Organic Chemistry: Functional Groups

  There are functional groups that fall under organic chemistry. Today, we will be learning how to name these functional groups and what differentiates them from each other.

Halides:

Halides, as the name suggests, contains elements from the halogen group (which is easy to remember considering they both begin with the same three letters). When naming a halide, one would apply the same rules in naming the parent carbon chain and then would proceed in naming the halogens as side chains.

CH3-CH2-CH2-CH-CH3

                                                                                         |
                                                                                         F

In the example above, start by:

 

1. Identifying & naming the parent chain which in this case, would be pentane.

 

2. Then count the side chain in the lowest possible numbering scheme (double or triple bonds would take priority over this rule). In this example, the number would be 2.

 

3. Finally take the the halogen, which would be flourine and make it floro (iodine=iodo, bromine=bromo, chlorine=chloro).

 

4. Put all the elements together: 2 floro pentane 

 

 

Amines:

 

  An amine includes not only carbons, but a nitrogen as well. Amines can be primary, secondary, or tertiary, depending on the number of carbon chains connected to it.

 

CH3-CH2-NH-CH3

In the example above, start by:

 

1. Identifying the side chains connected to the nitrogen which in the example, would be methyl and ethyl.

 

2. As the nitrogen is located in between the two carbon chains, no numbering is required.

 

3. Name the amine in alphabetical order: ethyl methyl amine

 

 

Amides:

 

  Amides, though also contain a nitrogen, differentiate from amines as they contain a double-bonded oxygen.

 

O=C-CH2-CH2-CH2-CH3

                                                                     |

                                                                   NH2

 

In the example above, start by:

 

1. Indentifying the parent carbon chain which, in the example, would be a chain of five carbons. Take the prefix of 'penta-' and add '-namide- to it.

 

2. Name the final product: Pentanamide.

 

 

Alcohols:

 

  The functional group of alcohols includes a single bonded '-OH'.

 

CH3-CH-CH2-CH-CH2-CH3

                                                                        |            |

                                                                      OH        OH

 

In the example above, start by:

 

1. Identifying the number of carbons in the parent chain which, in this case, would be described with the prefix, 'Hexa-'.

 

2. Number the '-OH's according to the lowest possible numbering scheme and use 'di' when creating the final name as there are two '-OH's.

 

3. Name the final product: 2, 4 hexadiol.

 

 

Aldehydes:

 

  Aldehydes include a carbon on either end, attached to a double-bonded 'O', and therefore, receiving the lowest number.

 

CH3-CH2-CH2-CH2-CH2-CH2-CH=O

 

In the example above, start by:

 

1. Identifying the number of carbons in the parent chain in the example can be described with the prefix of 'Hepta'.

 

2. Because the 'O' is found on the end of the carbon chain, there is no numbering needed.

 

3. Add to the prefix, '-nal' to complete the name: Heptanal.

 

Ketones:

 

  Ketones are similar to aldehydes in the way they have a double-bonded 'O' but differ in the way that the double-bonded 'O' is not found on either end.

 

CH3-C-CH3

                                                                                 ||

                                                                                 O

 

In the example above, start by:

 

1. Idenifying the number of carbons in the parent chain which in this case, can be described with the prefix 'Prop'.

 

2. Give the double-bonded 'O' the lowest possible number in the numbering sceme.

 

3. When you name the compound, and '-none' to it: 2 Propanone.

 

 

Carboxylic Acids:

 

  Carboxylic acids contain along with a carbon chain, a double-bonded 'O' as well as an '-OH'.

 

CH3-CH2-CH2-C=O

                                                                                          |

                                                                                        OH

 

In the example above, start by:

 

1. Identifying the number of carbons in the parent chain which, in this example, can be described with the prefix 'Buta-'.

 

2. Add '-noic acid' to the end to create the name: 'Butanoic Acid'.

 

Ethers:

 

  Ethers have carbon chains that branch off of an 'O'.

