How to be able to tell if a molecule will react via SN1 or SN2?

Question

I've been having a lot of issues with a question.

2 compounds, (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene react with EtOH in water.

It asks to classify the reaction mechanism for each and to then go on and explain why (1-chloro-2-methylpropyl)benzene has a much faster relative rate.

I am having major issues with the question, because I just cannot tell where there are any major differences between molecules which could make a difference. Both of them have a benzene ring, but this stabilises both the transition state and the carbocation intermediate so will favour both SN1 and SN2. Both reactions are done in the exact same solvent with the same nucleophile.

The only differences I can spot is that in (1-chloro-2,2-dimethylpropyl)benzene, there will be more stabilisation due to the inductive effect and hence it will favour SN1. But this does not help with determining the mechanism for (1-chloro-2-methylpropyl)benzene.

Any help will be much appreciated. Many thanks :)

Answer 1

Factors affecting S_N1 mechanism:

1. Weak nucleophile / base and Good leaving group.
2. More steric hindrance.
3. Polar protic solvent should be used.

Factors affecting S_N2 mechanism:

1. Strong nucleophile / base and Good leaving group.
2. Less steric hindrance.
3. Polar aprotic solvent should be used.

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Coming back to your problem. Actually, both of the compounds can go through S_N1 or S_N2 mechanism as the carbon linked with $\ce{Cl}$ is a secondary carbon. But, as you mentioned that you're using $\ce{EtOH}$, and you should know that $\ce{EtOH}$ is little basic, almost neutral, therefore, we can say that now, both of the compounds will go through S_N1 mechanism (Remember, the first point of S_N1 mechanism, i.e., weak base).

Now, the question is if both the compounds will follow S_N1 mechanism, then which will react faster. For this, we will have a look at the S_N1 mechanism:

Step 1: Leaving of the (-X) group

Step 2: Weak nucleophile attacks from either side

Step 3: Substitution product formed (Inversion or retention, based on the step 2)

So, our concern is only the step 2.

Note that, rate of reaction in step 2 depends on steric hindrance in your case. So, let's have a look at the structure of (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene:

The Carbon with red dot is the point where $\ce{EtOH}$ will attack after the (-Cl) will leave.

Now, on comparing the both the compounds, you will notice that in the first compounds there is greater steric hindrance as compared to the second compound. And, that's the reason why second compound will react faster than the first one.

Conclusion:

(1-chloro-2-methylpropyl)benzene will react faster than (1-chloro-2,2-dimethylpropyl)benzene due to less steric hindrance around the carbon that is directly attached to aromatic ring (i.e., the carbon with red dot).

And, both of the compounds will go through S_N1 mechanism due to presence of $\ce{EtOH}$.