Orgo: Nomenclature & Functional Groups

Question 1

functional group

Answer 1

A functional group is the part of a molecule where most of its chemical reactions occur. It is the part that effectively determines a compound’s chemical properties in addition to many of its physical properties.

Question 2

Alkyl groups

Answer 2

general, non-aromatic hydrocarbon groups that would be obtained by removing a hydrogen atom from an alkane:

Question 3

The symbol R

Answer 3

represent a generalized alkyl group. The group can be methyl, ethyl, propyl, or any of a multitude of others. You might think of R as representing the Rest of a molecule, which we aren’t bothering to specify because it’s not important

example: .

The R may be methyl, isopropyl, or ethyl:

Question 4

aryl group (Ar)

Answer 4

an aromatic hydrocarbon group derived from an arene by removal of a hydrogen atom.

a general class of primary carboxylic acids may be represented by:

(general class of carboxylic acids)

The Ar may be C6H4CH3, C6H3(CH3)2, or C6H2(CH3)3:

Question 5

phenyl group.

Answer 5

aryl group derived from the removal of a proton from benzene

Question 6

What is the name for the general fragment:

?

(RC=O) where C is connected to something else too

Answer 6

Systematic name:

alkanoyl (alkyl + carbonyl)

aka: acyl group

**Two common ones: **

Question 7

Major Nitrogen-, Oxygen-, and halogen-containing functional groups:

• nitrile
• imine
• amine
• amide
• nitro
• carbonyl
• aldehyde
• ketone
• alcohol
• carboxylic acid
• ester
• ether
• alkanol halide
• alkyl halide

Question 8

Carboxylic acids can act as hydrogen bond donors and acceptors. Think about the locations for the hydrogen bonds between the two carboxylic acid molecules shown below.

Answer 8

In carboxylic acids, a positively polarized -OH hydrogen atom from one molecule is attracted to a negatively polarized oxygen atom of another molecule. This results in a weak force between the molecules that holds them together. These intermolecular attractions must be overcome for a molecule to break free from the liquid and enter the vapor state, so carboxylic acids typically have higher boiling points than alkanes or ketones, both of which cannot act as hydrogen donor/acceptors.

Other functional groups capable of H-bonding include alcohols ( ), amides ( ), amines ( ), and imines ( ).

Question 9

Name the functional groups shown in the molecule below

Question 10

Oxididation Levels and Functional Groups

Answer 10

Functional groups which have the same oxidation level can interconvert by substitution reactions. To go from one level to another, an oxidizing or reducing agent is necessary.

For example, primary alcohols may be oxidized to aldehydes in the presence of the oxidizing agent PCC (pyridinium chlorochromate):

Ketones undergo reduction to secondary alcohols in the presence of NaBH4 (sodium borohydride):

Question 11

Bonds and Oxidation Levels

(arrange the common functional groups in order from least to most oxidized)

Answer 11

• The more bonds a carbon has to an electronegative element, the more oxidized it is. **Note: Halogens (X) count too as EN.
• The more bonds a carbon has to hydrogen, the more reduced it is.
• Hydrocarbons are in the lowest oxidation level.

Question 12

Nucleophiles v. Electrophiles, (vs Inert)

Answer 12

• Nucleophiles are literally “things that love nuclei”. Nuclei are positively-charged. Since positive charges attract them, nucleophiles are either negatively-charged, or at least have a negatively-polarized part that is reactive.
• Electrophiles “love electrons”. They are either positively-charged or have a positively-polarized part.

Nucleophiles react with Electrophiles!

Question 13

FG Reactivity Special Behavior #1:

Protonation of an Alcohol

Answer 13

Alcohols can react as either nucleophiles or electrophiles.

When an alcohol is protonated, it becomes positively-charged. It will then have a good leaving group (water) and may participate in a reaction as anelectrophile.

In the first step, the alcohol acts as a nucleophile. Because of the polarized C-O bond, the oxygen has a partial negative charge and will acquire a proton from HBr.

Once protonated, the molecule bears a good leaving group in H2O. The result is formation of a carbocation, which is electron-deficient. This carbocation will react electrophilically with Br-.

