OZ8 - Nucleophilic Substitution (What is the State of the Ozone Layer Now?)

Question 1

NOTE:

Reactions of haloalkanes involve breaking the carbon-hydrogen bond.

The bond can break homolytically or heterolytically (Topic OZ3).

Answer 1

NOTE:

Reactions of haloalkanes involve breaking the carbon-hydrogen bond.

The bond can break homolytically or heterolytically (Topic OZ3).

Question 2

What is homolytic fission?

Answer 2

Homolytic fission of haloalkanes occurs when haloalkanes reach the stratosphere, where they are exposed to intense UV radiation.

This is how ozone-depleting chlorine radicals are formed.

CH3–Cl + hv –> CH3* + Cl*

haloalkanes can be formed by a radical halogenation mechanism (Topic OZ3)

Question 3

What is heterolytic fission?

Answer 3

Heterolytic fission is more common under normal laboratory conditions.

The carbon-halogen bond is polar, and can break forming
a NEGATIVE HALIDE ION
and a POSITIVE CARBOCATION.

e.g.

• 2-chloro-2-methylpropane
  (C with 3 CH3 branches and one Cl)
  (C5H9Cl)

(red curly arrow on bond between central C and Cl, pointing to Cl)

–> breaks down to

• a carbocation
  (central C+ ion with 3 CH3 branches)
  (C+C4H9)
• and a chloride ion
  (Cl-)

____________

Sometimes, heterolytic fission is caused by a negatively charged substance reacting with the positively polarised carbon atom, causing a SUBSTITUTION reaction.

Question 4

What is SUBSTITUTION?

Answer 4

SUBSTITUTION:

A reaction in which one atom or group in a molecule is replaced by another atom or group

Question 5

What is a NUCLEOPHILE?

Answer 5

NUCLEOPHILE:

A nucleophile is a molecule or negatively charged ion with a lone pair of electrons that it can donate to a positively charged atom to form a covalent bond.

Question 6

Explain what happens in the nucleophilic substitution reaction of a haloalkane.

Use the reaction between hydroxide (OH-) ions and 1-bromobutane. The haloalkane is heated under reflux with ethanolic sodium hydroxide (this group is called the nucleophile)

Answer 6

1-bromobutane and ethanolic sodium hydroxide:

• the nucleophile (OH-) attacks the electron deficient carbon atom in the C-Br bond.
• the OH- donated two electrons to form a new dative covalent bond.
• the C-Br bond breaks heterolytically and the bromine atom receives two electrons, producing a bromide ion.
  In this case the bromide ion is called the leaving group.

Question 7

NOTE:

Some common nucleophiles…

Answer 7

COMMON NUCLEOPHILES
(in the structure (not formula), - is a bond, = is a double and =- is a triple)
(** is a lone pair of electrons, that are used in dative covalent bonds)
(^- is a negative charge symbol)

hydroxide ion
OH-
H - O (3 x **)^-

cyanide ion
CN-
N =- C^-

ethanoate ion
CH3COO-
CH3 - C (=O) - O (3 x **)^-

ethoxide ion
C2H5O-
CH3CH2 - O (3 x **) ^-

water
H2O
H - O (2 x **) - H

ammonia
NH3
H - N (2 x **) (-H) - H

Question 8

Using Nu- as a general symbol for any nucleophile and X as a symbol for a halogen atom, the nucleophilic substitution process can be expressed by…?

Answer 8

R-X + Nu- –> R-Nu + X-

The curly arrow moves from the lone pair of electrons on the Nu- to the electron deficient carbon in the haloalkane.

Another curly arrow moves from the bond between the 𝛿+ carbon and the 𝛿- halogen to the 𝛿- halogen.

Question 9

NOTE:
WATER AS A NUCLEOPHILE

A water molecule has a lone pair on the oxygen atom, so water can act as a nucleophile

– though this reaction is slower than the reaction with OH- ions.

When the two are heated together under reflux, water attacks the haloalkane.

e.g.:

H4H9Br + H2O –> C4H10OH + Br-

Initially, the C is 𝛿+ and the Br is 𝛿-.

There is a curly arrow from the C-Br bond to the Br atom in the mechanism, and a curly arrow from one of the lone e- pairs in the H2O nucleophile to the C 𝛿+ atom.

The Br- ion product is the leaving group, which has gained the lone e- pair from the nucleophile.

There is a positive charge on the oxygen (with a lone pair of e-) on the alcohol product.

Then the resulting ion (chain) loses H+ (that came from the H2O nucleophile) to form an alcohol (there is a curly arrow from the OH bond to the oxygen.

This results in
C4H9OH + H+

The general equation is known as a hydrolysis reaction:

R-X + H2O –> R-OH + H+ + X-

Answer 9

NOTE:
WATER AS A NUCLEOPHILE

A water molecule has a lone pair on the oxygen atom, so water can act as a nucleophile

– though this reaction is slower than the reaction with OH- ions.

When the two are heated together under reflux, water attacks the haloalkane.

e.g.:

H4H9Br + H2O –> C4H10OH + Br-

Initially, the C is 𝛿+ and the Br is 𝛿-.

