JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 09

Answer: (d) a, b, d

Explanation:
1. A secondary radical has its free valence on a carbon joined to two other carbons.
2. (a) sec-butyl, (b) isopropenyl (the carbon is joined to CH₂= and CH₃) and (d) isopropyl all qualify.
3. (c) vinyl has its free valence on a carbon joined to only one other carbon, so it is primary.

Teacher's Note:
1. Count the carbon neighbours of the radical carbon, counting a double bond partner once.
2. Vinyl is the standard example of a primary unsaturated radical.

Answer: (c) A is a tertiary alcohol while B is a primary amine

Explanation:
1. In A the OH is on a carbon joined to three other carbons, so A is a tertiary alcohol (alcohols are classified by the carbon).
2. In B the nitrogen carries only one alkyl group, so B is a primary amine (amines are classified by the nitrogen).
3. The tertiary butyl group is the same in both.

Teacher's Note:
1. Alcohols follow the carbon, amines follow the nitrogen.
2. tert-Butylamine is a primary amine.

Answer: (b) (CH₃)₃C– : neobutyl

Explanation:
1. (CH₃)₃C–CH₂– is neopentyl, HC≡C–CH₂– is propargyl and CH₂=CH–CH₂– is allyl.
2. (CH₃)₃C– is tertiary butyl (tert-butyl), and there is no group called neobutyl.

Teacher's Note:
1. Memorise the common radicals: isopropyl, isobutyl, sec-butyl, tert-butyl, neopentyl, allyl, vinyl, propargyl.
2. The prefix neo is used for the group with a quaternary carbon next to the free valence carbon.

Answer: (c) 1-Chloro-2-methylpropane

Explanation:
1. Isobutyl chloride is (CH₃)₂CH–CH₂Cl.
2. The longest chain is propane with the Cl on C1 and a methyl on C2.
3. The name is 1-chloro-2-methylpropane.

Teacher's Note:
1. Isobutyl is the group (CH₃)₂CHCH₂–.
2. Number from the end that gives the lowest locants, here the carbon bearing the chlorine.

Answer: (c) 2-Ethyl-2-butenoic acid

Explanation:
1. The principal functional group is –COOH, so the chain must include it as C1.
2. C1 = COOH, C2 = the carbon with the ethyl group and the double bond, C3 = CH, C4 = CH₃.
3. This gives 2-ethylbut-2-enoic acid.

Teacher's Note:
1. The acid carbon is always C1 and the suffix is -oic acid.
2. Locate the double bond and name the ethyl branch.

Answer: (c) 3-Ethyl-2,5-dimethylhexane

Explanation:
1. Taking the longest chain that includes the isopropyl group: CH₃–CH(CH₃)–CH(C₂H₅)–CH₂–CH(CH₃)–CH₃, six carbons.
2. Substituents: methyl at C2 and C5, ethyl at C3.
3. Name: 3-ethyl-2,5-dimethylhexane (the chain with the greater number of substituents is chosen when two chains have equal length).

Teacher's Note:
1. Several six-carbon chains exist; pick the one with the most substituents.
2. Number to give the lowest set of locants, {2,3,5}.

Answer: (d) Ethanamide, ethanoyl chloride

Explanation:
1. –CONH₂ on a two-carbon chain is named by replacing the -e of the alkane with -amide: ethanamide.
2. –COCl is an acid halide: replace -ic acid by -oyl and add the halide, ethanoyl chloride.

Teacher's Note:
1. Both groups are derivatives of carboxylic acids; the carbonyl carbon is C1.
2. Acetamide and acetyl chloride are the common names.

Answer: (b) 5

Explanation:
1. The principal chain must contain the principal functional group, –COOH, whose carbon is C1.
2. C2 is the quaternary carbon. From C2 the longest chain continues through the propyl group: C3, C4, C5.
3. The ethyl branch would give only 4 carbons.
4. The principal chain has 5 carbons (pentanoic acid with 2-ethyl-2-formyl).

Teacher's Note:
1. A carboxylic acid group outranks an aldehyde group, so CHO becomes a prefix (formyl).
2. The acid carbon is included in the count of the chain.

Answer: (c) 4,4-Dimethylpent-2-yne

Explanation:
1. The longest chain containing the triple bond has five carbons: CH₃–C≡C–C(CH₃)₂–CH₃.
2. Numbering from the end nearer the triple bond puts it at C2.
3. Two methyl groups are on C4: 4,4-dimethylpent-2-yne.

Teacher's Note:
1. Give the multiple bond the lowest locant.
2. The other three options are common or incorrect names.

Answer: (b) 4-Amino-3-methoxybut-2-enamide

Explanation:
1. The principal group is the amide, –CONH₂ (C1).
2. The chain C1 (CONH₂), C2 (=CH), C3 (C–OCH₃), C4 (CH₂NH₂) has the double bond at C2.
3. Substituents: 3-methoxy and 4-amino, so 4-amino-3-methoxybut-2-enamide.

Teacher's Note:
1. Amide outranks amine, ether and alkene in the priority list.
2. The other nitrogen becomes the prefix amino.

Answer: (c) N-Methylethanamine

Explanation:
1. Treat the longer alkyl group as the parent amine, ethanamine, and the smaller group on nitrogen as a substituent.
2. The methyl group on N gets the locant N.
3. Name: N-methylethanamine.

Teacher's Note:
1. The longer chain forms the parent name.
2. Locant N shows a substituent on the nitrogen atom.

