i can apply a knowledge of the rate equations for the hydrolysis of halogenoalkanes to deduce the...

I can apply a knowledge of the rate equations for the hydrolysis of halogenoalkanes to deduce the mechanisms for primary and tertiary halogenoalkane hydrolysis and to deduce the mechanism for the reaction between propanone and iodine (4.3h)

primary and tertiary halogenoalkane hydrolysis =

nucleophilic substitution reactions

Nucleophilic substitution mechanisms of halogenoalkanes

Two substitution mechanisms are possible when a bromoalkane reacts with aqueous

alkali:

R- Br + OH- → R-OH + BrOnly the experimental rate data can show which

mechanism actually takes place.

Nucleophiles are species that have a lone pair of electrons available to form a covalent bond with electrophiles. They will be found with either a full or a partial negative charge on one atom.

Nucleophiles

Lone pair forming a bond

Ammonia as a nucleophile reacting with chloromethane

d+d-

d+d-

Both molecules are polar because of the high electronegativity of N and Cl compared to H and C respectively.

This bond will break

This type of reaction is known as nucleophilic substitution. A nucleophile attacks an electrophile forming a bond. The carbon atom can only support 4 bonds so another is broken and a leaving group (here Cl) is substituted out.

:

© The Royal Society of Chemistry 2010-11

There are two major pathways a nucleophilic substitution can take…

Nucleophilic Substitution: SN2

HO-:

:

H3C Br

:

+ -HO CH3 Br-

Rate = k[H3CBr][OH-]Rate determining step contains both molecules*

NucleophileLone pair(s)Negative charge

ElectrophilePartial positive chargeBr leaving group

H3C BrHO-:

::

HO CH3 Br-

HO C

HH

H

Br

-

E

Transition State

SN2Bimolecular Nucleophilic Substitution

Animation is © 2001 by Daniel J. Berger

*only one step (correctly) assuming there are no faster steps after this one!

There are two major pathways a nucleophilic substitution can take…

Nucleophilic Substitution: SN1

Rate = k[(H3C)3CBr]Slow first step involving only (H3C)3CBr followed by fast addition of nucleophile

CarbocationPositively charged carbon

E

SN1Unimolecular Nucleophilic Substitution

HO-:

:

C Br

:

+ -

H3C

H3CH3C C+

CH3

CH3H3C

Br-HO-:::

C OH

H3C

H3CH3C

Br-

+

BrHO-:

::

HO-:

::

OH

Br-

Br-

Vacant p-orbital

Compare this animation to the previous one. Which type of mechanism will produce a pure product and which could produce a mixture of products?

Two steps, the first has a larger Ea so happens much more slowly

slow fast

Animation is © 2001 by Daniel J. Berger

SN1 – These reactions involve only one of the reactants in the rate determining step(SUBSTITUTION NUCLEOPHILIC FIRST ORDER)This applies to tertiary haloalkanes

SN2 – These reactions involve two reactants in the rate determining step(SUBSTITUTION NUCLEOPHILIC SECOND ORDER)

This applies to primary haloalkanes

As we move from a primary to a tertiary carbon two things happen...

H

HH

H

HH

H

HH

H

HH

H

HH

H

HH

Nucleophilic Substitution: Why does SN1 happen?

The carbon atom that is to be attacked becomes more sterically hindered by the surrounding atoms.

The carbocation becomes more stable due to inductive effects and hyperconjugation

Electrons in adjoining s-bonds to hydrogens that are b to the carbocation can stabilise it by forming an extended molecular orbital. This is called hyperconjugation

C+ C

H

HH C C

H+

H

H

R

R

R

R

Although not usually referred to as a polar bond, there is a difference in electronegativity between C + H which results in charge being pushed towards the vacant p-orbital to stabilise it. This is an inductive effect

© The Royal Society of Chemistry 2010-11

How attractive will the carbon be to the nucleophile? Bond Polarity – increases up the group

What factors affect the speed of nucleophilic substitution?

HO-

:

:

H3C Br

:

+ -HO CH3 Br-

How easy will it be to break the carbon-halogen bond? Bond Strength – increases up the group.

A simple experiment can be performed to determine which of these factors is the most important using silver nitrate solution which will form silver halide precipitates in the presence of the halide ions released in the reaction.

A precipitate will be formed from the reaction of CH3I before it will from CH3Cl indicating that bond strength is more important to the rate of the reaction.

AgCl precipitation

i.e. how effective is our leaving group?

Image by: Lupo

Image by: Dr. T

I can use kinetic data as evidence for SN1 or SN2 mechanisms in the nucleophilic substitution reactions of halogenoalkanes (4.3j)

How would they expect you to do this?

I can demonstrate that a proposed mechanism for the reaction between propanone and iodine is consistent with the data from the experiment in 4.3e (4.3i)

Look back at the results from the experiment for the propanone and iodine reaction that we carried out.

rate = k[CH3COCH3 ][H+](zero order with respect to iodine)

What mechanism would fit this reaction?

The mechanism must be consistent with the evidence:• if the reaction is second order overall it must involve two different species

The rate determining (slow) step must contain both chemical species. These are CH3COCH3(aq) and H+(aq). The mechanism will contain a step with these two species forming a new species. E.g.CH3COCH3 + H+ CH3COH+CH3

Proposed mechanism for the reaction between iodine and propanone

Any sensible species can be chosen as possible products, only further experiment can show if they really are present in the reaction

From a rate equation you can decide on one step in the mechanism. This will be the slow, rate determining step and must contain all species in the rate equation.