JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 21

Answer: (d) sp

Explanation:
1. The carbanionic carbon is part of a triple bond and has one sigma bond to carbon, so it is sp hybridised.
2. The lone pair occupies the remaining sp hybrid orbital.

Teacher's Note:
1. A triple-bonded carbon is sp hybridised.
2. The lone pair sits in an sp orbital, which has 50 percent s-character.

Answer: (a) 2-Amino-4-ethyl-5-methylhexan-1-ol

Explanation:
1. The main chain must contain the OH group and be as long as possible: HOCH2-CH(NH2)-CH2-CH(C2H5)-CH(CH3)-CH3 gives six carbons.
2. OH on C1 means amino on C2, ethyl on C4 and methyl on C5.
3. Substituents in alphabetical order: 2-amino-4-ethyl-5-methylhexan-1-ol.

Teacher's Note:
1. Number from the OH end.
2. Write the prefixes in alphabetical order.

Answer: (b) \( \ce{CH2=CH-C#CH} \)

Explanation:
1. In \( \ce{CH2=CH-C#CH} \) the first two carbons carry a double bond (sp²), and the last two carry a triple bond (sp).
2. The other options give sp,sp,sp,sp (a), sp² all (c) or sp³,sp²,sp²,sp³ (d).

Teacher's Note:
1. Write the hybridisation of each carbon in turn.
2. Match the pattern in order.

Answer: (b) 4-Bromo-2-phenylpent-2-ene

Explanation:
1. The longest chain containing the double bond has five carbons.
2. Numbering from the end that gives the double bond the lower locant puts it at C2, with phenyl on C2 and Br on C4.
3. The name is 4-bromo-2-phenylpent-2-ene.

Teacher's Note:
1. Number so that the double bond gets the lowest locant.
2. List the prefixes in alphabetical order.

Answer: (b) 5-Hydroxycyclohex-3-enone

Explanation:
1. The ketone is the principal group, so the C=O carbon is C1.
2. Numbering goes towards the double bond, which then lies at C3=C4, and the OH group is on C5.
3. The name is 5-hydroxycyclohex-3-en-1-one.

Teacher's Note:
1. In ring ketones the carbonyl carbon is C1.
2. Then give the double bond the lowest locant.

Answer: (b) 2-Formyl-5-methylhept-3-ene-1-nitrile

Explanation:
1. The nitrile has higher priority than the aldehyde, so the nitrile carbon is C1 and the CHO becomes a formyl prefix.
2. The longest chain is seven carbons: C1 (CN), C2 (CH-CHO), C3=C4, C5 (CH-CH3), C6, C7.
3. The name is 2-formyl-5-methylhept-3-enenitrile.

Teacher's Note:
1. The order of priority is nitrile, then aldehyde.
2. Count the nitrile carbon in the chain.

Answer: (d) 5, 13

Explanation:
1. Pi bonds: three in the benzene ring and two in the C#N triple bond, so 5.
2. Sigma bonds: six C-C ring bonds, five C-H bonds, one ring C to nitrile C, and one C-N sigma bond, so 13.

Teacher's Note:
1. Each double bond has one pi bond and a triple bond has two.
2. Remember the ring C-H bonds.

Answer: (c) Planar

Explanation:
1. All four carbons are sp² hybridised and trigonal planar, and the conjugated system keeps all atoms in one plane.
2. So the molecule is planar.

Teacher's Note:
1. sp² carbons give planar geometry.
2. Conjugation favours planarity.

Answer: (c) 3-Ethyl-2,3-dimethylhexane

Explanation:
1. The longest chain is six carbons, running through the propyl group, the central carbon and the isopropyl group.
2. This chain has three substituents (methyl on C2, and methyl and ethyl on C3), more than the other six-carbon chain.
3. Lowest locants give 3-ethyl-2,3-dimethylhexane, with prefixes in alphabetical order.

Teacher's Note:
1. Choose the longest chain and, when tied, the one with more substituents.
2. Cite the prefixes alphabetically.

Answer: (d) 2-Amino-2-ethyl-4-oxopentanoic acid

Explanation:
1. The chain contains COOH as C1 and has five carbons: C1 (COOH), C2 (C bearing NH2 and ethyl), C3 (CH2), C4 (C=O), C5 (CH3).
2. Prefixes in alphabetical order: amino, ethyl, oxo.
3. The name is 2-amino-2-ethyl-4-oxopentanoic acid.

Teacher's Note:
1. Carboxylic acid is the principal group and ketone becomes oxo.
2. Alphabetical order of prefixes: amino, ethyl, oxo.

Answer: (c) 1-Amino-2-buten-2-ol

Explanation:
1. The alcohol is the principal group, so the name ends with ol and the amino group is a prefix.
2. Numbering from the CH2NH2 end gives OH on C2 and the double bond at C2, while the other end gives OH on C3.
3. The lowest locant for OH gives 1-aminobut-2-en-2-ol.

