JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 43

Answer: (c) 3-Ethyl-4,4-dimethylheptane

Explanation:
1. Substituents are listed in alphabetical order (ethyl before methyl), ignoring multiplying prefixes like di.
2. Locants of the same substituent are separated by commas and locants are separated from names by hyphens.
3. So the correct name is 3-ethyl-4,4-dimethylheptane.

Teacher's Note:
1. Alphabetical order ignores di-, tri- and so on.
2. Do not use bis- for simple substituents.

Answer: (c) 4-oxopentanoic acid

Explanation:
1. The carboxylic acid group has the highest priority and is the principal group, with C1 as the COOH carbon.
2. The chain has 5 carbons and a ketone (oxo) at C4.
3. The name is 4-oxopentanoic acid.

Teacher's Note:
1. Carboxylic acid has priority over ketone.
2. The ketone is named as an oxo prefix.

Answer: (c) III

Explanation:
1. Electronegativity increases with the s character of the hybrid orbital: sp (50 per cent) > sp2 (33 per cent) > sp3 (25 per cent).
2. Only the carbon in (III) is sp hybridised, so it is the most electronegative.

Teacher's Note:
1. More s character means the electrons are held closer to the nucleus.
2. This explains the acidity of terminal alkynes.

Answer: (a) 3,4-dimethylhexane

Explanation:
1. The longest continuous chain goes through both ethyl groups: CH3CH2-CH(CH3)-CH(CH3)-CH2CH3, which has 6 carbons.
2. The two methyl branches are on carbons 3 and 4.
3. The name is 3,4-dimethylhexane.

Teacher's Note:
1. Look for the longest chain, which may pass through a substituent drawn as ethyl.
2. Number to give the lowest locants to substituents.

Answer: (d) \(\ce{F2CHCOO-}\)

Explanation:
1. Electron-withdrawing groups disperse the negative charge of the carboxylate ion by the -I effect and stabilise it.
2. Two fluorine atoms give the strongest -I effect, so \(\ce{F2CHCOO-}\) is the most stable.

Teacher's Note:
1. Stability of the conjugate base increases acid strength.
2. -I effect order: F > Cl, and more halogens increase the effect.

Answer: (a) 2 degree carbocation

Explanation:
1. H+ is an electrophile and adds to the terminal carbon (Markovnikov's rule).
2. This gives \(\ce{CH3-CH+-CH3}\), a secondary carbocation, which is more stable than a primary one.

Teacher's Note:
1. Electrophilic addition goes through carbocations.
2. More substituted carbocations are more stable.

Answer: (d) a pair of electrons to donate

Explanation:
1. Nucleophiles are electron-pair donors (Lewis bases) that attack sites of low electron density.
2. They need not be negatively charged; for example NH3 and H2O are neutral nucleophiles.

Teacher's Note:
1. Nucleophile means nucleus-loving.
2. Not all nucleophiles are anions.

Answer: (d) A is incorrect but R is correct

Explanation:
1. The resonance hybrid is more stable (lower in energy) than any single canonical form, so the assertion is false.
2. It is true that the hybrid cannot be represented by a single Lewis structure.

Teacher's Note:
1. The extra stability is called resonance energy.
2. Canonical forms do not have a real existence.

Answer: (d) \(\ce{RS-}\) is less basic but more nucleophilic than \(\ce{RO-}\)

Explanation:
1. Thiols are stronger acids than alcohols, so RS- is the weaker base.
2. Sulphur is larger and more polarisable and less strongly solvated, so RS- is a better nucleophile.

Teacher's Note:
1. Nucleophilicity increases down a group in protic solvents.
2. Basicity and nucleophilicity do not always parallel.

Answer: (d) \(\ce{CH3COO- < HC#C- < NH2- < CH3-}\)

Explanation:
1. The weaker the conjugate acid, the stronger the base.
2. Acidity of conjugates: CH3COOH (pKa 4.8) > HC≡CH (25) > NH3 (35) > CH4 (about 48).
3. So basicity increases as CH3COO- < HC≡C- < NH2- < CH3-.

Teacher's Note:
1. Use pKa values of the conjugate acids.
2. A carbanion on sp3 carbon is the strongest base here.

Answer: (c) 4

Explanation:
1. The molecule has one stereogenic centre (C bearing OH) and one double bond that can show cis-trans isomerism.
2. Number of stereoisomers = 2 x 2 = 4.

Teacher's Note:
1. Count each independent stereo-element.
2. The two ends of the double bond carry different groups.

Answer: (b) -COOH, -SO3H, -CONH2, -CHO

Explanation:
1. The IUPAC order of seniority is: carboxylic acid, sulphonic acid, anhydride, ester, acid halide, amide, nitrile, aldehyde, ketone, alcohol, amine.
2. So COOH > SO3H > CONH2 > CHO.

Teacher's Note:
1. Remember: acid, sulphonic, anhydride, ester, acyl halide, amide, nitrile, aldehyde, ketone, alcohol, amine.
2. The senior group becomes the suffix.

Answer: (b) \(\ce{RCH2X > R2CHX > R3CX}\)

Explanation:
1. SN2 reactions are sterically controlled because the nucleophile attacks from the back.
2. Less hindered halides react faster: primary > secondary > tertiary.

Teacher's Note:
1. Methyl > primary > secondary > tertiary for SN2.
2. The order is reversed for SN1.

Answer: (a) chemically reactive

Explanation:
1. A free radical has an odd electron which tends to pair up, so radicals are highly reactive.

Teacher's Note:
1. Radicals are neutral species with an unpaired electron.
2. They are intermediates in chain reactions.

Answer: (d) Optical

Explanation:
1. It has two identical stereogenic centres, so it exists as a pair of enantiomers and a meso form.
2. It shows optical isomerism.

Teacher's Note:
1. Two identical chiral centres lead to a meso form.
2. Geometrical isomerism needs a double bond or ring.

Answer: (b) Gaseous densities at the same temperature and pressure

Explanation:
1. Both have the same molar mass (46 g/mol).
2. For ideal gases at the same T and p, density depends only on the molar mass, so the densities are equal.
3. The other properties differ due to hydrogen bonding in ethanol.

Teacher's Note:
1. d = PM/RT for ideal gases.
2. Isomers have the same molecular mass.

Answer: (d) stereoisomerism

Explanation:
1. But-2-ene shows cis-trans (geometrical) isomerism and butan-2-ol shows optical isomerism.
2. Both are types of stereoisomerism.

Teacher's Note:
1. Stereoisomers differ only in the arrangement of atoms in space.
2. