JEE Chemistry Organic Chemistry Mixed Topics MCQs Set 01

Answer: (a) 2-(cyclohexylidene)cyclohexan-1-one

Explanation:
1. Cyclohexanone has alpha hydrogens, so in base one molecule forms an enolate that attacks the carbonyl of another.
2. This gives the aldol 2-(1-hydroxycyclohexyl)cyclohexanone.
3. On heating this beta-hydroxy ketone loses water to give the conjugated enone, 2-(cyclohexylidene)cyclohexan-1-one.

Teacher's Note:
1. Aldol addition gives a beta-hydroxy carbonyl compound; heat removes water to form an alpha,beta-unsaturated carbonyl compound.
2. The option with hydroxyl is the intermediate, not the final product.

Answer: (d) cyclohexanone

Explanation:
1. Hydrogenation with Pd/C at 1 atm is mild and reduces the C=C bond of the conjugated alkene easily.
2. The carbonyl group needs more forcing conditions, so it is left unchanged.
3. The product is the saturated ketone, cyclohexanone.

Teacher's Note:
1. Pd/C with hydrogen at low pressure reduces C=C faster than C=O.
2. To reduce the carbonyl one would use \(\ce{NaBH4}\) or \(\ce{LiAlH4}\).

Answer: (c) Primary aliphatic amines

Explanation:
1. Phthalimide is converted to potassium phthalimide, which reacts with an alkyl halide, and hydrolysis releases the primary amine.
2. The reaction goes by an SN2 path, so aryl halides do not react because their C-X bond is strong.
3. Hence only primary aliphatic amines are made.

Teacher's Note:
1. Gabriel synthesis gives pure primary amines with no secondary or tertiary amine contamination.
2. Aromatic primary amines need other methods, for example reduction of nitrobenzene.

Answer: (d) Sucrose

Explanation:
1. In sucrose the anomeric carbons of glucose and fructose are joined through the glycosidic link, so no free aldehyde or ketone group can form.
2. Maltose, lactose and glucose have a free hemiacetal group and reduce Tollens' and Fehling's reagents.

Teacher's Note:
1. Sucrose is the only common disaccharide that is non-reducing.
2. It is also called invert sugar after hydrolysis.

Answer: (b) 3-chlorocyclohex-1-ene

Explanation:
1. In 3-chlorocyclohexene, C-3 is bonded to four different groups: Cl, H, the vinyl carbon C-2 and the \(\ce{CH2}\) group C-4. It is a chiral centre, so the compound is optically active.
2. 1-Chlorocyclopentene has no sp3 stereocentre.
3. In 1-bromo-4-chlorocyclohexane and methylcyclohexane, the ring has a plane of symmetry through the two substituted carbons, so there is no chirality.

Teacher's Note:
1. Look for a carbon bonded to four different groups, with the two ring paths different.
2. A plane of symmetry rules out optical activity.

Answer: (d) 1,4,5-trichlorocyclohex-1-ene

Explanation:
1. But-2-yne is linear and but-1-ene has two hydrogens on one alkene carbon, so neither can show geometrical isomerism.
2. In a six-membered ring a double bond cannot be trans, so 1,2-dichlorocyclohexene has no isomers.
3. In 1,4,5-trichlorocyclohexene, the two Cl atoms on the adjacent sp3 carbons C-4 and C-5 can be on the same side or opposite sides of the ring, giving cis and trans isomers.

Teacher's Note:
1. Ring substituents on two different sp3 carbons give cis-trans isomerism.
2. A double bond in a small ring cannot have the trans form.

Answer: (d) \(\ce{CH3-CH=C(Cl)NH2}\) with \(\ce{CH3}\) and Cl on opposite sides

Explanation:
1. Assign CIP priorities on each carbon of the double bond.
2. In (a) F > H on one carbon and \(\ce{CH3}\) > H on the other, and the high-priority groups are together: Z.
3. In (b) ethyl > H and Cl > \(\ce{CH3}\), with ethyl and Cl together: Z.
4. In (c) \(\ce{CH3}\) > H and Cl > N, with \(\ce{CH3}\) and Cl together: Z.
5. In (d) the same priorities place \(\ce{CH3}\) and Cl on opposite sides: E.

Teacher's Note:
1. E (entgegen) means the higher priority groups are on opposite sides.
2. Z (zusammen) means they are on the same side.

Answer: (b) Diastereomers

Explanation:
1. The first projection, with the two Cl atoms on the same side, is the meso form (it has an internal plane of symmetry).
2. The second projection, with Cl atoms on opposite sides, is one of the chiral (optically active) forms.
3. They have the same connectivity but are not mirror images of each other, so they are diastereomers.

Teacher's Note:
1. Meso compound versus a chiral stereoisomer is always a diastereomer relationship.
2. Enantiomers must be non-superimposable mirror images.

