JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 28

Answer: (b) 4

Explanation:
1. Monochlorination (Cl2, light) of an alkane replaces one hydrogen by chlorine. The number of structural monochloro isomers equals the number of different (non-equivalent) types of hydrogen.
2. Types of H in 2-methylbutane: (i) the two equivalent CH3 groups on C2, (ii) the CH at C2, (iii) the CH2, (iv) the terminal CH3.
3. So 4 monochloro products.

Teacher's Note:
1. Count only chemically different hydrogens.
2. Methyl groups attached to the same carbon are equivalent.

Answer: (a) 1

Explanation:
1. All twelve hydrogen atoms of neopentane are equivalent (each is on a CH3 attached to the central quaternary carbon).
2. Hence only one monochloro product, neopentyl chloride, is possible.

Teacher's Note:
1. Symmetry makes all methyl groups identical.
2. A hydrocarbon giving a single monochloride has all H atoms equivalent.

Answer: (d) 3

Explanation:
1. In \( \ce{CH3-CH2-CH2-CH2-CH3} \) the C1 and C5 hydrogens are equivalent, as are those on C2 and C4.
2. Distinct sets: C1/C5, C2/C4 and C3, so 3 products: 1-, 2- and 3-chloropentane.

Teacher's Note:
1. Use the symmetry of the chain to pair the equivalent carbons.
2. The result is the same for any straight-chain with an odd number of carbons: (n+1)/2.

Answer: (c) 5

Explanation:
1. Types of hydrogen: the methyl hydrogens, the ring CH carrying the methyl (C1), the two C2/C6 hydrogens, the two C3/C5 hydrogens and the C4 hydrogens.
2. That is 5 types, hence 5 structural isomers.

Teacher's Note:
1. Ring carbons C2 and C6 are equivalent, as are C3 and C5.
2. Stereoisomers are not counted here.

Answer: (b) 2,3-Dimethylbutane

Explanation:
1. 2,3-Dimethylbutane \( \ce{(CH3)2CH-CH(CH3)2} \) has only two types of hydrogen: the 12 methyl hydrogens and the 2 tertiary hydrogens.
2. Hexane and 2,2-dimethylbutane give 3 products each; 2-methylpentane gives 5.

Teacher's Note:
1. Count the hydrogen types for all five isomers of C6H14: 3, 5, 4, 3, 2.
2. The most symmetrical isomer has the fewest products.

Answer: (a) 2

Explanation:
1. Hydrogen types in the five hexanes: hexane 3, 2-methylpentane 5, 3-methylpentane 4, 2,2-dimethylbutane 3, 2,3-dimethylbutane 2.
2. Exactly three monochloro products: hexane and 2,2-dimethylbutane.

Teacher's Note:
1. Work through all five isomers systematically.
2. Only symmetry makes counts smaller than the number of carbons.

Answer: (d) 2,2,3,3-Tetramethylbutane

Explanation:
1. For one monochloride, every hydrogen must be equivalent.
2. In \( \ce{(CH3)3C-C(CH3)3} \) all eighteen hydrogens are on equivalent methyl groups.
3. The other isomers contain at least two types of hydrogen.

Teacher's Note:
1. Look for a highly symmetrical structure with only methyl groups on quaternary carbons.
2. 2,2,4-Trimethylpentane gives four monochloro products.

Answer: (c) Ethylbenzene

Explanation:
1. In side-chain chlorination only the alkyl hydrogens are replaced; ring hydrogens are not.
2. Ethylbenzene has CH2 and CH3 hydrogens, giving C6H5CHClCH3 and C6H5CH2CH2Cl.
3. In each xylene all six side-chain hydrogens are equivalent, giving a single product.

Teacher's Note:
1. Ring hydrogens of aromatic compounds are not replaced under free-radical conditions.
2. Count only the types of side-chain hydrogen.

Answer: (b) 72 g/mol

Explanation:
1. Propane (2), butane (2), isobutane (2) and pentane (3) give fewer than four.
2. 2-Methylbutane (C5H12) gives four products.
3. M = 5 × 12 + 12 = 72.

Teacher's Note:
1. Check the smaller alkanes first, then the C5 isomers.
2. 2-Methylbutane is the first alkane with four types of hydrogen.

Answer: (a) 3

Explanation:
1. Each mole of pi bonds consumes one mole of H2.
2. One C=C contributes 1 and one C≡C (two pi bonds) contributes 2.
3. Total = 3 mol; the product is pentane.

Teacher's Note:
1. Hydrogenation cannot be stopped at an intermediate stage.
2. Moles of H2 = number of pi bonds (excluding the aromatic ring at room temperature).

Answer: (d) 1

Explanation:
1. Only the side-chain double bond is hydrogenated; the aromatic ring is stable at room temperature.
2. The product is ethylbenzene, so 1 mol of H2 is consumed.
3. At 100–150 °C the ring would also take 3 mol (total 4).

Teacher's Note:
1. Aromatic pi bonds need drastic conditions for hydrogenation.
2. Do not count benzene ring pi bonds at room temperature.

Answer: (c) 2

Explanation:
1. Hydrogenation gives butane, \( \ce{CH3-CH2-CH2-CH3} \).
2. Butane has two types of hydrogen (CH3 and CH2), so two monochloro products: 1-chlorobutane and 2-chlorobutane.

Teacher's Note:
1. The carbon skeleton is unchanged during hydrogenation and chlorination.
2. Saturated product first, then count the hydrogen types.

Answer: (b) cyclobutene

Explanation:
1. C4H6 has DU = 2.
2. Hydrogenation of cyclobutene gives cyclobutane, in which all eight hydrogens are equivalent: one monochloro product.
3. The other options give butane, which has two monochloro products.

Teacher's Note:
1. The product must be a symmetrical saturated compound.
2. Open-chain C4H6 isomers all give butane.

Answer: (a) 2

Explanation:
1. Propyne \( \ce{CH3-C#CH} \) takes 2 mol of H2 to give propane.
2. Propane has two types of hydrogen (CH3 and CH2), so 2 monochloro products.

Teacher's Note:
1. Propyne and propadiene both give propane, and propane is the lightest alkane with two monochlorides.
2. Skeleton unchanged.

Answer: (d) 2 mol of ethanal

Explanation:
1. Ozonolysis cleaves the C=C bond and puts a C=O on each carbon.
2. CH3–CH=CH–CH3 splits into two CH3CHO molecules.
3. Zn/H2O prevents oxidation of the aldehyde to acid.

Teacher's Note:
1. Ozonolysis tells the position of the double bond: rejoin the carbonyl carbons to get the alkene.
2. Reductive work-up gives aldehydes and ketones.

Answer: (c) methanal and propanone

Explanation:
1. (CH3)2C=CH2 is cleaved at the double bond.
2. The =CH2 end gives HCHO and the =C(CH3)2 end gives (CH3)2C=O.

Teacher's Note:
1. Terminal =CH2 gives methanal.
2. A disubstituted carbon gives a ketone.

Answer: (b) hexanedial

Explanation:
1. Cleavage of the ring double bond opens the ring and puts C=O at both ends of the chain.
2. Six carbons remain: OHC–(CH2)4–CHO, hexanedial.
3. Oxidative work-up would give hexanedioic acid.

Teacher's Note:
1. A cyclic alkene gives one open-chain dicarbonyl compound.
2. Reductive work-up gives aldehydes, not acids.

Answer: (a) hexa-2,4-diene

Explanation:
1. Join the carbonyl carbons of the fragments again with double bonds: the carbon of each CH3CHO and the two carbons of OHC–CHO give \( \ce{CH3-CH=CH-CH=CH-CH3} \).
2. This is hexa-2,4-diene; its cleavage gives CH3CHO, OHC–CHO and CH3CHO.

Teacher's Note:
1. Remove the oxygen atoms and join the former carbonyl carbons by double bonds.
2. Ethanedial comes from the central CH=CH unit.

Answer: (d) 2 mol of ethanoic acid

Explanation:
1. Alkynes are cleaved by ozone to give carboxylic acids.
2. CH3–C≡C–CH3 gives two molecules of CH3COOH.

Teacher's Note:
1. Alkene ozonolysis gives carbonyls; alkyne ozonolysis gives acids.
2. Each alkyne carbon becomes a COOH carbon.

Answer: (c) \( \ce{Fe4[Fe(CN)6]3} \)

Explanation:
1. Na fusion converts N to NaCN.
2. NaCN with FeSO4 gives Na4[Fe(CN)6]; on adding FeCl3 the ferric ferrocyanide Fe4[Fe(CN)6]3 forms, which is Prussian blue.

Teacher's Note:
1. Fe(SCN)3 is blood red (N and S together).
2. The nitroprusside violet colour indicates sulphur.

Answer: (b) both nitrogen and sulphur

Explanation:
1. With both N and S, sodium thiocyanate NaSCN forms during fusion.
2. NaSCN with ferric ions gives the blood-red ferric thiocyanate.

Teacher's Note:
1. N alone gives Prussian blue; S alone gives a purple colour with sodium nitroprusside.
2. No blood-red colour appears if only one of them is present.

Answer: (a) chlorine

Explanation:
1. AgCl is a white precipitate that dissolves readily in dilute ammonia.
2. AgBr is dull yellow and only partly soluble; AgI is yellow and insoluble.

Teacher's Note:
1. Acidify the extract with nitric acid first to remove CN⁻ and S²⁻.
2. The solubility in ammonia decreases from Cl to I.

Answer: (d) \( \ce{NH4NO3} \)

Explanation:
1. The test needs sodium fusion to form NaCN from carbon and nitrogen.
2. NH4NO3 contains no carbon, so no cyanide ion forms.
3. All the other compounds contain both C and N.

Teacher's Note:
1. Carbon is essential for forming the cyanide ion.
2. Compounds such as hydrazine and ammonium salts do not respond.

Answer: (c) immediate turbidity

Explanation:
1. Lucas reagent converts the alcohol to the alkyl chloride, which is insoluble and makes the solution turbid.
2. The rate follows the stability of the carbocation: tertiary reacts immediately, secondary in about 5 minutes and primary only on heating or after about 30 minutes.

Teacher's Note:
1. Order of reactivity: 3° > 2° > 1°.
2. The tertiary carbocation is the most stable.

Answer: (b) \( \ce{CH2=CH-CH2-OH} \)

Explanation:
1. Allyl alcohol ionises to the allyl carbocation, which is stabilised by delocalisation (it behaves like a tertiary carbocation).
2. Ordinary primary alcohols give no turbidity at room temperature for a long time.

Teacher's Note:
1. Delocalisation-stabilised carbocations (allyl, benzyl) form easily even from primary alcohols.
2. Benzyl alcohol also reacts quickly.