Organic Answers (2013 - 2014)

1

YORK MILLS COLLEGIATE INSTITUTE

SCIENCE DEPARTMENT

Grade 12 Chemistry, University Level, SCH4U

For the school year: 2013 - 2014

ORGANIC CHEMISTRY UNIT ANSWERS Table of Contents:

Organic Chemistry Unit Answers ...................................................................................................................................................... 1 Alkane Nomenclature ........................................................................................................................................................................ 2 Cycloalkane Nomenclature ............................................................................................................................................................... 2 Structural Isomers & Physical Properties of Aliphatics ..................................................................................................................... 3 Geometric Isomers ............................................................................................................................................................................ 4 Alkene Nomenclature ........................................................................................................................................................................ 4 Alkyne Nomenclature ........................................................................................................................................................................ 5 Aliphatic Compound Review ............................................................................................................................................................. 5 Aromatic Hydrocarbons .................................................................................................................................................................... 6 Alkyl Halides ..................................................................................................................................................................................... 7 Alcohols ............................................................................................................................................................................................ 8 Ethers................................................................................................................................................................................................ 9 Aldehydes ....................................................................................................................................................................................... 10 Ketones ........................................................................................................................................................................................... 11 Carboxylic Acids ............................................................................................................................................................................. 11 Esters .............................................................................................................................................................................................. 12 Amines ............................................................................................................................................................................................ 13 Amides ............................................................................................................................................................................................ 14 Nomenclature Review of Organic Derivatives ................................................................................................................................ 15 Substitution Reactions .................................................................................................................................................................... 16 Addition Reactions .......................................................................................................................................................................... 16 Elimination/Dehydration Reactions ................................................................................................................................................. 17 Hydrolysis Reactions ...................................................................................................................................................................... 18 Condensation Reactions ................................................................................................................................................................. 18 Oxidation Reactions ........................................................................................................................................................................ 19 Reduction Reactions ....................................................................................................................................................................... 20 Reactions of Organic Molecules Review ........................................................................................................................................ 20 Retrosynthetic Analysis ................................................................................................................................................................... 22 Polymer Questions .......................................................................................................................................................................... 23

Last revised: May 27, 2014 AWong

2

ALKANE NOMENCLATURE 1. CnH2n+2

2. Alkanes are saturated because they contain the maximum number of hydrogen atoms

3. Naming molecules: a. 2-methylbutane b. 2,3-dimethylpentane c. 3-ethyl-2-methylhexane

d. 2-methylhexane e. 2,2,4-trimethylpentane f. 5-ethyl-3-methyloctane

g. 3-ethyl-2-methylpentane h. 3,4,5-trimethylheptane

4. Drawing molecules:

a.

b.

c.

d.

e.

f.

g.

h.

i.

CYCLOALKANE NOMENCLATURE 1. CnH2n (assuming only one cyclic structure is present)

2. Cycloalkanes are considered saturated because they contain the maximum number of hydrogen atoms given they contain closed structures

3. Naming molecules: a. cyclohexane b. 1-ethyl-3-methylcyclopentane c. 1,2-diethylcycloheptane d. 1-methylcyclobutane

e. 1,2-dimethylcyclopropane f. 1-propylcyclopentane g. 1-cyclohexylcyclohexane h. 1-ethyl-2-methylcyclohexane


a.

b.

c.

d.

e.

f.

g.

h.

3

STRUCTURAL ISOMERS & PHYSICAL PROPERTIES OF ALIPHATICS 1. When a molecule is saturated, it means that the organic molecule has the maximum number of hydrogen atoms possible.

Alkanes are always saturated, in which case, the number of hydrogen atoms for an alkane can be determined by 2n + 2, where n represents the number of carbon atoms in the alkane.

2. Two molecules are structural isomers when they have the same molecular formula but the atoms are bonded in a different arrangement.

3. The molecules separate or come together during a change of state. This occurs when many of the intermolecular forces between the molecules are broken (as in the case of solid to liquid to gas), or when the intermolecular forces are formed (as in the case of gas to liquid to solid).

4. Pentane 36.1C, 2-methylbutane 27.8C & 2,2-dimethylpropane 9.5C. The linear structure of pentane allows for the molecules to closely approach each other, thus forming stronger intermolecular forces. On the other hand, 2,2-dimethylpropane, when they get close together, only part of the atoms are able to get close to the atoms of the other molecule, so the intermolecular forces are not as strong.

5. It turns out that 2,2-dimethylpropane should have the highest density of the three. The reason is because it is a small hydrocarbon & the most compact (spherical) of the three structures. All three molecules are very three-dimensional, so the only thing that can be of use in determining the density is the volume of space taken up by each of the three structural isomers. 2,2-dimethylpropane has the smallest volume of the three, so it has the highest density. (For longer chain hydrocarbons, straight chain hydrocarbons would have higher densities than their branched counterparts).

6. The number of carbon dioxide & water molecules would be the same because the molecular formulas are the same.

7. All three would release the same amount of energy, because all structural isomers have the same bonds, just arranged in a different way.

8. Small aliphatics tend to be gas state at room temperature because they have weak intermolecular forces. Weak intermolecular forces require little energy to break, & room temperature provides enough energy to break those intermolecular forces.

4

GEOMETRIC ISOMERS 1. If a molecule has a geometric isomer it means that the orientation of carbon atoms (or other functional groups) attached to

the double bonded carbons point in different directions.

2. Identifying geometric isomers:

a.

yes;

b. yes;

c. no

d. no

e. no

f. no

g. no

h.

yes;

ALKENE NOMENCLATURE 1. All alkenes have at least one carbon-carbon double bond

2. Alkenes are not saturated. This is because the double bond can be replaced with two hydrogen atoms & the carbon-carbon bonding arrangement is still intact

3. Naming molecules: a. trans-2-butene b. 1-cyclohexyl-cis-1-propene c. 1-pentene d. 3-methyl-trans-3-hexene

e. 1-trans-3-hexadiene f. 1,2-dimethyl-1,4-cyclohexadiene g. 3,4-dimethyl-trans-3-hexene h. 1-trans-3-pentadiene


a.

b.

c.

d.

e.

f.

g.

h.

i.

5

ALKYNE NOMENCLATURE 1. All alkynes have at least one carbon-carbon triple bond

2. Alkynes are not saturated. This is because the triple bond can be replaced with four hydrogen atoms & the carbon-carbon bonding arrangement is still intact

3. Naming molecules: a. 2-butyne b. 1-cyclohexyl-1-propyne c. 1-cyclopentyl-1-ethyne

d. 1-pentyne e. 1,3,5-hexatriyne f. 1-propyne

g. 4-methyl-2-pentyne h. 1,2-dicyclohexyl-1-ethyne


a.

b.

c.

d.

e.

f.

ALIPHATIC COMPOUND REVIEW 1. butane 2. 1,4-hexadiyne 3. 1,2-dicyclobutylethane 4. 2-methylpentane 5. 2,3,4,4-tetramethylheptane 6. 3,4-dimethylhexane 7. cis-2-trans-4-hexadiene 8. 1-butene 9. 4-methyl-trans-2-pentene 10. 1-cyclohexyl-1-ethyne

11. 4,6-dimethyl-1-heptyne 12. 1-cyclopentyl-1,3-pentadiyne 13. 2,3-diethyl-1,3-butadiene 14. 4-ethyl-2,3-dimethyl-2-trans-4-heptadiene 15. 3,6,6-trimethyl-trans-2-trans-4-octadiene 16. 3-ethyl-1-methyl-1-cyclopentene 17. 1-cyclohexene 18. 3,4,5,6-tetramethyl-1-cyclohexene 19. 1-ethyl-4-methylcyclohexane 20. 4-ethyl-3-methylheptane

a.

b.

c.

d.

e.

f.

g.

h.

i.

j.

k.

l.

m.

n.

o.

p.

q.

r.

s.

t.

u.

v.

w.

x.

y.

6

AROMATIC HYDROCARBONS 1. All aromatics (as far as grade 12 chemistry is concerned) contain at least one benzene ring. In general, if the cyclic

contains alternating carbon-carbon single & double bonds & is also flat, then it is aromatic. All aromatics obey Huckles rule.

2. Naming molecules: a. 1,2,3,4,5,6-hexamethylbenzene b. 1-ethyl-3-propylbenzene c. 1-cyclopentylbenzene d. 4-phenyl-1-butene e. 2-methyl-1,3-diphenylpropane

f. 2-ethyl-1-phenylpentane g. 1,3,5-trimethylbenzene h. 1,4-diphenyl-trans-2-butene i. 1-phenyl-1-ethyne


a.

b.

c.

d.

e.

f.

g.

h.