Question 14

FG Reactivity Special Behavior #2:

Carbonyls as Electrophiles

Answer 14

Carbonyls can also react as either nucleophiles or electrophiles.

In the presence of a strong acid, the carbonyl compound will become protonated thus becoming an electrophile.

In this example, the carbonyl double bond is polarized. That is, the effect of the electronegative oxygen is to leave the carbonyl carbon with a small positive charge. As a result, the carbonyl carbon is electron-deficient and is susceptible to attack by an electron-rich species (a nucleophile).

Once the carbonyl acquires a proton, it can react with an appropriate nucleophile to form the corresponding addition product.

By now you will have reviewed the basic structure and properties of functional groups that will appear on Test Day.

Let’s continue our review with a discussion of nomenclature.

Question 15

IUPAC system, the names of organic molecules

Answer 15

• Locants are the numbers that tell where principal functional group and substituents are located on the main chain or ring.
• Substituent Prefixes identify what substituents are located on the main chain or ring.
• The Parent is the part that identifies the size of the main chain or ring.
• The Suffix identifies the compound’s principal functional group.

Question 16

Identify the locant, substituent prefix, parent and suffix in 5-methyl-2-heptanol.

Answer 16

In 5-methyl-2-heptanol, the locant 5 indicates the methyl group is a substituent of the fifth carbon in the parent hydrocarbon chain, which is 7 carbons long (heptane).

The principal functional group in this molecule is an alcohol, as indicated by the ‘ol’ suffix.

Note: the longest carbon chain must be numbered such that the principal functional group receives the lowest possible number.

Question 17

IUPAC Suffixes for the Common Functional Groups

Question 18

IUPAC Suffixes and Prefixes for the Common Functional Groups

Question 19

How do you decide which suffix to use when there are two functional groups in a structure?

Answer 19

The more oxidized the functional group, the higher priority in the name.

This means:

• The suffix of the highest priority functional group is used as the ending, and this functional group gets the lowest possible number in the C-skeleton.
• Any other functional groups become substituents.
• On Test Day, you won’t have to worry too much about cases where both functional groups have the same oxidation level.

In naming compounds, functional groups are assigned priorities according to the following order:

Question 20

What is the name of the following compound?

Answer 20

The ketone functional group is more oxidized than the bromo- group. That is, the ketone group is the principal functional group in this molecule because it contains a C=O double bond.

The parent chain is pentane (5 carbon chain). The functional group with the highest priority is assigned the lowest possible number on the carbon chain. The parent chain contains bromo- and phenyl- substituents, both on C4.

Question 21

common names of common molecules

Answer 21

Basic:

_____________

For subsituents:

Question 22

What is the structure of acetic acid?

Answer 22

IUPAC: ethanoic acid

Formula: CH₃COOH, (also written as CH₃CO₂H or C₂H₄O₂)

Question 23

For each common name, what is the corresponding structure?

• acetylene
• ethylene

Question 24

Steps for formation of the systematic name for an organic compound

Answer 24

1. Identify the longest carbon chain containing the principal functional group.
2. Identify all substituents.
3. Number the carbon chain so that the principal functional group receives the lowest number. For alkanes, number the carbon atoms so that the substituents receive the lowest possible numbers.
4. The IUPAC name is obtained by arranging the substituent prefixes in alphabetical order and attaching them to the name of the longest carbon chain.

Question 25

IUPAC Nomenclature Rule #1:

Identify the longest carbon chain containing the principal functional group.

Answer 25

• When naming compounds, the longest continuous carbon chain within the compound is taken as the backbone.
• If there are two or more chains of equal length, the most highly substituted chain takes precedence.
• Number the chain from one end in such a way that the lowest set of numbers is obtained for the substituents.

Example (not always obvious):

Question 26

Straight chain alkanes have the general formula C_nH_{2n +2} (n is an integer)

Q: How many isomers exist with the formula C₅H₁₂?

Answer 26

Answer: 3

**Explanation: **

1. The first thing to recognize is that the formula C₅H₁₂ corresponds to a straight chain alkane (C₅H₂₍₅₎₊₂).
2. Start by drawing the simplest molecule, the one with the longest carbon chain and no branches.
   
3. Now be systematic. Decrease the length of the carbon chain to 4 and determine the number of possible isomers.
   
4. Now decrease the longest chain by one more carbon.
   

There are 3 isomers of C5H12.

Question 27

Naming alkanes

Answer 27

Simplest:

Longer chain: consist of prefixes derived from the Greek root for the number of carbon atoms, with the ending** -ane.**

Question 28

IUPAC Nomenclature Rule #2:

Identify all substituents.

Answer 28

Substituents are named according to their appropriate prefix with the ending -yl. More complex substituents are named as derivatives of the longest chain in the group.

• The prefix n- in the above example indicates an unbranched (“normal”) compound.
• If there are two or more equivalent groups, the prefixes di-, tri-, tetra-, etc. are used.
• Each substituent is assigned a number to identify its point of attachment to the principal chain. If the prefixes di-, tri-, tetra-, etc. are used, a number is still necessary for each individual group.

Question 29

Name the following molecule:

Answer 29

2-chloro-2,3-dimethylbutane

The compound is a butane because the longest carbon chain is 4 carbons long. The substituents are placed in alphabetical order and are assigned a number corresponding to their position on the carbon chain. We use ‘di’ to indicate there are 2 methyl group substituents.

Question 30

Identify the substituents in the following molecule:

Question 31

IUPAC Nomenclature Rule #4:

The IUPAC name is obtained by arranging the substituent prefixes in alphabetical order and attaching them together.

Answer 31

• List the substituents in alphabetical order with their corresponding numbers. Prefixes such as di-, tri-, etc., as well as the hyphenated prefixes (tert-sec-, n-) are ignored in alphabetizing.
• By contrast, cyclo-, iso-, and** neo- **are considered part of the group name and are alphabetized.
• Commas should be placed between numbers, and hyphens should be placed between numbers and words.
• Example: The longest carbon chain containing the principal functional group is 9 carbons long (nonane). The molecule contains methyl, ethyl, and tert-butyl substituents. The chain is numbered so the substituents receive the lowest possible numbers. Finally, we arrange the substituent prefixes in alphabetical order and attach them to the name of the longest carbon chain.

Question 32

Naming Cycloalkanes

Answer 32

Alkanes can form rings.

These are named according to the number of carbon atoms in the ring with the prefix cyclo-.

Substituted cycloalkanes are named as derivatives of the parent cycloalkane. The substituents are named, and the carbon atoms are numbered around the ring starting from the point of greatest substitution.

Examples:

• The compound on the left is a cyclobutane because it is a ring containing 4 saturated carbons. Its substituents are methyl and ethyl groups. We number the ring so as to give the principal functional group the lowest number, so the carbon bearing the substituents is C1. Putting the substituents in alphabetical order gives 1-ethyl-1-methylcyclobutane.*
• The compound on the right is a cycloheptane because it’s a ring made up of 7 saturated carbons. The substituents are an isopropyl group and two methyl groups. Numbering the ring so the substituents receive the lowest numbers gives: 4-isopropyl-1,1-dimethylcycloheptane.*

Question 33

Multiple Bonds

Answer 33

• Alkenes are compounds containing carbon-carbon double bonds.
• Noncyclic alkenes have the general formula C_nH_2n (n is an integer).
• The nomenclature rules are essentially the same as for alkanes, except that the ending **-ene **is used rather than -ane.
• When identifying the carbon backbone, select the longest chain that contains the double bond (or the greatest number of double bonds, if more than one is present).

*Example: *

• The backbone is numbered so that the double bond gets the lowest possible number.*
• Substituents are named as they are for alkanes, and their positions are specified by the number of the backbone carbon to which they are attached.*

Alkenes take priority over alkanes, so our longest carbon chain must contain the alkene, which is the principal functional group in this molecule. In this case, the only substituent is an ethyl group. Next, number the carbon chain so the principal functional group gets the lowest number.

• Finally, put it together. The 2 in front of ethyl indicates the ethyl group substitutes at the 2 position, while the 1 in front of pentene indicates the double bond is located between C1 and C2.*
• Notice we had to replace the “ane” ending with “ene” to indicate the alkene principal functional group.*