Final answer:
The hydrocarbon in question has a molecular ion with an m/z of 128, suggesting a molecular weight of 128 g/mol, with fragmentation peaks indicating an aliphatic structure with a propyl or isopropyl group. Considering the absence of peaks related to benzene ring fragments, the molecule is likely a branched alkane rather than an aromatic hydrocarbon or an alkene/alkyne.
Step-by-step explanation:
The question asks us to identify a hydrocarbon with a molecular ion with an m/z value of 128 and significant peaks with m/z values of 43, 57, 71, and 85 in its mass spectrum. To determine this, let's consider the mass and the fragmentation pattern suggested by the peaks.
A molecular ion with an m/z of 128 signifies a hydrocarbon with a molecular weight of 128 g/mol. Given the significant peak at an m/z of 43, which is commonly associated with a propyl group or an isopropyl group in mass spectrometry, we can surmise that the hydrocarbon likely contains this functional group as one of its fragments.
Now, to identify whether the hydrocarbon is aliphatic or aromatic and if it's an alkane, alkene, or alkyne, we consider the nature of the peaks. Aromatic hydrocarbons tend to show a molecular ion peak and often have stable fragment ions that relate to the benzene ring, whereas aliphatics show a different, often more complex fragmentation pattern.
Given the lack of high m/z value fragments consistent with an intact benzene ring structure, it is plausible that this molecule is an aliphatic hydrocarbon. With the molecular weight of 128 g/mol, a possible candidate could be a branched alkane since the peaks at m/z 43, 57, 71, and 85 could represent the loss of alkanes with increasing chain length from the parent ion.