 

CH3-CH2-CH2-O-CH3

 

In the example above, start by:

 

1. Identifying the carbon chains attached to the 'O'. They are propyl and methyl.

 

2. State the carbon chains in alphabetical order and add 'ether': Methyl Propyl Ether.

 

Esters:

 

  Esters are the product of a carboxylic acid and an alcohol. Therfore, they contain a double-bonded 'O' and another 'O' in addition to a carbon chain.

 

O=CH-O-CH2-CH2-CH3

 

In the example above, start by:

 

1. Identify how many carbons are attached to the carbon that is connected to the double-bonded 'O' and single 'O'. Since there is only one, use the prefix, 'Meth-'.

 

2. Count the carbons attached to the single-bonded 'O' as a side chain.

 

3. Name the compound and add '-anoate' to the prefix: Propyl Methanoate.

 

 

 

FOR MORE INFO ON FUNCTIONAL GROUPS, CHECK THESE LINKS OUT!

http://www.youtube.com/watch?v=96D7RHUQvMA

http://www.youtube.com/watch?v=CJRQ8gOjq4E&feature=fvwrel

 

Amines and Amides

-Amies are funtional groups that contain a Nitrogen compound bonded to either hydrogens or carbons.
-Primary amines have 1 carbon chain
-secondary amines have 2 carbon chains
-Tertiary amines have 3 carbon chains

ex. http://www.learnchem.net/orgo/images/namine2.gif

Amides
Amides are functional groups with CONH3
ex. http://www.ivy-rose.co.uk/Chemistry/Organic/molecules/amides/methanamide.gif

Aldehydes

An aldehyde is a compund that has a double bonded oxygen at the end of a chain
-the simplest is mathanal. (be careful not to get mixed up with alcohols)

ex. http://www.ivy-rose.co.uk/Chemistry/Organic/molecules/aldehydes/n-Propanal.gif
ex. http://www.ivy-rose.co.uk/Chemistry/Organic/molecules/aldehydes/n-Octanal_structure.gif

Keytone

-A keytone is a hydrocarbon with a double bonded oxygen that is not an ether
-the following standard rules and -one to the parent chain

ex. http://images-mediawiki-sites.thefullwiki.org/01/2/2/0/02730213444239629.png
ex. http://www.learnchem.net/orgo/images/nket2.gif

Ethers

An ether contains an oxygen group connected to two alkyl (carbon) chains.

ex. http://c0499862.cdn.cloudfiles.rackspacecloud.com/Ether-general-9263.png
ex. http://www.docbrown.info/page15/Image243.gif

Carboxylic acids

Carboxylic acids are formed by the function group
      double bonded O and an OH attached to it
-Use standard rules but change the parent chain ending to -oic acid
-the simplest carboxylic acid is methanoic acid
      O
       =
        c    - OH

http://upload.wikimedia.org/wikipedia/commons/thumb/b/b5/Carboxylic-acid.svg/748px-Carboxylic-acid.svg.png

More Functional Groups

Alcohols, halides, aldehyde, ketone
-organic compounds can contain elements on the than C and H
-those are known as functional groups
-carbon chains without functional groups are written as R-
1. Alcohols
2. halides
3.aldehyde
4.ketone
5.carboxylic acid
6.ethers
7.amines
8.amides
9.esters

Alcohols
an alcohol is a hydrocarbon with a -OH bonded to it
same naming rules apply but the parent chain ending to -ol

http://www.askiitians.com/iit-jee-chemistry/organic-chemistry/images2/isobutyl-alcohol.jpg

Alicyclics and Aromatics:

1. Carbon Chains can form two types of closed loops
2. Alicyclics are loops usually made with single bonds
3. If the parent chain is a loop standard naming rules apply with one addition:
    "cyclo" is added in front of the parent chain
 
   
Cyclopeptane:

1. There are three different ways to draw organic compounds

complete structural diagrams: 

condensed structural diagrams: 

2. Numbering can start anywhere and go C.W. or C.C.W on the loop but side chains need
    the lowest numbers possible

ex: trimethyl hexane : 

3. Loops can be a side chain
4. Same rules apply but the side chain is given a cyclo-prefix

ex: cyclobutyl octane 

Aromatics:


