Traite la pentane-2,4-dione par le bromure

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Pentane-2,4-dione is treated with isopropyl bromide in slightly basic anhydrous acetone, yielding two isomers, A and B, with respective yields of 28% and 72%. The major isomer, B, is characterized by its proton NMR spectrum, which includes a doublet at 1.28 ppm (6H), two singlets at 2.02 ppm and 2.20 ppm (3H each), a heptuplet at 4.40 ppm (1H), and a singlet at 5.42 ppm (1H). The methylene group between the carbonyls is noted to be fairly acidic and can be deprotonated, suggesting potential pathways for product formation. Identifying isomer B involves analyzing these spectral features and considering the deprotonation of the dione. Understanding the NMR data is crucial for determining the structure of compound B.
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on traite la pentane-2,4-dione par le bromure d'isopropyle en milieu faiblement basique dans l'acetone anhydre comme solvant .on obtient deux composés isoméres A et B,avec des rendements respectifs de 28% et de 72%.
Le composé B est caractérisé en R.M.N du proton par :
-un doublet à 1,28ppmd'intensité 6.
-deux singulets d'intensité 3 pour chacun et sortant à 2,02et2,20ppm.
-un heptuplet d'intensité 1 à 4,40ppm.
-un singulet d'intensité 1 à 5,42ppm.
identifier B.
 
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I believe you meant to say...
Treating pentane-2,4-dione with isopropyl bromide in slightly basic anhydrous acetone, two isomers (A and B) are obtained in yields of 28% and 72%. The compound B (the major isomer) is characterized by proton NMR. Examination of the proton NMR spectrum shows a doublet at 1.28ppm (integration gives 6H), a singlet at 2.02 ppm (integration gives 3H), a singlet at 2.20 ppm (integration gives 3H), a heptuplet at 4.40 ppm (integration gives 1H) and a singlet at 5.42ppm (integration gives 1H). What is B?

N'est-ce pas?
 


yes,ecxactly.
 


It will help you to know that the methylene group bewteen the two carbonyl groups is fairly acidic and can be deprotonated. Start with the deprotonated pentane dione and see what products are likely.
 

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