SCH4U
ORGANIC CHEMISTRY – Assignment (Chap 23-BH) *Chap 1&2-Nelson Chemistry 12
HYDROCARBONS - Composed of carbon chains and hydrogen.–(23.2 pg 1032) –table 1-pg 11-Chem12
–ALKanes: straight or branched chains of carbons with only single bonds (fully saturated with hydrogens)–pg 12-16
Group 1:
–ALKenes: straight or branched chains of carbons with one double bond (unsaturated hydrocarbons) - pg 16-19
* ALKadienes: 2 dble bonds, **ALKatrienes: 3 dble bonds,
–ALKynes: straight or branched chains of carbons with one triple bond - pg 16-19
–HaloALK’s:
contain one or more members of group VII (halogens)
–CycloALK’s: carbons are in a ring or cycle –Aromatics:contain at least one benzene ring (C6H6) – pg19-21
SIMPLE NOMENCLATURE: A review from SCH3U
# carbons
alkane
alkene
alkyne
1
methane
(CH4)
NA
NA
2
ethane
(C2H6)
eth ene
(C2H4)
eth yne
(C2H2)
3
propane
(C3H8)
prop ene
(C3H6)
prop yne
(C3H4)
4
butane
(C4H10)
but ene
(.…........)
…………
(....…......)
sketch
5-pentane (C5H12), 6-hexane(C6H14), 7-heptane(C7H16), 8-octane (C8H18), 9-nonane(C9H20), 10- decane (C10H22), X-(CxH2x+2).
STRUCTURAL ISOMERS:
When you draw butane (C4H10) you will notice that there are two possible correct structural diagrams.(see below or
see model on your desk) Both molecules exist as separate entities (ie different MP/BP) so can’t have the same name.
These molecules are said to be structural isomers.
CH3
CH3–CH2–CH2–CH3 *(n-butane)
or
CH3–CH–CH3
*(isobutane)
With few isomers known the naming problem was rectified by using the prefixes normal(n) for the straight chain and
iso for the branched chain. However as more isomers were found (octane has 18 , decane has 75) a better system had
to be developed so all chemists could communicate.
IUPAC RULES for NAMING ORGANIC MOLECULES:
1. a) Determine what is the LONGEST CONTINUOUS CHAIN of CARBONS and let this be the base name.
(eth = 2 carbons, prop = 3, hex = 6 etc.) * note: the longest chain does not have to be straight just continuous
b) Assign a number to each carbon such that any branch(es) etc will have the smallest possible #
2. Locate & identify any double (ene), triple (yne) bonds and/or if the longest chain is cyclic (prefix"cyclo").
*NOTE: dble/triple bonds are assigned the smallest possible number in the carbon chain
3. Locate & identify any FUNCTIONAL GROUPS. ( These change the end of the name)
** These will be covered in more detail later**
4. Locate & identify any BRANCHES (** must contain a carbon not in the main chain).
Branches (alkyls) are named by
a) identifying the number of carbons &
b) adding the suffix "yl"
(ex: methyl, ethyl, propyl, )
**if more than one - arranged alphabetically**
5. Locate & identify any SUBSTITUTED groups ( halogens etc.) . These do not contain a carbon but have been
substituted in place of a hydrogen. They are named by adding the suffix "o"
(Ex: fluoro-F, chloro-Cl, bromo-Br, iodo-I, nitro (-NO2), amino (-NH2) ), cyano (-CN) ) ** arranged alphabetically
Example # 1
7
8
ethyl branch
CH2–CH3
CH2–CH3
CH3–CH–CHBr–CH–CF=CH–CH3
methyl branch
6
5
4
3
2
1
IUPAC name = 5-bromo-3-fluoro-4-ethyl-6-methyl-2-octene
Note: 1) The longest chain of carbons does not have to be straight.
2) the double bond has the smallest possible # on the carbon chain, alpha arranged
Carbon #1
Example #2
Cl
CH2–CH2–CH3
1) The carbons in the ring(cycle) are not drawn – but are
assumed to be at at each point where the line bends
2) The double bond in the ring has to have the smallest #
which would always be #1 in a ring
branches
–CH2–CH3
3) the name is always organized
substituted group (Cl)  branches(alpha)base (ending)
Cl
IUPAC name
= 1,3-dichloro-4-ethyl-5-propyl-1-cyclopentene
GEOMETRIC ISOMERS
Carbon-carbon single bonds are created because of single orbital overlap and can rotate on each other a full 360°.
