Docstoc

chemistry 11th topic

Document Sample
chemistry 11th topic Powered By Docstoc
					                             Topic 11: Organic Chemistry

11.1 Homologous Series

11.1.1. Describe the features of a homologous series.
  Features include a general formula and neighboring members differing by CH2, with
         similar chemical properties and with a gradation in physical properties.

11.1.2. Predict and explain the trends in boiling points of members of a homologous
series.
   In a homologous series there is a gradual increase in boiling point as the number of
                   carbon atoms increases. Cross reference with 4.3.

11.2 Hydrocarbons

11.2.1. Draw structural formulas for the isomers of the non-cyclic alkanes up to C6.
Structural formulas should indicate clearly the bonding between atoms. For example, for
                                        pentane:
 CH3CH2CH2CH2CH3 or CH3-(CH2)3-CH3 or with the carbon drawn separately with all
                       hydrogens drawn attaching, but not C5H12.

11.2.2. State the names of alkanes up to C6.
  Name these using IUPAC rules. Consider both straight and branch-chained alkanes.

11.2.3. Explain the relative inertness of alkanes.
                             Refer to bond enthalpies. See 6.4.
Alkanes are relatively inert because the C-H bond is so strong, as well as C-C bond. It
takes an enormous amount of energy to break C-H bonds and C-C bonds, which would be
required to have a reaction with an alkane, so generally speaking it does not happen.

11.2.4. Draw structural formulas and state the names for straight-chain alkenes
        (CnH2n, where n is between 2 and 5)
                    Geometric (cis-trans) isomers are not required.

11.2.5. Describe complete and incomplete combustion of hydrocarbons.
     The formation of CO and C during incomplete combustion should be related to
                      environmental impacts and oxidation-reduction.
Complete combustion produces CO2 and H2O, incomplete combustion produces CO,C,
and H2O (usually occurs with saturated alkanes, where there is a lot of hydrogen, or
where there is a limited supply of oxygen.) C produces a ‘dirty’ flame leaving carbon
deposits on everything, CO is toxic and CO2 is a greenhouse gas. Incomplete combustion
is where the carbon is not being completely oxidized.

11.2.6. State that the combustion of hydrocarbons is an exothermic process.
                                    See 6.3 and 6.4
The combustion of hydrocarbons is an exothermic process; this is primarily what is
behind fossil fuels, we combust hydrocarbons. In your car, in powerplants, etcetera.

11.3 Other Functional Groups (7h…this is the longest subtopic ever, it’s bigger then
some entire topics combined).

Note: Along with alkanes and alkenes, compounds containing one or more functional
groups have been chosen to introduce students to:
     Interrelationship involving significant functional groups
     Important reaction types such as addition, substitution, oxidation, condensation,
        esterification and polymerization
In my humble opinion, this is the most ridiculous thing IB has ever done. There are years
of college courses dedicated to this stuff, yet they expect you to learn A LOT in very little
time. Oh well, this is really what makes IB harder then all the other syllabi out there like
AP.




11.3.1. Draw and state the names of compounds containing up to five carbon atoms
with one of the following function groups: aldehyde, ketone, carboxylic acid, alcohol,
amide, amine, ester and halogenoalkane.
             Functional groups in full and condensed forms are required, eh:
          Aldehyde: R-C=O (and another hydrogen bonded to the C) or RCHO.
Aldehydes and Ketones: Condensed form: RCHO Aldehyde. RCHOR Ketone.

Aldehydes have at least one hydrogen atom attached to the carbonyl atom. In ketones,
the carbonyl carbon atom is connected to two other carbon atoms.

