbon chain with no carbon-carbon double bonds. On the second carbon (counting the -COOH carbon as number 1) there is an amino group, -NH2.

What are the names of the following compounds?

a.

b.

c.

(CH3)2CHCH2CH2CH2CHCH2

d.

Solution:

a.

This consists of a six-membered ring, so the root is cyclohex. There are no double or triple bonds, nor any elements but carbon and hydrogen, so the suffix is ane, making cyclohexane. There is a branch with a one-carbon chain, which adds a methyl prefix. You would start numbering at the end as close to the branch as possible. However, since this is a cyclic compound, there is no end, so you would start at the branch. Since there is only one branch, the numbering is redundant. The best name is methylcyclohexane, but 1methylcyclohexane is acceptable. b. The longest chain has five carbons, making the root word pent. There are no carbon carbon double bonds, so pent is followed by an to make pentan. The functional group is a ketone, so the suffix is onemaking pentanone. The location of the functional group needs to be identified. If you start numbering the compound (at either end), the functional group is on carbon 3 making the full name 3-pentanone. c. This longest chain is seven carbons long. It branches at the end. There is a double bond at the other end. The seven-carbon chain makes it hept. The double bond adds hepten. Since there is no other functional group, this makes it heptene. The double bond starts with an end carbon, so that carbon is carbon number 1, and so 1-heptene. If you continue numbering the carbons, the branch is on carbon 6. The branch has one carbon, a methyl. The name is 6methyl-1-heptene.

d. The longest chain is six carbons long, although there are three ways to make a six-carbon chain. The root word is hex. The best choice is the chain that includes the functional groups. In this example, the only functional group is an alkene. Thus hexene. The chain is numbered, so that the functional group will have the lowest value. The longest chain, numbered in that manner, would be

. Either system will give the same name. The alkene is at carbon 2. There is one carbon branch (methyl) at carbon 2 and a two-carbon one (ethyl) at carbon 3. Therefore the name would be 1-methyl-2-ethyl-2-hexene.>> Naming Ethers

When naming ethers, name either side of the oxygen using the rules above and a yl suffix. String the two pieces together as separate words and add the third word, ether, at the end. If both pieces are the same, use the name only once with the prefix di. (e.g., dimethyl ether, not methyl methyl ether.)>> Naming Amines

Name each carbon chain attached to the nitrogen according to the preceding rules, ending with the suffix yl. String each piece together as one word ending with amine. If two pieces are the same, use the prefix di rather than repeating the name. If three pieces are the same, use the prefix tri rather than repeating the name (as with ethers).>> Example 2

Name the following compounds. a. b. c.CH3CH2OCH2CH2CH2CH3 (CH3CH2)2NH CH3NH2

d. Solution: a.

CH3OCH3

This is an ether. One side of the oxygen has two carbons (ethyl) and the other has four carbons (butyl). Therefore the name of the compound is ethyl butyl ether. b. This is a secondary amine (it has a nitrogen attached to two carbons). The carbon chains attached to the nitrogen are the same and have two carbons (ethyl), so the name of this compound is diethylamine. c. This is a primary amine. The one-carbon group is a methyl. Therefore the name of this compound is methylamine. d. This is an ether. The groups on either side of the oxygen are both methyl groups. Therefore the name of this compound is dimethyl ether.

>> back to the Top of the Page

B. Finding Structures from Names >> Example 3

What are the structures of the following compounds?a. b. c. d. e. f. 2,2-dimethylbutane 3-propyl-5-hexynal (hexanynal) 1-octanol tripropylamine 3-methylpentanoic acid 5-ethyl-2-heptyne

Solution: a. The butane is a four-carbon chain. There are two methyl (one-carbon) groups on the second carbon.

b. 3-propyl-5-hexynal has a six-carbon chain (hex), with a triple bond between the 5 and 6 carbons (yn) and an aldehyde (al) on the end (#1 carbon). Aldehydes must be on the end; otherwise they would be ketones. A threecarbon branch (propyl) is on the third carbon.

c. 1-octanol. There are eight carbons in the chain (oct), no double or triple bonds (an), and an alcohol (ol) on the first carbon (on either end). CH CH CH CH CH CH CH CH OH3 2 2 2 2 2 2 2

d. tripropylamine. The amine is a nitrogen that has three bonds. There are three propyl groups attached to the amine, so there is no space for other hydrogens. All the groups are the same (tri) and have three carbons (propyl). The structure could be written as (CH CH CH ) N or3 2 2 3

e.

3-methylpentanoic acid. There are five carbons in the chain (pent), no double or triple bonds (an), and a carboxylic acid at the end (oic acid). Carboxylic

acids can only be at the end, and the carbon of that group is carbon 1. On the third carbon from the carboxylic acid is a one-carbon group (methyl).

f. 5-ethyl-2-heptyne is a seven-carbon chain (hept) with a triple bond between the second and third carbons (yne) and a two-carbon group (ethyl) on the fifth carbon.

>> Geometric Isomers

This is only an issue for alkenes. In addition, if the groups attached to one of the carbons in the double bond are the same, there are no geometric isomers. The prefixes cis and trans are used when one group on the carbons opposite the double bond is the same. If those groups are on the same side, it is a cis isomer; if they are on opposite sides, a trans isomer.>> Example 4

Identify the following as cis and trans isomers. Name each compound.

a.

b.

c.

Solution: a. The group that is the same is the H group. It is on the same side (bottom) of the double bond; therefore it is a cis isomer. The longest carbon chain is from the upper left to the upper right, seven carbons (hept). The double bond (en) is between the 3 and 4 carbons (3). As above, it is the cis isomer. Consequently, the name is cis-3-heptene. b. The group that is the same on either side of the double bond is CH CH (ethyl). The group is on opposite sides of the double bond; it is therefore a trans isomer.3 2

The longest chain is again from the upper left to the upper right, and seven carbons long (hept). The double bond is between the 3 and 4 carbons (3- and ene). There is also a two-carbon branch from carbon 4. So the name is trans4-ethyl-3-heptene. c. The group that is the same is the ethyl group on the carbon to the right. Because those are the same carbon, this compound does not have a geometric isomer. The longest chain is from the lower left to the upper or lower right. It doesn't matter; thelength will be nine carbons (non). The double bond (en) is closer to the right end, so the numbering should start from there. That makes the double bond between carbons 3 and 4. There is also a two-carbon branch on carbon 3. The name is 3-ethyl-3-nonene.

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Education