SNS - Organic Chemistry - Alkanes

Question 1

Physical Properties

1. Density
2. Melting Point
3. Boiling Point

Answer 1

1. Generally increases with increasing molecular weight
2. As above. but also generally decreases with increased branching
3. As above

Question 2

Physical Properties

State

Answer 2

At room temperature:

1-4 C = gas

5-16 C = liquid

>17 C = solid

Question 3

Reactions

Answer 3

1. Combustion
2. Free radical halogenation
3. Pyrolysis
4. Substitution

Question 4

Reactions

Free Radical Halogenation

Answer 4

One or more hydrogens replaced by halogen atoms via free-radical substitution mechanism

Question 5

Reactions

Free Radical Halogenation

Mechanism

Answer 5

1. Initiation - diatomic halogens homolytically cleaved by either heat or light to form two free radicals

X₂ → 2X•

2, Propagation - a radical produces another radical that can continue the reaction. A free radical reacts with an alkane to form HX and an alkyl radical, or an alkyl radical reacts with X2 to form an alkyl halide and an alkyl radical

X• + RH → HX + R•

R• + X₂ → RX + X•

3. Termination - Two free radicals combine

2X• → X₂

RX• → R₂

X• + R• → RX

Question 6

X2 → 2X•

Answer 6

Alkanes

Free Radical Substitution

Initiation

Question 7

X• + RH → HX + R•

Answer 7

Alkanes

Free Radical Substitution

Propagation

Formation of Alkyl Radicals

Question 8

R• + X2 → RX + X•

Answer 8

Alkanes

Free Radical Substitution

Propagation

Reaction of Alkyl Radicals

Question 9

2X• → X₂

Answer 9

Alkanes

Free Radical Substitution

Termination

Formation of Halogens

Question 10

R• → R₂

Answer 10

Alkanes

Free Radical Substitution

Termination

Formation of Alkanes

Question 11

X• + R• → RX

Answer 11

Alkanes

Free Radical Substitution

Termination

Formation of Alkyl Halides

Question 12

Reactions

Free Radical Substitution

Bromination

Answer 12

Bromine radicals react fairly slowly

Primarily attack the hydrogen atoms on the carbon atom that can form the most stable free radical - the most substituted carbon atom. Thus a tertiary radical is most likely to be formed in a free radical bromination reaction

•CR₃ > •CR₂H > •CRH₂ > •CH₃

3º > 2ª > 1º > methyl

Question 13

Reactions

Free Radical Substitution

Chlorination

Answer 13

More rapid - thus depends not only on the stability of the intermediate but on the number of hydrogens present

Likely to replace primary hydrogens because of their abundance despite the relative instability of primary radicals

Question 14

Reactions

Combustion

Answer 14

Reaction of alkanes with molecular oxygen to produce carbon dioxide, water and heat

Often incomplete, producing significant quantities of CO rather than CO2.

Question 15

Reactions

Combustion

Mechanism

Answer 15

Very Complex

Believed to proceed through a radical process

Question 16

C3H8 + 5O2 → 3CO2 + 4H2O + heat

Answer 16

Alkanes

Combustion (propane)

Question 17

Reactions

Pyrolysis

Answer 17

Or cracking

Occurs when a molecule is broken down by heat. Most commly used to reduce the molecular weight of heavy oils and to increase the production of more desirable volatile compounds

Question 18

Reactions

Pyrolysis

Mechanism

Answer 18

C-C bonds are cleaved to produce smaller-chain alkyl radicals. These can recombine to form a variety of alkanes

CH₃CH₂CH₃ heat→ CH₃• + •CH₂CH₃

2CH₃• → + CH₃CH₃

2•CH₂CH₃ → CH₃CH₂CH₂CH₃

Alternatively, in process called disproportionation, a radical transfers a hydrogen atom to another radical to produce an alkane and an alkene

CH₃• + •CH₂CH₃ → CH₄ + CH₂=CH₂

Question 19

CH₃CH₂CH₃ heat→ CH₃• + •CH₂CH₃

Answer 19

Alkanes

Pyrolysis

Formation of smaller-chain alkyl radicals

Question 20

2CH₃• → + CH₃CH₃

2•CH₂CH₃ → CH₃CH₂CH₂CH₃

Answer 20

Alkanes

Pyrolysis

Formation of alkanes

Question 21

CH₃• + •CH₂CH₃ → CH₄ + CH₂=CH₂

Answer 21

Alkanes

Pyrolysis

Disproportionation

Question 22

Reactions

Substitution

Answer 22

Alkyl halides and other substituted carbon atoms can take part in nucleophillic substitution reactions

Two types:

SN1 - unimolecular nucleophilic substitution - so called as rate of reaction depends on only one species. Generally the rate determining step is the dissociation of this species to form a stable, positively charged carbocation

SN2 - bimolecular nucleophilic substitution - rate of reaction depends on two species, substrate and nucleophile. Involves a nucleophile simultaneously bonding with a compound and displacing the leaving group

