[sabre2th1]

how would you do this? I have no clue.. Haven't seen anything like this in my book :s

[s.aley]

What paper is this ?

For the first one, an OH- group attacks the carbon attached to the bromine, the bonding pair of electrons between c-br is donated to the br making br- ion, and simultaneously a lone pair of electrons from the OH- is donated to the carbon making a carbocation

This is heterolytic fission



Edit

Search sn1 / sn2 reactions on youtube, look for the khan academy videos

[mican]

(Original post by sabre2th1)


how would you do this? I have no clue.. Haven't seen anything like this in my book :s

i think because the tertiary haloalkane follow Sn1 mechanism and the primary follow Sn2...

[Waseef]

This is the first reactive intermediate:


Explanation:
<answer>To reduce the amount of repulsion, the OH^- ion attacks the carbon atom from the opposite side of the halogen atom and forms the reactive intermediate.</answer>

>As the C-OH bond gets stronger the C-Br bond gets weaker.
>This forms the reactive intermediate for a few moments and then it breaks to form the alcohol and bromide ion.

Since OH^- ion acts as a nucleophile in a substitution reaction with an intermediate formed using 2 reactants it is the SN2 mechanism (S= substitution, N= nuclephilic and 2 = bimolecular)


This is the second reactive intermediate:


Explanation:
<answer> Since attack on the Carbon atom is not possible due to repulsion from the bromine and CH₃ molecules, it forms a carbocation first.
> This carbocation then reacts with the OH^- to form the alcohol and bromide ion

Since OH^- ion acts as a nucleophile in a substitution reaction with an intermediate formed using 1 reactant it is the SN1 mechanism (S= substitution, N= nuclephilic and 1 = monomolecular)

Hope that answers the question
- Waseef

[charco]

(Original post by sabre2th1)


how would you do this? I have no clue.. Haven't seen anything like this in my book :s

Check this youtube video on nucleophilic substitution...

[sabre2th1]

(Original post by Waseef)
This is the first reactive intermediate:


Explanation:
<answer>To reduce the amount of repulsion, the OH^- ion attacks the carbon atom from the opposite side of the halogen atom and forms the reactive intermediate.</answer>

>As the C-OH bond gets stronger the C-Br bond gets weaker.
>This forms the reactive intermediate for a few moments and then it breaks to form the alcohol and bromide ion.

Since OH^- ion acts as a nucleophile in a substitution reaction with an intermediate formed using 2 reactants it is the SN2 mechanism (S= substitution, N= nuclephilic and 2 = bimolecular)


This is the second reactive intermediate:


Explanation:
<answer> Since attack on the Carbon atom is not possible due to repulsion from the bromine and CH₃ molecules, it forms a carbocation first.
> This carbocation then reacts with the OH^- to form the alcohol and bromide ion

Since OH^- ion acts as a nucleophile in a substitution reaction with an intermediate formed using 1 reactant it is the SN1 mechanism (S= substitution, N= nuclephilic and 1 = monomolecular)

Hope that answers the question
- Waseef

For the first reactive intermediate, the markscheme gives the following answer:



I understand what you are trying to say, but I am confused as to how I would draw the intermediate

Regarding the second intermediate, how many methyl groups have to be present in order to prevent an attack on a Carbon atom? There are 3 methyl groups in this example, but what if there were 2? Would they still prevent an attack (and therefore lead to a carbocation forming first?)

[sabre2th1]

(Original post by charco)
Check this youtube video on nucleophilic substitution...

In your video, the transition state for the SN2 reaction is quite simple:



However the transition state for the sn2 reaction in this question is quite complicated (as below):

[sabre2th1]

Thanks all btw

[Waseef]

(Original post by sabre2th1)
For the first reactive intermediate, the markscheme gives the following answer:



I understand what you are trying to say, but I am confused as to how I would draw the intermediate

Regarding the second intermediate, how many methyl groups have to be present in order to prevent an attack on a Carbon atom? There are 3 methyl groups in this example, but what if there were 2? Would they still prevent an attack (and therefore lead to a carbocation forming first?)

First Intermediate:
> Draw the carbon with the bromine atom as the central atom(I'll call it the central atom for ease of understanding, however, strictly speaking it is not the central atom).

> Draw all the other CH_{2 or 3} groups on one side of the central C atom and all the hydrogen connected to the central C atom on the other side of it

> Draw the Br on another side of the Central C atom and the OH group on the opposite side to it.

For example, when 1-bromopropane reacts with OH^- ions the following reactive intermediate forms before propan-1-al is formed:


As for the second question as to how many methyl groups need to be present to prevent attack from nucleophile:

> If it is a tertiary halogenoalkane (3 methyl/ethyl etc groups around the central C atom) then it will always be SN1 mechanism

> If it is a secondary halogenoalkane (2 methyl/ethyl etc groups around the central C atom) then it might form either SN1 or SN2 based on its structure and the conditions such as the solvent used.

-For example if it is a larger molecule it would prefer SN1 over SN2 since there is a greater repulsion from the larger chain on carbon atoms on one side.
-If a polar solvent such as water or ethanol is used SN1 mechanism is favored.

Why is that so?
- The other reason why the formation of carbocation may be favored is because alkyl groups bonded to the central carbon atom tends to stabilize the charge on the carbocation, so the greater number of alkyl groups, the greater the stability of the carbocation, thus the stability is tertiary>secondary>primary
- Polar solvents also help stabilize the charge on the carbocation, thus its formation is favored...

If there is any other confusion then please don't hesitate to ask.
- Waseef

[sabre2th1]

(Original post by Waseef)

If there is any other confusion then please don't hesitate to ask.
- Waseef

Thank you very much! I understand now!



The Edexcel markscheme uses wedges etc, it will be still okay if I don't use it right (so if I do it as above)?

[Waseef]

Yep... It will still work... but you could still use the wedges just to be safe