JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 38

Answer: (b) Tetrahedral

Explanation:
1. Carbon in \( \ce{CH4} \) is sp3 hybridised with four bonding pairs
2. The geometry is tetrahedral with H-C-H angles of 109.5 degrees

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) Tetrahedron

Explanation:
1. Carbon in \( \ce{CCl4} \) is sp3 hybridised
2. The four bonds point to the corners of a regular tetrahedron

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) sp2

Explanation:
1. Each ring carbon forms three sigma bonds (two C-C, one C-H)
2. The remaining p orbital forms the delocalised pi cloud, so the carbon is sp2

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) sp

Explanation:
1. s-character is 25% in sp3, 33% in sp2 and 50% in sp
2. So sp is the highest

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) sp

Explanation:
1. Each carbon of \( \ce{HC#CH} \) forms two sigma bonds
2. Two hybrid orbitals mean sp hybridisation

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) sp2 and sp3

Explanation:
1. The two doubly bonded carbons are sp2
2. The remaining carbons (\( \ce{CH3} \) and \( \ce{CH2} \)) are 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: (c) \( \ce{(CH3)3COH} \)

Explanation:
1. In \( \ce{(CH3)3C-OH} \) every carbon has four single bonds
2. In the others the carbonyl 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: (a) \( \ce{CH3-C#C-H} \)

Explanation:
1. In propyne the \( \ce{CH3} \) carbon (sp3) is joined to the sp carbon
2. The C-C bond is therefore sp-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: (c) sp and sp2

Explanation:
1. The nitrile carbon is sp hybridised
2. The next carbon is part of \( \ce{CH=CH2} \) and 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) 4

Explanation:
1. Each double bond has one pi bond (2 in all)
2. The triple bond has two pi bonds, giving 4

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) 10

Explanation:
1. Naphthalene has five C=C bonds drawn in its Kekule structure
2. Each carries two pi electrons, so 10

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) sp2 hybridised

Explanation:
1. Splitting \( \ce{CH3-CH3} \) evenly gives two methyl radicals
2. The methyl radical is planar, so the carbon is sp2 hybridised

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) 1 and 2 sp2; 3 and 4 sp3

Explanation:
1. The carbons of the C=C double bond in \( \ce{BrCH=CHBr} \) are sp2
2. After addition of \( \ce{H2} \) the product has only single bonds, so both carbons are 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: (d) (ii)

Explanation:
1. (i) has only sp3 carbons, (iii) only sp2 and (iv) only sp
2. Only (ii) contains both sp2 (C=C) and sp3 (\( \ce{CH3} \)) carbons

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) Tetrahedral

Explanation:
1. The carbon in \( \ce{CH2Cl2} \) is sp3 hybridised
2. Its shape is (slightly distorted) tetrahedral

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) It has a carbon which is not bonded to hydrogen

Explanation:
1. Isopentane is \( \ce{(CH3)2CH-CH2-CH3} \): three \( \ce{CH3} \), one \( \ce{CH2} \) and one \( \ce{CH} \)
2. Every carbon is bonded to at least one hydrogen, so (d) is false

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) 18

Explanation:
1. o-Xylene is \( \ce{C8H10} \) with 6 ring C-C, 2 ring-methyl C-C and 10 C-H bonds
2. The total is 18 sigma 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) 12

Explanation:
1. Benzene has 6 C-C sigma bonds and 6 C-H sigma bonds
2. The total is 12

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) 4

Explanation:
1. Two carbons of the C=C are sp2
2. The other four ring carbons are sp3

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) 3

Explanation:
1. The double bond has one pi bond and the triple bond has two
2. Total 3

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) sp and sp2 hybridised

Explanation:
1. The two carbons of the central C=C are sp2
2. The four carbons of the cyano groups are 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: (c) 3 sigma and 2 pi bonds

Explanation:
1. \( \ce{HC#CH} \) has one C-C sigma bond and two C-H sigma bonds
2. The triple bond also has 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) 6

Explanation:
1. Each sp2 carbon has one unhybridised p orbital (2 in all)
2. Each sp carbon has two (4 in all), so the total is 6

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. The greater the s-character of the carbon orbital, the shorter and stronger the C-H bond
2. Ethyne (sp, 50% s) is the strongest

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) All 6 sp2

Explanation:
1. Every carbon of the benzene ring has three sigma bonds
2. All six carbons are sp2 hybridised

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) only sp2 hybridised carbon atoms

Explanation:
1. All four carbons of \( \ce{CH2=CH-CH=CH2} \) are doubly bonded
2. Each has three sigma bonds and 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.