Chapter 16: Problem 32
The number of asymmetric carbon atoms in a molecule of glucose is (1) 6 (2) 4 (3) 5 (4) 3
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Chapter 16: Problem 32
The number of asymmetric carbon atoms in a molecule of glucose is (1) 6 (2) 4 (3) 5 (4) 3
These are the key concepts you need to understand to accurately answer the question.
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N)C(C)(C)C(=O)O (1)
geometrical isomerism. (2) optical isomerism. (3) geometrical and optical
isomerism. (4) tautomerism.
#
The structure shows
Lactic acid in which a methyl group, a hydroxyl group, a carboxylic acid group and a hydrogen atom arc attached to a central carbon atom shows optical isomerism duc to molecular gcometry at the (1) carbon atom of the methyl group. (2) carbon atom of the carboxylic acid group. (3) central carbon atom. (4) oxygen of the hydroxyl group.
Optical isomerism arises becausc of the presence of (1) an asymmetric carbon atom. (2) a centre of symmetry. (3) a line of symmetry. (4) a plane of symmetry.
\Lambdaddition of \(\mathrm{Br}_{2}\) on cis-butenc-2-gives (1) a racemic mixture of 2,3 -dibromobutenc. (2) mesoform of 2,3 -dibromobutane. (3) dextro form of 2,3 -dibromobutanc. (4) lacvo form of 2,3 -dibromobutanc.
An organic compound will show optical isomerism is (1) Four groups attached to C-atom are different. (2) Threc groups attached to \(\mathrm{C}\) -atom are different. (3) Two groups attached to C-atom are different. (4) Nll the groups attached to \(\mathrm{C}\) -atom are same.
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