I I know the answers, I just need explanation why and how. Thank you 1)(32 point

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I
I know the answers, I just need explanation why and how. Thank you

1)(32 points) For each problem, write the s letser of the compound froms the handout (at the g of the exam, page 1) which matches the described spectral property (4 ptsleach) G) An alkene that shows no IR stretching band for the carhon-carbon double bond A compound that will show 'H NMR signals for eactly two magnetically distinet protons G)y ADGL Acompound with a methykesegro A compound with a methylene group wbose protons show two separate signals for the individual methylene protons on an H NMR spectrum. A composmd that will undergo fragmentation in a mass spectrometer to form the allylic earbocation with mz-41 A compound that shows a large peak on a mass spectrometer at m/z-139 which is 127 mass units lighter than the parent ion (vi) A compound that shows a single peak on an 'H NMR spectrum obtained at 298 K and two peaks on the spectrum oblained at 100 K (vii) A compound that will show only two vinylic proton signals with coupling constants around 2.0 Hz vii) A compound that shows a peak for six equivalent hydrogens on an lI NMR spectrum that is split into a doublet. 3 of 12

Explanation / Answer

Matching statements with compound structures,

(i) alkene shows no C=C bond stretch in IR : E

as it is an aromatic ring which is different than an islated alkene

(ii) Compound having two distinct peaks in 1H NMR : B

Symmetrical structure, with tert-butyl and CH2 proton signals

(iii) A compound with methylene protons split into two signals in 1H NMR : Having an assymetric carbon next to methylene leads to splitting of protons of methylene into two signals.

(iv) allylic carbocations are very stable and thus give specific fragmentation in the mass spectrum.

(v) The most stable fragment forms in the mass spectrum by breaking of the C-C bond. the mass unit is found lower by 127 units.

(vi) All methylene appears same at room temperature whereas at lower temperature the acial and equatorial proton ring flip is restricted and thus two signals are seen.

(vii) Terminal alkene shows two vinylic signals.

(viii) The two CH3's (total 6H) is next to CH and thus appears as a doublet in 1H NMR for 6H integration.

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