A) 2-chlorobutane
B) 2, 3-dichlorobutane
C) 2, 3-dichloropentane
D) 2-hydroxypropanoic acid
Correct Answer: B
Solution :
[a]\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{*}{\mathop{C}}}\,H-C{{H}_{2}}C{{H}_{3}}\] One asymmetric carbon atom, forms \[d-\] and \[l-\]optical isomers. Meso due to internal compensation [b] Two asymmetric carbon atoms, forms \[d-,\,\,l-\]and meso forms. [c]\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{*}{\mathop{C}}}\,H-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{*}{\mathop{C}}}\,H-C{{H}_{2}}C{{H}_{3}}\] Two asymmetric carbon atoms but does not have symmetry. Hence, meso form is not formed. [d]\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{*}{\mathop{C}}}\,H-COOH\] One asymmetric carbon atom, meso form is not formed.You need to login to perform this action.
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