JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 47

Answer: (d) N-Deutero-N-phenylmethanamide

Explanation:
1. The group HC(=O)–N is an amide, so the parent is the one-carbon amide methanamide.
2. The substituents on nitrogen are a phenyl ring and a deuterium atom, so the locant N is used for both.
3. Name: N-deutero-N-phenylmethanamide.

Teacher's Note:
1. An amide is named from the acyl part (methan-) and the suffix -amide; groups on nitrogen get the locant N.
2. Isotopes such as D are cited like substituents in alphabetical order.

Answer: (a) sp²–sp³

Explanation:
1. C1 and C2 form the double bond, so C2 is sp².
2. C3 is a CH₂ group with four single bonds, so it is sp³.
3. The C2–C3 bond is therefore an sp²–sp³ sigma bond.

Teacher's Note:
1. Count the sigma bonds and lone pairs on a carbon: 4 means sp³, 3 means sp², 2 means sp.
2. Carbon atoms of a double bond are always sp².

Answer: (b) 3-Ethyl-2,4-dimethylhexane

Explanation:
1. Two six-carbon chains are possible. One through the isopropyl group gives 2,4-dimethyl with an ethyl on C3 (three substituents).
2. The other gives only two substituents (isopropyl and methyl).
3. The chain with the larger number of substituents is chosen, so the parent is hexane carrying methyl groups at 2 and 4 and ethyl at 3.
4. Alphabetical order puts ethyl first: 3-ethyl-2,4-dimethylhexane.

Teacher's Note:
1. When two chains have equal length, choose the one with more substituents.
2. Prefixes are listed alphabetically, ignoring di-, tri-.

Answer: (b) HC≡C–CH₂–CH₂–Br : 1-bromobut-3-yne

Explanation:
1. In option (b) the multiple bond must get the lowest locant, so the chain is numbered from the alkyne end: 4-bromobut-1-yne.
2. The name 1-bromobut-3-yne gives the triple bond the higher locant, so it is wrong.
3. Options (a), (c) and (d) follow lowest locants correctly (substituent set 2,2,4 beats 2,4,4).

Teacher's Note:
1. Rule order: principal group, then multiple bonds, then substituents for lowest locants.
2. Always test both numbering directions.

Answer: (b) Hex-1-en-5-yne drawn with the numbering started at the alkyne end: yne carbons 1,2 and ene carbons 5,6

Explanation:
1. For chains with both double and triple bonds, first make the set of locants for all multiple bonds as low as possible.
2. If that set is tied, the double bond gets the lower locant.
3. In option (b) the locant set is {1,5} either way, but the numbering starts at the alkyne end, so the alkyne gets 1 instead of the alkene. It is incorrect.
4. In (c) and (d) the locant set {1,3,6} is lower than the alternative {1,4,6}, so both numberings are correct.

Teacher's Note:
1. Ene gets preference over yne only when the locant sets are tied.
2. Write both numberings and compare locant sets before deciding.

Answer: (d) 3-Bromo-1-chlorocyclohex-1-ene

Explanation:
1. The double bond must be C1=C2, and the ring is numbered so that the substituents get the lowest locants.
2. The chlorine-bearing carbon is C1 (an alkene carbon) and the bromine-bearing carbon is then C3.
3. Name: 3-bromo-1-chlorocyclohex-1-ene (bromo is cited before chloro alphabetically).

Teacher's Note:
1. In cycloalkenes the double bond always takes locants 1 and 2.
2. Among equal choices the first substituent in alphabetical order gets the lower number, but only after the ring is fixed by the double bond and the lowest-locant set.

Answer: (b) 2,4-Dimethylpentan-3-one

Explanation:
1. A ketone is named with the suffix -one, never with the prefix oxo when it is the principal group, so options (a) and (c) are wrongly formed.
2. In 3,3-dimethylpentan-2-one both methyl groups sit on the same carbon, which breaks the given condition.
3. In 2,4-dimethylpentan-3-one the methyl groups are on C2 and C4, which are different carbons.

Teacher's Note:
1. Check each name against every condition in the stem.
2. The carbonyl carbon gets the lowest possible locant, but here the symmetric structure puts it at C3.

