JEE Chemistry Organic Chemistry MCQs Set 60

Answer: (a) Equanil

Explanation:
1. Equanil (meprobamate) acts on the central nervous system to relieve stress and anxiety: it is a tranquilizer.
2. Promethazine is an antihistamine; omeprazole and cimetidine reduce stomach acid secretion (antacids/anti-ulcer drugs).

Teacher's Note:
1. Tranquilizers include equanil, valium, luminal and seconal.
2. Antihistamines, antacids and tranquilizers are three different drug classes.

Answer: (c) 4-methylhex-2-yne

Explanation:
1. 2-Ethylpent-3-yne is \( \ce{CH3-CH(C2H5)-C#C-CH3} \).
2. The longest chain containing the triple bond has six carbons: \( \ce{CH3-C#C-CH(CH3)-CH2-CH3} \).
3. Numbering from the end nearer the triple bond gives the triple bond at C-2 and a methyl group at C-4: 4-methylhex-2-yne.

Teacher's Note:
1. Choose the longest chain that contains the multiple bond, not necessarily the one drawn as the main chain.
2. Give the lowest locant to the triple bond first.

Answer: (d) all of these

Explanation:
1. Tautomerism (keto-enol) requires a hydrogen on a carbon next to the carbonyl group (alpha-H) or, for conjugated ketones, on the gamma carbon.
2. (a) has two hydrogens on C-4 which, on enolisation, give phenol. (b) has alpha-hydrogens on the \( \ce{CH2} \) group (C-3); enolisation towards C-3 gives a double bond that is not at the bridgehead, so it is allowed.
3. (c) has no alpha-hydrogen on C-6 (quaternary), but the gamma-hydrogens on C-4 allow a dienol form. So all three show tautomerism.

Teacher's Note:
1. Bredt's rule forbids a double bond at the bridgehead, but the other alpha carbon of norbornanone is free to enolise.
2. In conjugated enones, gamma-hydrogens can also take part in keto-enol tautomerism.

Answer: (a) I > II > III > IV

Explanation:
1. Structure I is neutral and every atom has a complete octet: most stable.
2. Structure II carries a negative charge on oxygen and a positive charge on oxygen, but all atoms still have a full octet.
3. Structure III has a carbon with only six electrons (carbocation); structure IV places a positive charge on an oxygen with an incomplete octet and a negative charge on carbon, the least electronegative atom, so it is the least stable.
4. Order: I > II > III > IV.

Teacher's Note:
1. More covalent bonds, complete octets and less charge separation all increase stability.
2. A negative charge sits best on the more electronegative atom.

Answer: (c) 2,4,6-trinitrophenol (picric acid)

Explanation:
1. Phenoxide ion stability decides acidity of phenols.
2. Nitro groups at the ortho and para positions withdraw electrons strongly by \( -M \) and \( -I \) effects and delocalise the negative charge.
3. Three nitro groups (2,4,6-) stabilise the phenoxide most, so picric acid is the strongest acid (\( pK_a \approx 0.4 \)). The methyl group is electron-donating and weakens acidity.

Teacher's Note:
1. More electron-withdrawing groups at ortho and para positions give a stronger phenolic acid.
2. Electron-donating groups such as \( \ce{CH3} \) decrease acidity.

Answer: (a) methylenecyclohexane

Explanation:
1. In E2 the base removes a beta-hydrogen on the carbon next to the carbon holding bromine.
2. The only beta-carbon is the ring carbon attached to the \( \ce{CH2Br} \) group, and removal of its H gives \( \ce{C6H10=CH2} \): methylenecyclohexane.
3. The other two products would need a rearrangement (carbocation), which does not occur in a concerted E2 reaction.

Teacher's Note:
1. E2 occurs in one step without carbocation, so there is no rearrangement.
2. Count the beta-hydrogens: only one carbon qualifies here.

Answer: (d) \( \ce{C6H5-NO2} \)

Explanation:
1. \( \ce{-CH3} \), \( \ce{-OH} \) and \( \ce{-NH2} \) are activating groups: they increase the electron density of the ring, so substitution is faster than in benzene.
2. \( \ce{-NO2} \) is a strong deactivating group (\( -M, -I \)), so nitrobenzene reacts slower than benzene.

Teacher's Note:
1. Activating groups donate electrons; deactivating groups withdraw electrons.
2. Nitro is the classic strong deactivator and meta director.

Answer: (c) \( \ce{R1-CO-R2} \)

Explanation:
1. Ozonolysis cleaves the \( \ce{C=C} \) bond and puts a C=O group on each carbon.
2. The carbon \( \ce{R-CH=} \) gives \( \ce{R-CHO} \).
3. The carbon \( \ce{=CR1R2} \) gives the ketone \( \ce{R1-CO-R2} \).

