[Dolphinsiu]

Why the following reactions do not take place?

(a) CH3CH2CH3 + OH- --> CH3CH2CH2OH + H-
(b) CH3CH2C(CH3)2Br + CN- --> CH3CH2C(CH3)2CN + Br-
(c) CH3CHCH3OH + Cl- -->  CH3CHCH3Cl + OH-

[My attempt answers

(a) In this reaction, SN2 is favoured as primary substrate is used. Less steric hindrance favours concerted reaction in transition state. For SN2, polarity of nucleophiles is an important factor.
As nucleophile (OH-) is negative while the leaving group (H-) is neutral. The charge movement in transition state is dissipation. Hence the rate for more polar solvent is slight retardation. The above reaction, in turn, fails to take place.

(b) In this reaction, SN1 is favoured as tertiary substrate is used. There are three electron-donating groups to stabilize the carbocation formed by cleavage of leaving group. For SN1, CN- is a better leaving group than Br-. Therefore CN- as leaving group will heighten transition state energy and decrease the rate of reaction and this reaction is less likely to take place.

(c) In this reaction, Both SN1 and SN2 will undergo. For SN1, Cl- is a better leaving group than OH- so C – Cl is less energetically stable than C – OH Therefore OH- as leaving group will decrease the rate of reaction. For SN2, polarity of nucleophiles is a crucial factor.
As nucleophile (Cl-) is neutral while the leaving group (OH-) is negative. The charge movement in transition state is dissipation. Hence the rate for more polar solvent is slight retardation. In addition, more negative nucleophiles are more effective. OH- carries higher negative charge than Cl. Hence OH- is a strong nucleophile than Cl-, that is, Cl- is not strong enough to underogo nucleophilic attack on electrophilic substrate to replace OH-. As such, both SN1 and SN2 will fail to occur.]

Do my attempt answers in each part have any mistakes? If yes, please help me to correct it. Thank you!

[Dan]

You haven't been given a solvent for any of these reactions, so you can't argue on the basis of the nature of the solvent - you don't know what it is.

H^- and Cl^- are not neutral species! They are negative ions, they are written with a negative charge!

For Sn2 you must consider the nucleophile AND the leaving group. A poor leaving group (oe nucleophile) will result in poor Sn2.

[Dolphinsiu]

Actually I don't know what is the better way to explain the nucleophilicity of nucleophile and strength of leaving group. Would you take an example for better explaination?

[Dan]

Firstly, have you been given solvents for these reactions?

OK, for Sn2 to occur, you need the combination of a good nucleophile (solvent may come into play here - aprotic solvents do not solvate anions well, making them more nucleophilic), and a good leaving group (as well as an accessable C, ie. not a tertiary centre).
What makes a good leaving group? If the leaving group is stabilised is some way, then it is a good leaving group. eg the tosylate anion is stabilised by resonance, and is an excellent leaving group.
How can you tell if a leaving group is a good one? Consider the stability of the leaving group. A good way to think about it is to consider strength of the conjugate acid of the leaving group. I'll continue with the tosylate eg.

TsOH <----> TsO^- + H⁺       [Ts = F₃CC₆H₄SO₂]

TsOH is a very strong acid, which indicates that it's conjugate base (TsO^-) is relatively stable, and would make a good leaving group.
So, we would expect that

EtOTs + CN^- ----> EtCN + TsO^- via an Sn2 pathway.

If the leaving group is a strong base, this indicates that it is a poor leaving group.

eg NH₃ <----> NH₂^- + H⁺

Ammonia is a very very poor acid, ie. NH₂^- is a very strong base, and relatively unstable, so we would not expect Sn2 to occur on a primary amine:

EtNH₂ + OH^- ---------> no Sn2

For questions 1 and 3, extend this logic to the H^- and OH^- leaving groups.

Question 2 is different.

If substitution were to occur, it would be via Sn1, as you correctly stated. However, substitution may not be favoured at all, we could be looking at elimination. Solvent is important here. If the solvent is not polar enough to stabilise the carbocation intermediate in the Sn1 pathway, then we could see E2 instead. Cyanide can act as a base under the right conditions. Note that if Sn1 can't occur due to a lower polarity solvent, then nor can E1, as both reactions proceed via the same intermediate.

[Dolphinsiu]

Yes. The question has not given any solvent in the above three questions.

I only don't know why CN- can undergo SN1 although carbocation is stabilized by 3 methyl group.

Also, how can cyanide act as a base under the right conditions? Or in polar solvent?

[Dan]

I only don't know why CN- can undergo SN1 although carbocation is stabilized by 3 methyl group.

Also, how can cyanide act as a base under the right conditions? Or in polar solvent?

Cyanide is a good nuleophile, and also pretty basic. Sn2 at a tertiary centre can't happen. If the solvent is not polar enough to stabilise the carbocation, then Sn1 can't happen either (or E1). So the only thing that can happen is E2 or nothing. The carbocation would be stabilised by three alkyl groups, but it still needs a suitably polar solvent in order to form.
If the carbocation can form, I would expect Sn1 and E1 to occur, but the question states that a substitution product is not observed, so this leads me to thinking that it must be E2. Without info on the solvent it's not easy to say. 

[Dolphinsiu]

Thank you. In conclusion,

Solvent can affect nucleophilicity(SN2) and carbocation formation (SN1)!

but how can I know the solvent is helping SN1 or SN2 if polar solvent(e.g. water) is used?

[Dan]

but how can I know the solvent is helping SN1 or SN2 if polar solvent(e.g. water) is used?

Polar solvents help both, but for an anionic nucleophile, Sn2 is faster with polar aprotic solvents becuase the anion is not well solvated and is therefore more nucleophilic.
Sn1 is helped by polar protic solvents as the formaition of a carbocation and an anion in the rate determining step needs good solvation of both ions in order to happen.

However, when trying to decide if a reaction is Sn1/Sn2, first look at the C. Is it primary, sec, tert, neopentyl, allylic? Then think about the nucleophile (is it basic as well as nucleophilic?) and leaving group. If you still haven't made your mind up, look at the solvent (if it's given).

A very good, easy-to-read book for this is Mechanism in Organic Chemistry - Peter Sykes.

[Dolphinsiu]

Thank you very much, Dan!