In the IUPAC system, the characteristic ending for aldehydes is –al (from the first
syllable of aldehyde). So an aldehyde with four carbons and the last carbon is in an
aldehyde group would be called butanal. For substituted aldehydes, we number the chain
starting with the aldehyde carbon. So an aldehyde has priority over a double bond or a
hydroxyl group, not only in numbering, but also as the suffic. For cyclic aldehydes, the
suffic –carbaldehyde is used. i.e. ethanal

In the IUPAC system, the ending for ketones is –one (from the last syllable of ketone).
The chain is numbered so that the carbonyl carbon ahs the lowest possible number. i.e.
2-propanone.

Carboxylic Acid: Condensed: R-COOH

Carboxylic acids are organic acids that contain the carboxyl group. In acid derivatives,
the –OH group is replaced by other groups.

To obtain the IUPAC name of a carboxylic acid, we place the final e in the name of the
corresponding alkane with the suffix –oic and add the word acid. For substituted acids,
the chain is numbered beginning with the carboxyl carbon atom, and the substituents are
located in the usual way. i.e. ethanoic acid

Alcohol: Condensed: R-OH
Alcohol’s are similar to water but one of the hydrogen is replaced by an alkyl group. In
the IUPAC system, the hydroxyl group in alcohols is indicated by the ending –ol. i.e.
ethanol

Amine: Condensed: RNH2

In the IUPAC system, the amino group, -NH2, is named as a substituent. Recently,
Chemical Abstracts (CA) introduced a system for naming amines that is rational and easy
to use. In this system, amines are named as alkanamines. I don’t know which one they
will expect you to know. The two hydrogens on the N can be replaced by R groups to
give primary, secondary, and tertiary structures. i.e. ethylamine.

Amide: Condensed: RCONH2

Amides are carboxylic acid derivatives in which the –OH group is replaced by NH2,
NHR, or NR2. Amides are named by replacing the –ic or –oic ending of the acid name,
either the common or the IUPAC name, with the –amide ending. i.e. ethyamide.

Ester: Condensed: RCOOOR.

An ester is a carboxylic acid derivative in which the hydroxyl group is replaced by an
alkoxy (OR) group. This of it like an alkanoic acid with a carbon chain rather then a H.
The alkanoic acid type bit it -oate, this is preceded by the stem of the other half –ie
Ethylethanoate.

Halogenoalkane: R-X.

Name of halogen (fluro, chloro, bromo, iodo) followed by R name- i.e. Chloroethane.

11.3.2. Explain that functional groups can exist as isomers.
                                    Examples include:
              Ethanoic acid (CH3COOH) and methymethanoate (HCOOCH3)
                  Propanal (CH3CH2CHO) and propanone (CH3COCH3)
Functional groups can actually be isomers (although their properties are not generally
similar). For examples see above.

11.3.3. Outline the existence of optical isomers.
Restrict this to the fact that, if a carbon atom has four different substituents, the molecule
   exists in two enantiomeric forms that rotate the plane of polarized light in opposite
        directions. Students should be able to identify a chiral (asymmetric center).
Optical isomers result if a carbon atom has 4 different groups on each bond. If this is the
case, the compound exists in 2 enantiomeric forms (i.e. optical isomers). In general, they
react very similarly except in the presence of other optical isomers (also known as chiral
molecules-the chiral center is the carbon atom with four different groups). The two
enantiomers are mirror images of each other which cannot be superimposed on each
other. Biological systems commonly have a strong preference for one enantiomer over
the other (one can be bitter, the other sweet for example). The isomers can be identified
by their effect on polarized light (by a polarimeter). When polarized light passes through
one isomer it will be rotated to the left, while the other will rotate to the right.

11.3.4. Discuss the volatility, solubility in water and acid-base behavior of the
functional groups aldehyde, ketone, carboxylic acid, alcohol, amide, amine, ester and
halogenoalkane.
   For example, use functional groups to explain the higher boiling point of methanol
                     compared with methane. Cross reference with 4.3
O-H groups create hydrogen bonding (alcohols, alkanoic acids). Thus they are less
volatile and also soluble. Long chain molecules become less soluble since the non-polar
chain dominates the molecule.

