Halogenoalkanes Flashcards
(18 cards)
Free Radical Substitution
R/C: X2 in UV light
X2 + R-H –> R-X + HX
Topic: Alkanes
Electrophilic Addition (alkene –> halogenoalkane)
R/C: Dry HX at room temp
or
R/C: Halogen (X2) in organic solvent (Eg Br2 in CCl4)
Alkene –> 2 diff halogenoalkane products thru ES
WHY?
Unsymmetrical alkene
HOW is it determined?
Nucleophilic substitution (from alcohol to halogenoalkane)
- HX (g), heat
(Anhydrous ZnCl2 catalyst is used with HCl) - PX3, heat (X = Cl, Br, I)
PCl3, PBr3 (l) , PI3 (s) - PCl5 (s) - Also test for alcohol (white fumes)
- SOCl2 in pyridine, heat
thionyl chloride
- R–OH + HX –> R–X + H2O
- 3R–OH + PX3 –> 3R–X + H3PO3
- R–OH + PCl5 –> R–Cl + POCl3 + HCl (g)
- R–OH + SOCl2 –> R–Cl + SO2 (g) + HCl (g)
Miscellaneous: How is PCl3/PCl5 formed?
Pass dry chlorine gas into an excess of heated dry red phosphorous. Distill and collect the liquid PCl3 formed.
Excess dry chlorine gas is passed into liquid PCl3 and heated. –> solid PCl5 formed
PX3 can also be generated in-situ
eg reacting red phosphorous with iodine in the presence of an alcohol
Nucleophiles
Definition
Examples
Electron pair donors that are attracted to an electron deficient atom or region of low electron density
- Anions
- Neutral molecules with at least one lone pair of e-
Types of reactions halogenoalkanes can undergo
- Nucleophilic substitution
- Elimination
Why is C atom in C-X bond susceptible to attack by nucleophiles?
C-X bond in a halogenoalkane molecule is polarised by the EN halogen atom which draws the shared pair of electrons away from C.
C atom in C-X bond is associated with a partial positive charge, thus it is susceptible.
Halogen atom is displaced as a halide ion, known as a good leaving group.
Why is Sn1 mechanism favoured?
(unimolecular nucleophilis substitution)
Due to electronic considerations:
A tertiary halogenoalkane gives a stable tertiary carbocation intermediate as e- donating alkyl groups disperse the positive charge on the carbocation intermediate. Methyl carbocation and primary carbocation are less stabilised.
Sn1 mechanism, via a carbocation intermediate, is favoured for tertiary halogenoalkanes.
Why is Sn2 mechanism favoured? (bimolecular nucleophilic substitution)
Due to steric considerations:
A methyl or primary halogenoalkane has no or only one alkyl group, bonded to the delta+ carbon atom, which allows for easy approach of the nucleophile.
A tertiary halogenoalkane has three alkyl groups which can sterically hinder the approach of the nucleophile to the delta+ carbon atom.
Hence Sn2 mechanism, via a pentavalent transition state, is favoured for methyl and primary halogenoalkanes.
Traits of Sn2 mechanism
- No. of reactants
- How many steps
- Where nucleophile attacks from
- What is formed in the process
- What happens to nucleophile and X
- When does inversion of stereochemical configuration occur?
- 2 reactants
- One step
- attack from backside: nucleophile will attack from left (umbrella facing left) - tetrahedral in shape
- Pentavalent transition state (both nucleophile and X are bonded partially to the same C atom)
- forms trigonal bipyramidal shape
- nucleophile at left, X at the right
- have ki katakana sign at the side to represent transition state - nucleophile replaces X atom, umbrella will face right, forms X- ion
- number 3&5 –> inversion of stereochemical configuration as the nucleophile attacks the delta+ carbon atom from the opposite side of the halogen atom (only if chiral carbon)
In Sn2, why must the attack from the nucleophile be from the opposite side of the halogen atom?
Large bulky halogen atom sterically hinders the nucleophile from attacking the delta+ carbon from the same side of the halogen atom. There will be a great repulsion between the electron cloud of halogen atom and the nucleophile.
Traits of Sn1
1. No. of reacting species
2. No. of steps
3. What does the first step involve?
4. What hybridisation does the carbocation intermediate have?
5. What happens in step 2
6. Why is second step fast?
- Only 1 reaction species
- Involves two steps
- Slow step and first step (rate determing step) involves formation of carbocation - C-X bond breaks heterolytically (X takes both e- to become X-)
- Carbocation has sp2 hybridisation (trigonal planar with +ve charge)
- Nucleophile attacks carbocation (e- arrowed to C+) –> results in tetrahedral shape with nucleophile replace X)
- Second step is fast because it involves attraction between opposite charges of the cation and nucleophile (less energy is used)
Relationship between stability of carbocation and alkyl group
The electron donating alkyl groups help to stabilise the carbocation intermediate by dispersing the positive charge on the carbon
Sn1 reaction using a chiral reactant
- Why equal probability of nucleophile attack from top and bottom of the plane?
- What happens if carbocation becomes chiral carbon + what is the proportion of products formed?
- Since the carbocation is trigonal planar with respect to the electron deficient carbon, the nucleophile attacks from the top and bottom of the plane with equal probability
- If the carbon with the positive charge becomes a chiral carbon after the reaction, a racemic mixture (which is optically inactive), will be formed as both mirror image enantiomers are formed in equal quantities.
- Basically if attack from top
R1 OH (Nu:)
C R3 + OH : –> C
R2 R1 R2 R3
from trigonal planar to tetrahedral with OH on top. (umbrella shape facing down formed with the other 3 alkyl groups)
- use a mirror plane directly below to show attack from bottom
Differences between Sn1 and Sn2 mechanism
- Type of reaction / no. of steps
- Rate equation
- Relative rates
- Which R-X is more favoured
- Energy profile diagram
- Stereochemistry of products (chiral reactants)
- Non-elementary (2 step) vs elementary (1 step)
- rate = k [halogenoalkane given]^1
vs rate = k [nucleophile]^1 [halogenoalkane given] ^1 - tertiary > secondary > primary > methyl halides
vs
methyl halides > primary > secondary > tertiary - favoured by tertiary R-X vs favoured by primary RX/methyl halide
- two peaks for Ea (due to two steps) vs only one highest peak for Ea (both exothermic)
- racemic mixture obtained if chiral reactants are used vs inversion of stereochemical configuration
Exceptions to preference in mechanism (Sn1 instead of Sn2) Why?
- (CH3)3CCH2Cl
- C6H5CH2Cl (chloromethylbenzene)
- Steric hindrance due to bulky (CH3)3C- group makes it difficult for nucleophile to attack from opposite side of leaving group (halogen atom)
why exception: primary alkyl halide -> Sn2
- usually nucleophile would be sterically hindered by halogen atom (Cl2) due to large e- cloud therefore attacks opposite side
- Resonance stabilisation of benzyl carbocation by delocalisation of electrons from benzene ring
why exception: primary alkyl halide -> Sn2
- carbocation is stabilised (usually due to presence of 3 alkyl grps - tertiary reactant)
Practice qns - Context:
- Does not rotate plane polarised light (eg (+)-CH3CH(Cl)CH2CH3) + NaOH)
secondary alkyl halid + Sn1 (does not rotate –> racemic mixture)
Steps:
1. Formation of carbocation
2. Formation of two different enantiomers inverted across mirror plane (include 50% probability on top of arrow sign)
Explanation: Two enantiomers are formed in equal proportion and this gives a racemic mixture that is optically inactive
