[rleung]

Hi,  

My book asks what nucleophile should be used to react CH3CH2CH2CH2Br to form CH3CH2CH2CH2NHCH3, and the answer the solution manual gives is CH3NH2, which makes no sense to me.  Since this is a primary alkyl halide reactant, wouldn't this substitution occur thru SN2, which would mean that there would be no deprotonation of the nucleophile, hence, reaction of CH3CH2CH2CH2Br + CH3NH2 ---> CH3CH2CH2CH2-N(+)H2-CH3???  Therefore, my answer to the original question would be that CH3NH(-) would be the best nucleophile if you want to get CH3CH2CH2CH2NHCH3.  Please tell me I am not crazy  

Ryan

[AFFA]

Those textbooks focus on some ideal situation. If we have to use CH3NH2, its quantity should be in excess relative to the primary halide (or excess bulkyl amines such as DIPEA, DBU as proton scavenger). Also, try to choose some polar aprotic solvent like DMSO, DMF to favor SN2 pathway.

[rleung]

Thanks.  But saying that we do not have the capacity to meet those requirements, is CH3NH(-) just as good a nucleophile?  Thanks.

Ryan

[kkjc2]

The butyl methyl ammonium ion formed from the S_N2 reaction will be readily deprotonated by either the methyl amine, or a solvent molecule, or the bromide ion. The actual experiment will probably be done in a slightly basic condition/with excess amine for this purpose.
CH₃NH^- would be a very strong base and even if you were able to get that in a reagent form (which I don't think is available), it will readily deprotonate almost any protons available before doing any further reactions (eg. S_N2 etc) so at the end, it becomes methyl amine anyway.

[movies]

You can't just use MeNH^- because you need some kind of counterion for the minus charge.  Then you run into the same problem you have above where you have to lose that counterion somehow.  Imagine that counterion is H⁺ and you have the original case.

Nonetheless, as others said, there will always be some base around to deprotonate the S_N2 adduct.  In problems like these, you have to assume that you have all the reagents you need to get to the product.  You'd have a very tough time isolating the direct S_N2 adduct in this case.  That said, if you were to react a tertiary amine (e.g. Et₃N) with something like methyl iodide, you would make a quaternary ammonium salt (Et₃MeN⁺I^-) which could be isolated.  In that case, however, you can't just pull a proton off of the N though.