Download Descriptive Chemistry for the Final Exam

Chemistry 1051
Descriptive Chemistry
Intersession
Material on Final
Group 4 Elements
General comparison of chemical and physical properties.
PHHM (9th Edition): pp 906-909, 895-896
PHH (8th Edition): Section 22-5, 23-5 to top of 933.
Carbon and silicon are non-metals, germanium is a semi-metal and tin and lead are
metals.
Carbon
1.
2.
3.
4.
2 common allotropes, diamond (sp3 carbon) and graphite (sp2 carbon)
There are several other allotropes of carbon. Buckminster fullerene C60 and carbon
nanotubes are the most famous of the recently discovered allotropes. They contain
5 and 6 member rings of carbon atoms.
Forms covalent compounds with non-metals mostly with a valence of four.
It is the basis for organic chemistry. It forms molecules containing long chains of
carbon atoms (called catenation), both straight and branched structures. It forms
ring structures (5 and 6 carbon rings are common) and compounds with strong
double and triple bonds.
Silicon
1.
2.
3.
Pure form has a diamond-like structure.
Forms strong Si-O bonds found in silicates and related compounds such as
silicones which predominate over silanes which contain relatively weak Si-Si and
Si-H bonds.
Dominates the semi-conductor industry.
Germanium
exhibits semi-conductor behaviour
Tin
(a)
exists in 2 crystalline forms, α(gray) or non-metallic tin and β(white) or
metallic tin. Grey tin is stable below 13°C and white tin is stable above 13°C.
Conversion from white to grey tin at temperatures below 13°C causes tin
disease. The tin crumbles due to conversion to the lower density form.
(b)
production: high temperature reduction of SnO2, the chief tin ore
SnO2(s) + C(s)
(c)
(d)
∆
⎯
⎯→
Sn(l) + CO2(g)
main uses: 1. plating iron to make tin cans
2. in low melting alloys called solders.
compounds: exists with tin in + 2 oxidation state (5s2 electrons are an inert
pair) and +4 oxidation state.
chlorides:
oxides:
SnCl2 is a good reducing agent
tin from scrap tin plate is recovered as SnCl4
forms both SnO and SnO2
Page 1 of 14
Lead
similar to tin i.e. both are soft, malleable and low melting. Like tin, it also exists in
+2 and +4 oxidation states.
production: roasting of PbS is followed by reduction of PbO
∆
⎯
⎯→
2 PbO(s) + 2 SO2(g)
∆
⎯
⎯→ 2 Pb(l) + CO2(g)
2 PbS(s) + 3 O2(g)
2 PbO(s) + C(s)
uses:
(1)
(2)
to make lead-acid storage batteries used in cars and trucks
radiation shields, (e.g. to absorb x-rays in hospital labs)
oxides:
several, PbO, PbO2, Pb3O4
PbO2 is a strong oxidizing agent as Pb prefers the +2 oxidation state
Lead poisoning: ranges from mild to severe including damage to the brain. Due to
consumption of water and food exposed to lead-containing materials.
Questions
PHHM (9th Edition): Chapter 21: 41, 43, 51
PHH (8th Edition): Chapter 22: 7(c), 7(f), 51
Chapter 23: 57, 59
Extra Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
13.
14.
Explain why the Group 4A elements become increasingly more metallic in their
behaviour going from carbon to lead.
Explain why carbon compounds are more abundant than those of all other elements
combined. Hint: describe the versatility of carbon.
Define the term catenation.
Explain why silicon forms few compounds with Si-Si and Si-H bonds but many
compounds with Si-O bonds.
Which two Group 4A elements are found in semi-conductors?
Explain tin disease.
Write reactions for the reduction of PbO(s) and SnO(s) by C(s) at high temperature.
List 2 uses for the elements tin and lead.
Explain why it is important to recycle lead.
Write balanced reactions for the 2 steps (roasting to form the oxide followed by
reduction with carbon) in the production of lead from PbS.
