JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 68

Answer: (b) Hex-5-en-1-yn-3-ol

Explanation:
1. The compound is CH₂=CH–CH₂–CH(OH)–C≡CH. The chain is six carbons long and the suffix is the alcohol, so the OH must get the lowest locant.
2. Numbering from the alkyne end puts OH at C-3 (from the alkene end it would be C-4).
3. Name: hex-5-en-1-yn-3-ol (ene is cited before yne, ol last).
4. Answer: (b)

Teacher's Note:
1. The alcohol suffix gets the lowest locant first. Between double and triple bond, the double bond is named first (en before yn) but only gets the lower number when there is still a choice.

Answer: (d) 3-Hydroxybutane-1,2,4-tricarboxylic acid

Explanation:
1. The structure is HOOC–CH₂–CH(OH)–CH(COOH)–CH₂–COOH.
2. Taking the four-carbon chain with three COOH groups as carboxylic acid substituents, number from the end that gives the lowest locants to the COOH groups: 1,2,4 (not 1,3,4).
3. This places OH on C-3.
4. Name: 3-hydroxybutane-1,2,4-tricarboxylic acid.
5. Answer: (d)

Teacher's Note:
1. When the COOH group cannot all be part of one chain, the extra ones are named as carboxylic acid. Lowest locants go to the principal group first.

Answer: (c) 6-Chloro-4-ethyl-3-methylcyclohex-1-ene

Explanation:
1. In a substituted cycloalkene the double bond carbons are C-1 and C-2.
2. The ring has the double bond on one edge, CH₃ next to it, then C₂H₅, a CH₂, and Cl on the carbon next to C-1.
3. Numbering gives Cl = 6, C₂H₅ = 4, CH₃ = 3.
4. Name: 6-chloro-4-ethyl-3-methylcyclohex-1-ene.
5. Answer: (c)

Teacher's Note:
1. Fix the double bond at C-1 and C-2 first, then choose the direction that gives the lowest set of locants to the substituents.

Answer: (d) Acrylonitrile : But-2-enenitrile

Explanation:
1. Crotonaldehyde is CH₃CH=CHCHO, i.e. but-2-enal, and cinnamic acid is C₆H₅CH=CHCOOH, i.e. 3-phenylprop-2-enoic acid.
2. Lactic acid is CH₃CH(OH)COOH, i.e. 2-hydroxypropanoic acid.
3. Acrylonitrile is CH₂=CH–CN, which has three carbons: prop-2-enenitrile, not but-2-enenitrile.
4. Answer: (d)

Teacher's Note:
1. Learn the common names with their formulas. Acrylonitrile is the three-carbon nitrile used to make polyacrylonitrile.

Answer: (c) 1,1,1,2,3,3,3-Heptachloro-2-(trichloromethyl)propane

Explanation:
1. The central carbon carries one Cl and three CCl₃ groups.
2. The longest chain is three carbons (C-1 and C-3 are CCl₃ carbons, C-2 is the central carbon).
3. Substituents: Cl at 1,1,1,2,3,3,3 (seven in all) and one trichloromethyl group at C-2.
4. Name: 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane.
5. Answer: (c)

Teacher's Note:
1. Count every halogen on the chosen chain separately; the group left outside the chain is named as a substituent in brackets.

Answer: (d) Both (b) and (c)

Explanation:
1. Aniline is an accepted IUPAC name, so 4-bromoaniline is correct.
2. The systematic name benzenamine also gives 4-bromobenzenamine.
3. Both (b) and (c) are acceptable; the p- form is a common-system prefix.
4. Answer: (d)

Teacher's Note:
1. Aniline is a retained IUPAC name. Always compare the systematic and the retained form when a question asks for 'the' IUPAC name.

Answer: (b) 1-Phenylpropan-1-one

Explanation:
1. The compound is a ketone: a phenyl group and an ethyl group joined to C=O.
2. Chain: C-1 is the carbonyl carbon, then CH₂CH₃, so propan-1-one with phenyl on C-1.
3. Name: 1-phenylpropan-1-one (propiophenone).
4. Answer: (b)

Teacher's Note:
1. Ketones are named as alkanones; an aromatic ring becomes a phenyl prefix at the carbonyl carbon.

