Thomas Jefferson Snodgrass (Jeff1 found an unlabelled reagent bottle in his -80

Question

Thomas Jefferson Snodgrass (Jeff1 found an unlabelled reagent bottle in his -80 oC freezer. The bottle contained a liquid. He found that the compound (compound I) quickiy evaporated when warmed to room temperature. 'H-NMR of the compound showed. only two peaks, one, a triplet, was at 0.92 ppm and integrated to 6 protons and a second at 1.08 ppm that *u, u"., [(. [' quintet and integrated to 4 protons. Jeff treated the sample with Brz and light followed by sodium t-butoxide and isolated a new compound (Compound II which was also a gas at room temperature). NMR of compound II showed a doublet at 1.53 ppm that integrated to 6 protons and a quartet at 5.01 ppm that integrated to 2 protons. Treatment of compound 2 with Or followed by dimethylsulfide and water produced a single compound (compound IID. An IR spectrum of compound III had moderate peaks at2996,2836,2736, and 11 14 cm'l and a strong peak at l727cm'r. Use the clues in the story to identiff and draw the structures of compounds I, II, and III. You can figure out the chemistry without the spectroscopy.

Explanation / Answer

From the Molecular Formula : C5H10O This is . . . . CnH2nO . . . . 1 Unsaturation ( C=C or C=O ) __ or __ 1 Cyclic Ring From the Infrared Absorptions : 3600 - 3200 cm-1 ( Strong , Broad ) ----> O-H Stretching ( H-Bonded ) of Alcohol 1676 cm-1 ( Weak ) ----> Stretching of C=C Double Bond 965 cm-1 ----> Out of Plane Bending of Alkenes ; most probable of trans- R1-CH=CH-R2 From Proton NMR Spectrum : At ~ 1.1 ppm ; Doublet ( 6.3 Hz Coupling Constant ) ; 3 H ----> 3 Protons of Methyl Group adjacent to Carbon with 1 Hydrogen ; CH3 – ( C with 1 H ) – ; Signal Splitting by 1 Proton on Adjacent Carbon . [ This is Protons ( Hydrogens ) on C-1 of Final Concluded Molecular Structure ] At ~ 1.6 ppm ; Doublet ( 6.2 Hz Coupling Constant ) ; 3 H ----> 3 Protons of Methyl Group adjacent to C=C Carbon with 1 Hydrogen ; Allylic CH3 – ( CH=C ) – ; Signal Splitting by 1 Proton on Adjacent Carbon . [ This is Proton ( Hydrogen ) on C-5 of Final Concluded Molecular Structure ] Note : The values of Coupling Constant help to identify / classify the proton(s) provide the couplings ( signal splitting ) and also help in matching of the coupling pairs . ( Don't worry , there will be additional explanation how we utilize these information later . ) At ~ 2.4 ppm ; Singlet ( Disappears after shaking with D2O ) ; 1 H ----> 1 Proton Alcoholic –OH ; can be exchanged with D from D2O ; ( Signal Splitting by 1 Proton on Adjacent Carbon may not be observed because this proton can be exchanged with alcoholic proton in other molecules at a very fast rate ) . [ This is Proton ( Hydrogen ) of –OH Group attached on C-2 of Final Concluded Molecular Structure ] At ~ 4.1 ppm ; Quintet ( 6.3 Hz Coupling Constant ) ; 1 H ----> 1 Proton on Carbon with Alcohol ( –OH ) Group ; adjacent to C=C Carbon with 1 Hydrogen ; Allylic CH [ Methyl – CH(OH) – ( CH=C ) – ; Signal Splitting by 3 Protons on Adjacent Methyl Group + 1 Proton on Carbon of C=C Double Bond . ( Signal Splitting by 1 Proton from -OH group is not observed due to the same reason for the above signal at 2.4 ppm ) [ This is Proton ( Hydrogen ) on C-2 of Final Concluded Molecular Structure ] At ~ 5.4.and 5.6 ppm ; Multiplet ( with both 6.3 Hz and 6.2 Hz Coupling Constants shown ) ; 2 H ( Slightly Different 2 H ) ----> 2 Protons on C=C Double Bond ; Located on Different Carbons ; 1 Proton Signal ( nearer to ~ 5.4 ppm is coupled by 1 Proton at Carbon with Alcohol ( –OH ) Group with 6.3 Hz Coupling Constant + Another 1 Proton on C=C Double Bond ( C-4 ) ; 1 Proton Signal ( nearer to ~ 5.6 ppm is coupled by 3 Protons of Methyl Group ( C-5 Carbon ) with 6.2 Hz Coupling Constant + Another 1 Proton on C=C Double Bond ( C-3 ) ; [ These are 1 + 1 Protons ( Hydrogens ) on C-3 and C-4 of Final Concluded Molecular Structure ] From 13C NMR Signal Data : At 17.5 ppm ; ( with 3 H ) ----> Methyl Group ; Far away from –OH Group and C=C Double Bond . [ This is the C-1 of Final Concluded Molecular Structure ] At 23.3 ppm ; ( with 3 H ) ----> Methyl Group ; Nearer to C=C Double Bond ( Allylic Carbon ) [ This is the C-5 of Final Concluded Molecular Structure ] At 68.8 ppm ; ( with 1 H ) ----> Carbon with Alcoholic –OH Group [ This is the C-2 of Final Concluded Molecular Structure ] At 125.5 ppm ( with 1 H ) and At 135.5 ppm ( with 1 H ) ----> 2 Carbons of C=C Double Bond ( Most probable as R1-CH=CH-R ) [ These are the C-3 and C-4 of Final Concluded Molecular Structure ] From given information “ The Compound A is optically inactive , but it can be resolved into enantiomers . “ ---- > There is at least one chiral center __ ( one asymmetric carbon ) ; in this case the C-2 [ with –CH3 , –H , –OH , –CH=CH–R ] Final Concluded Molecular Structure trans- CH3 – CH = CH – CH(OH) – CH3 or CH3 – CH = CH – CH(OH) – CH3 ____ E Isomer trans- 3-Penten-2-ol ____ C-5 -- > C-4 & C-3 Double Bond -- > C-2* ( with –OH ) -- > C-1 (E) 3-Penten-2-ol ____ C-5 -- > C-4 & C-3 Double Bond -- > C-2* ( with –OH ) -- > C-1 C-2* is the chiral center ( asymmetric carbon ) Coupling ( Signal Splitting ) of Protons : C-1 Protons by C-2 Proton ; C-2 Protons by C-1 & C-3 Protons ; C-3 Protons by C-2 Proton are the ones with 6.3 Hz Coupling Constant Coupling ( Signal Splitting ) of Protons : C-4 Proton by C-5 Protons ; C-5 Protons by C-4 Proton are the ones with 6.2 Hz Coupling Constant

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