Name two allotropes of carbon in addition to graphite and diamond.
Explain the term roasting as it is applied to the industrial production of metals from
their ores. Why is the roasting process a major source of pollution?
Page 2 of 14
Organic Chemistry
PHHM (9th Ed): Chapter 26: Sections: 1, 2, 3, 5, 6, 7 (to page 1097)
PHH (8th Ed): Chapter 27: Sections 1, 2, 3, 5, 6, 7 (to pg 1084)
Topics not included in course: Nomenclature of organic compounds, conformations and
all reactions in text except those given on this handout.
Types of carbon atoms (described by the hybridization of the carbon)
1.
sp3 carbon: form 4 single bonds:
Tetrahedral geometry at the carbon
2.
sp2 carbon: form 2 single bonds + one double bond
Triangular planar geometry at the carbon
3.
sp carbon: form one single bond and one triple bond or 2 double bonds
Linear geometry at the carbon
hydrocarbons:
saturated:
contain only carbon + hydrogen
only C−H and C−C bond present, only sp3 carbons.
unsaturated:
contain at least one C=C or one C≡C i.e. at least one π bond.
Alkanes:
1.
aliphatic
formula CnH2n+2 where n = 1, 2, 3, etc.
normal:
branched:
2.
only one continuous chain of carbon atoms
at least two chains of carbon atoms
cyclic or alicyclic: at least one chain of carbon atoms joined to form a ring.
Smallest ring has three carbons.
formula: 2 less H per ring vs. normal alkane with same #
carbons e.g. one ring CnH2n.
Structural formula: shows all bonds and arrangement of atoms.
Alkenes: contain at least one C=C
General Formula: 2 less H per double bond vs. normal alkane with same #
carbons
Alkynes: contains at least one C≡C
General Formula: 4 less H per triple bond formed vs. normal alkane with
some # carbons
e.g.
H-C≡C-H
C2H2 vs. C2H6 (alkane)
Page 3 of 14
Functional Groups:
Functional groups are distinctive grouping of atoms. The rest of the molecule, usually an
alkyl group (alkane -H) is given the symbol R. A functional group can be converted to
other functional groups in reactions and is often very important in determining physical
properties such as solubility, melting point etc.
Types of sp3 carbons bonded to a functional group
type
# carbons directly bonded
# H’s directly bonded
primary
1(0)
2(3)
secondary
2
1
tertiary
3
0
Alcohol (R-OH): OH is called the hydroxyl functional group.
Alkyl halide (R-X): alkane with one or more H replaced by halogens (F, Cl, Br, I)
Ether (R-O-R′): like H2O with both H’s substituted by alkyl groups
Aldehyde (RCHO): contains the carbonyl functional group,
where R can be H or alkyl group e.g.
Ketone (RCOR′): also contains the carbonyl functional group.
R, R′ must be alkyl groups e.g.
Page 4 of 14
Carboxylic acid (RCOOH): contains the carboxyl functional group
R can be H or alkyl group e.g.
Ester (RCOOR′): combination of carbonyl and ether functional groups
R can be H, R′ cannot be H e.g.
Amine (RR’R”N): ammonia, NH3, with one or more H’s are replaced by alkyl groups.
Page 5 of 14
Isomers: different molecules with the same chemical formula
Constitutional (structural) isomers: molecules whose atoms are connected differently.
1.
skeletal: different carbon skeletons (normal vs. branched)
2.
functional: contain different functional groups.
3.
positional: different positions for a functional group for the same carbon skeleton.
Stereoisomers:
differ only in the arrangement of their atoms in space.
(several types)
e.g.
geometric: (cis-trans diastereomers) isomers which differ in the positioning
of functional groups relative to the double bond on an alkene (also found in
ring compounds).
NOTE:
In isomer A, the Cl’s are across (trans) the double bond and in isomer B, the
Cl’s are on the same side (cis) of the double bond. A and B are geometric
isomers. For geometric isomers, the two functional groups or substituents
must be on different sp2 carbons. For isomer C, the two Cl’s are on the same
sp2 carbon. Isomer C is a positional (structural) isomer of isomers A and
B.
