JEE Chemistry Organic Chemistry MCQs Set 61

Answer: (a) 3,4-dimethylenecyclobut-1-ene (a four-membered ring with one ring double bond and two exocyclic \(=\ce{CH2}\) groups on adjacent ring carbons)

Explanation:
1. In 3,4-dimethylenecyclobut-1-ene every ring carbon and both \(\ce{CH2}\) carbons are part of a C=C double bond, so all carbons are \(sp^2\).
2. 5-methylcyclohexa-1,3-diene has an \(sp^3\) ring carbon and a methyl group. Toluene has an \(sp^3\) methyl carbon. Pent-4-en-2-one has \(sp^3\) \(\ce{CH2}\) and \(\ce{CH3}\) carbons.

Teacher's Note:
1. A carbon is \(sp^2\) when it carries a double bond (or is part of an aromatic ring).
2. Any saturated carbon (CH3 or CH2 with single bonds only) is \(sp^3\).

Answer: (c) E

Explanation:
1. On C-1, \(\ce{NH2}\) (attached atom N) has priority over \(\ce{CH2OH}\) (attached atom C).
2. On C-2, compare \(\ce{CF3}\) with substituents (F,F,F) and \(\ce{CH2Cl}\) with (Cl,H,H). The first point of difference is the highest atom, Cl > F, so \(\ce{CH2Cl}\) has the higher priority.
3. \(\ce{NH2}\) and \(\ce{CH2Cl}\) are on opposite sides, so the configuration is E.

Teacher's Note:
1. Cis and trans are not used when each carbon carries two different groups; use the CIP rules instead.
2. Compare substituent atoms in order of decreasing atomic number at the first point of difference.

Answer: (c) (i), (iii) and (iv)

Explanation:
1. (i) A cumulated diene (C=C=C) with two different groups on each end carbon is chiral (axial chirality), so it is active.
2. (ii) 1,4-disubstituted cyclohexane has a plane of symmetry, so it is inactive.
3. (iii) Large groups in the ortho positions block rotation about the biphenyl bond, giving chiral atropisomers, so it is active.
4. (iv) The two ends of the molecule are different (COOH and COOCH3), so there is no meso symmetry and the compound is active.
5. (v) n-Hexane is achiral.
6. Optically active compounds are (i), (iii) and (iv).

Teacher's Note:
1. Look for a plane or centre of symmetry to decide that a compound is achiral.
2. Chirality does not need a chiral carbon: cumulated dienes and hindered biphenyls are chiral too.

Answer: (d) a < c < b < d

Explanation:
1. A nitro group stabilises the phenoxide ion, so all nitrophenols are stronger acids than phenol.
2. At the meta position only the \(-I\) effect acts (pKa about 8.4). At the ortho and para positions the \(-M\) effect also acts, giving pKa about 7.2 (ortho) and 7.1 (para).
3. The ortho isomer is slightly less acidic than the para isomer because of intramolecular hydrogen bonding in the neutral molecule.
4. Order: phenol < 3-nitrophenol < 2-nitrophenol < 4-nitrophenol.

Teacher's Note:
1. Both resonance (\(-M\)) and inductive (\(-I\)) effects work at ortho and para; only \(-I\) works at meta.
2. Intramolecular H-bonding holds the proton in the ortho isomer and lowers its acidity.

Answer: (c) prop-2-en-1-ol, \(\ce{CH2=CH-CH2-OH}\)

Explanation:
1. Resonance needs a lone pair, a p orbital or a double bond directly attached to a pi system.
2. Pyridine is aromatic. In 1-chlorocyclohexene the lone pair of Cl is conjugated with the double bond. In \(\ce{CH3-NH-CH2+}\), the lone pair on N is next to the carbocation.
3. In prop-2-en-1-ol, the \(sp^3\) \(\ce{CH2}\) group separates the double bond from the oxygen lone pair, so there is no conjugation and no resonance.

Teacher's Note:
1. Conjugation needs adjacent p orbitals; an \(sp^3\) carbon in between breaks it.
2. Allylic and vinylic positions differ: the lone pair must be on the vinylic carbon for resonance.

Answer: (c) 1-bromo-1,2,3,4-tetrahydronaphthalene (bromine on the benzylic carbon)

Explanation:
1. HBr adds to the double bond by electrophilic addition. \(\ce{H+}\) adds to the carbon that gives the more stable carbocation.
2. Proton addition to the non-benzylic carbon gives a benzylic carbocation, stabilised by the benzene ring.
3. Bromide then attacks this benzylic carbon, giving 1-bromo-1,2,3,4-tetrahydronaphthalene.

