Question

A compound with molecular formula C6H15N exhibits a singlet at d 0.9 (1H), a triplet at d 1.10 (3H), a singlet at d1.15 (9H), and a quartet at d 2.6 (2H) in its 1HNMR spectrum. Its IR spectrum shows one medium absorption band near 3400 cm-1. What is the structure of this compound?

Answer 1

Answer:

N-ethyl-2-methylpropan-2-amine

Explanation:

 

In this case, we have to start with the IR info. The signal on 3400 cm^-1 indicates the presence of a hydrogen bonded to the heteroatom. In this case, we have nitrogen in the formula, so we will have the amine group.

On the other hand, we have to analyze the NMR info:

a)  We have 2 singlets => This indicates the presence of 2 different hydrogens without neighbors.

b) We have a triplet => This indicates the presence of CH3 bonded to a CH2.

c) We have a quartet => This indicates the presence of CH2 bonded to a CH3.

From b) and c) we can conclude that we have the ethyl group bonded to a nitrogen.  

Finally, we have to add 4 more carbons in such a way that we only have a single signal. In this case the ter-butyl group.

In that way, we will have 2 singlets (from the CH3 groups in the ter-butyl and the H on the N). Also, we will have the quartet on the CH2 in the ethyl group and the triplet on the CH3 in the ethyl group