However carbon-carbon double bonds are no longer able to rotate. Thus if 2 groups are attached to these carbons they
are stuck in these positions. There are only two positions they can be stuck in - either on the same side of the double
bond ("cis") or on opposites sides of the dble bond (“trans"). Both molecules have the same formula & structure but a
different orientation in space hence are different molecules ( diff MP / BP etc. ) This type of isomer is called a
geometric isomer ( see below for an example of geometric isomers )
CH3
CH3
CH2
CH2
109°
CH3
120°
CH
CH3
CH
Br
n-b utane
Br
CH3
CH
Br
cis 2,3 - di b romob u tene
CH3
CH
Br
trans 2,3 - di b romob u tene
Any 2 groups attached across a double bond including carbons will exhibit cis-trans isomerism. The 2 groups DO
NOT have to be the same!( see page pg 1038)
Review– Organic Nomenclature
1. Are you ready ?
2. Naming alkanes:
3. Naming alkenes/alkynes:
*4. Naming Aromatics
1-9
practice 1&2
practice 3, 4, 5& 6
practice 7 & 8
*pg 4-5-Nelson Chem 12
*pg 15-16- Nelson Chem 12
*pg 18-19- Nelson Chem 12
*pg 21- Nelson Chem 12
5. Name the following organic compounds using the IUPAC system of nomenclature
a) C2H5–CCl2–CH=C(CH3)–CH2–CH3
b) CH2–CC–C(NO2)–CBr–CH3
|
|
|
C3H7
CH3
CH3
......................................................................
..........................................………...........................................
....................................................................
...........……………………….……………………….............
CH3 C2H5
|
|
C2H5–C(NH2)–CI–CH3
c)
d) CHI2–CBr2–CCl2–CH3
e) CF2=CF2
............................................................
........................................................................ ...................................................................
............................................................
........................................................................ .......………………….……………...........
f)
g) CH=C-CH(CH3)-CH3
CH3
NC-
Cl
h)
C3H7
Cl
CH-CH3
CH3
Br CH3
............................................................
........................................................................ ...................................................................
............................................................
........................................................................ .......………………….……………...........
i) CH3–CH(NO2)–CH(NO2)–CH(NO2)–C2H5 j) CH3–CH – CH–CH3 k) CH2=CH–CH=CH2
|
|
.................................................................................................
CH3–CH2 CH2–C2H5
......................................................... .......................................
l) H5C2 C2H5
|
|
CF=CF
m)
CH3
|
CH2=C–CH=CH2
.............................................................
…………………………………
..............................................................
...................................................
n) NH2–(CH2)6– NH2
....................................…….........................................................
*o) (CH3–CHI-CHCl)2–CH–CH3
.............................................. ...............................................
......................................................…..................................
.............................................. ...............................................
.........................................................…...............................
6. Draw correct structural diagrams for the following molecules
a)
c)
e)
g)
i)
k)
m)
o)
q)
1-bromo-1-chloro-2-methylpropane
1-bromo-2-nitropentane
3,3-diethyl-4-methy-5-isopropylloctane
2-methyl-1,1-dicyclopropyl-1-propene
trans-3,4-dimethyl-3-hexene
3-cyclopropyl-1-pentene
1,2-dicyclopropylethane
cis-3,4-diamino-1,3,5-heptatriene
3,3,3-trichloropropene
b)
d)
f)
h)
j)
l)
n)
p)
r)
3-ethyl-2-methylheptane
2,3,3,4-tetramethylhexane
cis-2,3-dimethyl-2-butene
4,4-di(2-butyl)nonane
3-methyl-1,4-pentadiene
1,2,3-trimethylcyclobutane
2,5-dichloro-2,5-difluoro-3-hexyne
trans-2-pentene
2,3-dinitro-1,3-butadiene
7.a) Draw and name the 5 different isomers of hexane b) Draw and name the 7 isomers of C3H4Cl2
8. Draw the structural diagram & name 2 different alkenes isomeric with cyclobutane
Group 2: CARBOHYDRATES - Composed of carbon chains where one or more carbons has been oxidized. These
oxygen atoms (and nitrogen atoms) make the carbon based molecule behave
differently the functional groups. Molecules that have the same groups of
Oxygens/Nitrogens all tend to behave similarly.
