Question

170 65 89 30 25 50 86 27 30 74 70 10 20 40 50 60 80 90 100 110 120 130 140 150 160 170 180 70 miz Values and Intensities: 15 Tr | 20 Trị 25 Tr | 26 7 27 16[ 28 11 32 Tr 36 1| 37 13 38 26 39 75 40 5 41 7 438 44 61 45 11! 46 Til 49 51 50 421 51 31 52 61 53 11 60 11 61 18 62 411 63 1011 64 231 €5137 66 71 73 41 74 1875 15 76 71 77 11 79 3180 1181 41 82 11 83 11 84 31 85 21 86 261 87 111 88 31 89 130 90 70 91 999 92 74 93 31 95 Tri 104 Tr 105 Tr | 106 Tr 107 Tr | 116 1| 11ỉ 8 118 1| 11B 8128 Tr | 129 Tr 130 Tr 131 Tr | 141 Tr | 143 3| 145 2| 169 29 170 238 | 171 47 | 172 229 | 173 16 | wnload Wwwwwwwwwwwwwwwwww.parom, hogy milyway Wowwwwwwwwwwwa wan 4000 3500 3000 3900 2000 PM 7.2 6.8 6.4 6.0 5.6 5.2 4.8 4.4 4.0 3.6 3.2 2.6 2.4 2.0 1.6 1.2 0.8 0.4 0.0 -0.4

What compound would these graphs describe?

I suspect the MS says Toluene, the NMR says its para substitued, and the IR is an aldehyde. Would this be 4-Methylbenzaldehyde?

Answer 1

Solution:

The structure of the compound is para-bromotoluene. Find the details below.

solution from mass spectra it is clear that there is two peak around 150 and both the peak heights are same means came intensities. So here is a M+2 peak and it is because of Br atom. Br isotopes are Br. and 81 Br, both isotopes present 49.5 and so.s. almost egnal. From 'H NMR region. That means position like 2 the doublet at substituent aromatic is in para- Br And also a singlet at ~ 2:3ppm, this is because of a -CH3 group. Also, it is confirmed from the mass spectra of the peak at 91. CH₂ T CH₂ 7+ -By m/2=91 Br Therefore, the structure is