Hydrogenation of the compound shown below in | vedclass

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(B) The given compound is $CH_3-C \equiv C-CH(CH_3)-CH=CH_2$. Hydrogenation with a poisoned palladium catalyst (Lindlar's catalyst) results in $syn$-a

Vedclass · Accepted Answer

An optically inactive compound

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(B) The given compound is $CH_3-C \equiv C-CH(CH_3)-CH=CH_2$. Hydrogenation with a poisoned palladium catalyst (Lindlar's catalyst) results in $syn$-addition of hydrogen to the triple bond,converting it into a $cis$-double bond. The resulting product is $CH_3-CH=CH-CH(CH_3)-CH=CH_2$. Due to the specific geometry and the presence of internal symmetry elements or the formation of a meso-like structure depending on the configuration,the product formed is optically inactive.

Hydrogenation of the compound shown below in the presence of a poisoned palladium catalyst (Lindlar's catalyst) gives:

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