Teacher's Note:
1. A benzylic carbocation is stabilised by resonance with the benzene ring.
2. Markovnikov's rule is really a rule about the more stable carbocation.

Answer: (c) \(\ce{C6H6}\)

Explanation:
1. Alkaline \(\ce{KMnO4}\) oxidises the side chain of toluene to a carboxylic group, giving benzoic acid (as its salt).
2. Heating with soda lime brings about decarboxylation: \(\ce{C6H5COONa + NaOH ->[CaO, \Delta] C6H6 + Na2CO3}\).
3. So B is benzene.

Teacher's Note:
1. Soda lime decarboxylation removes the carboxyl group as carbonate.
2. Any alkyl side chain with an alpha hydrogen is oxidised all the way to \(\ce{-COOH}\).

Answer: (d) N-(4-nitrophenyl)benzamide

Explanation:
1. The ring attached to nitrogen is activated by the lone pair on N (an \(-NHCOR\) group is an ortho, para director).
2. The ring attached to the carbonyl is deactivated by the \(-CO\) group.
3. So nitration occurs in the N-phenyl ring, mainly at the para position (the ortho position is hindered), giving N-(4-nitrophenyl)benzamide.

Teacher's Note:
1. In an amide, the nitrogen lone pair is shared with the C=O group, so the N-ring is only moderately activated, but it is still far more reactive than the benzoyl ring.
2. The para product dominates because of steric hindrance at ortho.

Answer: (c) \(\ce{I-} > \ce{Br-} > \ce{Cl-}\)

Explanation:
1. Iodide is the largest and most polarisable halide ion and the weakest base, so it is both the best nucleophile and the best leaving group.
2. The C-I bond is also the weakest, so it breaks most easily.
3. Leaving group order: \(\ce{I-} > \ce{Br-} > \ce{Cl-}\).

Teacher's Note:
1. A good leaving group is a weak base, because it can stabilise the negative charge on its own.
2. HI is the strongest acid among HCl, HBr and HI, so I- is the weakest base.

Answer: (b) Grignard reagent

Explanation:
1. Fehling's solution, Schiff's reagent and Tollens' reagent react with aldehydes only, so each of them separates ethanal from propanone.
2. Grignard reagent adds to both carbonyl compounds, giving an alcohol in each case, with no easily observable difference.
3. So the Grignard reagent cannot distinguish them.

Teacher's Note:
1. Aldehydes are easily oxidised, and these three reagents test this property.
2. Both aldehydes and ketones undergo nucleophilic addition, so addition reagents do not distinguish them.

Answer: (a) 4-methylbenzonitrile (p-tolunitrile)

Explanation:
1. Diazotisation converts the \(\ce{-NH2}\) group into the diazonium salt \(\ce{CH3C6H4N2+Cl-}\) (A).
2. In the Sandmeyer reaction \(\ce{CuCN/KCN}\) replaces \(\ce{-N2+}\) by \(\ce{-CN}\) with loss of \(\ce{N2}\).
3. So P is 4-methylbenzonitrile.

Teacher's Note:
1. Diazonium salts are stable only at 0-5 degrees Celsius.
2. Cu(I) salts (CuCl, CuBr, CuCN) in the Sandmeyer reaction replace the diazonium group.

Answer: (a) sorbitol

Explanation:
1. Na/Hg and water reduce the aldehyde (CHO) group of glucose to a primary alcohol (\(\ce{CH2OH}\)).
2. The product is the hexahydric alcohol sorbitol (glucitol).
3. Gluconic acid and saccharic acid are oxidation products, and fructose is an isomer.

Teacher's Note:
1. Reduction of an aldose gives an alditol.
2. Oxidation by bromine water gives gluconic acid, and by nitric acid gives saccharic acid.

Answer: (d) All of these.

Explanation:
1. Saccharin is the cyclic imide of 2-sulphobenzoic acid (ortho-sulphobenzimide), as in statement (a).
2. It is about 550 times sweeter than sucrose and it is used as an artificial sweetener, especially for diabetic patients.
3. So all the statements are correct.

Teacher's Note:
1. Artificial sweeteners such as saccharin are not metabolised, so they add almost no calories.
2. Their sweetness is compared with cane sugar on a weight basis.