Question

A compound, C11H12O2, has an IR spectrum showing a peak at 1710 cm-1. Its 1H NMR spectrum has peaks at delta 1.3 (3 H, triplet), 4.3 (2 H, quartet), 6.5 (1 H, doublet), 7.4-7.6 (5 H, multiplet), and 7.7 (1 H, doublet). Draw its structure in the window below. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.

Answer 1

Degree of unsaturation would be 6 and we can draw the structure from given information

Step-by-step explanation:

Degree of unsaturation represents the number of double bond and number of rings in a compound. If a compound have 1 degree of unsaturation it would have 1 double bond or 1 ring and if a compound have 2 degree of unsaturation it would have 2 double bond, 1 ring and 1 double bond, 2 ring or 1 triple bond and so on.

From the given information we can calculate the degree of information by the following formula.

number of Carbon atoms+ 1- [number of hydrogen atoms + No of chlorine]/2

By putting the values

11+ 1 (-12)/2

11-11/2 =6

SO according to DOU rules the resultant compound should contain 1 ring and five double bond. Kindly find the attachment for structure.

Answer 2

Step-by-step explanation:

Degree of unsaturation is mathematically evaluated as

Degree of unsaturation
`= No \ of carbon +1 - ((No\ of\ H + No\ of\ Cl ) )/(2)`

Looking at the molecular formula of the given compound we see

that the number of carbon is = 11

the number of hydrogen is = 12

The number of Cl is = 0

So

Degree of unsaturation
`= 11 +1 - (12)/(2)`

`= 6`

This number tells us that there are five double bond present and one ring

This mean this compound contain a benzene

The structure of the compound is shown on the first uploaded image