4. Compared to a 6 carbon straight chain alkane (without branches): a. Benzene should have the higher melting/boiling point because it is more flat, & thus able to get closer to another

benzene molecule, so the intermolecular forces should be stronger & harder to break. Therefore the MP/BP of benzene should be higher than hexane.

b. Benzene is more flat, so it can stack with other benzene molecules more closely, so more molecules can fit in the same volume of space. Hexane is more three dimensional, meaning it cannot pack so well in the same volume of space.

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ALKYL HALIDES

1. All alkyl halides contain at least one halogen atom: F, Cl, Br or I.

2. Naming molecules: a. 1-bromo-2-methylpropane b. 1-bromo-2-chlorobenzene c. 2-ethyl-4-iodo-1-pentene d. 3,5-dibromo-1-cyclopentene

e. 5-fluoro-1,3-heptadiyne f. 5-fluoro-3-iodo-1-hexyne g. 2-bromo-1-cyclohexylpropane h. 4,4-diiodo-2,3-dimethylheptane


a. Br

I

b.

Cl

Cl

Cl

F

F

F

c.

Br

Br

d. Cl

e. C

Br

Cl

F

I

f. I

g. Cl

h. F

4. All four molecules have the same molecular shape. This means the only difference between them is the effects of the halogen on the overall polarity of the molecule. 1-fluoropropane should have the higher boiling point because out of all the carbon-halogen bonds, C-F displays the most polar covalent character. This means that 1-fluoropropane is the most polar out of the four molecules. As the most polar molecule, it would have the strongest intermolecular forces, which in turn requires the most energy to break & thereby separate the molecules away from each other.

5. Without looking up the boiling point data for 1-chloropentane & 3-chloropentane, provide a valid argument supporting why: a. 1-chloropentane should have the higher boiling point because the structure is mostly linear (without branches), which

means that the molecules can get really close to each other, thereby increasing the strength of the intermolecular forces. Increased intermolecular forces means it needs more energy to break them, hence higher boiling point.

b. 3-chloropentane should have the higher boiling point because the C-Cl bond is at the middle of the molecule, where

the 2 carbon tails are not long enough to shield (or block) the C-Cl bond from forming intermolecular forces with other molecules.

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ALCOHOLS 1. All alcohols contain at least one hydroxyl group (OH).

2. Naming molecules: a. 3,4-dimethyl-1-pentanol OR 1-hydroxy-3,4-dimethylpentane b. 1,2-cyclohexanediol OR 1,2-dihydroxycyclohexane c. 2-propanol OR 2-hydroxypropane d. 1-cyclopentyl-1,2-butanediol OR 1-cyclopentyl-1,2-dihydroxybutane e. 1-phenyl-1-ethanol OR 1-hydroxy-1-phenylethane f. 1-ethene-1,1,2,2-tetrol OR 1,1,2,2-tetrahydroxy-1-ethene g. 2,2-propanediol OR 2,2-dihydroxypropane h. cis-1-ethene-1,2-diol OR 1,2-dihydroxy-cis-1-ethene


a. OH

b.

OH

OH

c. OH

d. HO OH

e.

OH

f. OH

g. OH

h. OH

4. 1 alcohols have the OH group at the end of the molecule. The C with the OH group is attached to one other carbon. 2 alcohols have the OH group in the middle of the molecule. The C with the OH group is attached to two other carbons. 3 alcohols have the OH group in the middle of the molecule. The C with the OH group is attached to three other carbons.

5. Alternate naming systems arent useful because if no one understands what tert means, then the trivial name is useless.

6. Alcohols with more than one OH should be more soluble in water than those with fewer OH groups. This is because of the solubility rule of like dissolves like. The more similar the solvent & solute molecules are to each other, the better they will dissolve each other. Since water (the solvent) contains two OHs, then alcohols with more OHs would dissolve more readily in water than those with one OH group.

9

ETHERS 1. All ethers have the ether linkage: C-O-C.

2. Naming molecules: a. phenoxybenzene b. 2-cyclopropoxybutane c. 1-(2-methylpropoxy)-2-methylpropane d. butoxycyclopentane e. ethoxypentane

f. 2-(2-propoxy)pentane g. ethoxycyclopropane h. methoxy-2-methylbutane i. 2-methoxy-3-methylbutane

3. Drawing molecules

a. O

b. O

c. O

Br

d. O

e. O

f. O

g. O

h. O

4. Alcohols should have the higher boiling point because alcohols have more intermolecular forces (LF, dd & Hb) while ethers have fewer intermolecular forces (LF & dd). The more intermolecular forces (& stronger intermolecular forces), the higher the boiling point because the more energy that is needed to break the intermolecular forces.

