Identify the major product that is obtained when 2-hexyne is treated with H2 and Lindlar's catalyst.
1 answer:
Answer:
<em>cis</em>-2-hexene
Explanation:
Hello.
In this case, since a Lindlar's catalyst is mainly composed by palladium, which is a transition metal with a very exceptional surface area which makes it a suitable catalyst for hydrogenation reactions by which hydrogen is added to the insaturation in order to make a double bond where the triple bonds was in an alkyne, based on the attached picture we can see that the major product of that reaction is <em>cis</em>-2-hexene.
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Answer 1) : According to the complete question attached in the answer,
The radial wave function which is denoted by shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.
Also, the radial probability distribution curve denoted as shown in blue color is observed to touch zero, and shows the place of radial node.
Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by <em>'n'</em>.
It is observed that for any atomic orbital, the total number of nodes will be n-1 .
Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.
Hence, there will be only radial nodes, which is
(n−1 = total number of radial nodes in s orbitals)
According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)
This represents the 5s orbital.
Answer 2) The radial nodes are observed in I'm seeing radial nodes at
,
,
and
.
where represents the hydorgen bohr atomic radius = 0.0529177 nm
Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.
