Describe the number of signals and their splitting in the 1H NMR spectrum of (CH3)2CHOCH3.
a. 4 signals: 2 doublets, a singlet, and a septet
b. 2 signals: a doublet and a septet
c. 3 signals: a doublet, a quartet, and a septet
d. 3 signals: a singlet, a doublet, and a septet
e. 3 signals: 2 doublets and a septet

In this case, we can start with the structure of [tex](CH_3)_2CHOCH_3[/tex] . When we draw the molecule we will obtain 2-methoxypropane (see figure 1).

In 2-methoxypropane we will have three signals. The signal for the [tex]CH_3[/tex] groups in the left, the [tex]CH[/tex] and the [tex]CH_3[/tex] in the right. Lets analyse each one:

-) [tex]CH_3[/tex] in the right

In this carbon, we dont have any hydrogen as neighbors. Therfore we will have singlet signal in this carbon.

-) [tex]CH[/tex]

In this case, we have 6 hydrogen neighbors ( the two methyl groups in the left). So, if we follow the n + 1 rule (where n is the amount of hydrogen neighbors):

[tex]multiplicity~=~n+1~=~6~+~1~=~7 [/tex]

For this carbon we will have a septet.

-) [tex]CH_3[/tex] in the left

In this case we have only 1 hydrogen neighbor (the hydrogen in [tex]CH[/tex] ). So, if we use the n+1 rule we will have:

[tex]multiplicity~=~n+1~=~1~+~1~=~2 [/tex]

We will have a doublet

With all this in mind the answer would be:

d. 3 signals: a singlet, a doublet, and a septet

See figure 2 to further explanations