Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

DAT > SNS - Organic Chemistry - Alkanes > Flashcards

SNS - Organic Chemistry - Alkanes Flashcards

1
Q

Physical Properties

  1. Density
  2. Melting Point
  3. Boiling Point
A
  1. Generally increases with increasing molecular weight
  2. As above. but also generally decreases with increased branching
  3. As above
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Physical Properties

State

A

At room temperature:

1-4 C = gas

5-16 C = liquid

>17 C = solid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Reactions

A
  1. Combustion
  2. Free radical halogenation
  3. Pyrolysis
  4. Substitution
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Reactions

Free Radical Halogenation

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Reactions

Free Radical Halogenation

Mechanism

A
  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•

  1. Termination - Two free radicals combine

2X• → X2

RX• → R2

X• + R• → RX

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

X2 → 2X•

A

Alkanes

Free Radical Substitution

Initiation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

X• + RH → HX + R•

A

Alkanes

Free Radical Substitution

Propagation

Formation of Alkyl Radicals

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

R• + X2 → RX + X•

A

Alkanes

Free Radical Substitution

Propagation

Reaction of Alkyl Radicals

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

2X• → X2

A

Alkanes

Free Radical Substitution

Termination

Formation of Halogens

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

R• → R2

A

Alkanes

Free Radical Substitution

Termination

Formation of Alkanes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

X• + R• → RX

A

Alkanes

Free Radical Substitution

Termination

Formation of Alkyl Halides

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Reactions

Free Radical Substitution

Bromination

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Reactions

Free Radical Substitution

Chlorination

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Reactions

Combustion

A

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

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Reactions

Combustion

Mechanism

A

Very Complex

Believed to proceed through a radical process

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

C3H8 + 5O2 → 3CO2 + 4H2O + heat

A

Alkanes

Combustion (propane)

17
Q

Reactions

Pyrolysis

A

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

18
Q

Reactions

Pyrolysis

Mechanism

A

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

19
Q

CH3CH2CH3 heat→ CH3• + •CH2CH3

A

Alkanes

Pyrolysis

Formation of smaller-chain alkyl radicals

20
Q

2CH3• → + CH3CH3

2•CH2CH3 → CH3CH2CH2CH3

A

Alkanes

Pyrolysis

Formation of alkanes

21
Q

CH3• + •CH2CH3 → CH4 + CH2=CH2

A

Alkanes

Pyrolysis

Disproportionation

22
Q

Reactions

Substitution

A

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

23
Q

Nucleophiles

Basicity

A

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

24
Q

Nucleophiles

Basicity

Put in order of increasing strength:

ROH, H2O, RCO2- HO- RO-

A

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

25
Q

Nucleophiles

Size and Polarity

A

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

26
Q

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-

A

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

27
Q

Nucleophiles

Leaving Groups

A

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

28
Q

Nucleophiles

Leaving Groups

Put in order of increasing suitability as a leaving group

CL-, Br-, F-, I-

A

I- > Cl- > Br- > F-

Opposite to order of base strength

29
Q

Protic Solvent

A

Capable of hydrogen bonding

30
Q

Reactions

Substitution

SN1

Mechanism

A

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
31
Q

Reactions

Substitution

SN1

Rate

A

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
32
Q

Reactions

Substitution

SN2

Mechanism

A

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
33
Q

Intermediate vs Transition State Definition

A

Intermediate - well-defined species with a finite lifetime

Transition state - theoretical structure used to define a mechanism

34
Q

Reactions

Substitution

SN2

Rate

A

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

35
Q

Reactions

Substitution

SN1

Stereochemistry

A

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

36
Q

Reactions

Substitution

SN2

Stereochemistry

A

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

37
Q

Reactions

Substitution

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

A
  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

DAT (53 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions