JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 51

Answer: (c) 3

Explanation:
1. The carbon skeleton is unchanged, so we need alkenes with the 2-methylbutane skeleton.
2. 2-Methylbut-1-ene, 2-methylbut-2-ene and 3-methylbut-1-ene: three alkenes.

Teacher's Note:
1. Count positions where a double bond can be placed without making a carbon pentavalent.
2. The quaternary carbon is absent, so any CH-CH or CH-CH2 bond can host a double bond.

Answer: (a) 3

Explanation:
1. C3 carries two methyl groups and has no hydrogen, so no triple bond can include C3.
2. Possible triple bonds: C1-C2, C4-C5 and C5-C6, giving three alkynes.

Teacher's Note:
1. Hydrogenation keeps the carbon skeleton.
2. A triple bond needs each carbon to have at most one other carbon substituent.

Answer: (d) methylcyclopentane

Explanation:
1. M = 84 means C6H12.
2. Methylcyclopentane has four ring CH2 groups and gives 4 isomers (CH3, C1, C2, C3 chlorinated). Cyclohexane gives 1, ethylcyclobutane gives 5, 1,1-dimethylcyclobutane has only 3 secondary carbons.

Teacher's Note:
1. Count distinct hydrogens: Cl can replace one of each kind.
2. The number of monochloro structural isomers equals the number of non-equivalent hydrogen types.

Answer: (b) 2-methylbut-2-ene and 3-methylbut-1-ene

Explanation:
1. Both 2-methylbut-2-ene and 3-methylbut-1-ene give 2-methylbutane.
2. The other pairs give 2-methylhexane/3-methylhexane, 3-ethylhexane/2-methylheptane and ethylcyclopropane/1,2-dimethylcyclopropane, all different.

Teacher's Note:
1. Same skeleton after saturation means same product.
2. Check the carbon skeleton of each name.

Answer: (b) 2,3-dimethylbutane

Explanation:
1. 2,3-Dimethylbutane has two types of H: the 12 methyl hydrogens and the 2 tertiary hydrogens.
2. n-Pentane has 3 types and 2,4-dimethylpentane has 3; toluene gives one side-chain product.

Teacher's Note:
1. Count non-equivalent H atoms.
2. Aromatic hydrogens are inert to free radical chlorination.

Answer: (d) 1

Explanation:
1. All 18 hydrogens are in six equivalent methyl groups, so only one monochloro derivative is possible.

Teacher's Note:
1. Highly symmetrical molecules have few isomers.
2. Count the symmetry-distinct H atoms.

Answer: (a) 2-methylbut-2-ene

Explanation:
1. Hydrogenation products: pent-2-ene gives pentane (3 monochloro), 2-methylbut-2-ene gives 2-methylbutane (4), 3-methylhex-2-ene gives 3-methylhexane (7), 2,3-dimethylbut-2-ene gives 2,3-dimethylbutane (2).

Teacher's Note:
1. First find the alkane, then count its H types.
2. 2-Methylbutane has CH3 (two kinds), CH2 and CH.

Answer: (c) 3-methylhex-3-ene

Explanation:
1. The ketone is butanone, from =C(CH3)CH2CH3; the acid is propanoic acid, from CH3CH2CH=.
2. Joining them gives CH3CH2CH=C(CH3)CH2CH3, 3-methylhex-3-ene.

Teacher's Note:
1. Join the two fragments through a double bond.
2. A =CH- end gives an acid in oxidative work-up; a disubstituted =C end gives a ketone.

Answer: (d) a terminal alkyne

Explanation:
1. Only a terminal alkyne has an acidic sp C-H and forms a white silver acetylide. An aldehyde would give a silver mirror.

Teacher's Note:
1. Alkyne-H is acidic.
2. Internal alkynes do not react.

Answer: (b) but-1-yne

Explanation:
1. Cuprous acetylide forms only with terminal alkynes (RC≡CH).

Teacher's Note:
1. But-1-yne is HC≡C-CH2CH3.
2. Pent-1-en-3-yne has an internal triple bond.

Answer: (a) but-1-yne

Explanation:
1. B has an acidic terminal C-H (but-1-yne); A is but-2-yne, which has no such hydrogen.

Teacher's Note:
1. Terminal alkynes react with sodium.
2. Position isomers: triple bond at C1 or C2.

Answer: (c) ceric ammonium nitrate

Explanation:
1. Ceric ammonium nitrate, a yellow solution, turns red with alcohols (and phenols green to brown).

Teacher's Note:
1. Schiff's detects aldehydes, Molisch carbohydrates.
2. The colour change is yellow to red.

Answer: (c) acetophenone

Explanation:
1. 2,4-DNP shows a carbonyl group, iodoform a CH3C=O group. C6H5COCH3 is C8H8O and fits both.

Teacher's Note:
1. Benzophenone has no methyl ketone group.
2. Check the formula.

Answer: (a) butanoic acid

Explanation:
1. Phenolphthalein turns pink when NaOH is in excess, so X is acidic: CH3CH2CH2COOH (C4H8O2). The esters hydrolyse only slowly and the dione is C4H6O2.

Teacher's Note:
1. A carboxylic acid titrates with NaOH.
2. Esters are neutral.

Answer: (d) 2,4-dimethylaniline

Explanation:
1. Only primary amines give the isocyanide (foul smell). 2,4-Dimethylaniline is a primary aromatic amine; the others are secondary or tertiary.

Teacher's Note:
1. CHCl3 + KOH + RNH2 gives RNC.
2. Count N-H bonds.

Answer: (b) separation of amine mixtures

Explanation:
1. Benzenesulphonyl chloride reacts differently with 1°, 2° and 3° amines, allowing their separation and identification.

Teacher's Note:
1. Primary products dissolve in alkali, secondary do not.
2. Tertiary amines do not react.

Answer: (b) \( \ce{(CH3CH2)3N} \)

Explanation:
1. A tertiary amine has no N-H hydrogen, so it cannot form a sulphonamide.

Teacher's Note:
1. The primary and secondary amines form sulphonamides.
2. Hinsberg's reagent is benzenesulphonyl chloride.

Answer: (d) thiourea, NH\(_2\)CSNH\(_2\)

Explanation:
1. Fusion of a compound containing both N and S with sodium gives NaSCN, which gives the blood red [Fe(SCN)]2+.

Teacher's Note:
1. Both nitrogen and sulphur are required.
2. Thiourea contains both.

Answer: (a) hydrazine, NH\(_2\)-NH\(_2\)

Explanation:
1. Prussian blue needs NaCN, which needs carbon and nitrogen. Hydrazine contains no carbon.

Teacher's Note:
1. Without carbon no cyanide forms.
2. The other three contain C and N.

Answer: (c) NaCN

Explanation:
1. Carbon, nitrogen and sodium combine to form sodium cyanide.

Teacher's Note:
1. Lassaigne's extract contains NaCN.
2. Isocyanide is not formed.

Answer: (d) sulphur

Explanation:
1. S2- gives black PbS: \( \ce{Pb^{2+} + S^{2-} -> PbS} \).

Teacher's Note:
1. Acetic acid is used because H2SO4 would precipitate PbSO4.
2. Black PbS identifies sulphur.