Question

Butanone is formed by oxidation of Butan-2-ol using acidified potassium dichromate (VI) as shown by the following equation. CH:CH-CH(OH)CH3 + [0] → CH3CH,COCH3 + H2O (a) Is butan-2-ol Primary, secondary or tertiary alcohol? Explain your answer. (2) (b) Look at the infrared spectrum below and see whether the compound is butanone or butan-2-ol. Use your knowledge of spectroscopy to identify and explain your answer(3) Table A Infrared absorption data Bond Wavenumber Icm-1 N-H 3300-3500 (amines) O-H 3230-3550 (alcohols) C-H 2850-3300 O-H 2500-3000 (acids) CON 2220-2260 Co 1680-1750 C=C 1620-1680 -O 1000-1300 750-1100 100 Transmittance/ % pon 50 - 0+ 4000 3000 1000 500 2000 1500 Wavenumber/cm!

Answer 1

a) In butan-2-ol hydroxyl group (OH) is attached to secondary carbon atom, hence it is a secondary alcohol.

ОН butan-2-ol

b) The given IR spectra belongs to butanone because the characteristic IR absorption peak is at 1700 cm-1 which corresponds to C=O stretching

For alcohol, OH stretching at 3300-3600 cm-1 is not there in IR spectra.

c) Tertiary alcohol are not recats with oxidizing agent due to absent of proton on carbon containing OH group.

ОН 2-methylpropan-2-ol

d) C4H9OH is a Tertiary alcohol because hydroxyl group (OH) is attached to tertiary carbon atom.

Example of types of alcohol for your reference OH ОН ОН secondary alcohol tertiary alcohol primary alcohol