Question

o
o and π bond in
Indicate the no. of
atoms
CH₂ = CH - CH2 -C=CH
CH₂=CH-CN​

Answer 1

Answer:

Don't know sorry siso sorry

Answer 2

Explanation:

One of the first things you learn about alkenes is that rotation about the C-C pi (π) bond does not occur. For instance, at normal temperatures and pressures., trans-2-butene (shown below left) is never observed to spontaneously convert to cis-2-butene (right) . They’re separable compounds, with different melting and boiling points. You can buy each of them separately from Aldrich. This wouldn’t be possible if there was free rotation about the double bond.

This has other physical consequences besides the rotation barrier: it influences molecular geometry as well. Since Pi-bonding is a phenomenon exclusive to p-orbitals, that means that each pi bond an atom participates in will leave one fewer p-orbital available for “hybridization” with the s orbital [and remaining p orbital(s)] on the atom. This results in the familiar “trigonal planar” (sp2 ) geometry for typical alkene carbons and “linear” geometry (sp) for alkyne carbons.

Hence, alkenes are “flat”, as opposed to alkyl carbons, which adopt a tetrahedral geometry.

2. The Importance Of Orbital Overlap For Pi Bonding

A vivid illustration of the importance of orbital overlap is presented by a case where we might naively think a double bond “should” form – but does not. Bredt observed in 1924 that alkenes tend not to form on “bridgehead” positions, such as in the molecule at bottom left, an observation that came to be called “Bredt’s rule“.

Why not? If you make the model, you’ll see that the geometry of the bicyclic ring forces those p orbitals to be oriented at right angles. There’s no overlap between the p orbitals. Therefore, it resembles a carbon with two adjacent radicals more than it does a real pi bond!

pi bonding is not possible in bridgehead alkenes since p orbitals are at right angles to each other not properly aligned bredts rule

Three dimensional drawings on a flat surface don’t really do justice to glory of the 3-dimensional structure. Here’s a model and accompanying video. [RIP Vine, which was awesome for short organic chemistry videos]

Those pink things are supposed to be the p-orbitals. See how they’re at right angles to each other? That means they don’t overlap.

a half-filled orbital (e.g. a radical)

We call this “building up” of p orbitals into larger “pi systems”, “conjugation”. In each of the middle molecules below, the alkene (pi bond) is conjugated with an adjacent p orbital.

examples of conjugation pi bond with carbocation lone pair conjugation with other pi bond conjugation with radical example of no conjugation

The “conjugation killer” to watch out for is an atom lacking lone pairs connected to only single bonds, such as CH2 in the example below-right (1,4-pentadiene). These two pi bonds are not conjugated.

[A note on the second example. We usually think of the geometry of a nitrogen with three single bonds as trigonal pyramidal (e.g. as in NH3). But in the presence of an adjacent pi bond, there is a slight “re-hybridization” of the nitrogen from sp3 to sp2 (trigonal planar) such that the lone pair is in a p orbital, not an sp3 orbital. This is a tradeoff: the slightly increased strain of the eclipsed N-H bonds is made up for by a decrease in overall energy due to better overlap of a p orbital with the pi bond. We usually think of this as “resonance energy“.]