10

ALDEHYDES 1. All aldehydes have at least one carbonyl group located at the end of a chain.

2. Naming molecules: a. propanal b. 2-methylbutanal c. trans-2-butenal

d. 2-cyclohexylethanal e. 2-hydroxypropanal f. phenylmethanal

g. 3,4-dimethylpentanal h. butanedial i. heptanal


a. O

b. O

c. O

d. O

e. O

f. O

g.

O

h. O

4. The boiling points of aldehydes do not increase drastically as the number of carbons increase because as the number of carbons increase, only the London forces increase, which happens to be the weakest of all the intermolecular forces. Aldehydes also have dipole-dipole forces which are much stronger than London forces, but in aldehydes, the dipole forces remain the same as the number of carbons increase. This means that the increase in London forces is relatively insignificant compared to the stronger dipole forces that already exist in aldehydes. For alkanes, the only intermolecular force present is London forces, so when the London forces increase, it can drastically increase the boiling point.

5. 1 alcohols should have the higher boiling point because alcohols have hydrogen bonding in addition to the London forces & dipole forces that they have in common with aldehydes.

6. The carbonyl group in an aldehyde is always in the first carbon in the chain, so it is redundant to write the 1 in the name.

11

KETONES 1. All ketones have the carbonyl group located in the middle of the molecule. In other words, no terminal carbonyl group.

2. Naming molecules: a. 2,3-butanedione b. 1,3-dicyclopropyl-2-propanone c. 1-phenylethanone d. 1-hexene-3-one

e. 4-methyl-2-pentanone f. 3-ethyl-4-methyl-2-hexanone g. 3,4-dimethyl-1-cyclopentanone h. 5-methyl-2-hexanone


a. O

b. O

c. O O

d.

OO

O

e. O

f. O

g.

O

h. O

i. O

4. The name ethanone can exist if a cyclic or phenyl group is located next to the carbonyl group.

5. It is necessary to include the location number of the carbonyl group in ketones because the carbonyl group can potentially be located anywhere in the middle of the molecule, particularly if it has more than 2 non-terminal carbon atoms in the parent chain.

CARBOXYLIC ACIDS 1. All carboxylic acids contain at least one carboxyl group. The carboxyl group consists of a hydroxyl group attached to the

carbon of a carbonyl group.

2. Naming molecules: a. butanedioic acid b. pentanedioic acid c. 2,3-dimethylbutanoic acid d. 4-pentenoic acid e. 4-cyclopentyl-3-methylbutanoic acid

f. cyclobutylmethanoic acid g. cis-2-butenedioic acid h. trans-2-butenedioic acid i. 2-phenylethanoic acid


a.

O

HO

OH

b.

OH

OO

OH

c.

O

OH

d. OH

O

e. O

HO

f. OH

O

g. OH

O

h. O

HO

O

OH

4. Like aldehydes, it is not necessary to write the location numbers for the carboxyl group, as it is always at the end of the molecule.

5. Small carboxylic acids tend to be solids at room temperature because they can get close to other molecules due to their small size. As molecules get closer together, the strength of the intermolecular forces increase, causing the molecules to get closer together. If they are close enough, they form the solid state.

12

ESTERS

1. All esters have the ester linkage. The ester linkage consists of an oxygen atom single bonded to the carbon of a carbonyl group.

2. Naming molecules: a. 2-methylpropyl propanoate b. 3-methylbutyl ethanoate c. octyl ethanoate d. methyl butanoate

e. ethyl butanoate f. pentyl butanoate g. 2-methylpropyl methanoate h. propyl ethanoate


a.

O

O

b. O

O

c.

O

O

d.

O

O

e. O

O

f. O

O

4. Small esters would most likely be in the solid state as they are very polar, with the carbonyl group & the C-O groups. If the molecule is very polar, then its intermolecular forces would be at their strongest, which causes molecules to be closest together, which results in the solid state. (You can also argue that larger esters have stronger London forces, but the increased number of carbons can shield the polar part of the ester molecule, thus the polarity goes down a little bit).

