JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 26

Answer: (b) 2,3-Dimethylbutane

Explanation:
1. Longest continuous chain has 4 carbons, so the parent is butane.
2. Two methyl groups sit on carbons 2 and 3 (numbering from either end gives the same locants).
3. Name: 2,3-dimethylbutane.

Teacher's Note:
1. Count the longest chain first, then list every branch with its locant.
2. Molecular formula C6H14: the name must describe six carbons in total (4 + 2).

Answer: (a) 3-Ethyl-2-methylpentane

Explanation:
1. Two five-carbon chains are possible: one through the isopropyl methyl group and one through the ethyl branch.
2. When chains are equally long, choose the one with more substituents. The first chain carries two (methyl and ethyl); the second carries only one (isopropyl).
3. Number to give the lowest locants: methyl on C2, ethyl on C3. Substituents are listed alphabetically (ethyl before methyl): 3-ethyl-2-methylpentane.

Teacher's Note:
1. Equal chain length: the more substituted chain is the parent.
2. Prefixes are written in alphabetical order, not in order of locant.

Answer: (d) the greater number of substituents

Explanation:
1. Rule 1: choose the longest chain.
2. Rule 2: if several chains have equal length, choose the chain carrying the maximum number of substituents.
3. Only if the number of side chains is also equal do we compare locants.

Teacher's Note:
1. The order of rules is: length, then number of substituents, then lowest locants.
2. Branch size and atomic mass are never used to select the parent chain.

Answer: (c) 2-Bromo-4-chloropentane

Explanation:
1. The parent chain is pentane. Numbering from either end gives the locant set {2,4}.
2. When the locants are the same from both ends, the substituent cited first alphabetically gets the lower number.
3. Bromo precedes chloro alphabetically, so Br is at C2 and Cl at C4: 2-bromo-4-chloropentane.

Teacher's Note:
1. Tie in locant sets is broken by alphabetical order of the prefixes.
2. Prefixes are always written alphabetically, so bromo comes before chloro.

Answer: (b) 4-Bromo-2,2-dichloropentane

Explanation:
1. Numbering from the CCl2 end gives locants {2,2,4}; from the other end {2,4,4}.
2. At the first point of difference 2 is lower than 4, so the CCl2 end is carbon 1.
3. Name (alphabetical, 'di' ignored): 4-bromo-2,2-dichloropentane.

Teacher's Note:
1. Compare locant sets term by term; the first difference decides.
2. 'Di' and 'tri' are not counted when arranging prefixes alphabetically.

Answer: (a) 5

Explanation:
1. A primary carbon is attached to only one other carbon.
2. The primary carbons are the five methyl groups: C1 and C5 of the pentane chain, two methyl groups on C2 and one methyl group on C3.
3. So there are 5 primary carbons (and 1 secondary, 1 tertiary, 1 quaternary).

Teacher's Note:
1. Every CH3 group of an alkane is a 1° carbon.
2. Check: 5 + 1 + 1 + 1 = 8 carbons, matching C8H18.

Answer: (d) 5

Explanation:
1. Use DU = (x + 1) − (y − w)/2 for CxHyNw.
2. Here x = 9, y = 12, w = 2, so DU = 10 − (12 − 2)/2.
3. DU = 10 − 5 = 5.

Teacher's Note:
1. Nitrogen counts with a negative sign in the hydrogen term; halogens count with a positive sign; oxygen and sulphur do not matter.
2. Each DU is one double bond or one ring.

Answer: (c) 8

Explanation:
1. Oxygen and sulphur (divalent) are ignored.
2. Formula: x = 10, y = 8, halogens s = 2, nitrogen w = 4.
3. DU = (10 + 1) − (8 + 2 − 4)/2 = 11 − 3 = 8.

Teacher's Note:
1. Replace each halogen by H and delete each divalent atom; add one H for each N removed.
2. Ignoring divalent atoms is the usual trap in such formulas.

Answer: (b) 0

Explanation:
1. Oxygen is ignored; chlorine counts like hydrogen.
2. DU = (5 + 1) − (11 + 1)/2 = 6 − 6 = 0.
3. A DU of zero means the compound is saturated and acyclic.

Teacher's Note:
1. DU 0 means no double bond, no triple bond and no ring.
2. Check with the alkane formula C5H12: replace one H by Cl and insert O into a C–H or C–C bond.

Answer: (a) 10

Explanation:
1. x = 6, y = 0, w = 6.
2. DU = (6 + 1) − (0 − 6)/2 = 7 + 3 = 10.

Teacher's Note:
1. Nitrogen subtracts from the hydrogen count, so with no hydrogen the term becomes positive.
2. DU can exceed the number of carbon atoms when many nitrogens are present.

Answer: (d) isobutyl

Explanation:
1. It is obtained by removing a hydrogen from the CH3 end of 2-methylpropane (isobutane).
2. The free valency is on a CH2 group attached to a CH(CH3)2 group, which is the isobutyl group.

Teacher's Note:
1. Isobutyl: the free valency is on a primary carbon of isobutane.
2. tert-Butyl comes from the tertiary hydrogen of isobutane instead.

Answer: (c) sec-butyl

Explanation:
1. Remove one hydrogen from the second carbon of n-butane.
2. The valency is on a secondary carbon, giving sec-butyl (1-methylpropyl).

Teacher's Note:
1. sec- indicates a secondary carbon bearing the free valency.
2. n-Butyl has the valency at the chain end instead.

