JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 55

Answer: (d) 2-vinylphenol

Explanation:
1. 2-Vinylphenol has an acidic phenolic OH (reacts with EtMgBr to give ethane) and a C=C that decolourises bromine water.
2. 2-Methoxystyrene has no acidic hydrogen; the ester and the keto nitrile have no alkene.

Teacher's Note:
1. Need both an acidic H (or C=O) and an alkene.
2. Bromine water tests C=C.

Answer: (d) A, B and C only

Explanation:
1. 2-Phenylbutan-2-ol has Ph, CH₃ and C₂H₅ on the carbinol carbon.
2. A, B and C all supply those three groups (one from the Grignard, two from the ketone).
3. D would give a secondary alcohol, 1-phenylbutan-1-ol.

Teacher's Note:
1. A tertiary alcohol comes from a ketone.
2. Check the three groups on carbinol carbon.

Answer: (a) (III) (ii) (P)

Explanation:
1. (III) cyclohexyl methyl ketone + NaBD₄ adds D- to the carbonyl (nucleophilic addition, ii) to give CH₃-CD(OH)-cyclohexyl, which contains the CH₃CH(OH)- group needed for the iodoform test; the ketone reactant gives the 2,4-DNP test (P).
2. The other products have no CH₃CH(OH)- or CH₃CO- group.

Teacher's Note:
1. Iodoform test needs CH₃CO- or CH₃CH(OH)-.
2. Methyl ketone reduction keeps CH₃CH(OH)-.

Answer: (a) RCOOH + Cl₂

Explanation:
1. SOCl₂, PCl₅ and PCl₃ replace OH by Cl in carboxylic acids.
2. Chlorine gas alone does not convert the acid to the acid chloride (it would chlorinate the α carbon with a catalyst, as in the HVZ reaction).

Teacher's Note:
1. Typical reagents contain P-Cl or S-Cl bonds.
2. SOCl₂ is preferred as the by-products are gases.

Answer: (d) 3

Explanation:
1. Two Cl atoms are replaced one after the other, giving acetyl chloride and then acetone.
2. Acetone adds a third CH₃MgCl to give tert-butyl alkoxide: 3 moles in all.

Teacher's Note:
1. Replace each Cl, then add to the carbonyl.
2. Final product is (CH₃)3COH.

Answer: (d) 2-methylbutan-2-ol

Explanation:
1. CH₃- adds to the carbonyl carbon of CH₃CH₂COCH₃.
2. The alkoxide formed on hydrolysis gives the tertiary alcohol CH₃CH₂C(CH₃)2OH, 2-methylbutan-2-ol.

Teacher's Note:
1. Ketones give tertiary alcohols with Grignard reagents.
2. The carbonyl carbon becomes the carbinol carbon.

Answer: (a) C₆H₅CH₂C(CH₃)2OH

Explanation:
1. The Grignard reagent first replaces OCH₃ to give the ketone C₆H₅CH₂COCH₃.
2. The ketone reacts faster than the ester and adds a second CH₃-, so the tertiary alcohol C₆H₅CH₂C(CH₃)2OH forms (two identical CH₃ groups).

Teacher's Note:
1. Esters with excess Grignard give tertiary alcohols with two identical new groups.
2. The ketone cannot be isolated.

Answer: (b) 4

Explanation:
1. I is addition (hydration); II, VII and VIII are nucleophilic substitution.
2. III, IV, V and VI each remove H and a leaving group (HBr, HCl or H₂O) from adjacent carbons to form C=C: elimination.
3. Total = 4.

Teacher's Note:
1. Elimination forms a double bond.
2. Lucas reagent and HI convert alcohols by substitution.

Answer: (a) a mixture of diastereomers

Explanation:
1. The acid contributes a stereocentre of fixed R configuration.
2. The racemic alcohol contributes (R) and (S) centres, so the esters are (R,R) and (R,S): diastereomers, a mixture with different physical properties.

Teacher's Note:
1. Fixed R acid + (R/S) alcohol gives R,R and R,S.
2. Diastereomers are not mirror images.