Question 23

Nucleophiles

Basicity

Answer 23

If two nucleophiles have the same attacking atom, for example oxygen, nucleophilicity is roughly correlated with basicity - the stronger the base, the stronger the nucleophile

Question 24

Nucleophiles

Basicity

Put in order of increasing strength:

ROH, H2O, RCO2- HO- RO-

Answer 24

RO- > HO- > RCO2- > ROH > H2O

Question 25

Nucleophiles Size and Polarity

Answer 25

If attacking nucleophiles differ, nucleophilic ability doesn't necessarily correspond to basicity. For example, in a protic solvent, large atoms tend to be better nucleophiles as they can shed their solvent molecules and are more polarizable In aprotic solvents, however, the nucleophiles are 'naked' - not solvated. Nucleophilic strength is then related to basicity

Question 26

Nucleophiles Size and Polarity Put these molecules in order of increasing strength for a (a) protic solvent (b) non-protic solvent Cl, Br, F, I, H2O, RO-, HO-, CN-

Answer 26

a) CN- \> I- \> RO- \> HO- \> Br- \> Cl- \> F- \> H2O b) F- \> Cl- \> Br- \> I.

Question 27

Nucleophiles Leaving Groups

Answer 27

The ease with which nucleophilic substitution occurs is also related to the leaving group. The best are those that are weak bases, as can accept an electron pair and dissociate to form a stable species

Question 28

Nucleophiles Leaving Groups Put in order of increasing suitability as a leaving group CL-, Br-, F-, I-

Answer 28

I- \> Cl- \> Br- \> F- Opposite to order of base strength

Question 29

Protic Solvent

Answer 29

Capable of hydrogen bonding

Question 30

Reactions Substitution SN1 Mechanism

Answer 30

Two steps: 1. Dissociation of a molecule into a carbocation and a good leaving group. Carbocations are stabilised by polar solvents that have lone electron pairs to donate (eg water, acetone) or by charge delocalisation. More highly substituted carbocations are therefore more stable 2. Combination of the carbocation with a strong nucleophile. To get the desired product, the original substituent should be a better leaving group than the nucleophile so that at equilibrium RNu is the main product

Question 31

Reactions Substitution SN1 Rate

Answer 31

Rate limiting step is the dissociation of the molecule to form a carbocation - energetically unfavourable Therefore a first-order reaction Rate can be increased by anything that accelerates carbocation formation. Most important factors are: 1. Structural - highly substituted alkylhalides allow for distribution of the positive charge over a greater number of carbon atoms and thus form the most stable carbocations 2. Solvent effects - highly polar solvents are better at surrounding and isolating ions than are non-polar solvents. Polar protic solvents such as water work best as solvation stabilises the intermediate state 3. Nature of the leaving group - weak bases dissociate more easily from the alkyl chain and thus make better leaving groups, increasing the rate of carbocation formation

Question 32

Reactions Substitution SN2 Mechanism

Answer 32

Nucleophile actively displaces the leaving group. For this to occur, nucleophile must be strong and the reaction can't be sterically hindered 1. The nucleophile attacks the reactant from the backside of the leaving group to form a trigonal bipyramidal transition state 2. As the reaction progresses the bond to the nucleophile strengthens while the bong to the nucleophile weakens 3. The leaving group is displaced as the bond to the nucleophile becomes complete

Question 33

Intermediate vs Transition State Definition

Answer 33

Intermediate - well-defined species with a finite lifetime Transition state - theoretical structure used to define a mechanism

Question 34

Reactions Substitution SN2 Rate

Answer 34

The single step for this reaction involves two species - the molecule with the leaving group and the attacking nucleophile. Therefore second order, follows second order kinetics

Question 35

Reactions Substitution SN1 Stereochemistry

Answer 35

Involve highly carbocation intermediates which are approximately planar and therefore achiral If the original compound is optically active because of the reacting chiral centre, then a racemic mixture will be produced - SN1 reactions therefore result in a loss of optical activity

Question 36

Reactions Substitution SN2 Stereochemistry

Answer 36

Single-step reaction involving a chiral transition state. Since the nucleophile attacks from one side of the central carbon and the leaving group departs from the opposite side, the reaction 'flips' the bonds attached to the carbon If the reactant is chiral, optical activity is usually retained, however an inversion of configuration occurs

Question 37

Reactions Substitution (a) SN1 vs (b) SN2 1. Steps, 2. Solvents, 3. Carbon attacked, 4. Rate, 5. Optical activity, 6. Nucleophiles

Answer 37

1. (a) two, (b) one 2. (a) favoured in polar, protic solvents, (b) favoured in polar aprotic solvents 3. (a) 3º \> 2º \> 1º \> methyl, (b) 3º \> 2º \> 1º 4. (a) first order =k[RX], (b) second order =k[Nu][RX] 5. (a) Racemic products, loss of optical activity if existed previously, (b) Optically active, inverted products 6. (a) Favours by use of bulkly nucleophiles, (b) Strong, non-bulky, no steric hindrance