See over for examples
A) ALCOHOLS:
General formula R-OH where R = any group of carbons (pg 38-42-Nelson Chem 12)
NAME: name the group of carbons (R) then add the suffix "ol" instead of “e”
CH3OH  methanol
CH3–CH2–OH ethanol
CH3-CHOH-CH3
2-propanol
wood alcohol
Rubbing alcohol
grain alcohol -booze
CH2–CH2 –CH2
CH3-CH2-CH2–OH
1-propanol
(1° alcohol)
|
|
|
CH3-CHOH-CH3
2-propanol
(2° alcohol)
OH OH OH
CH3-C(CH3)OH-CH3 2-methyl 2-propanol (3° alcohol)
1,2,3-propantriol  glycerol
9. Using the IUPAC rules name the following organic alcohols
a) CH3-CH2-CH2–OH
…………………………………………………………………….………….
b) CH3-CHOH-CH3
…………………………………………………………………….………….
c) CH3-C(CH3)OH-CH3
…………………………………………………………………….………….
d) CH3–CHOH–CH2CH3
…………………………………………………………………….………….
e) CH3–C(CH3)2–CH2OH
…………………………………………………………………….………….
f) CH2OH–CH2OH
…………………………………………………………………….………….
10. Using the IUPAC rules draw the following organic alcohols
a) cyclobutanol
b) 2,3-dimethyl-2-hexanol
c) 4,4-dimethyl-1,2-pentandiol
11. i) Write the correct name for each alcohol & ii) Which alcohol is not an isomer of all the others ? & iii) Which alcohols are identical ?
a) CH3-C(CH3)2-CH2–OH
..............………....................................................................
b) C2H5-C(CH3)2–OH
..............………....................................................................
c) C3H7-CHOH-CH3
..............………....................................................................
d) CH3-CH2-CH2-CH2-OH
..............………....................................................................
e) C2H5-CH(C2H5)-OH
..............………....................................................................
f) C4H9-CH2-OH
..............……….....................................................................
g) CH3-CHOH-CH2-CH2-CH3
..............………....................................................................
h) CH3-(CH2)3-CH2OH
..............………....................................................................
B) ETHERS: General formula R-O-R' where R & R' = any alkyl branch (pg 45-46-Nelson Chem 12)
NAME: Name R then R' (alphabetically) then add "ether”
CH3–O–C2H5 ethylmethylether
CH3
O
#1
C3H7–O–C4H9 butylpropylether
#1
#1
CH3 -CH 3-CHCl-CH 3
3-chlorobutyl-3methylcyclohexylether
O
CH3
CH-CH 2 -CH2 -CH3
2-cyclobutenyl-2-pentylether
C) ALDEHYDES: General formula RCH=O where R = any alkyl branch (pg 49-52-Nelson Chem 12)
NAME: This time the carbon is part of the functional group and must be at the end of the carbon
chain(terminal) and is always #1. add the suffix " al " instead of “e”.
CH2– CH– CH2–CH=O
H–CH=O
methanal / formaldehyde
CH3–CH=O
ethanal / acetaldehyde
CH2Br–CBr=CH–CH2–CH=O 4,5 dibromo-3-pentenal
4-phenyl-3-cyclopropylbutanal
D) KETONES: General formula RCOR’ where R& R’ = any alkyl branch (pg 49-52-Nelson Chem 12)
NAME: This has the same functional group as aldehydes except it is never at the end and is
never carbon #1. add the suffix "one " instead of “e”.
CH3–CO –CH3
propanone / acetone
CH3–CH2–CH2–CO –CH3
2-pentanone
CH3–CH2–CO–CH=C(CH3)-CH2–CH3
3-methyl-3-hepten-5-one
O
O2N
CH2 -CH 3
2-nitro-2-ethylcyclopentanone
E) CARBOXYLIC ACIDS: General formula RCOOH where R= any alkyl branch (pg 58-61-Nelson Chem 12)
NAME: The fully oxidized carbon is usually carbon #1. add the suffix "oic acid” instead of “e”.