Answer: (b) Cyclohexanecarbonyl chloride

Explanation:
1. When –COCl is attached to a ring, the carbonyl carbon is not part of the ring.
2. The suffix is -carbonyl chloride added to the ring name.
3. Name: cyclohexanecarbonyl chloride.

Teacher's Note:
1. Use -carbonyl chloride for a ring and -oyl chloride for a chain.
2. The same pattern gives cyclohexanecarboxylic acid and cyclohexanecarbaldehyde.

Answer: (b) 3-Methyl-3-penten-1-yne

Explanation:
1. The chain has five carbons and contains both multiple bonds.
2. Numbering from the triple bond end gives the locants {1,3} (yne at 1, ene at 3); from the other end they would be {2,4}.
3. So the name is 3-methylpent-3-en-1-yne.

Teacher's Note:
1. Multiple bonds together get the lowest locants.
2. In names with both, -ene is written before -yne.

Answer: (d) 3-Ethyl-2,5,5-trimethylheptane

Explanation:
1. Two seven-carbon chains are possible: one runs through the isopropyl group's carbon and one through the ethyl group.
2. The chain that goes through the isopropyl CH carries four substituents (2-methyl, 3-ethyl and 5,5-dimethyl), the other only three.
3. The chain with more substituents is chosen, and numbering gives the lowest locants {2,3,5,5}.
4. Name: 3-ethyl-2,5,5-trimethylheptane.

Teacher's Note:
1. When two chains are of equal length, choose the one with the greater number of side chains.
2. Then number to get the lowest locants at the first point of difference.

Answer: (b) 1-Chloro-3-ethyl-2-methylcyclopentane

Explanation:
1. Number the ring so the substituents get the lowest locants: {1,2,3} in either direction.
2. At the tie, the first-cited substituent in alphabetical order (chloro, ethyl, methyl) gets the lowest locant.
3. So chloro is C1, methyl is C2, ethyl is C3: 1-chloro-3-ethyl-2-methylcyclopentane.

Teacher's Note:
1. Alphabetical order, not size, decides ties in numbering.
2. Substituents are cited in alphabetical order: chloro, ethyl, methyl.

Answer: (b) 5-Ethyl-3-methylcyclohex-1-ene

Explanation:
1. The double bond carbons are C1 and C2 (the double bond gets the lowest locants).
2. Of the two directions, number so that the substituents get the lowest set: C1 is the lower left carbon, C2 the upper left, C3 the top carbon (methyl), C4 upper right, C5 lower right (ethyl).
3. This gives 5-ethyl-3-methylcyclohex-1-ene.

Teacher's Note:
1. In cycloalkenes the double bond always occupies C1 and C2.
2. Then choose the direction that gives the lowest substituent locants {3,5}.

Answer: (b) 2-Oxoethanenitrile

Explanation:
1. The nitrile (–CN) has higher priority than the aldehyde, so the compound is named as a nitrile.
2. The chain has two carbons (the CN carbon is C1) and the CHO group is the 2-oxo substituent.
3. Name: 2-oxoethanenitrile.

Teacher's Note:
1. Priority order: acid > anhydride > ester > acid halide > amide > nitrile > aldehyde > ketone.
2. The higher-priority group takes the suffix.

Answer: (c) Pent-4-en-2-one

Explanation:
1. The ketone is the principal group; the chain has five carbons.
2. Numbering from the end that gives the carbonyl the lowest locant puts C=O at C2 and the double bond at C4.
3. Name: pent-4-en-2-one.

Teacher's Note:
1. The principal group takes the lowest locant, even over a double bond.
2. The suffix -one replaces the final -e.

Answer: (a) 3-Pentyne

Explanation:
1. Pentyne must be numbered from the end that gives the triple bond the lowest locant, so the correct name is pent-2-yne, not 3-pentyne.
2. The other three names follow the rules.

Teacher's Note:
1. Always number from the end nearest to the multiple bond.
2. A locant above half the chain length is a warning sign.

Answer: (c) 4-Methoxyaniline

Explanation:
1. The ring carries –NH₂ and –OCH₃ para to each other.
2. The amino group gives the parent name aniline (benzenamine) and the methoxy group is a substituent at C4.
3. Name: 4-methoxyaniline.

Teacher's Note:
1. Amine has higher priority than ether.
2. Benzenamine and aniline are accepted parent names.

Answer: (d) 3-Hydroxymethylpentane-1,2,5-triol

Explanation:
1. Choose the longest chain containing all three OH groups: HOCH₂–CH(OH)–CH(CH₂OH)–CH₂–CH₂OH, five carbons.
2. Numbering from the end that gives {1,2,5} (not {1,4,5}) to the OH groups.
3. The CH₂OH branch is at C3: 3-(hydroxymethyl)pentane-1,2,5-triol.

Teacher's Note:
1. The chain must carry the maximum number of principal groups, then be the longest.
2. Lowest locants go to the suffix group.

Answer: (c) Propane-1,2,3-tricarbonitrile

Explanation:
1. The three –CN groups are on a three-carbon chain, one on each carbon.
2. When the CN carbons are not counted in the chain, the suffix -carbonitrile is used.
3. Name: propane-1,2,3-tricarbonitrile.

Teacher's Note:
1. Three identical nitrile groups on a chain are named with -tricarbonitrile.
2. The nitrile carbon is outside the chain in this style.