5. Using the NMR and IR data, along with the molecular formula, provide structures forcompounds A and B, along with the reagents needed to perform the transformation.→Compound AC5H10NMR:1.70 ppm, 3H, s1.82 ppm, 3H, s2.05 ppm, 3H, d5.20 ppm, 1H, qIR:3050 cm¹¹, m2900 cm¹¹, s1650 cm¹¹, mCompound BC5H₁1BrNMR:0.90 ppm, 3H, t1.75 ppm, 2H, q1.84 ppm, 6H, sIR:2950 cm¹¹, s2900 cm¹¹, s1450 cm¹, s

Double bond equivalence (DBE) of the molecular formula can be calculated by using the formula DBE = x +1 - y / 2 where, x= number of carbon atoms and y = number of hydrogen atoms. If the molecule contains the halogen then DBE formula is DBE = x +1 - (y + z) / 2 where, x= number of carbon atoms, y = number of hydrogen atoms and z = number of halogen atoms. DBE value tells the unsaturation in the molecule. DBE for C5H10 is Here x = 5 and y = 10 Then DBE = 5 + 1 - 10 / 2 = 6 -5 = 1 DBE value of C5H10 is one means it may contains one double bond. Now DBE for C5H11Br is (x = 5, y = 11 and z = 1) DBE = 5 + 1 - (11 + 1) / 2 = 6 - (12 / 2) = 6 - 6 = 0 DBE value is zero means it has no any unsaturation. For the 1H-NMR chemical shift and IR values of compound A it seems that compound A is The IR value at 3050 cm-1 is due to sp2, =C-H stretching. Value at 2900 cm-1 is due to sp3, C-H stretching and the value at 1650 cm-1 is due to C=C stretching. The protons of the 2-methylbut-2-ene are labelled as below. Proton labelled as A appeared as the quartet of one hydrogen at 5.20 ppm. The multiplicity of the signal can be calculated by using the formula N + 1 where, N = number of neighboring hydrogens. The multiplicity of the signal A is quartet because it couples with three B type hydrogens i.e. 3 + 1 = 4. Proton labelled as B appeared as the doublet of three hydrogen at 2.05 ppm. The multiplicity of the signal B is quartet because it couples with three A type hydrogens i.e. 1 + 1 = 2. Proton labelled as C appeared as the singlet of three hydrogen at 1.82 ppm. Proton labelled as D appeared as the singlet of three hydrogen at 1.70 ppm. Compound B is IR values at 2950 and 2900 cm-1 is due to the asymmetric and symmetric sp3, methyl stretching respectively. The value at 1450 cm-1 is due to the C-C stretching. The protons of the 2-bromo-2-methylbutane are labelled as below. Proton labelled as A are two equivalent methyl groups appeared as the singlet of six hydrogens at 1.84 ppm. Proton labelled as B appeared as the quartet of two hydrogens at 1.75 ppm. The multiplicity of the signal B is quartet because it couples with three C type hydrogens i.e. 3 + 1 = 4. Proton labelled as C appeared as the triplet of three hydrogens at 0.90 ppm. The multiplicity of the signal C is triplet because it couples with two B type hydrogens i.e. 2 + 1 = 3.This reaction is the addition of HBr to unsymmetrical alkene. This addition takes place according to Markovnikov rule. According to Markovnikov rule the negative part i.e. bromide ion attached to that carbon which has less number of hydrogens. The reaction is given below.

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5. Using the NMR and IR data, along with the molecular formula, provide structures for compounds A and B, along with the reagents needed to perform the transformation. Compound A CsH10 NMR: 1.70 ppm, 3H, s 1.82 ppm, 3H, s 2.05 ppm, 3H, d 5.20 ppm, 1H, q IR: 3050 cm¹¹, m 2900 cm¹¹, s 1650 cm¹¹, m Compound B CsH₁1Br NMR: 0.90 ppm, 3H, t 1.75 ppm, 2H, q 1.84 ppm, 6H, s IR: 2950 cm¹¹, s 2900 cm¹, s 1450 cm¹¹, s

5. Using the NMR and IR data, along with the molecular formula, provide structures for compounds A and B, along with the reagents needed to perform the transformation. Compound A CsH10 NMR: 1.70 ppm, 3H, s 1.82 ppm, 3H, s 2.05 ppm, 3H, d 5.20 ppm, 1H, q IR: 3050 cm¹¹, m 2900 cm¹¹, s 1650 cm¹¹, m Compound B CsH₁1Br NMR: 0.90 ppm, 3H, t 1.75 ppm, 2H, q 1.84 ppm, 6H, s IR: 2950 cm¹¹, s 2900 cm¹, s 1450 cm¹¹, s

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

NMR Spectroscopy

NMR stands for Nuclear Magnetic Resonance in which the study of different molecules is done based on the interaction of radiofrequency electromagnetic radiations with the nuclei of molecules placed in a strong magnetic field. A precise study of spectra of visible light and all bands of electromagnetic radiation is called spectroscopy.

Question

5. Using the NMR and IR data, along with the molecular formula, provide structures for
compounds A and B, along with the reagents needed to perform the transformation.
→
Compound A
C5H10
NMR:
1.70 ppm, 3H, s
1.82 ppm, 3H, s
2.05 ppm, 3H, d
5.20 ppm, 1H, q
IR:
3050 cm¹¹, m
2900 cm¹¹, s
1650 cm¹¹, m
Compound B
C5H₁1Br
NMR:
0.90 ppm, 3H, t
1.75 ppm, 2H, q
1.84 ppm, 6H, s
IR:
2950 cm¹¹, s
2900 cm¹¹, s
1450 cm¹, s

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