I have absolutely no idea how to answer these questions and how to make the dedu

Question

I have absolutely no idea how to answer these questions and how to make the deductions of how to draw these diagrams. Please explain step by step the easy way of how and why I should draw the molecules the way that they are drawn as if I you were explaining it to a 2 year old so that I can finally get.

Part i 120 points Draw the structure of the following compounds. Your structure will be graded as correclor not. Partial credit only will be given for structural fragments (3rd box, points deducted if these exceed the molecuiar formuia) and for constraints and the lHD (2nd box) Hint Walch ine Integralion values! Compound A: Ca H12C IR 3035, 2975. 1740 cm 1, 1H NIMR 66.9 (s, 2H), 2 (q, 2H), 17 (q, 2H). 1 3 (t, 3-) 1 1 (t, 3H), NMR 208, 129, 53, 35, 29, 12.8 1 Structure 2 interpretation and constraints IHD 2 3. Structural fragments (don't exceed formula) Ore. CEO are CHz CH

Explanation / Answer

Molecular formula: C8H12O.

So Index of Hydrogen Deficiency (IHD) is calculated as follows:

For an aliphatic chain with 8 carbons it should have 2x8 + 2 = 18 Hydrogens.

Difference of (18 - 8) = 10 Hydrogens. That is equal 10/2 = 5 double bonds.

IR data: 3035, 2975 and 1740 CM-1:

3035 and 2975 indicate presence of a double bond in the structure which are C=C stretching and =CH stretching.

1740 indicates presence of an isolated keto group but not an alpha beta unsaturated ketone, CH=CHCO

1H NMR data: Delta values: 6.9 (s, 2H), 2.4 (q, 2H), 1.7 (q, 2H), 1.3 (t, 3H), 1.1 (t, 3H)

2.4 (q, 2H), 1.3 (t, 3H).................indicates CH2CH3 fragment. The CH2 is next to CO as in COCH2CH3

1.7 (q, 2H), 1.1 (t, 3H)................indicates CH2CH3 fragment

6.9 (s, 2H)..................................indicates presence of two equivalent olefin protons as in CH=CH. From IHD and IR data we know there's a possibility of double bond in the compound.

13C NMR data: 208, 129, 53, 35, 29, 12, 8

208...........SP2 Carbonyl Carbon as in C=O

129...........SP2 Carbons as in CH=CH

53, 35..........SP3 Carbons next to Carbonyl carbon CH2COC

29, 12, 8...........SP3 Carbons as in CH2 and CH3

The structure drawn is reasonable from the given data.

    

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