Answer: (b) CH₃CH₂CH(CH₃)COOH : isovaleric acid

Explanation:
1. In the first compound Br is on the beta carbon of butyraldehyde, so the name is correct.
2. CH₃CH₂CH(CH₃)COOH is 2-methylbutanoic acid (alpha-methylbutyric acid). Isovaleric acid is (CH₃)₂CHCH₂COOH, 3-methylbutanoic acid.
3. o-Toluidine and allylamine are correct.
4. Answer: (b)

Teacher's Note:
1. Isovaleric acid has the isopropyl group at the end of the chain. Do not confuse it with 2-methylbutanoic acid.

Answer: (c) 3-Methylenehex-1-en-5-yne

Explanation:
1. The compound is CH₂=CH–C(=CH₂)–CH₂–C≡CH.
2. The longest chain containing both multiple bonds has six carbons: CH₂=CH–C–CH₂–C≡CH, with a =CH₂ (methylene) on C-3.
3. Locants {1,5} are the same in both directions, so the double bond gets the lower number: hex-1-en-5-yne.
4. Name: 3-methylenehex-1-en-5-yne.
5. Answer: (c)

Teacher's Note:
1. Choose the chain with the maximum number of multiple bonds; when locants tie, the double bond takes the lower number.

Answer: (a) Chloromethyl 3-methyl-2-oxobutanoate

Explanation:
1. The compound is (CH₃)₂CH–CO–COO–CH₂Cl.
2. The acid part is a four-carbon acid: C-1 (COO), C-2 (C=O, the oxo group), C-3 (CH with a methyl).
3. The alkyl part is chloromethyl.
4. Name: chloromethyl 3-methyl-2-oxobutanoate.
5. Answer: (a)

Teacher's Note:
1. In an ester the alkyl group (here chloromethyl) is named first as a separate word. Number the acid chain from the carboxyl carbon.

Answer: (c) Hex-4-en-1-yne

Explanation:
1. The chain has six carbons with one double and one triple bond.
2. Locants: from the triple-bond end the set is {1,4}; from the other end it is {2,5}. The lowest set is {1,4}.
3. Cite ene before yne: hex-4-en-1-yne.
4. Answer: (c)

Teacher's Note:
1. Apply the IUPAC rules in order: principal group, longest chain, lowest locants, then alphabetical order of prefixes.

Answer: (b) Pent-2-en-4-yne

Explanation:
1. In 'pent-2-en-4-yne' the numbering is wrong: from the other end the locants are {1,3} (pent-3-en-1-yne), which is lower than {2,4}.
2. The other three names follow the rules (alphabetical prefixes, lowest locants, principal group suffix).
3. Answer: (b)

Teacher's Note:
1. Give the lowest set of locants to double and triple bonds taken together; only then break a tie in favour of the double bond.

Answer: (b) 5

Explanation:
1. The principal group is COOH, so the parent chain must include the COOH carbon (C-1).
2. Chain: C-1 (COOH), C-2 (=C), C-3 (=C), then through the ethyl group on C-3: CH₂ and CH₃. That gives five carbons.
3. The chain through CHO or through the ethyl on C-2 is not longer.
4. Answer: (b)

Teacher's Note:
1. The principal chain must carry the principal functional group; among such chains pick the longest one.

Answer: (b) 1,3-Dimethylbutyl

Explanation:
1. Number the substituent chain from the carbon with the free valency as C-1.
2. The longest chain from C-1 is four carbons: C-1 (CH, with a methyl), C-2 (CH₂), C-3 (CH with a methyl), C-4 (CH₃).
3. Name: 1,3-dimethylbutyl.
4. Answer: (b)

Teacher's Note:
1. In an alkyl substituent the carbon carrying the free valency is always C-1.

Answer: (c) 1,2-Epoxypropane

Explanation:
1. The ring is an epoxide (oxirane) formed from propene.
2. In substitutive nomenclature of this type the epoxy bridge is a prefix on the hydrocarbon chain: 1,2-epoxypropane.
3. 'Propylene oxide' is only the common name.
4. Answer: (c)

Teacher's Note:
1. Epoxy is a bridging prefix; the two carbons joined by oxygen get the locants.

Answer: (d) 5-Methylcyclopent-2-en-1-ol

Explanation:
1. The OH-bearing carbon is C-1 (principal group).
2. Numbering must then give the double bond the lowest locants: it falls at C-2=C-3.
3. The methyl group is on the other neighbour of C-1, so it is at C-5.
4. Name: 5-methylcyclopent-2-en-1-ol.
5. Answer: (d)

Teacher's Note:
1. In a cyclic alcohol the OH carbon is C-1; the double bond then takes the lowest locants.