5. Small esters would most likely be soluble in water. Small esters are more polar than larger esters. Since water is small & polar, small esters would be more soluble than large esters.

13

AMINES 1. All amines contain the nitrogen atom in a 1, 2 or 3 situation.

2. Naming molecules: a. N-methylethylamine b. 2,4-diaminopentane c. N,N-dimethylethylamine d. 1-amino-trans-2-butene e. 2-amino-4-methylpentane

f. 1-aminopropane (or propylamine) g. N,N-dimethylphenylamine h. 1-amino-2-propene i. 1,4-diamino-trans-2-butene


a. N

b. H2N

NH2

c. NH2

d. H2N NH2

e. NH2

f. NH

4. 1 amines have one alkyl group attached to the N, while 2 amines have two alkyl groups attached to the N, & 3 amines have three alkyl groups attached to the N.

5. Assuming all are structural isomers, the 1 should be more soluble in water than the others due to the hydrogen bonding the N would have. While some of the Hs would be shielded by the single long carbon chain, the presence of hydrogen bonding from two N-H bonds is more than enough to increase the solubility of the amine in water.

6. Assuming all are structural isomers, the 3 amine would be more soluble in benzene than the others. The reason is because it contains no hydrogen bonding as it does not have any N-H bonds in the structure. (1 & 2 amines have hydrogen bonding). The lack of hydrogen bonding means the 3 amine is less polar than its 1 & 2 counterparts.

7. An argument could be presented which can support both of the amine & alcohol as having the higher boiling point. a. If 1-aminopropane has the higher boiling point, it would be due to the fact that there is two sources of hydrogen

bonding, whereas 1-hydroxypropane has only one source of hydrogen bonding. b. If 1-hydroxypropane has the higher boiling point, it would be due to the fact that the O-H bond is more polar covalent

than the N-H bond. The more polar covalent the bond, the stronger the dipole force, which can be stronger than the intermolecular forces generated by two weaker N-H bonds.

14

AMIDES 1. All amides have the amide linkage. The amide linkage consists of a nitrogen atom bonded to the carbon of a carbonyl

group.

2. Naming molecules: a. 2,N-dimethylpropanamide b. 1,N-dicyclopentylmethanamide c. 2,N-dimethyl-N-phenylpropanamide d. N -ethylpentanamide e. N,N-dimethylpropanamide

f. N-cyclopentylbutanamide g. N-phenyl-trans-2-pentenamide h. 1-phenyl-N-propylmethanamide i. methanamide


a. O

NH2

b.

O

NH

c.

O

NH

d.

O

NH

e.

O

N

f.

O

H2N

4. Amides are rarely gases at room temperature because they are relatively polar. This means the intermolecular forces between molecules are relatively strong, which causes molecules to be close enough to be either in the liquid or solid state.

15

NOMENCLATURE REVIEW OF ORGANIC DERIVATIVES 1. 2-amino-3-ethyl-3,4-dimethyl-6-heptene 2. 1-bromo-3-fluorobenzene 3. 4-ethyl-2,3,5-trimethylhexanoic acid 4. 5-methyl-2,3-hexanedione 5. butoxypentane 6. 6-methyl-3-propylheptanal 7. N,N-diethylpropanamide 8. 3,8-dimethyl-4,5-decanedione 9. propyl ethanoate 10. N-propylpentylamine 11. 2,3-dimethylbutanoic acid 12. 2-butoxyhexane 13. N-butyl-N-ethylpropanamide 14. 3,3-dimethyl-4-phenylheptanal

15. 4-ethyl-5,6-dimethyl-2,3-heptanediol 16. 2-methyl-5-propyl-4-nonanone 17. phenyl hexanoate 18. 2,5,6-triamino-3,4-diethyloctane 19. 3-methyl-4-propyl-1,5-heptanediol 20. 2,3-dimethyl-2-butene 21. 1,1-diphenylmethane 22. propanedial 23. 2-methyl-3-pentanone 24. 1,2-dichloro-trans-1-ethene 25. N-ethyl-N-methylbutylamine 26. 3-(2-butoxy)pentane 27. 3,5-dimethylheptanal

a.

OHOH

b.

O

c. N

d. O

e.

O

OO

f. O

HO

g.

O

NH

h.

O O

O

i. N

j. O

k. O

l.

O

N

m.

OH OH

n.

Cl Cl

Cl

o. NH2

NH2

NH2

p. O

q.

r.