Answer: (c) 3-Chloro-5-hydroxybenzenecarbonyl chloride

Explanation:
1. The acid chloride group –COCl is the principal group, so the parent is benzenecarbonyl chloride with that carbon at C1.
2. The other substituents are at C3 and C5, either way round, so the first cited alphabetically (chloro) takes the lower locant 3.
3. Name: 3-chloro-5-hydroxybenzenecarbonyl chloride.

Teacher's Note:
1. Acid derivatives outrank alcohols in priority.
2. When locant sets tie, alphabetical order decides.

Answer: (b) 5-Amino-2-bromocyclohexane-1-carboxylic acid

Explanation:
1. The carboxylic acid is the principal group, so its carbon is C1 of the ring.
2. Number towards the bromine so that the locant set is {1,2,5} rather than {1,3,6}.
3. Alphabetical order: amino before bromo, giving 5-amino-2-bromocyclohexane-1-carboxylic acid.

Teacher's Note:
1. Compare locant sets only after fixing the principal group at C1.
2. Option (b) has the correct locants but is not in alphabetical order.

Answer: (a) N-Ethyl-N-methylethanamine

Explanation:
1. The longest alkyl group on nitrogen (ethyl) gives the parent ethanamine.
2. The other groups on nitrogen are ethyl and methyl, both given the locant N.
3. Alphabetical order: N-ethyl-N-methylethanamine.

Teacher's Note:
1. Choose the longest carbon chain on N as the parent amine.
2. Substituents on nitrogen use the italic N locant.

Answer: (c) Ethyl 2-bromopropanoate

Explanation:
1. The alkyl part on oxygen is ethyl and the acid part is a three-carbon chain including the carbonyl carbon.
2. The bromine is on C2 (the carbon next to C=O), so the acid part is 2-bromopropanoate.
3. Name: ethyl 2-bromopropanoate.

Teacher's Note:
1. Name an ester as alkyl alkanoate; the alkyl group is a separate word.
2. Number the acid chain starting from the carbonyl carbon.

Answer: (a) Metamers

Explanation:
1. Ethyl benzenecarboxylate is C₆H₅COOC₂H₅; phenyl propanoate is C₂H₅COOC₆H₅.
2. Both are C₉H₁₀O₂ esters with the same functional group.
3. The alkyl groups on the two sides of the ester group are exchanged, which is metamerism.

Teacher's Note:
1. Metamers have the same functional group and formula but different alkyl groups on either side of the polyvalent atom or group.
2. Functional isomers must have different functional groups.

Answer: (c) 4-Methoxy-2-nitrobenzaldehyde

Explanation:
1. The aldehyde is the principal group, so the ring carbon bearing CHO is C1 (benzaldehyde).
2. The nitro group is ortho (C2) and the methoxy group is para (C4).
3. Alphabetical order: methoxy before nitro, giving 4-methoxy-2-nitrobenzaldehyde.

Teacher's Note:
1. Benzaldehyde is an accepted parent name, and its CHO carbon is C1.
2. Choose the direction giving the lower locant set {2,4}, not {4,6}.

Answer: (b) sp² and sp³

Explanation:
1. The smallest ester is methyl methanoate, HCOOCH₃.
2. The carbonyl carbon forms a C=O double bond and is sp².
3. The methyl carbon has four single bonds and is sp³.

Teacher's Note:
1. Every ester contains one sp² carbonyl carbon.
2. The smallest ester has just one other carbon.

Answer: (d) 3

Explanation:
1. The parent chain is heptane with one extra carbon as a methyl branch.
2. A methyl group can be placed on C2, C3 or C4 (C5 and C6 repeat these by symmetry).
3. Isomers: 2-methylheptane, 3-methylheptane and 4-methylheptane, so 3.

Teacher's Note:
1. An ethyl branch would make the longest chain 8 carbons, so only methyl is possible here.
2. Use the chain symmetry to avoid counting the same isomer twice.

Answer: (d) 6

Explanation:
1. Place three methyl groups on a cyclohexane ring and remove duplicates by symmetry.
2. The distinct substitution patterns are 1,1,2; 1,1,3; 1,1,4; 1,2,3; 1,2,4 and 1,3,5.
3. Total: 6 (stereoisomers are not counted).

Teacher's Note:
1. Fix one methyl at C1 and move the others systematically.
2. Check each pattern against its mirror numbering.

