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

Answer:

The possible structures are ketone and aldehyde.

Explanation:

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

$N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}$

$N_{db}$=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 - $C_{4}H_{8}O$

$N_{db}= 4+1-\frac{8+0-0}{2}=1$

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.