Free rotation about C-C bonds
Holding one carbon in a fixed position, the other carbon with its three bonding positions
can freely rotate in a circle. This can occur because the σ molecular orbital containing the
bonding electrons is not distorted. In general, little energy is needed. As a consequence,
all three positions for a functional group on a primary sp3 carbon or two positions on a
secondary sp3 carbon are interconvertible and hence identical.
Note: Free rotation cannot occur about C−C bonds in rings.
Page 6 of 14
Rotation about C=C bonds
Holding one carbon in a fixed position and rotating the other sp2 carbon with its two
bonding positions, breaks the π bond formed by the 2pz orbitals of the carbons. This
requires a lot of energy. Hence, cis isomers cannot be converted into trans isomers
except at high temperatures.
How to find some or all constitutional and geometric isomers for a given
chemical formula.
Step 1:
Calculate the number of double bond equivalents (DBE). One DBE accounts
for 2 less H atoms in the formula compared to the n−alkane formula with
same number of carbon atoms.
# DBE =
# H (n - alkane) - # (H + halogens)
2
NOTE: oxygens are ignored in calculating DBE’s.
A ring or a double bond is one DBE. A triple bond is two DBE’s
e.g.
C4H8O vs. C4H10
#DBE = (10−8)/2 = 1
This means all isomers must contain either one ring, or one C=O or one C=C
Step 2:
determine all unique (normal and branched) carbon skeletons.
Draw all constitutional isomers of C5H12 (no DBE’S, only alkane skeletal isomers are
possible). There are only three unique carbon skeletons for five carbons.
Step 3:
Q1.
determine all unique positions for functional groups on each unique carbon
skeleton.
Draw all alkene isomers of C4H8 (1 DBE, 2 unique carbon skeletons)
Page 7 of 14
Q2.
Draw all structural isomers of C3H7Cl (no DBE, only one unique carbon skeleton)
Only positional isomers possible:
Q3.
Draw 2 aldehydes and one ketone with the formula C4H8O (1 DBE, 2 unique carbon
skeletons)
aldehydes:
ketone:
Q4.
Draw an ether and two alcohols with the formula C3H8O(no DBE, one unique
carbon skeleton)
ether
CH3-CH2-O-CH3 (only one unique place to insert O into skeleton)
alcohols
CH3-CH2-CH2-OH (primary)
(secondary)
Sample questions on isomers (Answers can be found on Dr. Flinn’s website)
1.
Draw all possible aldehydes (4) and ketones (3) for C5H10O.
2.
Draw all possible ethers (3) and alcohols (4) with the formula C4H10O. Label each
alcohol as primary, secondary, or tertiary.
3.
Draw all possible alkenes (6) and cyclic alkanes (4) with the formula C5H10. Label
any cis and trans isomers.
4.
Draw all possible amines (7) with the formula C4H11N. Label each as primary,
secondary, or tertiary.
5.
Draw two esters and one carboxylic acid with the formula C3H6O2.
6.
Draw all possible isomers of C3H5Cl (4).
Page 8 of 14
Reactions of organic compounds:
Mainly involve conversion of one type of functional group to another.
Some reactions of amines:
1.
all are weak bases and hydrolyse to form their conjugate acid and OH⎯
e.g. (CH3)3N(aq) + H2O(l) ⇌ (CH3)3NH+(aq) + OH⎯(aq)
2.
They are neutralized by strong acid to form a salt
e.g. CH3NH2(aq) + HCl(aq) → CH3NH3Cl(aq)
Some reactions of carboxylic acids: RCOOH
1.
all are weak acids and hydrolyse to form their conjugate base and H3O+
e.g.
2.
They are neutralized by strong base to form a salt and H2O
e.g.
3.
HCOOH(aq) + H2O(l) ⇌ HCOO⎯(aq) + H3O+(aq)
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
They react with sodium to form H2(g) and a salt
e.g.