1. Benzene [ C6H6 ] is a cyclic hydrocarbon with unique bonds between the carbon atoms
2. Structurally it can be drawn with alternating double bonds
3. Careful analysis shows that all 6 C-C bonds are identical and really represent
     a  1.5 bond
4. This is due to e- resonance
5. E- are free to move all around the ring

Aromatic Nonmenclature:


1. A benzene molecule is given a special diagram to show its unique bond structure
2. Benzene can be a parent chain or side chain "phenyl"

ex:  pentamethyl benzene 

Alkenes & Alkynes

  Since Carbon can form multiple bonds, it is possible for it to form double and triple bonds with other carbon atoms. When multiple bonds are created, fewer Hydrogens are needed to attach to the Carbon atoms. The naming rules are very similar to that of the naming rules for Alkanes.

 1. The position of the double or triple bonds must always have the lowest numbering scheme and is put in front of a parent chain. They take priority over the side chains.

CH3

|

CH2

|

                                                             CH2 = CH2 - CH2 - CH3

                                                               1        2        3       4

3 ethyl 1 butene

CH3

|

                                                                        CH3 - C - CH2 - CH = CH

|

CH3

                                                                          5     4       3       2       1

4, 4 diethyl 1 pentyne  

2. Double bonds end in '-ene' and are known as Alkenes. In diagrams, they are represented by a double line.

3. Triple bonds end in '-yne' and are known as Alkynes.In diagrams, they are represented by a triple line.

Check out this PDF for more practice problems:

http://www.mouatonline.com/Teachers/BHutchinson/Chemistry/Chem11/Unit%207/PS%207.2%20-%20Alkenes%20etc%28ans%29.pdf

Try these videos out too and learn more about Alkenes & Alkynes!

http://www.youtube.com/watch?v=KWv5PaoHwPA

http://www.youtube.com/watch?v=TBQjbGcUcUw&feature=related

Organic Chemistry

  Essentially, organic chemistry is the study of many different carbon compounds as carbon can form multiple covalent bonds. The number of organic compounds escalate to approximately 17,000,000 and can form chains, rings, or branches. The simplest organic compounds are composed of Carbon (C) and Hydrogen (H):

 EX.1: Above is an example of an organic compound called ethane.

  Saturated compounds exist with no double or triple bonds. Therefore, compounds with only single bonds are called Alkanes and always end in '-ane'. There are three categories of organic compounds:

  1. Straight Chains

2. Cyclic Chains

3. Aromatics

But today, we will only look at the composition of and naming of the straight chains:

1. Circle the longest continuous chain (this means the longest number of carbons in a row without going back on the chain). This will become the BASE or PARENT chain and can be determined by choosing a prefix and then adding the ending '-ane'.

meth=1 carbon atom

eth=2  carbon atoms

prop=3 "

but=4 "

prop=5 "

hex=6 "

hept=7 "

oct=8 "

non=9 "

dec=10 "

*IN THIS CASE, THE PARENT CHAIN WILL BE HIGHLIGHTED AND WE WILL BE USING A STRUCTURAL DIAGRAM.

                                                                                CH3 CH3

                                                                                  |      |

                                                                      CH3 - CH - CH - CH3

butane

2. Number the base chain in a way so that the side chains (anything not circled as your longest continuous chain), have the lowest possible numbering scheme.

*NUMBERING COULD START FROM LEFT OR RIGHT SIDE IN THIS CASE AS IT MAKES LITTLE DIFFERENCE.

     CH3 CH3

                                                                                  |      |

                                                                      CH3 - CH - CH - CH3

                                                                        1       2      3      4

butane

3. Name each side chain using one of the following prefixes with their '-yl' ending.

methyl=1

ethyl=2

propyl=3

butyl=4

propyl=5

hexyl=6

heptyl=7

octyl=8

nonyl=9

decyl=10

     CH3 CH3

                                                                                  |      |

                                                                      CH3 - CH - CH - CH3

                                                                        1       2      3      4

                                                                           -methyl -methyl butane

4. Give each side chain the appropriate number. If there are more than one of a certain side chain, you would have to number and label accordingly with added prefixes (di for 2, tri for 3, etc.)

