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.

Explanation:

   Degree of unsaturation is mathematically evaluated as

       Degree of unsaturation

      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

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  

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

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: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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