Two isomers with the formula C_5H_9N are shown. How might they be distinguished by IR...


1) Two isomers with the formula {eq}C_5H_9N {/eq} are shown. How might they be distinguished by IR Spectroscopy?

2) Propose a structure for one compound which meets the following descriptions:

i. {eq}C_4H_8O {/eq} with IR absorptions at 2700 and 1710 cm{eq}^{-1} {/eq}

ii. {eq}C_5H_8 {/eq} with IR absorptions at 3100 cm{eq}^{-1} {/eq} and 2150 cm{eq}^{-1} {/eq}

iii. {eq}C_5H_{10}O {/eq} with a strong, broad IR absorption at 3300 cm{eq}^{-1} {/eq} and one at 1680 cm{eq}^{-1} {/eq}

Infrared Spectroscopy

Infrared spectroscopy (IR) involves the interaction of infrared radiation with matter. Functional groups will behave (vibrate, stretch, flex, wiggle, basically move around) at different ranges on the spectra based on the type of functional group. This can be used to identify and study chemical substances.

The width and location of the peak in an IR is indicative of what functional group caused it. Alcohol and carboxylic acid peaks are very broad verses carbonyl peaks which are very narrow and sharp. Substituted benzene rings have peaks that correspond to the substitution pattern (Mono, para, meta, etc) in the fingerprint and overtone regions of the IR.

Answer and Explanation:

1) A two strong peak in the 3500-3300{eq}cm^{-1} {/eq} region caused by the two N-H will differentiate the primary amine from the nitrile. There are...

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Infrared Spectroscopy & Molecule Identification
Infrared Spectroscopy & Molecule Identification

from AQA A-Level Chemistry: Practice and Study Guide

Chapter 2 / Lesson 13

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