HCOOH
methanoic acid / formic acid
CH3COOH
ethanoic acid / acetic acid
CH3CH2CH=CHCH2CH(CH3)COOH
2-methyl-4-heptenoic acid
HOOC(CH2)4COOH
1,6-hexandioic acid
CH3 -CH-CH 3
|
3,3-difluoroCF2
4-methyl
|
pentanoic
CH
2
acid
|
HO– C=O
F) ESTERS: General formula RCOOR’ where R&R’ = any alkyl branch (pg 64-66-Nelson Chem 12)
NAME: The fully oxidized is now NOT at the end of the carbon chain name R’ first (as a branch)
then the acid radical with a suffix “oate” for esters
**Note: esters are made by combining an alcohol with a carboxylic acid.
CH3–COO–CH3 methyl ethanoate / methyl acetate CH3CH2CH2COO–CH2CH3 ethyl butanoate / pineapple
acid radical
branch
CH3COO–(CH2)7CH3
octyl ethanoate / oranges
CH2=C(CN)COO–CH3 methyl -2-cyano-2-propenoate / crazy glue monomer
HCOO–CH2CH(CH3)CH3
C2H 5
#1
2-methylpropyl methanoate / raspberry
#1
COO-CH2CH2CH2CCl2CH3
4,4-dichloropentyl-3-ethy lcyclohexy lmethanoate
COO-CH2CH3
ethy l benzoate
G) AMINES: General formula R–NH2, R–NH - R’ & R–NR’ R’’ (pg 69-72-Nelson Chem 12)
NAME: The functional group is now a nitrogen. The new method modifies the alkane name by
replacing the “e” with “-amine”. IUPAC labels the “N” as an amino branch
CH3–NH2 methanamine / aminomethane CH3CH2– NH2 ethanamine/aminoethane *primary (1°) amines
(CH3)2–NH dimethanamine
CH3– NH–CH2CH3 ethylmethanamine *secondary (2°) amines
(CH3)3– N trimethanamine *teretiary (3°) amines
NH2–(CH2)6– NH2 1,6-hexandiamine
CH(CH3)2– NH2
2-propanamine
H) AMIDES: General formula R–CO–NH2, R–CO-NH-R’ & R–CO–NR’ R’’ (pg 74-76-Nelson Chem 12)
NAME: Carboxylic acids can be converted to amides by replacing the -OH on the carbon with
NH2. They are named by replacing the “oic acid” with “-amide”. Substituted alkyl
groups(R’ & R’’) are named as branches preceded by “N” or “N,N”.
CH3–CO- NH2 ethanamide / from ethanoic acid CH3CH2CH2–CO– NH2 butanamide / from butanoic acid
CHCl2–CH(CH3)–CO– NH2
3,3-dichloro-2-methylpropanamide
CH3–CO- NH–CH3 (N-methyl)ethanamide CH3CH2–CO- N(CH3)–CH2CH3 (N-ethyl-N-methyl)propanamide
Aromatic Hydrocarbons (pg 19-21-Nelson Chem 12)
This group of hydrocarbons is based on the benzene molecule –C6H6 – a
planar hexagonal (bond angles=120°) structure that is very unsaturated with
all carbon-carbon bonds of equal strength & length. At first thought this
structure must have alternating double bonds but this model did not correlate
to the physical & chemical behavior of this molecule. So the current model
has single(σ) carbon bonds with the π electrons in a cloud above & below
the ring adding increased stability Because of this stability a very large
group of compounds contain this ring structure – the aromatics.
The early structure had
alternating dble bonds.
Modern structure has a
cloud of 6 delocalized
electrons above & below
the ring ( the oval)
Naming the aromatics
Because of the stability of the ring, aromatic compounds simply have one or more of the H's replaced with a
substitute ( any halogen, amino, nitro, another hydrocarbon etc. ).
There are also several different ways to name some aromatics, although IUPAC would like us all to use their
method. *NOTE. Benzene can also be named as a branch ( C6H5-R )=PHENYL
As well some aromatics have special names (from the past ) that are still very popular ex:
Br
NO2
1-bromobenzene
1-nitrobenzene
Disubstituted Aromatics
CH3
1-methylbenzene
** toluene **
benzoic acid
** Phenol**
Br
Br
Br
OH
COOH
CHO
**Benzaldehyde
Br
Br
ortho(o) position
Br
meta (m) position
Br
1,2-dibromobenzene
1,3-dibromobenzene
1,4-dibromobenzene
Para (p) position
o-dibromobenzene m-dibromobenzene p-dibromobenzene
12. Draw the structure of the following benzene derivatives.
a) chlorobenzene
b) o-iodotoluene
c) 3-chlorophenol
d) m-nitrobenzaldehyde
e) 1,3,5-triethylbenzene
f) benzyl alcohol (C6H5CH2OH)
g) aminobenzene (aniline)
h) 1,4-dimethylbenzene (p-xylene) i) 2,4,6-trinitrotoluene
j) 1,3-dibromo-2-phenyl-2-pentene
k) 3,5-difluoro-4-isobutylbenzoic acid
l) 1-phenylethene (styrene)
m) 1,1-di(4-chlorophenyl)-2,2,2-trichloroethane
n) 1-phenyl-2-aminopropane ( benzedrine/speed)
13. Using the IUPAC rules name the following organic molecules each containing a functional group
a) CH3–CHOH–CH2CH3
b)
-OH
e) CH(OH)2-CCl3
F-
-F
d) CH3–C(CH3)2–CH2OH
O
f) C5H11–O–C4H9
g)
CH-CH3
CH3
c) CH2CHClCH2OH
h) CH3(CH2)2CHOHCH(C2H5)CH2OH
i) CH3CHBr–O–CH2CH2CH2Cl
j) CH2=CH-CH=CH-CH=CH-CH2–O–CH3
k) C3H7–O–CH2CH=CH–CHF-CH3
l)
-CH-CHO
CH3
m) CH3CH(CH3)–CH2CHO
p)
CH=C-CH2-CO-CH2-CH3
n) CH3CH2CH(C2H5)CH(CH3)CHO
o)
CH2-CO-CH2
q) CH3-CHF-CH2-CO-CH2-CH3
Br
r) CH2Br-CHBr-CH(C6H5)-CO-CH3
s) CH3-C(CH3)2-CH2-COOH
t) CH3CHFCH(CH3)CH2COOH u) HOOC–COOH v) H2N
COO-CH3 w) CH3-COO-I
x) CH3(CH2) 3–COO–(CH2) 3CH3
y) CH3-CH2-COO-CH2-CH(CH3)-CH2-CH3
z) C4H9–NH2
aa) C3H7–N(CH3)–C2H5
ab) CH(CH3)2–NH–CH2-CH2Cl
ac) CH3CH2CO–NH2
ad) CH3-CH2-CH2CO–NH–C5H11
ae) NH2CH2COOH (glycine)
14. Using the IUPAC rules draw the following organic molecules each containing a functional group
a) cyclopentanol
b) 3-isopropyl-2,3-dimethyl-2-hexanol
c) 4,4-dimethyl-1,2-pentandiol
d) pentachlorophenol
e) 1,2,3-trichloro-4,5,6-trihydroxybenzene
f) 1-phenylethanol
g) dicyclobutylether
h) methyl-2-methylpentylether
i) methylphenylether
j) 2,4-difluoro-3,4-dimethylpentanal
k) 2-ethylhexanal
l) 4-hydroxy-3-methoxybenzaldehyde
m) 3-cyclopropyl-4-methyl-2-hexanone (vanillin)
n) 3-cyclopentenone
o) 4,5-dibromo-3-ethyl-2-heptanone
p) trichloroethanoic acid
q) 2-iodo-6-methyloctanoic acid
r) 2-hydroxypropanoic acid (lactic acid)
s) cis-butendioic acid
t) 3-butenoic acid
u) propyl-2-methylbutanoate
v) pentyl butanoate
w) 3-methylbutyl-3-methylbutanoate
x) ethyl-4-aminobenzoate (benzocaine)
y) 2-chloroethyl ethanoate
z) methyl-2-aminophenylmethanoate
aa) ethylbutanamine
ab) ethyl,propyl,pentanamine
ac) N,N-ethyl,butylpropanamide
ad) N,N-diethyl-metatoluamide (DEET)
ae) 2-aminopropanoic acid (alanine)
af) 2-amino-3-phenylpropanoic acid (phenylalanine)
ag) 2-hydroxy-1,2,3-propantricarboxylic acid (citric acid)
Reactions involving organic molecules
This section deals with the variety of chemical reactions that involve the organic molecules you have just learn to identify
1) They all combust- forming oxides ie C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
or H2NC5H10COOH + 9¼ O2 → 6 CO2 + 6½ H2O + NO2
15.a) 3-ethyl-2-methylheptane + O2 (g)
b) 2,5-diammino-4-isopropylhexanoic + O2 (g)
2) The alkANEs (fully saturated) tend to undergo SUBSTITUTION reactions producing an alkyl halide ie
halogenation with Cl2 or Br2 etc (see pg 24 CHEM 12)
Energy
CH4 + Cl2 
 CH3Cl + HCl
hf
or
Energy
C2H6 + Br2 
 C2H5Br + HBr
hf
*The MECHANISM is“FREE RADICAL” (light sensitive). A “FREE RADICAL” is a high energy
unstable molecule because of an unpaired electron. ex: Cl. or .CH3
A)
Cl2 → 2 Cl.
