Isomerism MCQ Questions & Answers in Organic Chemistry | Chemistry
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1.
Given below are the structures of few compounds with molecular formula $${C_4}{H_{10}}O.$$ Select metamers from these structures.
\[\begin{align}
& \left( \text{i} \right)C{{H}_{3}}-O-C{{H}_{2}}C{{H}_{2}}C{{H}_{3}} \\
& \left( \text{ii} \right)C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH \\
& \left( \text{iii} \right)C{{H}_{3}}-C{{H}_{2}}-O-C{{H}_{2}}-C{{H}_{3}} \\
& \left( \text{iv} \right)C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
OH\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}-C{{H}_{3}} \\
\end{align}\]
A
(i) and (ii)
B
(ii) and (iii)
C
(i) and (iii)
D
(ii) and (iv)
Answer :
(i) and (iii)
$$C{H_3}OC{H_2}C{H_2}C{H_3}$$ and $$C{H_3}C{H_2}OC{H_2}C{H_3}$$ are metamers. (ii) and (iv) i.e., $$C{H_3}C{H_2}C{H_2}C{H_2}OH$$ and \[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
OH\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}-C{{H}_{3}}\] are position isomers.
2.
Indicate whether each of the following pairs are identical or not ?
A
I - enantiomers; II - diastereomers; III - enantiomers
B
I - identical; II - enantiomers; III - enantiomers
C
I - enantiomers; II - diastereomers; III - identical
D
I - enantiomers; II - identical; III - identical
Answer :
I - enantiomers; II - diastereomers; III - enantiomers
3.
Which of the following compounds will exhibit geometrical
isomerism ?
A
1 - Phenyl - 2 - butene
B
3 - Phenyl - 1 - butene
C
2 - Phenyl - 1 - butene
D
1,1 - Diphenyl - 1 - propene
Answer :
1 - Phenyl - 2 - butene
$$PhC{H_2}CH = CHC{H_3}$$ will exhibit geometrical isomerism because in others one of the doubly bonded carbon atom has two similar groups.
4.
Allyl isocyanide has
A
$$9\,\sigma \,\,{\text{and}}\,\,4\,\pi {\text{ - bonds}}$$
B
$$8\,\sigma \,\,{\text{and}}\,\,5\,\pi {\text{ - bonds}}$$
C
$$9\,\sigma ,3\,\,\pi \,\,{\text{and}}\,\,2\,\,{\text{non - bonded electrons}}$$
D
$$8\,\sigma ,3\,\pi \,\,{\text{and}}\,\,4\,\,{\text{non - bonded electrons}}$$
Allyl isocyanide has the structure
$${H_2}C = CH - C{H_2} - N\mathop = \limits^ \to C$$
it has $$9\sigma $$ and $$3\pi $$ bonds along with 2 nonbonded electrons
5.
Which one of the following acids does not exhibit optical isomerism ?
A
Lactic acid
B
Tartaric acid
C
Maleic acid
D
$$\alpha $$ - amino acids
Answer :
Maleic acid
Optical isomers contain an asymmetric ( chiral ) carbon atom ( a carbon atom attached to four different atoms or groups ). Therefore all except maleic acid exhibit optical isomerism.
6.
Racemic mixture is formed by mixing two
A
isomeric compounds
B
chiral compounds
C
meso compounds
D
enantiomers with chiral carbon.
Answer :
enantiomers with chiral carbon.
A mixture of equal amount of two enantiomers is called a racemic mixture. A racemic mixture does not rotate plane - polarized light. They are optically inactive because for every molecule in a racemic mixture that rotate plane of
polarized light in one direction, there is a mirror image molecule that rotates the plane in oppsite direction.
7.
Out of the following, the alkene that exhibits optical
isomerism is
A
3 - methyl - 2 - pentene
B
4 - methyl - 1 - pentene
C
3 - methyl - 1 - pentene
D
2 - methyl - 2 - pentene
Answer :
3 - methyl - 1 - pentene
For a compound to show optical isomerism, presence of chiral carbon atom is a necessary condition.
8.
Following types of compounds ( as I, II )
are studied in terms of isomerism in :
A
chain isomerism
B
position isomerism
C
conformers
D
stereoisomerism
Answer :
stereoisomerism
TIPS/Formulae :
Stereoisomerism is of two types i.e., geometrical isomerism and optical isomerism
Both the structures shows sterioisomerism. Structure I shows geometrical isomerism as it contains two different atoms $$\left( H \right)$$ and groups $$\left( {C{H_3}} \right)$$ attached to each carbon
Structure II shows optical isomerism as it contains a chiral carbon (attached to four different group) atom.
9.
Maleic acid and fumaric acids are
A
Chain isomers
B
Functional isomers
C
Tautomers
D
Geometrical isomers
Answer :
Geometrical isomers
Maleic acid and fumaric acids are geometrical isomers.
10.
The structure shows :
A
geometrical isomersism
B
optical isomerism
C
geometrical & optical isomerism
D
tautomerism.
Answer :
optical isomerism
NOTE :
A compound which consists of at least one asymmetric carbon atom is capable of showing the phenomenon of optical isomerism.
The structure cannot show geometrical isomerism as one of the carbons along the double bond has identical group (methyl). Tautomerism is not possible because of the absence of $$ - CO,$$ group. It shows optical isomerism because it has chiral $$C$$ atom with four different groups, $$H,C{H_3},COOH\,{\text{and}}\,{\left( {C{H_3}} \right)_2}C = CH.$$