Answer: (c) 4

Explanation:
1. A primary amine is R–NH₂ with R = C₄H₉.
2. There are four butyl groups: n-butyl, sec-butyl, isobutyl and tert-butyl.
3. So the four amines are butan-1-amine, butan-2-amine, 2-methylpropan-1-amine and 2-methylpropan-2-amine.

Teacher's Note:
1. The number of primary amines equals the number of alkyl groups of that carbon count.
2. C₄H₉ has four isomeric alkyl groups.

Answer: (a) 6

Explanation:
1. Ethers have the form R–O–R'. Total carbon atoms in R and R' is five.
2. Methyl ethers: methyl with n-butyl, sec-butyl, isobutyl or tert-butyl gives 4.
3. Ethyl ethers: ethyl with n-propyl or isopropyl gives 2. Total = 6.

Teacher's Note:
1. Split five carbons as 1+4 and 2+3.
2. Count the isomeric alkyl groups for each partner.

Answer: (b) 1

Explanation:
1. A tertiary alcohol has the OH on a carbon carrying three other carbon atoms.
2. With only five carbons the only possibility is a C₄ skeleton with an extra methyl at that carbon: 2-methylbutan-2-ol.
3. So the number is 1.

Teacher's Note:
1. Write the tertiary carbon first: C(OH)(C)(C)(C).
2. The remaining carbon is added to one of the methyl groups.

Answer: (c) CH₃CH(OH)CH(CH₃)CH₃ : 2-methyl-3-butanol

Explanation:
1. Compound (c) has an OH group, so numbering starts from the end nearer to OH: the OH is on C2 and the methyl on C3.
2. Correct name: 3-methylbutan-2-ol. The given name numbers from the wrong end.
3. The other three names follow the rules (the old style of writing locants before the parent is accepted).

Teacher's Note:
1. The principal group gets the lowest locant before any substituent.
2. Always try numbering from both ends.

Answer: (c) Structure (c)

Explanation:
1. Picric acid is a phenol with three nitro groups at the two ortho positions and the para position.
2. Picric acid is 2,4,6-trinitrophenol, which is structure (c).
3. The other options are benzoic acid derivatives with a single substituent at C3.

Teacher's Note:
1. Picric acid is a strong organic acid because three nitro groups stabilise the phenoxide ion.
2. Phenol is the parent, so the OH carbon is C1.

Answer: (d) carboxylic acid

Explanation:
1. A saturated open-chain compound with one C=O double bond has one degree of unsaturation, giving C_nH_2nO₂ for a carboxylic acid (or an ester).
2. Diketones and dialdehydes have two C=O groups: C_nH_2n-2O₂.
3. Saturated diols are C_nH_2n+2O₂.

Teacher's Note:
1. Each double bond or ring removes two hydrogens from the alkane formula.
2. Esters share the formula of the carboxylic acids.

Answer: (c) 3,3-Dimethyl-1-cyclohexanol

Explanation:
1. The alcohol is the principal group, so the carbon with OH is C1.
2. Counting to the carbon with two methyl groups gives C3 by the shorter route.
3. Name: 3,3-dimethylcyclohexan-1-ol (written as 3,3-dimethyl-1-cyclohexanol in older style, option c).

Teacher's Note:
1. The suffix -ol is used for alcohols, not a hydroxy prefix.
2. Hydroxy is a prefix only when a higher priority group is present.

Answer: (d) Acetonitrile

Explanation:
1. Acetone, acetic acid and acetamide each have a carbonyl carbon (C=O), which is sp².
2. Acetonitrile CH₃–C≡N has an sp³ methyl carbon and an sp nitrile carbon.
3. So acetonitrile has no sp² carbon.

Teacher's Note:
1. A triple bond means sp hybridisation.
2. Carbonyl carbons are always sp².

Answer: (d) –COOH, –SO₃H, –CONH₂, –CHO

Explanation:
1. The seniority order of these classes is carboxylic acid, then sulfonic acid, then amide, then aldehyde.
2. Only option (d) lists them in this order.

Teacher's Note:
1. Remember: acids, anhydrides, esters, acid halides, amides, nitriles, aldehydes, ketones, alcohols, amines.
2. Sulfonic acids rank just below carboxylic acids.