Answer: (b) 2,3-Dioxopropanoic acid

Explanation:
1. The carboxylic acid is the principal group, so the chain is propanoic acid (C-1 = COOH).
2. C-2 is a ketone C=O and C-3 is the CHO carbon, both named as oxo prefixes.
3. Name: 2,3-dioxopropanoic acid.
4. Answer: (b)

Teacher's Note:
1. When an acid is present, the aldehyde and ketone groups become oxo prefixes on the chain.

Answer: (a) 5

Explanation:
1. Cyclic C₅H₁₀ has one ring.
2. Ring of five: cyclopentane (1). Ring of four with CH₃: methylcyclobutane (1). Ring of three with two carbons added: ethylcyclopropane, 1,1-dimethylcyclopropane and 1,2-dimethylcyclopropane (3).
3. Total structural isomers = 5 (the cis/trans forms of 1,2-dimethylcyclopropane are stereoisomers, not structural).
4. Answer: (a)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (a) Only the enol (II)

Explanation:
1. Tautomers are interconvertible by shifting a proton and a double bond.
2. Cyclohexanone has an alpha-hydrogen, so it forms the enol cyclohex-1-en-1-ol (II).
3. Cyclohexanol (III) differs in the oxidation level and cannot be formed by a proton shift alone.
4. Answer: (a)

Teacher's Note:
1. Keto-enol tautomerism needs an alpha-hydrogen. The enol is the tautomer, not the saturated alcohol.

Answer: (a) CH₃–CHO (I)

Explanation:
1. In simple aldehydes and ketones the keto form is far more stable than the enol form.
2. The C=O bond is stronger than C=C, so the equilibrium for CH₃CHO lies almost entirely on the keto side.
3. Answer: (a)

Teacher's Note:
1. The keto form dominates unless the enol is stabilised by conjugation or hydrogen bonding, as in acetylacetone.

Answer: (d) II > III > I

Explanation:
1. The keto form (II) is the most stable.
2. Of the two enols, III is the more substituted alkene (CH₃–C(OH)=CH–CH₃), so it is more stable than the terminal enol I.
3. Order: II > III > I.
4. Answer: (d)

Teacher's Note:
1. A more substituted C=C is more stable by hyperconjugation, so it also decides the order among enols.

Answer: (c) Configurational isomers

Explanation:
1. Geometrical isomers (cis-trans) differ in the arrangement of groups around a double bond or ring, and they cannot be interconverted without breaking a bond.
2. Such isomers are configurational isomers (and also diastereomers).
3. Answer: (c)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (b) CH₂=CH–CH=CH–CH₃

Explanation:
1. Geometrical isomerism needs both carbons of a double bond to carry two different groups.
2. In CH₃–CH=CH–CH=CH₂ the double bond between C-2 and C-3 has CH₃ and H on one carbon, and H and vinyl on the other, so cis and trans forms exist.
3. Every terminal =CH₂ carbon has two identical H atoms, so the other options show no geometrical isomerism.
4. Answer: (b)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (c) Both (a) and (b)

Explanation:
1. Both CH₃ groups are on the same side of C=C, and the higher-priority groups (CH₃) are together.
2. Hence it is cis-2-butene, which is the same as (Z)-2-butene.
3. R and S describe chirality, not a double bond.
4. Answer: (c)

Teacher's Note:
1. Z means the higher-priority groups are on the same side; for 2-butene cis and Z coincide.

Answer: (b) Anti (staggered)

Explanation:
1. In the anti staggered conformation the two CH₃ groups are 180° apart.
2. This minimises both torsional and steric strain, so it has the lowest energy.
3. Answer: (b)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (b) CH₂(OH)CH₂OH

Explanation:
1. Conformers arise from rotation about a C–C single bond.
2. CD₃–H and CH₄ have only one carbon, so there is no C–C bond.
3. Ethylene glycol has a C–C bond and shows gauche and anti forms.
4. Answer: (b)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (a) 4

Explanation:
1. Write the aldehydes C₄H₉CHO: C₄H₉ can be n-butyl, isobutyl, sec-butyl or tert-butyl.
2. They give pentanal, 3-methylbutanal, 2-methylbutanal and 2,2-dimethylpropanal.
3. Total = 4 structures.
4. Answer: (a)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.

Answer: (a) Positional isomers

Explanation:
1. Both have the same carbon skeleton (a five-carbon chain including the CHO carbon) and the same groups (CN and CHO).
2. Only the position of the CN group changes: C-3 in the first and C-4 in the second.
3. They are positional isomers.
4. Answer: (a)

Teacher's Note:
1. First draw the possible structures systematically (chain, position, functional group), then count each distinct one only once.