looking at this reaction (first image), The second image shows the 13C NMR spectrum of the 2-acetnaphthone which is the starting material and an 13C NMR spectrum obtained for the reaction product:Based a comparison of the two 13C NMR spectra, is product A or product B the major product of the reaction? - Explain your reasoning. 2-AcetnaphthoneNaBH4CH3OHProduct AOHorProduct B? 20013C NMR spectrum of the product:2001501501001005050[ppm]0 [ppm]F25 [rel201510G014 [rel]12ОТ89420

NaBH4 reduces aldehyde/ketone to primary/secondary alcohol

Answered: 2-Acetnaphthone NaBH4 CH₂OH Product A…

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Chemistry

2-Acetnaphthone NaBH4 CH₂OH Product A OH or Product B ?

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter16: Amines

Section: Chapter Questions

Problem 16.39P

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Concept explainers

Alkanes and Cycloalkanes

Alkanes are saturated organic compounds which are made up of C and H atoms. Alkanes have the general formula of CnH2n+2. Alkane contains sp3 hybridized carbon atoms with four sigmas (σ) bonds & every hydrogen atom is connected with one of the carbon atoms. The size of the alkane could be determined by the number of carbon atoms located in it.

Question

looking at this reaction (first image),

The second image shows the ¹³C NMR spectrum of the 2-acetnaphthone which is the starting material and an ¹³C NMR spectrum obtained for the reaction product:

Based a comparison of the two ¹³C NMR spectra, is product A or product B the major product of the reaction? - Explain your reasoning.

2-Acetnaphthone
NaBH4
CH3OH
Product A
OH
or
Product B
?

200
13C NMR spectrum of the product:
200
150
150
100
100
50
50
[ppm]
0 [ppm]
F
25 [rel
20
15
10
G
0
14 [rel]
12
ОТ
8
9
4
2
0

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