Answer: (b) CH₃CHO + CH₃CH₂MgBr, then H₃O+

Explanation:
1. Butan-2-ol is CH₃CH(OH)CH₂CH₃, a secondary alcohol, so it comes from an aldehyde other than HCHO.
2. CH₃CHO + CH₃CH₂MgBr gives it directly.
3. The other routes give pentan-2-ol (butanal + CH₃MgBr), tert-butyl alcohol (acetone + CH₃MgBr) and, from HCHO + isopropyl MgBr gives the primary alcohol 3-methylbutan-1-ol.

Teacher's Note:
1. Count carbons on each side of the carbinol carbon.
2. HCHO always gives a primary alcohol.

Answer: (d) A, B and C only

Explanation:
1. Excess Grignard adds two identical R groups from the Grignard reagent, so the carbinol carbon carries two identical groups.
2. A (from ethyl acetate + PhMgBr), B (propanoate + C₂H₅MgBr) and C (ethyl formate + isopropyl MgBr) fit.
3. D has three different groups and cannot form.

Teacher's Note:
1. Two of the three substituents on the carbinol carbon must be identical.
2. Formate esters give secondary alcohols.

Answer: (b) A, B, C and D only

Explanation:
1. NaH is a strong base that reacts with any compound having an acidic hydrogen and releases H₂.
2. The O-H of an acid and an alcohol, the N-H of an amide and the terminal alkyne C-H are all acidic enough.

Teacher's Note:
1. Compounds with reactive hydrogen give H₂.
2. H- + H+ → H₂.

Answer: (c) DMSO

Explanation:
1. DMSO, (CH₃)2S=O, is polar but has no hydrogen on O or N.
2. NH₃, H₂O and CH₃COOH can all donate hydrogen bonds.

Teacher's Note:
1. Aprotic means no acidic hydrogen on a heteroatom.
2. Polar aprotic solvents leave anions poorly solvated.

Answer: (a) cleavage of a pi bond and formation of two new sigma bonds

Explanation:
1. In an addition to C=C or C=O, the pi bond breaks and each of the two atoms forms a new sigma bond with the adding groups.
2. The reverse process (two sigma bonds broken, a pi bond formed) is elimination.

Teacher's Note:
1. Addition converts a pi bond into two sigma bonds.
2. Elimination does the opposite.

Answer: (d) 5

Explanation:
1. Each acylation replaces H by COCH₃, adding 42 mass units.
2. Increase = 390 - 180 = 210 = 5 × 42, so 5 amino groups.

Teacher's Note:
1. Mass change per acetyl = 43 - 1 = 42.
2. Divide the total change by 42.

Answer: (b) 2

Explanation:
1. CN- attacks the planar carbonyl carbon from either face, creating a new stereocentre (C-1).
2. Because C-3 is already a stereocentre of fixed configuration, the two products are diastereomers: 2 products.

Teacher's Note:
1. A new stereocentre forms with an existing one.
2. Diastereomers, not enantiomers, result.

Answer: (c) 88

Explanation:
1. 56 mL of CH₄ at STP = 56/22400 = 0.0025 mol.
2. One mole of alcohol (one OH) gives one mole of CH₄, so moles of alcohol = 0.0025.
3. Molar mass = 0.22/0.0025 = 88 g mol^-1 (e.g. pentan-1-ol).

Teacher's Note:
1. Each active hydrogen gives one CH₄.
2. Molar mass = mass / moles.

Answer: (b) 2-methylpropan-2-ol

Explanation:
1. CH₃COBr + CH₃MgI gives acetone.
2. Acetone + a second CH₃MgI gives (CH₃)3COMgI.
3. NH₄Cl solution hydrolyses it to (CH₃)3COH, 2-methylpropan-2-ol.

Teacher's Note:
1. Excess Grignard goes on to the tertiary alcohol.
2. NH₄Cl is a mild proton source.

Answer: (b) CH₃NH₂ + NaOH

Explanation:
1. A reaction proceeds if the stronger acid gives up its proton to the stronger base (forming weaker acid and weaker base).
2. CH₃NH₂ is a far weaker acid than water, so NaOH cannot deprotonate it. The other three are feasible: sulphonic acid > carbonic acid; phenol > NH₃; terminal alkyne > H₂.

Teacher's Note:
1. Compare the acidity of the proton donor with the conjugate acid of the base.
2. Amines are very weak acids.

