Chapter S: INTEGRATED Srscraoscory PROBLEMS 129 Compound 71 is a liquid (boiling
ID: 304604 • Letter: C
Question
Chapter S: INTEGRATED Srscraoscory PROBLEMS 129 Compound 71 is a liquid (boiling point 89 readily prepared by the free radical halegenation of propyne. The Mass, IR, and H NMR below, Long-range coupling with J-2 He is visible in the 'H NMR. Elementa a: C which adds two moles of hydrogen gas in the presence of PuC. The compound is spoctra, along with C NMR data, are given Analysis: C, 30.29;H,1% contains halogen. Mass Spectirm Iafrared Spectrum Wave Number, om - VT Wavelength, microns HNMR ppm & 0 "C Spectral Data: doublet, 68.1 ppm; singlet, 80.3 ppm; triplet, 22.6 ppm Structure: C2000 Beooks (Cole Publishing CompanyExplanation / Answer
Structure: It should be 3-bromopropyne (propargyl bromide): Br-CH2-C(triplebond)CH (C3H3Br)
Mass: The compounds molecular formula is C3H3Br acoounting to a molar mass of 36+3+79 = 118
The peask at 118 confirms the structire and the peak at 120 is the M+2 peak due to the second isotope of Br with molar mass of 81.
The peak at 79 and 81 correspond to the Bromide isotopes
Strong peak at 39 corresponds to the propargyl cation (+CH2C(triplebond)CH)
IR Spectrum: Peaks and correspoding funtional groups:
3200 cm-1 alkyne CH stretch (from CH group) - Confirms alkyne presence
2900 cm-1 sp3 CH stretch (from the CH3 group) - Confirms alkane presence
2100 cm-1 CtriplebondC stretch - Confirms CtriplebondC
1H NMR: Br-CH2(doublet at 3.8)-C(triplebond)CH(triplet at 2.3 ppm)
Triplet at 2.3 ppm can be assigned to the CH group attached to a triple bond. Since there is only 1 proton on the CH group, the intensity is 1. This peak is expected between 2-2.5 ppm due to the alkyl group next to it. And it undergoes long range coupling with the CH2 group and gives rise to a triplet due to the 2 neighboring protons from CH2. n+1 rule, 2+1 = 3, leading to a triplet.
Doublet at 3.8 ppm can be assigned to the CH2 group attached to Br. Since there are 2 protons on the CH2 group, the intensity is 2. This peak is expected to show up around 3.5 -4 ppm due to the electronegative element (Br) next to it. And it undergoes long range coupling with the CH group and gives rise to a doublet due to 1 neighboring proton from CH. n+1 rule, 1+1 = 2, leading to a doublet.
13C NMR: Br-CH2(singlet at 80.3)-C(doublet at 68.1)(triplebond)CH(triplet at 22.6 ppm)
Triplet at 22.3 ppm can be assigned to the C of CH group attached to a triple bond. This peak is expected between 20-25 ppm due to the alkyl group next to it. Doublet at 68.1 ppm can be assigned to the C group that has no Hs attached to it. The singlet at 80.3 ppm is from the CH2 group attached to Br, the down field indicating the presence of electronegative element next to it.