Chem10004 Question

[MelonBar]

Was just revising some earlier stuff and came across a question asking for the mechanism of the reaction between
2-chloro-2-methylpentane and NaOH.

This is question 1.5 from the tutorial workbook. anyways i checked the answers and apparently all that happens is the OH group from the NaOH substitutes the chloro - but I thought, because the NaOH is a strong base we should get an alkene?

coming from chem dud so would appreciate an easy to understand response

thanks so much

[nerdgasm]

Hi MelonBar!

I'm a 2nd year chemistry student at Monash, and I studied mechanisms and organic reactions last semester (though my chemistry skills aren't all that flash), but it seems to me this question is a little unclear.

It looks like we're trying to decide between a substitution and elimination reaction here. As you've correctly pointed out, NaOH (or more specifically OH-) is a strong base, and so this would tend to favour elimination reactions over substitution reactions, due to OH- being good at taking protons from the neighbouring carbon to the leaving group.

However, I suppose it's worth considering that in addition to being a strong base, OH- is also a good nucleophile (it has a negative charge and O is quite electronegative), and it is also a relatively small molecule. These properties make it good for attacking a carbocation, in an SN1 reaction (I'm assuming you've touched on Sn1, Sn2, E1 and E2 reactions).

Another thing to consider is that Cl- is a good leaving group (a good way to judge this is to look at the strength of the conjugate acid, in this case HCl), and would leave to form a tertiary carbocation. This suggests to me that it wouldn't take much at all for the Cl to leave and form a carbocation.

So assuming we DO get Cl- leaving, we have a stable carbocation. I'm guessing what happens is that the OH- will prefer to attack the carbocation, rather than the neighbouring hydrogen (due to being a good, small nucleophile), thus leading to a substitution reaction of the form Sn1.

Personally, I don't think there's enough in this question to definitively conclude whether there's substitution or elimination occurring. It genuinely looks like more information on the reaction (solvents, for example) is required. I've just tried to outline some arguments on why substitution (more specifically Sn1) is possible, but I'll be more than happy to be proven wrong!

So, on that note, I'll leave it there, and hope some of the pro chem people on this forum can help out!

[LeviLamp]

Just going to type my thoughts out and hopefully arrive at an answer :v

For a tertiary alkyl halide, E2 predominates when a strong base is present (yes), and SN1 and especially E1 will occur more happily and in competition under neutral or acidic conditions. However, (and this is kind of mean of them) SN1 reactions like to take place in hydroxylic solvents such as ethanol and water, which happen to be what sodium hydroxide is usually dissolved in. I think this is because the electron-rich OH groups on the solvent molecules can orient themselves around the carbocation and help to stabilise the positive charge and therefore lower the energy of the carbocation intermediate? Elimination is favoured when the temperature is high, big bulky reactants are used, the basicity is high and the nucleophile is bad. I guess since the base here is a strong nucleophile and the carbocation is surrounded by electron-rich stabilising groups, and since we're assuming a fairly low temperature (like room temperature??) it's not unreasonable to assume that the SN1 reaction will predominate despite the high basicity of OH-.
(Side note; Wikipedia says substitution reactions predominate by default, so if you're ever completely unsure just go for the SN1/SN2 pathway, it's worked for me so far)

http://en.wikipedia.org/wiki/Elimination_reaction <- read the little bit at the bottom, it isn't in our course but it really really helps

(Also, amalgam, does nucleophile size impact upon SN1 reaction preference? I thought since carbocations were planar the nucleophile could come from either side unhindered, which is why SN1 is favoured over SN2 for tertiary halides?)

[lzxnl]

Was just revising some earlier stuff and came across a question asking for the mechanism of the reaction between
2-chloro-2-methylpentane and NaOH.

This is question 1.5 from the tutorial workbook. anyways i checked the answers and apparently all that happens is the OH group from the NaOH substitutes the chloro - but I thought, because the NaOH is a strong base we should get an alkene?

coming from chem dud so would appreciate an easy to understand response

thanks so much

OH- is small. It is a small negative charge that has a choice of species to attack: a positive carbon, or a hydrogen which won't have as large a charge. An E1 reaction would occur if we had a large base like LDA (lithium diisopropylamide) that wouldn't physically fit; it can only then abstract a hydrogen.

LeviLamp, to my understanding, that is the case. If the nucleophile is large, the SN2 reaction is sterically hindered. Alternatively, if there are many bulky substituents, there won't be the space for the SN2 reaction. It's not as big of a problem with the SN1 reaction for the reason you've given.

A reason why the SN1 reaction would be preferred over the SN2 reaction here is because the first slow step, the step where the chlorine drops off, the carbocation intermediate is stabilised by the alkyl groups; carbon 2 has three other carbons bonded to it.

And yeah, read the bottom of http://en.wikipedia.org/wiki/SN1_reaction . Water itself has a relatively high dielectric constant (due to its high polarity), but a sodium hydroxide solution would be even more polar, so it would be great at stabilising the intermediate.

[nerdgasm]

Just as an addendum, I'd like to say that I agree with LeviLamp and nliu1995 on their points here. I agree steric factors are less important in Sn1 compared to Sn2  (due to the different mechanism of reaction), but I still believe that if we had a really bulky basic nucleophile, there would still be a bit of hindrance, even for the Sn1 reaction, simply because it's still a tiny bit harder to access the carbon atom than the hydrogen atoms.

Regardless, I think all the points mentioned above make a very good case for why Sn1 can work for this reaction.