JEE Chemistry Isomerism MCQs Set 06

Answer: (a) Tetrahedral nature of carbon

Explanation:
1. The four bonds of an sp3 carbon point to the corners of a tetrahedron
2. With four different groups, the arrangement is non-superimposable on its mirror image

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (c) E/Z isomerism

Explanation:
1. \( \ce{ClCH=CH-NO2} \) has two different groups on each doubly bonded carbon
2. So E and Z forms exist

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (b) \( \ce{CH3CHBrCl} \)

Explanation:
1. \( \ce{CH3-CHBrCl} \) has a carbon with H, Br, Cl and \( \ce{CH3} \)
2. The others have two identical atoms on the carbon

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (c) \( \ce{(CH3)3C-Cl} \)

Explanation:
1. SN1 rate depends on carbocation stability
2. The tertiary chloride gives the most stable carbocation

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (d) One

Explanation:
1. The products are 1-chloro-2,3-dimethylbutane (chiral, one enantiomeric pair) and 2-chloro-2,3-dimethylbutane (achiral)
2. So there is one pair

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (a) \( \ce{CH3CH2-C(CH3)(OH)-CH2CH3} \)

Explanation:
1. Tertiary alcohols dehydrate most easily because they form the most stable carbocation
2. (a) is a tertiary alcohol, (b) secondary and (c), (d) primary

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (c) \( \ce{p-O2N-C6H4-CH2^-} \)

Explanation:
1. A benzyl carbanion is resonance stabilised, and the nitro group delocalises the charge further by -M and -I effects
2. The methyl group in (d) is electron donating and destabilises it

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (b) o-Nitrophenol

Explanation:
1. The phenoxide ion of o-nitrophenol is stabilised by the -M and -I effects of the nitro group
2. Phenol and o-cresol are weaker, and the alcohol in (a) is much weaker still

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (d) \( \ce{CH2=CH2 + Br2 -> CH2BrCH2Br} \)

Explanation:
1. The pi electrons of ethene attack the electrophile \( \ce{Br^+} \) end of \( \ce{Br2} \)
2. (a) is radical substitution, and (b) and (c) are nucleophilic additions

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (c) 8

Explanation:
1. The number of optical isomers is 2^n for n dissimilar stereocentres
2. For n = 3 this gives 8

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (a) 2

Explanation:
1. Butane and 2-methylpropane
2. Two isomers

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (b) 3

Explanation:
1. The isomers are \( \ce{CH3-CBrCl2} \), \( \ce{CH2Br-CHCl2} \) and \( \ce{CH2Cl-CHClBr} \)
2. Three

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (c) Optical rotation and is derived from D(+)-glyceraldehyde

Explanation:
1. The sign (+) shows dextrorotation, which is measured experimentally
2. The prefix D shows that its configuration is related to D(+)-glyceraldehyde, the reference compound

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (b) 2-Butene

Explanation:
1. Geometrical isomerism needs two different groups on each doubly bonded carbon
2. Only 2-butene meets this

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (b) 2

Explanation:
1. The two carbons bearing OH each have H, OH, a different carbon group on each side
2. Both are asymmetric, so the answer is 2

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (a) 2-Butene and 1-phenylpropene

Explanation:
1. 2-Butene and 1-phenylpropene have two different groups on each doubly bonded carbon
2. Propene and 2-methyl-2-butene have two identical groups on one carbon

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (c) 2^n

Explanation:
1. Each dissimilar stereocentre can have two configurations
2. The total is 2^n

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (b) 5

Explanation:
1. Hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane
2. Five isomers

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (a) 2

Explanation:
1. 1-Chloropropane and 2-chloropropane
2. Two isomers

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. All twelve hydrogens of \( \ce{C(CH3)4} \) are equivalent, so one monosubstituted product forms
2. Neopentane is indeed an isomer of pentane, but that fact does not explain the single product

Teacher's Note:
1. Enantiomers differ only in optical rotation; diastereomers differ in all physical properties.
2. Meso compounds have stereocentres but are optically inactive because of internal symmetry.

Answer: (c) A is true, but R is false.

Explanation:
1. Anti-addition of bromine to the trans alkene gives the meso dibromide, so the assertion is true
2. The addition is anti (through a bromonium ion), not syn, so the reason is false

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Cyclohexanone has alpha hydrogens and forms the enol by hydrogen shift
2. The reason names the two tautomeric forms, which is exactly what keto-enol tautomerism is

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The trans isomer is usually more stable because the similar groups are far apart and steric strain is less
2. The reason is a correct definition, but it does not explain the stability

Teacher's Note:
1. Look for a carbon with four different groups to spot chirality, and for two different groups on each doubly bonded carbon to spot geometrical isomerism.
2. Check that the isomers have the same molecular formula before naming the type.

Answer: (d) A is false, but R is true.

Explanation:
1. In allene (\( \ce{CH2=C=CH2} \)) each end carbon has two identical H atoms, so the molecule is achiral and optically inactive
2. The molecule does have planes of symmetry, so the reason is true

Teacher's Note:
1. Write out the structures of all isomers systematically (skeleton first, then position of groups) when counting.
2. Remember that the number of stereoisomers is 2^n only when the stereocentres are all different.