Question

An alkyne with the molecular formula c6h10 was reduced with h2 and lindlar's catalyst. Ozonolysis of the resulting alkene afforded only one aldehyde product. What is the iupac name of the alkene that was formed by reduction of the starting alkyne?.

Answer 1

The alkene formed by the reduction of the alkyne is 1,2-dimethylcyclohexene.

Step-by-step explanation:

When an alkyne with the molecular formula C6H10 is reduced with H2 and Lindlar's catalyst, it forms an alkene. In this case, the alkene formed is most likely cyclohexene. The reduction of the alkyne removes two hydrogen atoms from each carbon in the triple bond, resulting in the formation of double bonds between the carbon atoms.

Ozonolysis of this alkene would lead to the formation of one aldehyde product. The IUPAC name for this alkene would be 1,2-dimethylcyclohexene.The presence of two methyl groups (dimethyl) on adjacent carbon atoms in the cyclohexene ring suggests a symmetrical substitution pattern.

Answer 2

Upon reduction with H2 and Lindlar's catalyst, the symmetrical alkyne C6H10 forms the alkene 3-hexene. This conclusion is based on ozonolysis yielding a single type of aldehyde, implying a symmetrical structure of the original alkyne.

Step-by-step explanation:

The alkyne with molecular formula C6H10 upon reduction with H2 and Lindlar's catalyst followed by ozonolysis yielding only one aldehyde product suggests that the alkyne had to be symmetrical. Therefore, the starting alkyne must have been 3-hexyne since its reduction would lead to the alkene, 3-hexene. Upon ozonolysis of 3-hexene, only one type of aldehyde, propanal, would be formed, supporting the evidence provided by the single aldehyde product.

When naming the resulting alkene from the reduction of the original alkyne by IUPAC rules, the longest carbon chain is identified and numbered so that the double bond has the lowest possible number. Since the double bond starts on the third carbon atom in the chain of six carbon atoms, the IUPAC name of the alkene formed is 3-hexene.