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
Arrange the following alkenes in the decreasing order of stability:
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The reactions which cannot be applied to prepare an alkene by elimination, are
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Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
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