Q24P Chapter-9 covered a synthesis of... [FREE SOLUTION]

Chapter 12: Q24P (page 639)

Chapter-9 covered a synthesis of alkynes by a double dehydrohalogenation of dihalides. A student tried to convert trans-2,5-dimethylhex-3-yne by adding bromine across the double bond and then doing a double elimination. The infrared and mass spectra of the major product are shown here.
Do the spectra confirm the right product? If not, what is it?
Explain the important peaks in the IR spectrum.

Short Answer

Expert verified

The mass spectrum is consistent with the formula of the alkyne $C_{8} H_{14}$ m/z 110.The IR is not consistent with the alkyne. Often, symmetrically substituted alkynes have a miniscule $C 鈮� C$ peak, so the fact that the IR does not show this peak does not prove that the alkyne is absent.

The important evidence in the IR is the significant peak at 1620 ${cm}^{- 1}$ and the $= CH$ absorption above 3000 ${cm}^{- 1}$ and this absorption is a characteristic of a conjugated diene. Instead of the alkyne being formed, the reaction must have been a double elimination to the diene.

Step by step solution

Step-1. Mass spectrum and IR spectrum:

A mass spectrum is simply the m/z ratios of the ions present in a sample plotted against their intensities. Each peak in a mass spectrum shows a component of unique m/z in the sample, and heights of the peaks denotes the relative abundance of the various components in the sample.

An IR spectrum is a plot of percentage transmittance of the radiation through the molecule versus wave number of the radiation. IR spectroscopy is a measurement of interaction of infrared radiation with matter by absorption, emission, or reflection.

Step-2. Explanation of part (a) and (b):

The mass spectrum is consistent with the formula of the alkyne, $C_{8} H_{14}$ as the molecular ion peak is observed at m/z 110. The IR is not consistent with the alkyne. Often, symmetrically substituted alkynes have a minisculepeak, so the fact that the IR does not show this $C 鈮� C$ peak does not prove that the alkyne is absent.

The important evidence in the IR is the significant peak at 1620 ${cm}^{- 1}$ and the $= C - H$ absorption above 3000 ${cm}^{- 1}$ and this absorption is a characteristic of a conjugated diene. Instead of the alkyne being formed, the reaction must have gone through a double elimination to form the conjugated diene. The IUPAC name of the product 2,5-Dimethyl-2,4-hexadiene.