initiation step
B) Cl. + CH4 → .CH3 + HCl
chain step 1
C) .CH3 + Cl2 → CH3Cl + Cl.
chain step 2
D)
2 Cl. → Cl2
termination step
Energy
CH4 + Cl2 
 CH3Cl + HCl
hf
overall reaction
**NOTE: Stability of the carbonyl free radicals is 3° > 2° > 1° . this means that if propane undergoes this
type of reaction the only product formed would be 2-chloropropane.
CH3– CH2– CH3 + Cl2 → CH3– CHCl – CH3 + HCl
16.(complete & name the product) a) butane + Br2 b) 2-methylbutane +Cl2 c) hexane + I2 d) 3-methylpentane + F2
3) The alkyl halides can undergo an elimination reaction forming an alkene (see pg 36 - CHEM 12)
CH3– CHCl – CH3 + OH- (a base) → CH2=CH– CH3 + H2O + Cl4) The alkENEs & alkYNEs (unsaturated) are much more reactive because of the π cloud of electrons on top of the
double bond and tend to undergo ADDITION reactions across this dble bond. (see pg 25 CHEM 12)
C3H6 + Br2 → C3H6Br2 halogenation
C3H6 + H2 → C3H8
hydrogenation
C3H6 + HCl → C3H7Cl hydrohalogenation
C3H6 + H2O → C3H7OH
hydration
*This MECHANISM is“ELECTROPHYLLIC” (hungry for electrons) – ΔG is very negative
H2C = CH2 + Cl:Cl  H2C -CH2 + Cl-  H2C-CH2 + Cl-  H2C -CH2
\ /
| +
| |
attacks the π cloud
Cl
Cl
Cl Cl
seeking the 2 electrons
**NOTE: Stability of the carbonium ion C+ is 3° > 2° > 1° . this means that if 1-propene undergoes this
type of reaction with a hydrogen halide like HCl the only product formed would be 2-chloropropane as
the 2° carbon(in the middle) would become C+ and that is where the negative Cl- will attach. This is
known as Markovnikov’s Rule (the H attaches to the carbon that already has the most H’s)
CH2=CH–CH3 + HCl → CH3– CHCl – CH3
17. i) Do practice exercise #1 (pg 27) , understanding concepts # 2,3,4,5 & 6 ( pg 31)
ii) (complete & name the product)
a) cyclohexene + Cl2
c) 2,2-dimethyl-4-octene + F2
b) 3-methyl-3-hexene +HBr
d) 2-methyl-3-pentene + HI
5) The alcohols (R-OH) are a reactive group of molecules because of the -OH of electrons and tend to undergo
different reactions around this active site (see pg 42 CHEM 12)
a) Preparation of alcohols (Markovnikov) CH3CH2CH=CH2 + H2O → CH3-CH2-CHOH-CH3
hydration
b) Preparation of alkenes from alcohols: CH3-CHOH-CH3 + cat → CH3CH=CH2 + H2O dehydration
c) Oxidation of alcohols
i) primary alcohols:
CH3-CH2OH + ox agent
→ CH3-CHO
→
CH3-COOH
ethanol
ethanal(aldehyde)
ethanoic acid
ii) secondary alcohols: CH3-CHOH-CH3 + ox agent → CH3-CO-CH3
2-propanol
2-propanone (ketone)
iii) tertiary alcohols: CH3-C(CH3)OH-CH3 + ox agent → NO RX
2-methyl-2-propanol
6) The ethers (R-O-R’) are generally not very reactive (they do burn well) (see pg 47 CHEM 12)
a) Preparation of ethers from alcohols
CH3-OH + CH3-OH → CH3-O-CH3 + H2O condensation rx
7) Changing Carboxylic acids to organic salts (ESTERS) – esterification rx (see pg 64 CHEM 12)
Ex: R-COOH (carboxylic acid) + R”-OH (alcohol) → R-COO-R’ (ester)
CH3-CH2-CH2COOH
+ CH3-CH2OH
8) Changing esters back to acids & alcohols – hydrolysis rx
9) Preparation & reactions of amines
(see pg 73 CHEM 12)
10) Preparation & reactions of amides (see pg 74-78 CHEM 12)
11) *Flowchart of organic reactions
(see pg 83 CHEM 12)
→ CH3-CH2-CH2COO-CH2-CH3
ethyl butanoate (sweet smell in pineapples )
(see pg 67 CHEM 12)
12) POLYMERS: Many repeating units of simpler molecules called monomers into endless strings of larger units !