O

O

OHHO

s.

t.

O

u. OH

v.

Cl

Cl

w. NH

O

x. O

16

SUBSTITUTION REACTIONS

Note: only the main organic molecule is/are listed & by-products such as HBr, HCl, H2O are not included.

1.

+ Br2 3FeBr

Br

2. CH3Br + H2O CH3OH

3.

+ CH3Cl 3AlCl

+

+

4.

+ Cl2 UVlight Cl

5.

+ Cl2 UVlight Cl

6. Br

+ HF F

7. Br + NH3 H2N

8.

+ HNO3 42SOH NO2

9.

+ Cl2 UVlight Cl

+

Cl

+ Cl

+

Cl

a. + Cl2 UVlight Cl + HF F

b. + Cl2 UVlight Cl +

3AlCl

c. + Cl2 UVlight

Cl + H2O

OH

d. + Cl2 UVlight Cl + NH3 NH2

ADDITION REACTIONS

1. + H2 Pt

2.

+ HCl peroxide Cl

17

3.

+ HBr Br

4. + H2 Pt

5.

+ Br2 Br

Br

6.

+ HCl Cl

a. + H2O OH

b. + H2 Pt

c. + HCl peroxide

Cl + 3

AlCl

d. + HF peroxide

F

e. + HCl Cl

+

3

AlCl

f. + HCl Cl

+ NH3

NH2

g. + Cl2 Cl

Cl

ELIMINATION/DEHYDRATION REACTIONS

1. OH 42SOH

2. Cl

NaOH +

3. OH 42SOH

4. OH

42SOH

5. Br

NaOH

6. OH

42SOH

18

a. OH 42SOH

b. OH 42SOH + Cl2

Cl

Cl

+ NaOH

c. OH 42SOH + H2O

OH

d. OH )(bstepsfrom + HBr

Br

+ HBr

Br

Br

HYDROLYSIS REACTIONS

1. NH2

O + H2O

H

O

OH

+ NH3

2. O

+ H2O

H

OH

+ HO

3.

O

N

+ H2O H

O

OH

+ HN

4. O

O + H2O

H

O

OH + HO

5. O

+ H2O H

OH

+ HO

6.

O

O

+ H2O H

O

OH

+ HO

a. O + H2O H

OH + NH3 NH2 (disregarded the other products)

b. O

O + H2O

H OH + NH3 NH2 (disregarded the other products)

c. NH

O

+ H2O

H NH2 (disregarded the other products)

CONDENSATION REACTIONS

1.

O

OH

+ NH

42SOH

O

N

19

2. O

OH + OH 42

SOH

O

O

3. OH + OH 42SOH O

4. O

HO + OH

42SOH

O

O

a. OH + HO 42SOH O

b. OH + O

HO 42SOH

O

O

c. OH + NH3 NH2 + O

OH 42SOH

NH

O

OXIDATION REACTIONS

OH

][O O

1. O

][O

O

OH

2. OH

][O

O

3.

OH

][O

O

4. OH

][O

O

5. O

][O NR

6. OH ][O

O

7. HO OH

OH

][O HO OH

O

or

O OH

OH

a. OH ][O O

20

b. OH 42SOH + H2O

OH

][O

O

c. OH ][O O

][O O

OH

REDUCTION REACTIONS

1. HO OH ][R

NR

2. OH ][R

NR

3. O

][R

OH

4. NH2

O ][R

NH2

HO

5. O

OH ][R

O

6. OH

][R NR

7.

O

OH

OHO

HO

O

][R

O

OHO

HO

O

a. O

O + H2O

H OH

O

][R O

][R OH

b. O

O + H2O

H HO

O ][R O

][R OH

REACTIONS OF ORGANIC MOLECULES REVIEW

1. combustion C6H14 + + 9.5 O2 6 CO2 + 7 H2O

2. addition

(halogenation) 2

Cl

Cl

Cl 2,3-dichloropentane

21

3. condensation OH

O + OH

42SOH O

O

ethyl pentanoate

4. oxidation OH ][O

O

heptanal

5. reduction O

][R

OH

1-butanol

6. substitution Cl

+ Cl2 UVlight

Cl

Cl

1,2-dichloro-2-methylpropane

7. addition

(halohydrogenation) + HF F

2-fluorobutane

8. condensation O

HO +

H2N

O

42SOH

O

NH

O

(dont worry about this one)

