Question

Describe the number of signals and their splitting in the 1H NMR spectrum of (CH3)2CHOCH3.

Answer 1

Answer:

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

Explanation:

In this case, we can start with the structure of $(CH_3)_2CHOCH_3$ . 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 $CH_3$  groups in the left, the $CH$  and the $CH_3$  in the right. Lets analyse each one:

-) $CH_3$  in the right

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

-) $CH$

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):

$multiplicity~=~n+1~=~6~+~1~=~7$

For this carbon we will have a septet.

-) $CH_3$  in the left

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

$multiplicity~=~n+1~=~1~+~1~=~2$

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