Teacher's Note:
1. Each alkene carbon keeps its own substituents and gains a double-bonded oxygen.
2. A carbon bearing two alkyl groups gives a ketone.

Answer: (c) \( \ce{CHI3} \) and cyclohexanone

Explanation:
1. The methyl ketone \( \ce{-COCH3} \) undergoes the iodoform reaction: \( \ce{-COCH3 -> -COO- + CHI3} \).
2. Acidification gives the beta-keto acid, 2-oxocyclohexanecarboxylic acid.
3. Beta-keto acids decarboxylate on heating, giving cyclohexanone and \( \ce{CO2} \).

Teacher's Note:
1. Beta-keto acids lose \( \ce{CO2} \) easily through a six-membered cyclic transition state.
2. The iodoform reaction needs a \( \ce{CH3CO-} \) group.

Answer: (a) 7-chlorocyclohepta-1,3,5-triene (chlorocycloheptatriene)

Explanation:
1. \( S_N1 \) reactivity depends on the stability of the carbocation formed.
2. Chlorocycloheptatriene gives the cycloheptatrienyl (tropylium) cation, which is aromatic (6 pi electrons, planar, cyclic, fully conjugated) and exceptionally stable.
3. Benzyl cation is stabilised by delocalisation but less than tropylium; \( \ce{CH3+} \) is unstable; vinyl chloride has a strong C-Cl bond and an unstable vinyl cation.

Teacher's Note:
1. Aromatic carbocations such as tropylium are very stable.
2. Vinyl halides are almost inert to substitution.

Answer: (c) \( \ce{CH3I + C2H5OH} \)

Explanation:
1. HI protonates the ether oxygen. The iodide ion then attacks the less hindered alkyl group by the \( S_N2 \) mechanism.
2. The methyl group is less hindered than the ethyl group, so \( \ce{CH3I} \) forms and \( \ce{C2H5OH} \) is released (with excess HI the alcohol becomes \( \ce{C2H5I} \)).

Teacher's Note:
1. For primary alkyl groups, \( S_N2 \) occurs at the smaller alkyl group.
2. Mixed ethers with a tertiary group follow \( S_N1 \) instead.

Answer: (a) 2-oxocyclohexane-1-carbaldehyde (ring ketone and \( \ce{-CHO} \))

Explanation:
1. PCC is a mild oxidising agent: it oxidises primary alcohols to aldehydes (without going on to acids) and secondary alcohols to ketones.
2. The ring \( \ce{CH-OH} \) becomes a ketone and the \( \ce{-CH2OH} \) becomes \( \ce{-CHO} \), giving 2-oxocyclohexane-1-carbaldehyde.

Teacher's Note:
1. Unlike \( \ce{KMnO4} \) or chromic acid, PCC stops at the aldehyde stage.
2. Both alcohol groups are oxidised, so excess PCC gives a keto-aldehyde.

Answer: (a)

Explanation:
1. With concentrated KOH, a dialdehyde with no alpha-hydrogens undergoes an intramolecular Cannizzaro reaction: one \( \ce{-CHO} \) is oxidised to \( \ce{-COOK} \) and the other reduced to \( \ce{-CH2OH} \) (A).
2. On acidification the carboxylate gives the hydroxy acid, which on heating loses water intramolecularly (esterification) to form a cyclic ester (lactone).
3. The new ring contains the two ring carbons, the two aryl carbons, \( \ce{CH2} \), O and C=O: a seven-membered lactone (B).

Teacher's Note:
1. Aldehydes without alpha-hydrogen give the Cannizzaro reaction; two groups in one molecule react intramolecularly.
2. A hydroxy acid with the groups suitably placed cyclises to a lactone on heating.

Answer: (c) 4-methylbenzonitrile

Explanation:
1. Diazotisation gives 4-methylbenzenediazonium chloride (D).
2. In the Sandmeyer reaction \( \ce{CuCN/KCN} \) replaces the diazonium group by \( \ce{-CN} \) with loss of \( \ce{N2} \).
3. E is 4-methylbenzonitrile (p-tolunitrile).

Teacher's Note:
1. Sandmeyer reaction replaces \( \ce{N2+} \) by Cl, Br or CN using the cuprous salt.
2. The methyl group stays unchanged.

Answer: (d) both (a) and (b)

Explanation:
1. Tollens' reagent, \( \ce{[Ag(NH3)2]+} \), is a mild oxidising agent that oxidises all aldehydes (aliphatic and aromatic) to carboxylates and is reduced to silver (silver mirror).
2. Acetaldehyde and benzaldehyde both give the test; ketones such as acetone do not.

Teacher's Note:
1. Tollens' test distinguishes aldehydes from ketones.
2. Fehling's solution is not given by benzaldehyde, but Tollens' reagent is.