JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 42

Answer: (d) 2, 3

Explanation:
1. SN2 is bimolecular and its rate depends on both substrate and nucleophile, so (1) is false
2. Backside attack (2) and a concerted bond change (3) are correct

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (b) IV > II > I > III

Explanation:
1. Toluene is activated, benzene is the reference, chlorobenzene is mildly deactivated and the anilinium ion is strongly deactivated
2. Order: IV > II > I > III

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Answer: (d) Free radicals

Explanation:
1. Light or heat splits \( \ce{Br2} \) into bromine atoms
2. The reaction proceeds by a free radical chain mechanism

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (a) Chlorobenzene

Explanation:
1. The C-Cl bond of chlorobenzene has partial double-bond character (resonance) and the carbon is sp2
2. So it does not react under ordinary conditions, unlike the alkyl chlorides

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

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

Explanation:
1. The +I effect increases with the number of alkyl groups on the carbon
2. tert-Butyl has the most

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Answer: (c) I > II > III

Explanation:
1. The methoxy group is activating, so anisole is more reactive than benzene
2. The nitro group is strongly deactivating, so nitrobenzene is the least reactive

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (c) \( \ce{C6H5CH2^+} \)

Explanation:
1. The benzyl cation is stabilised by delocalisation of the charge into the benzene ring
2. The alkyl cations only have inductive and hyperconjugative stabilisation, and methyl has the least

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Answer: (c) SN2

Explanation:
1. Inversion shows backside attack by \( \ce{OH^-} \) on the carbon while \( \ce{Br^-} \) leaves
2. This single-step bimolecular nucleophilic substitution is SN2

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (b) \( \ce{(CH3)3C^+} \)

Explanation:
1. The tertiary cation has the most hyperconjugation and +I stabilisation
2. Stability: 3 degrees > 2 degrees > 1 degree > methyl

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Answer: (a) -COOH

Explanation:
1. A meta-directing group is electron-withdrawing and deactivating
2. -COOH is meta-directing, while -NH2, -OH and -Cl are ortho/para-directing

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Answer: (d) Benzoic acid > phenol > n-propylbenzene

Explanation:
1. Phenol (strongly activated) is more reactive than benzoic acid (deactivated)
2. So the order in (d) is wrong, while (a), (b) and (c) are correct orders

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (b) \( \ce{p-CH3O-C6H4-CH2^+} \)

Explanation:
1. The methoxy group donates electrons by +M to the carbocation through the ring
2. The nitro and chloro groups withdraw electrons and destabilise it

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Answer: (c) 3 degree > 2 degree > 1 degree

Explanation:
1. The +I effect increases with the number of carbon atoms attached to the carbon bearing the group
2. So tertiary > secondary > primary

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (a) \( \ce{CH3NH2} \), \( \ce{RO^-} \) and \( \ce{CH3MgBr} \)

Explanation:
1. Nucleophiles have a lone pair or a carbanion-like carbon to donate
2. \( \ce{CH3NH2} \), \( \ce{RO^-} \) and \( \ce{CH3MgBr} \) (a carbanion source) donate, but \( \ce{AlCl3} \) is an electron-pair acceptor, an electrophile

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Answer: (c) \( \ce{CH3^-} \)

Explanation:
1. Across the second period nucleophilicity (and basicity) falls as electronegativity rises
2. Carbon is the least electronegative, so \( \ce{CH3^-} \) is the strongest

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Answer: (c) \( \ce{CH3CN} \)

Explanation:
1. The nitrogen lone pair of \( \ce{CH3CN} \) is nucleophilic
2. The nitrile carbon is electron-poor (C#N is polarised) and is attacked by nucleophiles

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Answer: (c) sp3 < sp2 < sp

Explanation:
1. A carbanion is more stable when the lone pair is in an orbital with more s-character (closer to the nucleus)
2. So sp3 < sp2 < sp

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Answer: (b) \( \ce{C2H2} \) < \( \ce{C2H4} \) < \( \ce{C6H6} \) < \( \ce{C2H6} \)

Explanation:
1. Bond lengths: C-C in ethane 1.54, benzene 1.39, ethene 1.34 and ethyne 1.20 angstrom
2. So the order is \( \ce{C2H2} \) < \( \ce{C2H4} \) < \( \ce{C6H6} \) < \( \ce{C2H6} \)

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Answer: (d) 120 degrees

Explanation:
1. Equal bond lengths between single and double values indicate a delocalised, aromatic ring such as benzene
2. Each carbon is sp2, so the C-C-C angle is 120 degrees

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

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

Explanation:
1. Aniline has a lone pair on nitrogen to donate
2. In the anilium ion this pair is used to hold a proton, so no lone pair is left and the positive charge makes it electron-poor

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

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

Explanation:
1. Water replaces bromide, giving benzyl alcohol
2. This is nucleophilic substitution, which explains the formation of the alcohol

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

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

Explanation:
1. Catenation is the ability of carbon to link with itself in chains and rings
2. Forming multiple bonds is true of carbon but is not the reason for catenation, which comes from strong C-C single bonds

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

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

Explanation:
1. Olefins (alkenes) have the formula \( \ce{CnH2n} \), so the assertion is false
2. The reason is a correct description of an alkene

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

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

Explanation:
1. m-Dinitrobenzene dissolves in ether but glucose, being highly polar, does not
2. Glucose is indeed water soluble, but that is not the reason it stays out of the ether

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

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

Explanation:
1. Both statements are true
2. Polarity of the bond helps heterolysis, but the assertion is the definition of the process, so the reason is not an explanation of it

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

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

Explanation:
1. Resonance gives each C-O bond a bond order of 4/3, so all three are equal
2. The reason is true but it is the resonance, not the trend of bond length with multiplicity, that explains the equality

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.