Which of the following compounds is optically | vedclass

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(B) For an allene $(R_1R_2C = C = CR_3R_4)$ to be optically active,the groups attached to each terminal carbon must be different,i.e.,$R_1 
eq R_2$ a

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$CH_2 = C = CH_2$

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(B) For an allene $(R_1R_2C = C = CR_3R_4)$ to be optically active,the groups attached to each terminal carbon must be different,i.e.,$R_1 
eq R_2$ and $R_3 
eq R_4$.$(A)$ $CH_3 - CH = C = CH - CH_3$: Here,$R_1 = CH_3, R_2 = H$ and $R_3 = CH_3, R_4 = H$. Since $R_1 
eq R_2$ and $R_3 
eq R_4$,it is optically active.$(B)$ $CH_2 = C = CH_2$: Here,$R_1 = R_2 = H$. Since the terminal carbons have identical groups,it possesses a plane of symmetry and is optically inactive.$(C)$ $Cl(Br)C = C = C(Cl)Br$: Here,$R_1 = Cl, R_2 = Br$ and $R_3 = Cl, R_4 = Br$. Although the terminal carbons have different groups,the molecule has a plane of symmetry (the plane containing the central carbon and the substituents),making it optically inactive.$(D)$ $Me(Et)C = C = C(Cl)Br$: Here,$R_1 = Me, R_2 = Et$ and $R_3 = Cl, R_4 = Br$. Since $R_1 
eq R_2$ and $R_3 
eq R_4$,it is optically active.Note: In many standard contexts,$CH_2 = C = CH_2$ is the simplest example of an optically inactive allene. However,option $C$ is also optically inactive due to its symmetry. Given the typical nature of such questions,$B$ is the most fundamental answer.

Which of the following compounds is optically inactive?

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