Identify each of the following compounds from its molecular...

Key Concepts

Degree of Unsaturation

This concept involves calculating the number of rings and multiple bonds in a molecule based only on its molecular formula. It is crucial because it helps narrow down the structure by indicating the presence of unsaturated units such as double bonds, triple bonds, or aromatic rings.

¹H NMR Spectroscopy

This analytical technique is used to determine the number, types, and environments of hydrogen atoms in a molecule. The method provides data through chemical shifts, signal integration, and splitting patterns, all of which reveal details about the molecular structure.

Chemical Shifts

Chemical shifts indicate the resonance position of hydrogen atoms in the ¹H NMR spectrum. These positions are highly dependent on the electronic environment surrounding each proton, allowing the identification of functional groups and the local structure of the molecule.

Spin-Spin Coupling and Splitting Patterns

This concept refers to the interaction between non-equivalent neighboring protons which splits NMR signals into multiplets. The pattern, such as doublets or triplets, along with the coupling constants, gives insight into the proximity and number of neighboring hydrogen atoms.

Molecular Formula Analysis

Analyzing the molecular formula provides initial clues about the possible structures by outlining the atom count and calculating the degree of unsaturation. This step is essential to guide further structural elucidation using spectroscopic data like NMR.

Transcript

00:05 Question number 71, four molecular formula are given along with proton nmr the structure of the compound is to be determined the first molecule is c -8 h -8 the nmr 5 .5 .5 .2 a doublet 5 .7 a doublet and then 6 .7 4 signals caught it and 7 to 7 .2 to 7 .2 to 7 .5 and 5 proton multiply.

01:30 5 proton multiply.

01:33 So now these 3 signals corresponds to double bonded protons tons since they are quite downfield and each of this signal correspond to 1 h by integration and this is 5 h so this is double bonded compounds of the molecule is a monosubstit benzene ring 7 .2 to 7 .5 is benzene ring since there are 5 protons we can assume it is a monos substituted benzene this is a c we can write this cw1c so this problem this has to be the other proton and here we have c c h and finething so this problem so this molecule should be styling it should be the structure of the molecule the second compound c6 h12 o with two signals c6h12 4 you can see unsaturation index equals 6 minus 6 plus 1 1 that is 1 double bond or ring we see the number values 1 .2 2 .2 2 .2 3 is to 1 .2 delta 2 .2 in the ratio 3 is to 1 .2 delta in the ratio 3 is to 1.

04:32 If we divide them in the ratio 3 is to 1, there should be 9 protons.

04:37 There should be 3 protons.

04:40 So a 9 proton synchlet typically is a t -buttal group.

04:47 Then the delta value of 2 .2 of the methyl group, the delta value clear indicates the presence of a co group over there.

05:07 So this is the structure.

05:09 The next molecule, molecular formula c9h184 with a singlet c9h18o with one single at 1 .2, single at 1 .2.

05:52 Now this molecule, the single at 1 .2 means all the 18 propons are developed.

06:13 So 18 protons means two sets of nine proton each and nine proton in a single it corresponds to a tbutary loop so there are two t -butyl groups in this molecule so we will draw two t -butyter tubes now this should be connected now we have one two three four four plus four eight so we have one more carbon and that should be attached to equity you this will be the structure it will give only a singlet the fourth the fourth one c4 h8 oh c4 h8o now looking at the nature of the enumar the 3 .8 atriplet 3 .8 a triplet of 2 .8 a triplet of two programs 2 doublets at 4 and 4 .2 2 .2 2 doublets so this is the triplet is 2 protons and each of this is one proton so clearly indicates as you have seen in the first problem it indicates a ch2 double border cs now we have 6 .8 6 .6 .5 again a quartet four signals 6 .5 there are four signals you can say it's a quartet and the signal at 3 .7 is not a triplet three points but there is a signal at 1 .3 a triplet and this is a quartet 1 .3...

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