Answer: (b) 2,3-Dimercaptopropan-1-ol

Explanation:
1. The principal functional group is OH, so the chain is named as an alcohol (propan-1-ol) with the OH on C-1.
2. The two SH groups (mercapto substituents) are on C-2 and C-3.
3. Name: 2,3-dimercaptopropan-1-ol (also called 2,3-bis(sulfanyl)propan-1-ol).

Teacher's Note:
1. The alcohol group gets the lowest locant, and SH is cited as a prefix.
2. Option (d) is wrong because the suffix must take the lowest locant.

Answer: (d) All of these

Explanation:
1. The longest chain including the nitrile carbon has three carbons, with a methyl group on C-2: 2-methylpropanenitrile (IUPAC name).
2. It can also be named as isopropyl cyanide (alkyl cyanide naming) and isobutyronitrile (common name).
3. All three names describe the same compound.

Teacher's Note:
1. Nitrile carbon is C-1 in IUPAC naming.
2. Cyanide names treat the CN group as a separate word.

Answer: (d) Butyric acid

Explanation:
1. The series runs formic (1 carbon), acetic (2), propionic (3), butyric (4), valeric (5).
2. The 4th member is butyric acid, \(\ce{CH3CH2CH2COOH}\).

Teacher's Note:
1. Members differ by a \(\ce{CH2}\) unit.
2. Memorise the order formic, acetic, propionic, butyric, valeric.

Answer: (c) Crotonamide

Explanation:
1. The corresponding acid \(\ce{CH3CH=CHCOOH}\) is crotonic acid, so the amide is crotonamide.
2. Butyramide would be the saturated amide, and the IUPAC name is but-2-enamide.

Teacher's Note:
1. Replace "-ic acid" by "-amide" to get the common name of an amide.
2. Crotonic acid is the common name of but-2-enoic acid.

Answer: (d) Acetonitrile

Explanation:
1. The carbonyl carbon in acetone, acetic acid and acetamide is attached to three atoms (one double bond) and is \(sp^2\).
2. In acetonitrile the nitrile carbon is triple bonded and is sp, while the methyl carbon is \(sp^3\).
3. So acetonitrile has no \(sp^2\) carbon.

Teacher's Note:
1. Count the sigma bonds plus lone pairs: 3 gives sp2, 2 gives sp.
2. A triple bond means an sp carbon.

Answer: (b) 12, 3

Explanation:
1. C-H bonds: 3 + 1 + 3 = 7 sigma.
2. C-C bonds: \(\ce{CH3-CH}\), \(\ce{CH=C}\) (one sigma), \(\ce{C-CH3}\), \(\ce{C-CN}\) give 4 sigma.
3. The \(\ce{C#N}\) bond has 1 sigma. Total sigma = 7 + 4 + 1 = 12.
4. Pi bonds: one in \(\ce{C=C}\) and two in \(\ce{C#N}\) = 3.

Teacher's Note:
1. A double bond is one sigma and one pi; a triple bond is one sigma and two pi.
2. Count all single bonds as sigma.

Answer: (b) 3

Explanation:
1. A secondary amine has the form R-NH-R'.
2. Splitting four carbons between the two groups: methyl + n-propyl, methyl + isopropyl, and ethyl + ethyl.
3. That gives 3 isomers: N-methylpropan-1-amine, N-methylpropan-2-amine and N-ethylethanamine.

Teacher's Note:
1. Make a systematic list by the size of the two alkyl groups.
2. Do not forget the branched propyl group.

Answer: (a) \(\ce{CHCl3}\)

Explanation:
1. The carbylamine reaction is \(\ce{RNH2 + CHCl3 + 3KOH -> RNC + 3KCl + 3H2O}\).
2. It needs chloroform, which gives dichlorocarbene with alcoholic KOH.
3. The isocyanide formed has a foul smell and is the positive test for a primary amine.

Teacher's Note:
1. Only primary amines (aliphatic or aromatic) give this test.
2. Secondary and tertiary amines do not.

Answer: (b) phthalide (isobenzofuran-1(3H)-one)

Explanation:
1. Phthalaldehyde has no alpha hydrogen, so with NaOH it undergoes an intramolecular Cannizzaro reaction: one CHO is reduced to \(\ce{CH2OH}\) and the other oxidised to \(\ce{COO-}\).
2. On acidification the 2-(hydroxymethyl)benzoic acid is formed.
3. On heating, the \(\ce{OH}\) and \(\ce{COOH}\) groups are close together and lose water to give the five-membered lactone, phthalide.

Teacher's Note:
1. Aldehydes without alpha hydrogen give the Cannizzaro reaction.
2. A hydroxy acid whose OH and COOH groups can form a five-membered ring cyclises easily to a lactone.