There is a curly arrow from the C-Br bond to the Br atom in the mechanism, and a curly arrow from one of the lone e- pairs in the H2O nucleophile to the C 𝛿+ atom.

The Br- ion product is the leaving group, which has gained the lone e- pair from the nucleophile.

There is a positive charge on the oxygen (with a lone pair of e-) on the alcohol product.

Then the resulting ion (chain) loses H+ (that came from the H2O nucleophile) to form an alcohol (there is a curly arrow from the OH bond to the oxygen.

This results in
C4H9OH + H+

The general equation is known as a hydrolysis reaction:

R-X + H2O –> R-OH + H+ + X-

Question 10

NOTE:
AMMONIA AS A NUCLEOPHILE:

Ammonia (NH3) can act in a similar way to water with the lone pair of electrons on the nitrogen atom attacking the haloalkane.

The haloalkane is heated in a sealed tube with concentrated ammonia solution. The product is an amine with an NH2 group.

The overall equation is:

R-X + NH3 –> R-NH3 + X-
⇌
R-NH2 + H+ = X-

Amines have the general formula R-NH2.
They are nitrogen analogues of alcohols (R-OH).

Examples of amines:

methylamine
CH3NH2

ethylamine
C2H5NH2

propylamine
C3H7NH2

Answer 10

NOTE:
AMMONIA AS A NUCLEOPHILE:

Ammonia (NH3) can act in a similar way to water with the lone pair of electrons on the nitrogen atom attacking the haloalkane.

The haloalkane is heated in a sealed tube with concentrated ammonia solution. The product is an amine with an NH2 group.

The overall equation is:

R-X + NH3 –> R-NH3 + X-
⇌
R-NH2 + H+ = X-

Amines have the general formula R-NH2.
They are nitrogen analogues of alcohols (R-OH).

Examples of amines:

methylamine
CH3NH2

ethylamine
C2H5NH2

propylamine
C3H7NH2

Question 11

NOTE:
USING NUCLEOPHILIC SUBSTITUTION TO MAKE HALOALKANES

You can use the REVERSE of a hydrolysis reaction to produce a haloalkane from an alcohol. This time the nucleophile is a halide ion (X-).

The reaction is done in the presence of a strong acid, and the first step involves bonding between H+ ions and the oxygen atom on the alcohol.

e.g. C4H9OH + H+ –> C4H9OH2

With a curly arrow from one of the lone e- pairs on the O (nu) to the H+ ion.

This gives the carbon atom to which the oxygen (which has an e- pair but is -ve charged) is attached to a higher partial positive charge.
It is now more readily attacked by halide ions:

C4H9OH2 + Br- –> C4H9Br + H2O

With a curly arrow from one of the Br- lone e- pairs to the end carbon atom, and another curly arrow from the C-O bond to the e- lone pair on the O.

The Br in the product has 3 remaining e- lone pairs, and started with 4 as an ion.

The overall equation for the reaction is:

CH3CH2CH2CH2OH + H+ + Br-
–>
CH3CH2CH2CH2Br + H2O

Answer 11

NOTE:
USING NUCLEOPHILIC SUBSTITUTION TO MAKE HALOALKANES

You can use the REVERSE of a hydrolysis reaction to produce a haloalkane from an alcohol. This time the nucleophile is a halide ion (X-).

The reaction is done in the presence of a strong acid, and the first step involves bonding between H+ ions and the oxygen atom on the alcohol.

e.g. C4H9OH + H+ –> C4H9OH2

With a curly arrow from one of the lone e- pairs on the O (nu) to the H+ ion.

This gives the carbon atom to which the oxygen (which has an e- pair but is -ve charged) is attached to a higher partial positive charge.
It is now more readily attacked by halide ions:

C4H9OH2 + Br- –> C4H9Br + H2O

With a curly arrow from one of the Br- lone e- pairs to the end carbon atom, and another curly arrow from the C-O bond to the e- lone pair on the O.

The Br in the product has 3 remaining e- lone pairs, and started with 4 as an ion.

The overall equation for the reaction is:

CH3CH2CH2CH2OH + H+ + Br-
–>
CH3CH2CH2CH2Br + H2O

Question 12

IMPORTANT NOTE:

THE BOND ENTHALPY / STRENGTH IS THE OVERRIDING FACTOR IN DETERMINING REACTIVITY.

NOT POLARITY.

Bond strength decreases down the group of halogens.
On this basis, you would expect the bond C-I to be hydrolysed most easily because it is the weakest.

It is the weakest because iodine is bigger, and so the radius is bigger so the electrons are further away from the nucleus (lower electrostatic attraction between nucleus and electron makes it easier to be ‘given up’)

THIS IS AN EXPERIMENTAL FINDING.

(more info on back of this card)

Answer 12

The strength of the C-F bond makes it difficult to break, so fluoro compounds are very unreactive.

As you go down the halogen group, the carbon-halogen bond gets weaker, so the compounds get more reactive.

Bromo- and iodo-compounds are fairly reactive, so they are useful intermediates in organic synthesis.