     CH3 CH3

                                                                                  |      |

                                                                      CH3 - CH - CH - CH3

                                                                        1       2      3      4

                                                                         2, 3  dimethyl butane

5. When providing the final name for the compound, you must list the side chains alphabetically.

2, 3  dimethyl butane 

For more practice, try out these worksheets: 

http://www.arps.org/users/hs/thompsom/chemcom/unit_3/Naming_Alkanes_Worksheet_1.pdf

& while you're at it, check out this video on naming Alkanes!

http://www.youtube.com/watch?v=KKAD-OOOHxg

 

Polar Molecules

  In polar molecules, there is an overall separation in charges. An important aspect in determining whether a molecule is polar or non-polar is to look at its diagram and see if there is any symmetry. If a molecule is unsymmetrical, it is usually polar and if it is symmetrical, it would usually be non-polar. If one was to figure out that the molecule was non-polar, that would mean that the pull of the electrons is usually balanced. After determining if a molecule is polar, that would mean that the molecule has either different atoms or a different amount of atoms. One would then use signs to show which side is partially positive or partially negative.

EX.1:

ION CONCENTRATION:

HCl---> H+1 + Cl-1 [dissociation]

Dissociation:
1. Ionic compounds are made up of two parts:
    - cation: positively charged particle
    - anion: negatively charged particle

2. When ionic compounds are dissolved in water, the cation and anion separate from each other

3. This process is called dissociation

4. When writing dissociation equations, the atoms and charges must balance

5. The dissociation of sodium chloride is:
     NaCl---> Na+ + Cl-
ex: 1. Fe[OH]2=---> Fe+2 + 20H-1
      2. Na3PO4=---> 3Na+1 + PO4-3
      3. Fe3[PO4]2= 3Fe+2 + 2PO4-3
      4. If the volume doesn't change then the concentration of individual ions depends
          on the balanced coefficients in the dissociation equation
          a. Determine the {Na+} and {PO4}-3 in a 1.5M solution of Na3PO4
          Na3PO4 [1.5M]---> 3Na+[4.5M] + PO4-3[1.5M]
1.5mol/L x 1/1= 1.5mol/L

6. Determine both ion concentrations when a 2.5M solution of lithium sulfate dissociates
    Li2SO4[2.5M]---> 2Li+ + SO4-2
    {Li+}= 5.0M
    {SO4-2}=2.5M
     ZnCl2[0.376M]---> Zn+2 + 2Cl-1[0.752M]
7. A 100ml solution of 0.500M PcCl2 is added to 200ml solution of 0.100M
    NaOH. Determine the final concentration of each ion
    PbCl2[0.500M]--->Pb+2[0.500M] = 2Cl-1[1.00M]

c1v1= c2v2
[0.500M][0.100L]= c2[0.300L]
c2= 0.167M= [Pb+2]
c1v1=c2v2
[1.0M][0.1L]= c2[0.300L]
c2=0.333M=[Cl-1]
c1v1=v1v2
[0.100][0.200L]= c2[0.300L]
c2=0.0667M= [Na+] = [OH-1]
        

Titrations

-A titration is an experimental technique used to determine the concentration of an unknown solution.

Terms and Equipment
- Buret contains the known solution. Used to measure how much is added.
- Stopcock valve used to control the flow of solution from the the buret .
- Pipet is used to accurately measure the volume of unknown solution.
- Erlenmejer Flask contains the unknown solution
- Indicator is used to identify the end point of the titration.
- Stock solution is the known solution.
http://www.youtube.com/watch?v=9DkB82xLvNE

Dilutions

  It is evident that when two solutions are mixed, the concentration (measured in mol/L) changes. In the process of dilution, the concentration is decreased by adding a solvent, or any substance that dissolves another substance (usually water.) Similarly to how one adds one cup of water to one can of juice concentrate, or how one adds one cup of water to one can of soup, the substance that is dissolved (solute) does not change in amount. This could be explained with the equation:

n1 = n2    ->    c1v1 = c2v2

 Where c=Concentration, n=number of moles, & v=volume.

