
Which compound is optically active?
(A) $ Butan - 1 - amine$
(B) $ Butan - 2 - amine$
(C) $ 2 - methylpropan - 1 - amine$
(D) $ 2 - methylpropan - 2 - amine$
Answer
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Hint: The chemical compounds consisting of at least one chiral carbon is optically active. Chiral carbon is the carbon attached to the four different groups. When all the carbon atoms in the compound are attached to two or more same groups, then that compound can be known as optically inactive.
Complete answer:
The presence of chiral carbon gives optical activity to the molecule.
Option A consists of $ Butan - 1 - amine$ , in which the carbon chain consists of four carbon atoms. The amine group is at $ {1^{st}} $ position. The structure of $ Butan - 1 - amine$ will be $ C{H_3}C{H_2}C{H_2}C{H_2}N{H_2} $ . None of the carbon is chiral. So, it is not optically active.
Option B consists of $ Butan - 2 - amine$ . The structure of $ Butan - 2 - amine$ will be $ C{H_3}C{H_2}CH\left( {N{H_2}} \right)C{H_3} $ . The carbon to which an amine group is attached is chiral carbon. So, it is optically active.
Option C consists of $ 2 - methylpropan - 1 - amine$ . The structure of $ 2 - methylpropan - 1 - amine$ will be $ C{H_3}CH\left( {C{H_3}} \right)C{H_2}N{H_2} $ . None of the carbon is chiral carbon. Thus, it is optically inactive.
Option D consists of $ 2 - methylpropan - 2 - amine$ . The structure of $ 2 - methylpropan - 2 - amine$ will be $ C{H_3}C\left( {C{H_3}} \right)\left( {N{H_2}} \right)C{H_3} $ . None of the carbon is chiral carbon. Thus, it is optically inactive.
Thus, in the given compounds, $ C{H_3}C{H_2}C{H_2}C{H_2}N{H_2} $ is an optically active compound.
Option A is the correct one.
Note:
When any chemical compound is optically active, it can be classified into two forms namely d-form and l-form. These two forms rotate the plane-polarized light in different directions. d-form rotates the plane polarised light towards the right side which can be known as Dextro and the l-form rotates the plane polarised towards the left side which can be known as leavo.
Complete answer:
The presence of chiral carbon gives optical activity to the molecule.
Option A consists of $ Butan - 1 - amine$ , in which the carbon chain consists of four carbon atoms. The amine group is at $ {1^{st}} $ position. The structure of $ Butan - 1 - amine$ will be $ C{H_3}C{H_2}C{H_2}C{H_2}N{H_2} $ . None of the carbon is chiral. So, it is not optically active.
Option B consists of $ Butan - 2 - amine$ . The structure of $ Butan - 2 - amine$ will be $ C{H_3}C{H_2}CH\left( {N{H_2}} \right)C{H_3} $ . The carbon to which an amine group is attached is chiral carbon. So, it is optically active.
Option C consists of $ 2 - methylpropan - 1 - amine$ . The structure of $ 2 - methylpropan - 1 - amine$ will be $ C{H_3}CH\left( {C{H_3}} \right)C{H_2}N{H_2} $ . None of the carbon is chiral carbon. Thus, it is optically inactive.
Option D consists of $ 2 - methylpropan - 2 - amine$ . The structure of $ 2 - methylpropan - 2 - amine$ will be $ C{H_3}C\left( {C{H_3}} \right)\left( {N{H_2}} \right)C{H_3} $ . None of the carbon is chiral carbon. Thus, it is optically inactive.
Thus, in the given compounds, $ C{H_3}C{H_2}C{H_2}C{H_2}N{H_2} $ is an optically active compound.
Option A is the correct one.
Note:
When any chemical compound is optically active, it can be classified into two forms namely d-form and l-form. These two forms rotate the plane-polarized light in different directions. d-form rotates the plane polarised light towards the right side which can be known as Dextro and the l-form rotates the plane polarised towards the left side which can be known as leavo.
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