above is the experiment please help explain the C NMR, H NMR, And IR analysis. f

Question

above is the experiment
please help explain the C NMR, H NMR, And IR analysis. for dimethylbenylamine and hexane.
The IR for hexane was not produced (so it is not needed to explain but welcomed) but please explain the h nmr and C NMR.

19 EXIERIMENT ALKYLATION OF THE AMINO GROup USING THE ESCHWEILER- CLARK METHYLATION PROCEDURE 0 In this experiment a primary amine will be converted into a terntiary amine using the Eschweiler -Clark methy lation reaction. The procedure presented herein is a modified procedure Irom a report in the Journal of Chemical Education that appeared in 1968. Download this article from htp llpubs acs org (from the USM net- work) and read it prior to coming to lab. Notice that the scale that was originaly utilized is quite large com- pared to the procedure reported herein The amino functional group can serve as a good nucleophile in typical Sy2 type reactions with alkyl halides Unfortunately, the amino group has a habit of over-alkylating ulimately giving quatermary ammonium salts as a significant product. This is of course perlectly fine il the chemist wishes to prepare quaternary ammo- nium salts. However, it is frequently the case that tertiary amines are the desired product and having the reac tion stop at the tertiary amine typically proves quite dilficult under typical Sn2 conditions. Scheme 19,1 illustrates the problem inherent in the alkylation of amines using alkyl halides as allylating agents,

Explanation / Answer

For 1H nmr analysis, first count the different types of protons available in the molecule. In case of dimethylbenzylamine, you have 3 different types of protons on phenyl ring as ortho meta and para. The two methyl groups on N are also in different environment. So finally you have 5 different types of protons and hence 5 nmr signals were expected. The 2 aliphatic methyl protons will show signal at 2.2 and 3.4 ppm. One of the methyl group is oriented in such a way with the phenyl ring so that it is in the vicinity of pi electrons and that is why these protons are showing peak at 3.4 ppm. Aromatic protons are visible near 7.5 ppm. From integration of the peaks, it is obvious that 3 protons corresponds to 1 integration factor. Therefore for 5 aromatic protons, it should be 1.65 but it is 2.05 because of the disturbance from the peak of CDCl3 (contains trace amount of CHCl3) at 7.26 ppm.

For the analysis of 13C nmr, we will count the different types of carbons. Here also, we have 5 different types of carbons. Those 2 aliphatic carbons can be seen at 45 and 65 ppm. And the range for aromatic carbons is 110-140ppm. The aromatic carbon which is next to N atom is a tertiary carbon and hence will be most down fielded (140ppm in the soectra). As you know that amines exert positive mesomeric effects on the ortho and para positions in aromatic rings and this effect up fielded the chemical shift in 13C nmr. Therefore, among all 5 secondary carbons, the two meta carbons will be least up fielded because +Mesomeric effect doesn't affect meta position, and hence will show peaks at 129 and 128 ppm. After that ortho carbons will be most affected and will be most up fielded and para will be slightly less affected. Never get confused from the peak at 78 ppm as they arises from the CDCl3 solvent.

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