9. reduction O

][R OH 2-hydroxybutane

10. hydrolysis O

O

+ H2O

H OH

O

+ HO hexanoic acid & 1-propanol

11. reduction OH

O ][R

OH

OH 1,3-pentanediol

12. oxidation OH

][O O 2-butanone

13. Cl + Cl2 No reaction

14. substitution + Cl2 UVlight

Cl

(any #)-chloropentane

15. O

O NaOH No reaction

16. combustion C9H18 + 13.5 O2 9 CO2 + 9 H2O

17. oxidation O

][O

O

OH hexanoic acid

22

18. addition

(halohydrogenation) + HCl Cl

2-chlorohexane

RETROSYNTHETIC ANALYSIS

1. + Cl2

Cl

Cl

2.

+ H2O

OH

3. + H2O OH (50% yield)

][O

O

4. + H2O OH

+ H2O peroxide HO

OH + HO 42SOH O

5. + H2O OH

+ NH3

NH2

6.

+ H2O peroxide OH

][O

O ][O

O

OH

7. + H2O peroxide

OH ][O

O ][O

O

OH

+ H2O HO + NH3 H2N

O

OH

+ H2N

42SOH O

NH

8. + H2O peroxide OH

+ H2O peroxide

OH ][O

O ][O

O

HO

OH + OHO

42SOH

O

O

a. + Cl2 UVlight

Cl

+ H2O

OH

(do this

23

twice)

OH

+

OH

42SOH

O

b.

+ Cl2 3FeCl Cl

+ NH3

NH2

c. C C C + H2 Pt

CC

C

(need 3 of

these)

H

C

C C C

C

C C

C

C

d. CH4 + Cl2 UVlight CH3Cl

+

3AlCl

+ Cl2 UVlight

Cl

+ H2O

OH

OH

][O

O

][O

O

OH

e. CH4 + Cl2 UVlight CH3Cl

+ NH3 CH3NH2

CH3NH2 + Cl2 UVlight ClCH2NH2 + NH3

H

NH2H2N C

H

H

NH2H2N C

H

+ Cl2 UVlight

H

NH2H2N C

Cl

+ NH3

H

C NH2

NH2

H2N

f. OH ][O

O

][O

O

OH + NH3

O

NH2

g. + H2O OH

][O

O

h. + Cl2 UVlight

Cl

NaOH

+ Cl2

Cl

Cl

NaOH

Cl

NaOH

POLYMER QUESTIONS 1. Defining terms:

a. Monomer a small, repeating unit within an organic molecule chain b. Dimer two monomer units together c. Polymer an organic molecule chain consisting of many monomer units d. Polyester an organic molecule chain consisting of carboxylic acid (with at least 2 carboxyl groups) & alcohol

monomers (with at least 2 hydroxyl groups)

24

e. Polyamide an organic molecule chain consisting of carboxylic acid (with at least 2 carboxyl groups) & amine (1 or 2 only) monomers

2. Identifying monomers:

a. C

CH3

CH2 b.

Cl

CH2C

H

3. Out of the above monomers, 2(a) would have the lower MP/BP because the monomer is non-polar. Non-polar molecules generate only LF as intermolecular forces, which are easy to break.

4. Three repeating units of tetrafluoroethene:

C

F

F F

F F

F F

F F

F

C C C C C

F

F

5. Forming polymers from monomers:

a. Not possible

b. C

C C C

C

C C

C

C

c. HO O

O

O

O O

O

O

OH

d.

O

NHHO

O

O

O

NH

O

O

NH2

e.

6. Identifying monomers: a. 1,2-ethanediol & butanedioic acid b. 4-aminobutanoic acid

7. Cross-linking: a. This molecule will form hard cross-linking because the second double bond can form extra addition polymerization

between polymer strands b. This molecule will not form a polymer, let alone create cross-links c. This molecule will form hard cross-linking because the extra H atoms will form condensation polymerization reactions

to connect different polymer strands together d. This molecule will form soft cross-linking because there arent any opportunities for extra addition polymerization or

condensation polymerization. The one double bond it has must be used to form a polymer

8. Identifying what groups must be present in monomers for the following: a. Addition polymerization at least one double or triple bond b. Condensation polymerization the amino or hydroxyl or carboxyl group c. Hard crosslinking more than one double or triple bond d. Soft crosslinking no more than one double or triple bond must be present

Documents

Organic Answers (2013 - 2014)