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Explain optical activity of 2-chlorobutane. - Chemistry

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प्रश्न

Explain optical activity of 2-chlorobutane.

रासायनिक समीकरणे/रचना
स्पष्ट करा
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उत्तर १

  1. The property of a substance by which it rotates the plane of polarization of incident plane-polarized light is known as optical activity.
  2. 2-Chlorobutane has one chiral carbon atom and thus exists as a pair of enantiomers, which are non-superimposable mirror images of each other.
  3. These mirror images are referred to as optical isomers, which differ from each other in terms of a measurable property called optical activity. The structure of 2-chlorobutane and its mirror image can be represented as,
  4. Thus, 2-chlorobutane will show optical activity, and in accordance with the direction of optical rotation, one isomer will be dextrorotatory, and the other will be laevorotatory.
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उत्तर २


          Enantiomers of 2-Chlorobutane
(with structural and Fischer’s projection formula)

  1. 2-Chlorobutane contains an asymmetric \[\begin{array}{cc}\ce{Cl}\phantom{.......}\\|\phantom{........}\\\ce{CH3 - \underset{∗}{C}H - CH2 - CH3}\end{array}\] carbon atom (the starred carbon atom) which is attached to four different groups, i.e. ethyl (– CH2 – CH3), methyl (CH3), chloro (Cl) and hydrogen (H) groups.
  2. As seen in the image, these groups can be arranged around the carbon atom in two different ways. It therefore exists in two enantiomers. The two enantiomers cannot be superimposed; they are mirror images of one another.
  3. The laevorotatory isomer (l-isomer) is one of the enantiomers that will rotate the plane of plane-polarized light to the left. The second enantiomer, known as the dextrorotatory isomer (d-isomer), will rotate the plane of plane-polarized light to the right.
  4. The racemic mixture, also known as the racemate (dl-mixture), is an equimolar mixture of the d- and l-isomers that is optically inactive. External compensation is the cause of the racemic mixture's optical inactivity.
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Optical Isomerism in Halogen Derivatives
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Halogen Derivatives - Short Answer Questions (Type-Ⅰ)

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एससीईआरटी महाराष्ट्र Chemistry [English] 12 Standard HSC
पाठ 10 Halogen Derivatives
Short Answer Questions (Type-Ⅰ) | Q 4

संबंधित प्रश्‍न

Explain optical isomerism in 2-chlorobutane.


Optical activity of a molecule is associated with _______________


Which of the following shows optical isomerism?


Identify the INCORRECT statement.


Which among the following compound is not optically active?


Nicol's prism is made of ____________.


How many asymmetric carbon atoms are present in a molecule if it has 16 optical isomers?


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


(+) 2-methylbutan-1-ol (-) 2-methylbutan- 1-ol have different values for which property?


Identify the chiral molecule from the following.


Identify chiral molecule/s from the following.

a.

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

b.

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

\[\ce{CH3 - CH2 - CH2 - CH2Br}\]

d.

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


Identify the chiral molecule from the following.


Identify chiral molecules from the following

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.

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


Identify the chiral molecule from the following.


What is dextrorotatory substance?


Identify the chiral molecule from the following.


Identify chiral molecules from the following.

a.

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

b.

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

c.

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

d.

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


Identify the chiral molecule from the following.


Identify the chiral molecule from the following:


Identify the chiral molecule from the following.


Identify the chiral molecule from the following.


Identify chiral molecule/s from the following.


Identify the chiral molecule from the following.


Identify the chiral molecule from the following.


Identify chiral molecule/s from the following.


Identify the chiral molecule from the following.


Identify chiral molecule/s from the following.


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