EBK ORGANIC CHEMISTRY
EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 9781260475685
Author: SMITH
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 12C, Problem 67P

Reaction of ( CH 3 ) 3 CCHO with ( C 6 H 5 ) 3 P = C ( CH 3 ) OCH 3 , followed by treatment with aqueous acid, affords R ( C 7 H 14 O ) . R has a strong absorption in its IR spectrum at 1717 cm 1 and three singlets in its 1 H NMR spectrumat 1 .02 ( 9 H ) , 2 .13 ( 3 H ) , and 2 .33 ( 2 H ) ppm . What is the structure of R? We will learn about this reaction in Chapter 21.

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A compound (C7H14O) has a strong peak in its IR spectrum at 1710 cm–1. Its 1H NMR spectrum consists of three singlets in the ratio 9:3:2 at δ 1.0, 2.1, and 2.3, respectively. Identify the compound.
Compound X (molecular formula C10H12O) was treated with NH2NH2, −OH to yield compound Y (molecular formula C10H14). Based on the 1HNMR spectra of X and Y given below, what are the structures of X and Y?
Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.

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NMR Spectroscopy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=SBir5wUS3Bo;License: Standard YouTube License, CC-BY