Chapter 21, Problem 123CP

Reaction of aldehyde D with amino alcohol E in the presence of NaH forms F (molecular formula C11H15NO2). F absorbs at 1730 cm−1 in its IR spectrum. F also shows eight lines in its 13C NMR spectrum, and gives the following 1H NMR spectrum: 2.32 (singlet, 6 H), 3.05 (triplet, 2 H), 4.20 (triplet, 2 H), 6.97 (doublet, 2 H), 7.82 (doublet, 2 H), and 9.97 (singlet, 1 H) ppm. Propose a structure for F. We will learn about this reaction in Chapter 18.

Reaction of aldehyde D with amino alcohol E in the presence of NaH forms F (molecular formula C11H15NO2). F absorbs at 1730 cm−1 in its IR spectrum. F also shows eight lines in its 13C NMR spectrum, and gives the following 1H NMR spectrum: 2.32 (singlet, 6 H), 3.05 (triplet, 2 H), 4.20 (triplet, 2 H), 6.97 (doublet, 2 H), 7.82 (doublet, 2 H), and 9.97 (singlet, 1 H) ppm. Propose a structure for F. We will learn about this reaction in Chapter 16.

a) Draw the structures of the products in the following conversions. Draw intermediates where necessary. NaBH, Ph,P-CH, THE (V) CH,OH 1. CH,MgBr 2. Н.о 1. KMNO,„H*, heat 2. SOCI, Benzaldehyde (iv) 1. HCN 2. Н,О, Н- (ii)

Gamma(y)-amino butyric acid (GABA) is a neurotransmitter (a chemical that is used to send signals from one neuron to another) of the mammalian central nervous system. In order to understand how GABA works, conformationally restricted analogues, such as compound 1, have been made. During a synthesis of compound 1, compound 2 was subjected to allylic bromination using NBS and a radical initiator (AIBN) instead of light (Aust. J. Chem. 1981, 34, 2231-2236). H₂N. CH3 GABA CO₂H CH3 dddd CH3 H₂N Modify the structures given below to draw all eight possible allylic bromides that can be formed when compound 2 undergoes allylic bromination, considering all possible regiochemical and stereochemical outcomes. You can use the single bond tool to add/remove pi bonds. CH3 CO₂H CH3 solddd CH3 CH3 2 CH₂ CO₂Et

Treatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppm

The treatment of isoprene [CH2=C(CH3)CH=CH2] with one equivalent ofmCPBA forms A as the major product. A gives a molecular ion at 84 in itsmass spectrum, and peaks at 2850–3150 cm−1 in its IR spectrum. The 1HNMR spectrum of A is given below. What is the structure of A?

Compound A (C9H18O) forms a phenylhydrazone, but it gives a negative Tollens’ test. The IR spectrum of A has a strong band near 1710 cm−1 . The broadband carbon NMR spectrum of A is given below. a)Propose a structure for A and explain how your structure fits the spectral data. b) Propose a mechanism for this reaction.

Structure created

(d) (i) Identify products B-F in the following reaction scheme. (ii) Provide a mechanism for the formation of B from benzene. (iii) Provide a mechanism for the formation of C from B. H2SO4 Brz B C FeBr3 Brz FeBrз H2SO4 D E & F