C=O bonds in alkanoic acids and alkanals are a polar bond, thus allow dipole forces to
act between molecules which increase the boiling point (this is more significant in small
molecules). Small molecules are soluble due to polarity (effect decreases with long
chains.)

Esters have no hydrogen bonding, thus they are very volatile and have a low boiling
point. They are polar though due to the double bonded oxygen and thus are water soluble
when they are short.

Amides have an N-H bond which is polar and is very good at hydrogen bonding. Thus
they are highly water soluble and all amides have higher boiling points then the alkanes
they are compared to.

Amines have hydrogen bonding present in primary and secondary situations, so they are
soluble when short and have a higher boiling point then alkanes. Tertiary a mines are
very similar to alkanals but are branched, thus less dense packing.

Halogenoalkanes when they are short will be soluble due to polar bonds and the boiling
point will be somewhat higher.

Acid Base properties: Alkanoic (carboxylic) acids are acidic. Alcohols are generally not
due to the donating effect of the R group. Amines are derivatives of ammonia, and so are
basic (though stronger due to the donating effect of the R groups). Amides are not due to
withdrawing effect of the C=O group. The others, in general, are neutral (however
alkanols can, in acidic conditions act as a base and accept a proton, though the electron
donating effect of the alkyl groups generally stop any acidic action.)
Note: Mechanisms are not required for 11.3.5. to 11.3.10. Conditions are also not
required unless specified.

11.3.5. Outline the reaction of symmetrical alkenes with hydrogen, bromine, hydrogen
halides and water.
      A double bond is relatively reactive, therefore molecules such as H2C=CH2 are
                      important starting materials in organic synthesis.
A double bond in an alkene is relatively reactive and fairly easy to break, so it is a
starting point for a reaction. When this reacts with hydrogen, the double bond will break
and one of the electrons will go to each carbon, which will share with a new hydrogen,
thus creating a saturated hydrocarbon. When it reacts with bromine (and all halogens),
the double bond breaks and each carbon forms a bond with one of the two molecules in
the diatomic halogen. With hydrogen halides, a halide goes to one carbon and the
hydrogen goes to the other. With water, a hydrogen goes to one carbon and an OH group
goes to the other, creating an alcohol. As you can see, double bonds allow for all sorts of
creations.

11.3.6. Outline the use of reactions of alkenes.
 Hydrogenation is used in the production of margarine, hydration of ethene is used in the
manufacture of ethanol, and bromination can be used to distinguish between alkanes and
                                          alkenes.
The reactions of alkenes are used to make ethanol (an alcohol), hydrogenation of
margarine (breaking double bonds and replacing them with hydrogen) is used in the
production of margarine (remember the name saturated fat?). And bromination can be
used to distinguish between alkanes and alkenes. Bromine when it is diatomic is a
brownish red color, and when it is added to a solution, if the color disappears that is
because the bromine split and formed with an alkene to make a halogenoalkane, thus the
original substance was an alkene. But if the brownish red color remains, the substance
was already an alkane.

11.3.7. Outline the polymerization of alkenes.
   Polyethene and polyvinyl chloride should be used as examples of addition polymers.
 Students should be able to draw the structures of the monomer and the repeating unit of
                                         the polymer.
This is a little too long to type, maybe take a look at what Matthew Sheppard has on his
webpage, or look at page 99 in the O-Chem book mentioned in bibliography.

What he has: Polymerisation of alkenes (by a free radical mechanism)...

Initiation ... 2R2 -> 2R° (R° is a free radical with a lone electron)

R° + CH2=CH2 -> R-CH2-CH2°

Propagation ... R~~~~° + CH2=CH2 -> R~~~~~°.

Termination ... R~~~~° + °~~~~~R -> R~~~~~~~~R.

