Chapter 14: Q.21558-14-16P. (page 546)
Sketch the NMR spectrum of CH3CH2Cl , giving the approximate location of each NMR signal.
Short Answer

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Chapter 14: Q.21558-14-16P. (page 546)
Sketch the NMR spectrum of CH3CH2Cl , giving the approximate location of each NMR signal.

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Label each statement as True or False.
a. When a nucleus is strongly shielded, the effective field is larger than the applied field and the absorption shifts downfield.
b. When a nucleus is strongly shielded, the effective field is smaller than the applied field and the absorption is shifted upfield.
c. A nucleus that is strongly deshielded requires a lower field strength for resonance.
d. A nucleus that is strongly shielded absorbs at a larger δ value.
Question: As we will learn in Chapter 20, reaction of (CH3)2CO with followed by affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600–3200, 3303, 2938, and 2120cm-1 . D shows the following 1 H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?
Question: How many different types of protons are present in each compound?
Question: Treatment of 2-methylpropanenitrile [(CH3)2CHCN] with CH3CH2CH2MgBr, followed by aqueous acid, affords compound V, which has molecular formula C7H14O. V has a strong absorption in its IR spectrum at 1713 cm-1, and gives the following 1 H NMR data: 0.91 (triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and 2.60 (septet, 1 H) ppm. What is the structure of V? We will learn about this reaction in Chapter 22.
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