Teacher's Note:
1. The principal group takes the lowest locant.
2. Amino is only a prefix when OH is present.

Answer: (b) 3-Chloro-2-butanol

Explanation:
1. The alcohol is the principal group and gets the lowest locant, 2.
2. Numbering from the other end, Cl moves to C3.
3. The name is 3-chlorobutan-2-ol (3-chloro-2-butanol).

Teacher's Note:
1. Principal group first for the lowest locant.
2. Halogen is only a prefix.

Answer: (c) 3-Vinyl-2-hepten-6-ynal

Explanation:
1. The aldehyde carbon is C1. The longest chain through the CHO carbon is seven carbons, containing both the double bond (C2=C3) and the triple bond (C6#C7).
2. The vinyl group is a substituent on C3.
3. The name is 3-ethenylhept-2-en-6-ynal (3-vinyl-2-hepten-6-ynal).

Teacher's Note:
1. Choose the longest chain that contains the principal group and multiple bonds.
2. Double bond gets the lower locant than the triple bond when there is a choice.

Answer: (a) Ethyl-2-oxocyclopentane carboxylate

Explanation:
1. The ester is the principal group, so the compound is named as an ester of cyclopentanecarboxylic acid, with the alkyl group (ethyl) first.
2. The ketone is the prefix oxo, on C2 next to C1 bearing the ester group.
3. The name is ethyl 2-oxocyclopentane-1-carboxylate.

Teacher's Note:
1. Ester outranks ketone, which becomes an oxo prefix.
2. The ring carbon carrying the ester group is C1.

Answer: (a) 3,4-Dimethylpent-1-en-3-ol

Explanation:
1. The alcohol is the principal group, and the chain should contain the OH carbon and the double bond: C1=C2 (vinyl), C3 (OH, CH3), C4 (CH with CH3), C5.
2. Numbering from the double bond end gives OH locant 3, against 3 from the other end. The double bond then gets 1 instead of 4.
3. The name is 3,4-dimethylpent-1-en-3-ol.

Teacher's Note:
1. When the OH locant ties, the double bond decides the numbering.
2. List the methyl groups with their locants.

Answer: (a) 3,4-Dihydroxybenzoic acid

Explanation:
1. The carboxylic acid group is the principal group, so the ring carbon bearing COOH is C1.
2. The OH groups are on the carbons that are meta and para to it, C3 and C4.
3. The name is 3,4-dihydroxybenzoic acid.

Teacher's Note:
1. Carboxylic acid outranks alcohol and phenol.
2. The ring carbon carrying the principal group is C1.

Answer: (c) \( \ce{CH3-CH(OH)-CH(CH3)-CH3} \) : 2-Methyl-3-butanol

Explanation:
1. In \( \ce{CH3-CH(OH)-CH(CH3)-CH3} \) the OH is the principal group and must get the lowest locant.
2. The chain is numbered from the CH3 next to the CHOH carbon, so OH is on C2 and methyl on C3: 3-methylbutan-2-ol.
3. The name given in (c) is wrong. The other names are correct.

Teacher's Note:
1. The alcohol gets the lowest locant, not the substituent.
2. Check the principal group first.

Answer: (b) \( \ce{CH3-C(Br)(Cl)-COO-CH2CH3} \)

Explanation:
1. Propanoate has a three-carbon acid part, with C1 as the carbonyl carbon.
2. Both Br and Cl are on C2, which is next to the carbonyl carbon.
3. The ethyl group is on the oxygen of the ester, giving \( \ce{CH3-C(Br)(Cl)-COOC2H5} \).

Teacher's Note:
1. Build the acid part first, then add the alkyl group on oxygen.
2. Both halogens sit on C2.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. The ketone is at C3 and the two alkoxy groups are at C2 and C4 either way.
2. The tie is broken by alphabetical order, so ethoxy gets the lower locant 2, and the assertion is true.
3. The reason is false, since the size of a substituent does not decide the locants.

Teacher's Note:
1. When locants tie, the prefix first in alphabetical order gets the lower number.
2. Size of a group has no role.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Both numberings give the locant set {1, 4}.
2. The double bond gets the lower number, so the correct name is pent-1-en-4-yne, and the assertion is false.
3. The reason is also false, because the rule favours the double bond.

Teacher's Note:
1. On a tie, the double bond is numbered first.
2. Do not give the preference to the triple bond.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. All twelve hydrogens of neopentane are equivalent, so only one monosubstituted product forms, and the assertion is true.
2. Neopentane has two types of carbon (four primary and one quaternary), so the reason is false.

Teacher's Note:
1. Equivalent hydrogens give one substitution product.
2. Primary and quaternary carbons are different types.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Acetic acid is a saturated carboxylic acid, so the assertion is false.
2. It has only one double bond (C=O), so the reason is false.
3. Both are false.

Teacher's Note:
1. Acetic acid has a single C=O only.
2. It is a saturated carboxylic acid.