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Identify the chiral molecule from the following. - Chemistry

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प्रश्न

Identify the chiral molecule from the following.

Which of the following is an example of chiral molecule?

पर्याय

  • 1-Bromobutane

  • 1,1-Dibromobutane

  • 2,3-Dibromobutane

  • 2-Bromobutane

MCQ
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उत्तर

2,3-Dibromobutane and 2-Bromobutane

Explanation:

At first, we need to make the structure of all compounds to say which is chiral or not.

(a) 

\[\begin{array}{cc}
\phantom{........}\ce{H}\\
\phantom{........}|\\
\ce{C - C - C - C - Br}\\
\phantom{........}|\\
\phantom{........}\ce{H}
\end{array}\]

(b)

\[\begin{array}{cc}
\phantom{.........}\ce{Br}\\
\phantom{........}|\\
\ce{C - C - C - C - Br}\\
\phantom{........}|\\
\phantom{........}\ce{H}
\end{array}\]

(c)

\[\begin{array}{cc}
\phantom{..}\ce{Br}\phantom{..}\ce{Br}\phantom{}\\
\phantom{}|\phantom{....}|\phantom{}\\
\ce{C - C - C - C}
\end{array}\]

(d)

\[\begin{array}{cc}
\phantom{......}\ce{Br}\\
\phantom{.....}|\\
\ce{C - C - C - C}\\
\phantom{.....}|\\
\phantom{.....}\ce{H}
\end{array}\]

Now, we can see that 2,3-Dibromobutane and 2-Bromobutane has a Chiral carbon (that carbon which has 4 different attachments).

Hence, It's a chiral molecule.

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Optical Isomerism in Halogen Derivatives
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पाठ 10: Halogen Derivatives - Exercises [पृष्ठ २३१]

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बालभारती Chemistry [English] Standard 12 Maharashtra State Board
पाठ 10 Halogen Derivatives
Exercises | Q 1. v. | पृष्ठ २३१

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While assigning R, S configuration, the correct order of priority of groups attached to chiral carbon atom is ______


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a.

\[\begin{array}{cc}
\ce{CH3 - CH - CH2 - CH3}\\
|\phantom{.........}\\
\ce{OH}\phantom{.......}
\end{array}\]

b.

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH - CH2 - CH3}\\
|\phantom{..}\\
\ce{Br}
\end{array}\]

c.

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH2 - CH2Br}
\end{array}\]

d.

\[\begin{array}{cc}
\ce{CH3 - CH - CH3 - CH3}\\
|\phantom{.........}\\
\ce{CH3}\phantom{.......}
\end{array}\]


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a.

\[\begin{array}{cc}
\ce{CH3 - CH - CH2 - CH3}\\
|\phantom{.........}\\
\ce{OH}\phantom{.......}
\end{array}\]

b.

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH - CH2 - CH3}\\
|\phantom{.}\\
\phantom{.}\ce{Br}
\end{array}\]

c.

\[\ce{CH3 - CH2 - CH2 - CH2 Br}\]

d.

\[\begin{array}{cc}
\ce{CH3 - CH - CH3 - CH3}\\
|\phantom{.........}\\
\ce{CH3}\phantom{......}
\end{array}\]


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