The monomers can be polymerized via “ADDITION” type rx’s (very active dble bond) (see pg 100 CHEM 12)
or by removing a water molecule –> “CONDENSATION” type rx’s. (see pg 108 CHEM 12)
Ex 1
 [ –CH2-CH2-CH2-CH2–]x
CH2=CH2 
ethene
Ex 3
energy / catalyst
polyethylene
chloroethene
 [ – CH(CH3)–CH2–]x
CH(CH3)=CH2 
energy / catalyst
propene
Ex 5
 [–CH2-CHCl-CH2-CHCl–]x
Ex 2 CH2=CHCl 
energy / catalyst
Ex 4
polypropylene
H2N-(CH2)6-NH2 +
1,6-hexandiamine
HOOC-(CH2)6-COOH
CF2=CF2
7. Saran
energy / catalyst


tetrafluoroethene
[ –CF2-CF2–]x
teflon
→ [ – HN-(CH2)6-NH -OC-(CH2)6-CO –]x + H2O
1,6-hexandioic acid
Ex 6 A polyester type molecule
polyvinyl chloride
6 6 nylon
8. orlon
9. polystyrene
10. Lucite
(pressure sensitive glue)
O=C-O-C4H9
|
[-CH-CH2-]x
[ -CH2-CCl2-]x
11. silcone
[-O-Si(CH3)2-]x-
[ -CH2-CH(CN)-]x
12. rubber
[ -CH2-CH-]x
|
13. neoprene
[ -CH2-C(CH3)=CH-CH2-]x
[ -CH2-C(Cl)=CH-CH2-]x
15. Kevlar see pg 110
16. Proteins ( polyamides) see pg 117
17. polypeptides see pg 118
18. Starches/cellulose sugar polymers see pg 125
[ -CH2-C(CH3)-]x
|
O=C-O-CH3
14. instant glue
[ -CH2-C(CN)-]x
|
O=C-O-CH3
Assignment: ORGANIC Chemistry
1. Complete the following Organic rx equations by drawing & naming the product (or reactant ) in each case
a) 4-ethyloctane (l) + oxygen →
b) CH3–CH2–CH(CH3)–CH3
+ Cl2
En / Pt  catalyst


c) CH3–CH2–CH=CH–CH3
+ Cl2
→
d) CH3–C(C2H5)=CH–CHCl–CH2–CH3
+ HCl →
acid / H 

 CH3–CH–(CH3)–COO–CH–(CH3)2
e)
f) CH2 – CH2
+
|
|
CH2 – CH–COOH
CH3–CHOH–CH3
acid / H 


g) CH3–CHCl–C(CH3)2–COO–CH(C2H5)–CH3
h)
-CH2OH
i)
-COOH +
j)
l)
m)
+
ox  agent / H 


-OH
acid / H 


Fe  catalyst


Br2
-CH2-CHOH-CH2-
k)
+ Br2
→
ox  agent / H 


acid / H 

-OH + CH3-CH2-CH2–CH(CH3) -CH2-COOH 
n)
o)
base / OH 


+
-OH
CH3-CHCl-CH2-CH3
AlCl 3


acid / H 


2. Do questions #3, #8, # 9, & #11 on pg 96/97 as well as # 2, #13 & # 19 on pg 146/147