Answer: (b) tert (as in tert-butyl)

Explanation:
1. Iso and neo are treated as part of the name and are considered.
2. The multiplying prefixes di, tri, tetra are ignored for simple groups but are considered when they are part of a complex substituent name.
3. The italicised prefixes sec- and tert- are ignored.

Teacher's Note:
1. Ignored: di, tri, tetra (simple groups), sec, tert.
2. Considered: iso, neo, and di/tri inside a complex radical name.

Answer: (a) isohexane

Explanation:
1. Iso is used when two methyl groups are at the end of an unbranched chain and there is no other substituent.
2. 2-Methylpentane has six carbons, so its retained name is isohexane.
3. Neohexane is 2,2-dimethylbutane; isopentane is 2-methylbutane.

Teacher's Note:
1. The retained 'iso' name applies to 4 to 6 carbon alkanes of the type (CH3)2CH–R.
2. Count the total carbons to select the correct prefix.

Answer: (d) 2,2,4-Trimethylpentane

Explanation:
1. Iso-octane has the structure \( \ce{CH3-C(CH3)2-CH2-CH(CH3)-CH3} \).
2. Longest chain: pentane. Methyl groups at 2,2 and 4 (the alternative numbering gives 2,4,4, which is higher).

Teacher's Note:
1. Choose the numbering giving the lower first point of difference: {2,2,4} beats {2,4,4}.
2. The formula is C8H18.

Answer: (c) 2,2-Dimethylpropane

Explanation:
1. Neopentane is \( \ce{C(CH3)4} \), a central carbon bonded to four methyl groups.
2. The longest chain is propane (3 carbons) with two methyl groups on C2: 2,2-dimethylpropane.

Teacher's Note:
1. Neo means 2,2-dimethyl at the end of the chain.
2. C5H12 has three isomers: pentane, 2-methylbutane (isopentane) and 2,2-dimethylpropane (neopentane).

Answer: (b) prop-1-enyl

Explanation:
1. Number the carbon bearing the free valency as C1.
2. The double bond starts at C1, giving prop-1-enyl.

Teacher's Note:
1. The valency carbon is always numbered 1 in a substituent name.
2. Prop-2-enyl is allyl, \( \ce{CH2=CH-CH2-} \).

Answer: (a) 1-methylethenyl

Explanation:
1. The valency carbon is C1 of a two-carbon chain (ethenyl) that carries a methyl group at C1.
2. Hence 1-methylethenyl.

Teacher's Note:
1. The longest chain starting at the valency carbon and containing the double bond is only two carbons here.
2. Prop-1-enyl and prop-2-enyl are unbranched three-carbon groups.

Answer: (d) \( C_nH_{2n-3} \)

Explanation:
1. Removing one H atom from \( C_nH_{2n-2} \) leaves \( C_nH_{2n-3} \).
2. Alkane gives alkyl \( C_nH_{2n+1} \), alkene gives alkenyl \( C_nH_{2n-1} \), alkyne gives alkynyl \( C_nH_{2n-3} \).

Teacher's Note:
1. Subtract one hydrogen from the parent formula.
2. Memorise the three series: 2n+1, 2n−1, 2n−3.

Answer: (c) benzyl

Explanation:
1. Phenyl is \( \ce{C6H5-} \); benzyl has an additional CH2 between the ring and the free valency.
2. Benzal is \( \ce{C6H5-CH=} \) and tolyl is \( \ce{CH3-C6H4-} \).

Teacher's Note:
1. Benzyl: Ph–CH2–; benzal: Ph–CH=; benzo (benzotrichloride type): Ph–C≡.
2. Tolyl has a methyl on the ring (o-, m-, p-).

Answer: (b) Hexanedioic acid

Explanation:
1. Both COOH carbons are part of the parent chain, so the chain has six carbons.
2. With the suffix 'dioic acid' the name is hexanedioic acid (adipic acid).

Teacher's Note:
1. The 'e' of the primary suffix is retained before a consonant such as d in 'dioic'.
2. 'Carboxylic acid' as a suffix would exclude the COOH carbons from the chain.

Answer: (a) 3-Methylbutanenitrile

Explanation:
1. The nitrile carbon is part of the chain and is C1.
2. Chain: C1 (CN), C2 (CH2), C3 (CH), C4 (CH3) so butane with a methyl on C3.
3. Name: 3-methylbutanenitrile.

Teacher's Note:
1. Nitrile is named as alkanenitrile; the CN carbon is counted in the parent chain.
2. The 'e' of alkane is retained before the consonant n of nitrile.

Answer: (d) N-Methylethanamide

Explanation:
1. The acyl part has two carbons (ethanoyl), so the parent is ethanamide.
2. The methyl on nitrogen is shown with the locant N: N-methylethanamide.

Teacher's Note:
1. Substituents on the nitrogen of an amide are given the locant N.
2. Identify the acyl chain by the carbonyl carbon: CH3–CO– means ethanamide.

Answer: (c) N-Ethylpropan-1-amine

Explanation:
1. The longer carbon group on nitrogen is propyl (3 carbons), so the parent is propan-1-amine.
2. The ethyl group on nitrogen is shown with locant N.

Teacher's Note:
1. Choose the longer chain on N as the parent amine.
2. Secondary amines are named as N-alkylalkanamines.

Answer: (b) 3-Oxobutanal

Explanation:
1. The aldehyde is senior to the ketone, so the chain is numbered from the CHO carbon (C1).
2. The ketone carbonyl at C3 is expressed as a prefix, oxo.
3. Name: 3-oxobutanal.

Teacher's Note:
1. Seniority: aldehyde > ketone.
2. The senior group takes the suffix; the junior one becomes a prefix (oxo).