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Molecular formula

LO82

 

SQ1

The Peak integration Give H =1+2+3+(6+1)+9 =22 It has O=5 and 2 un-saturation so CnH(2n-2)O5

So calculating n we get n=12 so the molecular formula is C12H22O5

Now The molecule show 2 tautomeric forms since it involve oxygen So it would be Keto enol tautomer.

Further the molecule Has Most of the H atom in Upfield So ((6+1)+ 9))=16 H atom in the upfield and likely to be CH2 and CH3 Hydrogen.

Further investigation of NMR shows that Multiplicity of upfield peaks are singlet for 9H So 3 –CH3 gr. Are their again 3.2 ppm has one 6 H peak and closely following one 1H peak so There are 3 equivalent CH2-CO grs methelenic H (-CH2-) again we have one 3H and one 2H peak given total 4 methyl gr. in molecule and 4 (-CH2-) gr. In the molecule.

So The Coumpud is likely 1,3-diethyl 2-(2-hydroxypentan-2-yl)propanedioate .

The Single H pertain to Hydroylic and in alpha Carbon H in between the Conjugate Keto groups.

Now Compare The IR so see the proof of the Str.

We have 1160-1310 /cm brad peak for the conjugate carbonyl

We have 3000-3100 and 2800-3000 broad peaks for CH3 and Ch2 groups

We have small peak at 3450 /cm for OH

We have Peak broadening and from 1000 to 1750 /cm due ester interaction on carbonyl and Tautomerism since conjugate Easter Carbonyl Has this Charteristic So –O-CO-CH(R)-CO-O- str is satisfied.

Now Compare The IR so see the proof of the Str.

So we can Say The Molecule is 1,3-diethyl 2-(2-hydroxypentan-2-yl)propanedioate

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