Answer: (d) CH₃Cl and methyl benzoate

Explanation:
1. The carbinol carbon carries Ph and two CH₃ groups.
2. Two CH₃ groups come from CH₃MgCl (P = CH₃Cl) adding to the ester carbonyl of methyl benzoate PhCOOCH₃ (Q).
3. Chlorobenzene + ethyl acetate would give 1,1-diphenylethanol instead.

Teacher's Note:
1. Two identical groups come from the Grignard reagent.
2. The third group is already on the ester carbonyl.

Answer: (b) 4

Explanation:
1. Ketones and esters give tertiary alcohols: (a), (f), (h). Formaldehyde gives a primary alcohol: (c).
2. Aldehydes other than HCHO give secondary alcohols: (b), (d), (e).
3. Formyl chloride with excess RMgX gives R₂CHOH, a secondary alcohol: (g).
4. Total = 4 (b, d, e, g).

Teacher's Note:
1. Formic acid derivatives give secondary alcohols.
2. Count the carbinol carbon's substituents.

Answer: (b) (B) and (D)

Explanation:
1. The Grignard reagent is destroyed by acidic hydrogens (COOH in B and CH₂OH in D), so addition does not occur with the equivalent amount.
2. Benzaldehyde and 4-methoxybenzaldehyde have no acidic hydrogen.

Teacher's Note:
1. Look for O-H groups.
2. Grignard reagents are protonated first.

Answer: (b) Ph₂CD(OH)

Explanation:
1. The hydride (here deuteride) ion from LiAlD₄ adds to the carbonyl carbon, so D is bonded to carbon.
2. The oxygen picks up an ordinary proton from H₃O+, giving Ph₂CD(OH).

Teacher's Note:
1. Deuteride goes to carbon, proton from the workup goes to oxygen.
2. H₃O+ supplies H, not D.

Answer: (a) HCHO + PhCH(CH₃)CH₂MgX

Explanation:
1. The target is a tertiary alcohol with Ph, CH₃ and C₂H₅ on the carbinol carbon.
2. HCHO always gives a primary alcohol, so the first route cannot give it; the others each supply the three required groups.

Teacher's Note:
1. Formaldehyde gives primary alcohols.
2. Check each route for the groups on carbinol carbon.

Answer: (a) (ii) > (iv) > (i) > (iii)

Explanation:
1. Electron withdrawing groups on the ring increase the positive charge on the carbonyl carbon and speed up attack by water.
2. NO₂ is the strongest withdrawing group, then CHO; CH₃ is electron donating and slows the reaction.
3. Order: (ii) > (iv) > (i) > (iii).

Teacher's Note:
1. Withdrawing groups raise electrophilicity.
2. The methyl group is electron releasing.

Answer: (a) CH₃S-

Explanation:
1. Nucleophilicity increases with polarisability and with size of the donor atom down a group.
2. S is large and highly polarisable, and CH₃S- is only weakly solvated by protic solvents, so it is the best nucleophile here.

Teacher's Note:
1. Compare donor atoms: S is more polarisable than O or F.
2. Sterically crowded Et₃N is poorer than expected.

Answer: (d) 3

Explanation:
1. Secondary alcohols come from ketones.
2. Ketones of formula C₅H₁₀O: pentan-2-one, pentan-3-one and 3-methylbutan-2-one: 3.

Teacher's Note:
1. Aldehydes would give primary alcohols.
2. Count the ketone isomers.

Answer: (c) both C₆H₅COOMgI and CH₄

Explanation:
1. The Grignard reagent is a very strong base and removes the acidic hydrogen of the carboxylic acid.
2. CH₃MgI + C₆H₅COOH → CH₄ + C₆H₅COOMgI, so both products form together.

Teacher's Note:
1. Grignard reagents react with any compound having an acidic hydrogen.
2. The methyl group takes up H+ to form methane.

Answer: (b) 4-nitrobenzenesulphonate

Explanation:
1. The leaving group ability rises as the negative charge is better stabilised.
2. The strongly electron withdrawing nitro group stabilises the anion most, so 4-nitrobenzenesulphonate (nosylate) is the best leaving group; the methyl group destabilises the anion.

Teacher's Note:
1. Electron withdrawing groups stabilise anions.
2. Order: NO₂ > Br > H > CH₃.