2 HCOOH(aq) + 2 Na(s) → 2 HCOONa(aq) + H2(g)
Some reactions of esters: R′COOR
Esterification
The reversible formation of an ester by reaction of an alcohol (ROH) with a carboxylic
acid (R′COOH) using strong acid as catalyst. The hydroxyl H of the alcohol combines with
the OH of the carboxylic acid to form water. The hydroxyl oxygen of the alcohol bonds to
the carboxyl carbon to form the ester.
General: ROH + R′COOH
⎯acid
⎯→
catalyst
R′COOR + H2O
Ester hydrolysis:
Decomposition of an ester. This is the reverse of esterification reaction. The ester breaks
apart at the bond between the carboxyl carbon and the singly-bonded oxygen. H and OH
from a water molecule are inserted to form the original alcohol and carboxylic acid.
+
(a)
acid catalyzed hydrolysis R′COOR + H2O
Page 9 of 14
H
⎯⎯→
R′COOH + ROH
(b)
base hydrolysis (saponification)
Excess strong base is used to decompose the ester. The original alcohol and the
salt containing the conjugate base of the carboxylic acid are formed. The free
carboxylic acid can be generated by making the solution acidic.
Soap making: Fats are esters of glycerol (a trialcohol) and long chain (C14-C22)
carboxylic acids called fatty acids. Heating fat with strong base was once used to
make soap (salt of fatty acids).
General
R′COOR
⎯NaOH
⎯⎯→ R′COO⎯ Na+ + ROH
R′COO⎯ Na+
∆
H+
⎯⎯→ R′COOH
examples:
Oxidation reactions
One oxidation step: one carbon to hydrogen bond breaks and a new carbon to oxygen
bond forms.
1.
Primary alcohols
Acidified Cr2O72⎯ as oxidizing agent. Initial product is an aldehyde (one oxidation
step) which readily oxidizes to form a carboxylic acid (two oxidation steps). When
the solution is heated to increase reaction rate, some of the more volatile aldehyde
escape into the gas phase and is not further oxidized.
Page 10 of 14
Oxidation with pyradinium chlorochromate (PCC). This milder oxidizing agent,
in a non−aqueous solvent such as CH2Cl2(l) oxidizes a primary alcohol to the
aldehyde but no further.
Oxidation with basic KMnO4(aq). The product is potassium salt of the carboxylic
acid (two oxidation steps). The free carboxylic acid can be generated by making
the solution acidic.
2.
Secondary alcohols
Oxidation with acidified Cr2O72⎯: Product is a ketone (one oxidation step).
General
e.g.
3.
Tertiary alcohols
DO NOT OXIDIZE because a C−C bond must be broken.
Page 11 of 14
4.
Aldehydes
Oxidation with basic KMnO4(aq). Product is potassium salt of carboxylic acid
(one oxidation step). The free carboxylic acid can be generated by making the
solution acidic.
General
e.g.
5.
Ketones
DO NOT OXIDIZE because a C−C bond must be broken.
Page 12 of 14
Questions on reactions
Complete by determining the organic products of each reaction. Write NR if no reaction:
Answers available on the web at http://www.chem.mun.ca/~cgflinn/chem1051/
Page 13 of 14
General physical properties of organic compounds
Hydrocarbons: non-polar, dispersion forces only.
Compounds with polar functional groups: alcohols, carboxylic acids, amines, ethers,
esters, aldehydes, ketones.
Compounds which can form hydrogen bonding: alcohols, carboxylic acids, primary and
secondary amines.
Effect of R group on properties: the longer the non-polar R group, the more the
compound behaves like a non-polar compound.
e.g. CH3OH is completely soluble in water C4H9OH is only slightly soluble in water
PHHM (9th Ed.): Chapter 26: 1, 4, 7, 9,49
PHH (8th Ed.): Chapter 27: 19, 20, 23-25, 27
Developed by Dr. Chris Flinn
Page 14 of 14