JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 39

Answer: (c) Tertiary (3 degree) C-H bond

Explanation:
1. Breaking a C-H bond gives an alkyl radical
2. The tertiary radical is the most stable, so the tertiary C-H bond is the weakest

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) 7 sigma, 3 pi

Explanation:
1. Sigma bonds: 3 C-C and 4 C-H = 7
2. Pi bonds: 1 from the double bond and 2 from the triple bond = 3

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) One sigma bond and two pi bonds

Explanation:
1. The head-on overlap of two sp orbitals gives one sigma bond
2. Two sideways p-p overlaps give two pi 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) 15 sigma and 3 pi bonds

Explanation:
1. Toluene \( \ce{C6H5CH3} \) has 6 ring C-C, 1 ring-methyl C-C and 8 C-H bonds = 15 sigma
2. The three ring double bonds give 3 pi bonds

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) Carbon tetrachloride

Explanation:
1. Only a molecule with four identical atoms on a central sp3 carbon has perfectly equal angles
2. \( \ce{CCl4} \) is perfectly tetrahedral, while the others are distorted by unlike substituents

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) sp2

Explanation:
1. sp2 uses one s and two p orbitals in a plane at 120 degrees
2. This is the trigonal planar arrangement

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) sp, sp2 and sp3

Explanation:
1. The two end carbons of the allene unit (C-1 and C-3) are sp2 and the central C-2 is sp
2. The \( \ce{CH3} \) carbon is sp3

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) Buta-1,3-diene

Explanation:
1. Conjugation needs alternating single and double bonds
2. Buta-1,3-diene fits; buta-1,2-diene has cumulated bonds and penta-1,4-diene has an sp3 carbon between its double 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: (a) 109 degrees 28 minutes

Explanation:
1. Methane is a regular tetrahedron with sp3 carbon
2. The angle is 109 degrees 28 minutes

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) sp-sp2

Explanation:
1. The carbon of the triple bond that is joined to the next carbon is sp
2. The next carbon is part of the C=C and is sp2, so the bond is sp-sp2

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) Trigonal planar

Explanation:
1. Each carbon is sp2 hybridised with three sigma bonds at about 120 degrees
2. The whole molecule is planar

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) sp2

Explanation:
1. Each carbon of graphite is bonded to three others in a layer
2. The fourth electron is delocalised, so carbon is sp2

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) 5

Explanation:
1. Naphthalene (\( \ce{C10H8} \)) has five C=C bonds in its Kekule structure
2. Each has one pi bond, so there are 5

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) sp3

Explanation:
1. Every carbon of diamond is bonded to four others in a tetrahedral network
2. This is sp3 hybridisation

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) 10 and 3

Explanation:
1. Sigma bonds: 4 C-C and 6 C-H = 10
2. Pi bonds: 2 from the triple and 1 from the double bond = 3

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) Graphite is in planar form while diamond is in tetrahedral form

Explanation:
1. Graphite consists of planar layers held by weak forces, which slide over each other
2. Diamond has a rigid three-dimensional tetrahedral network

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) sp2, sp

Explanation:
1. The end carbon is doubly bonded and is sp2
2. The central carbon has two double bonds and is sp

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) \( \ce{C2H5OH} \)

Explanation:
1. Only ethanol has an O-H bond to act as hydrogen-bond donor and acceptor
2. The others can only accept hydrogen 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) 5

Explanation:
1. The heptane chain has two end \( \ce{CH3} \) groups
2. Two methyl substituents and one \( \ce{CH3} \) of the ethyl group make 5

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) Acetonitrile

Explanation:
1. In \( \ce{CH3CN} \) the carbons are sp3 and sp
2. Acetic acid, acetone and acetamide each have a C=O carbon, which is sp2

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) \( \ce{CH3OH} \)

Explanation:
1. The order of acidity is \( \ce{CH3OH} \) (pKa about 15.5) > \( \ce{HC#CH} \) (25) > \( \ce{C6H6} \) (43) > \( \ce{C2H6} \) (50)
2. Methanol is the strongest acid

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) Ethyne

Explanation:
1. Ethyne is \( \ce{HC#CH} \)
2. Ethene has a double bond, and butane and glycerine only single bonds

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) 15

Explanation:
1. 6 ring C-C, 1 ring-methyl C-C and 8 C-H bonds
2. Total 15

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) 12 sigma and 3 pi

Explanation:
1. 6 C-C and 6 C-H sigma bonds make 12
2. The Kekule structure has 3 pi bonds

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) 9 sigma bonds, 1 pi bond and 2 lone pairs

Explanation:
1. Sigma bonds: 2 C-C, 5 C-H, 1 C-O and 1 O-H = 9
2. One pi bond (C=C) and two lone pairs on oxygen

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) trans-2-Butene

Explanation:
1. In trans-2-butene the two C-CH3 bond moments lie on opposite sides and cancel
2. The others have a net dipole because the moments do not cancel

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.