Allene is the name given to propdiene, \(_{ \, \, \, \, \, \, \, \, Allene}^{H_2C = C = CH_2}\)
Hybridisation of an atom is determined by determining the number of hybrid orbitals at that atom which is equal to the number of sigma \((s)\) bonds plus number of lone pairs at the concerned atom.
Pi\((p)\) bonds are not formed by hybrid orbitals, therefore, not counted for hybridisation.
Here, the terminal carbons have only three sigma bonds associated with them, therefore, hybridisation of terminal carbons is \(sp^3\). The central carbon has only two sigma bonds associated, hence hybridisation at central carbon is sp.
So, the correct answer is (B): sp and sp2
In the compound, the central atom is combined with other atoms with the help of two sigma bonds and two pi bonds, the hybridization for the compound is sp. The terminal carbon on the other hand, is attached to the carbon atom and also hydrogen by 3 sigma and 1 pi bond, the hybridization will be sp2.
So, the correct answer is (B): sp and sp2
The reactions which cannot be applied to prepare an alkene by elimination, are
Choose the correct answer from the options given below:
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
In organic chemistry, an alkene is a hydrocarbon containing a carbon-carbon double bond.[1]
Alkene is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds.
Read More: Ozonolysis
Read More: Unsaturated Hydrocarbon