Question

Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at ≥ 160 ppm. Then draw the structure of an isomer with no rings of molecular formula C4H8O that has all of its 13C NMR signals at < 160 ppm.

Structure that has a signal ≥ 160 ppm: draw structure ...

Structure with no rings that has all signals < 160 ppm: draw structure ...

Answer 1

The possible structures are ketone and aldehyde.

Step-by-step explanation:

Number of double bonds of the given compound is calculated using the below formula.

`N_(db)=N_(c)+1-(N_(H)+N_(Br)-N_(N))/(2)`

`N_(db)`=Number of double bonds

`N_(c)` = Number of carbon atoms

`N_(H)` = Number of hydrogen atoms

`N_(N)` = Number of nitrogen atoms

The number of double bonds in the given formula -
`C_(4)H_(8)O`

`N_(db)= 4+1-(8+0-0)/(2)=1`

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound
`C_(4)H_(8)O` having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.