EX.1:

Determine C when 200 mL of 0.30 M of Sodium Nitrate (NaNO3) is diluted to a final volume of 600 mL.

1. Remember to first convert mL to L as M is measured in mol/L. After doing so, write out your equation that explains the conservation of the solute before and after dilution.

200 mL  x  1 L    =    0.200 L                     600 mL  x  1 L    =    0.600 L

              1000 mL                                                  1000 mL

c1v1 = c2v2

2. With the information you are given in the question, plug in the numbers into your equation and isolate the variably you are looking for.

(0.30)(0.200)  =  C2(0.600)

                                                                    (0.600)           (0.600)

3. Once you have isolated your variable, calculate & answer the question with correct amount of significant digits.

(0.0600)  =  C2

                                                                                                (0.600)

                                                                       0.10 mol/L = C2

In order to answer other questions similar to this one, you have to consider what is it that the question is asking you to find? In most cases you will have to isolate the variable of which you are solving for.

For more examples on this topic, check out these links:
http://misterguch.brinkster.net/PRA025.pdf

http://www.youtube.com/watch?v=ygr0CGdkAws

Titration Lab

Hello fellow subscriber to this blog!

Today we did a lab on titration. 

We used Vinegar CH3COOH

 and as well as Sodium Hydroxide NAOH

We tested to see the concentration of the vinegar using sodium hydroxide and also 2 drops of Phenolphthalein

We carefully measured how much sodium hydroxide can be poured into vinegar to make it fully "pink". It was cool how every drop it shows a bit of pink then goes away. 

BUT!

To be the MASTER OF TITRATION [kinda sounded cooler when Mr.Doktor said it]

It only takes one drop...

Yes one drop to make it all pink.

Like this :)

Well that is all that we did so stay tuned for our next lesson (: 

Solution Stoicheiometry

heyo subscriber!

Today we just finished basic stoichiometry! Now we are going deeper in to stoichiometry, we are now dealing with solutions.

What is a solution? Well it is a mixture of two different types of substances. One is the solvent which dissolves something called the solute. One type of common solvent is water which makes compounds or elements aqueous when dissolved in water.

Take salt for example:
NaCl + H2O => NaCl(aq)

So yeah, we did basic stoichiometry solutions like this.
Want something harder?

check it out, part 2 and part 3 is in the related videos :)

Thank you! 

Mole to Mole Conversions

- Coefficients in balanced equations tell us the number of mole reacted or produced
- They can be used as conversion factors
- What you need over what you have

ex. N2 + 3H  -----> 2NH3

3.5 x  2  =7 mols
          1 

Percent Yield

  One way to think of percent yield is as a measure of success. To find this number you will need two things: the theoretical yield and the experimental yield. The theoretical yield is the amount of products that should be formed and the experimental yield refers to the amount you're given based on the experiment. Because you are usually given the experimental yield, you will need to find the theoretical yield first...

EX.1:

If you burn 12 g of carbon (C) to produce 48.0 g of carbon dioxide (CO2) what is the percent yield?

1. The first and one of the most important steps is to make sure you have a correct and balanced equation.

C + O₂ --- > CO₂

2. You will then have to find the theoretical yield of carbon dioxide because you are given the experimental value. Take what you have and convert it into moles as it is the most common conversion factor.

12 g of C    x    1 mol of C

                      12.0 g of C

3. Next, use your mole to mole ratio: what you need over what you have. Cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2

                      12.0 g of C            1 mol of C

4. After having your moles of Carbon Dioxide, use molar mass as your conversion factor and cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2    x    44.0 g of CO2

                      12.0 g of C            1 mol of C            1 mol of CO2

5. Multiply and divide your equation and you will now have your theoretical yield. Plug it into the percent yield formula.

% YIELD    =    Experimental     x 100

                           Theoretical

% YIELD    =    48.0 g    x    100

                          44.0 g

% YIELD    =    109%

For more examples check out this link:

 http://www.youtube.com/watch?v=TKNxdL7DN1I