Polyethene: Monomer is CH2=CH2, the general polymer is –[-CH2-CH2-]n-
Polyvinyl chloride: Monomer is CClH=CH2 (chloroethene), the general polymer is –[-
CH2-CHCl-]n-
11.3.8. Outline the condensation reaction of an alcohol with a carboxylic acid to form
an ester, and state the uses of esters.
    Esters are used as flavouring agents, in plasticizers, as solvents and in perfumes.
This is an addition elimination reaction (or addition-dehydration, since we’re eliminating
water, could also be called a condensation reaction).
CH3CH2OH + HOOCCH3 (in the presence of H2SO4 and warming)  CH3CH2OOCCH3
+ H2O

Esters are used as flavoring agents, in plasticizers, as solvents and in perfumes.

11.3.9. Describe the partial and complete oxidation of ethanol.
     A suitable oxidizing agent is acidified potassium dichromate (VI). Both oxidation
    products (ethanal and ethanoic acid) can be obtained by altering the conditions, eg
  ethanal by distilling off the product as it is formed, and ethanoic acid by heating under
                                             reflux.
This process requires a primary alcohol (which ethanol is), or otherwise the reaction is
stopped because the intermediate formed is a ketone rather than an aldehyde.

CH3CH2OH is oxidized by K2Cr2O7 (in the presence of an acid)  CH3CHO which is
then oxidized again by K2Cr2O7 (once again in the presence of an acid)  CH3COOH.

Ethanal is an intermediate. If it were poured off once it was formed and not allowed to
mix with more potassium dichromate (VI) then it would not go to be ethanoic acid.
Secondary alcohols only give ketones. Tertiary alcohols, having no hydrogen atom on
the hydroxyl-bearing carbon, do not undergo this type of oxidation. Notice that as an
alcohol is oxidized to an aldehyde or ketone and then to a carboxylic acid, the number of
bonds between the reactive carbon atom and oxygen atoms increases from one to two to
three. In other words, we say that the oxidation state of that carbon increases as we from
an alcohol to an aldehyde or ketone to a carboxylic acid.

11.3.10. Deduce the condensation polymers formed by amines and by carboxylic acids.
            Emphasize the need for two functional groups on the monomers:
           Polyamides (nylons) – eg hexanedioic acid and 1,6-diaminohexane.
            Polyesters- eg benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Nylon ... hexane -1,6-diamine + hexanedioic acid

H2N-(CH2)6-NH2 + HOOC(CH2)4COOH ->

NH2(CH2)6-NH-CO-(CH2)4-CO-NH-(CH2)6-NH-CO-(CH2)4-CO...-NH-(CH2)6NH2.

As each new group is created, a water molecule is eliminated...

Polyester ... benzene-1,4-dicarboxylic acid + ethane-1,2-diol

HOOC-Benzene-COOH + HOCH2CH2OH ->
HOOC-Benzene-COOCH2CH2OOC-Benzene-...COOH (once again, water is eliminated
each time)

Notice the fact that two functional groups are required on each monomer...Otherwise the
reaction would stop without producing a long chain.


11.3.11. Outline the formation of peptides and proteins from 2-amino acids.
 All 2-amino acids (alpha-amino acids), except aminoethanoic acid (glycine), can show
optical activity. Peptides are formed from amino acids, and two functional groups allow
 for the formation of macromolecules. Students should be familiar with simple primary
  structures (order of amino acids) for peptides containing up to three amino acids, eg:

Amino Acids ... H2N-CHR-CO-H ... carbon atom (asymmetric) is connected to 4
different species -> optically active (except glycene).

H2N-CHX-CO-H + H2N-CHY-CO-H + H2N-CHZ-CO-H -->

-NH-CHX-CO-NH-CHY-CO-NH-CHZ-CO- (poly peptide) This group can join to other
peptides to form a protein.

				
DOCUMENT INFO
Shared By:
Stats:
views:34
posted:4/22/2012
language:English
pages:8
Description: chemistry 11th topic about organic chemistry
Ali siddique Ali siddique Techsol http://xpertcosmetics.com/
About i'm a Student of Bs.cs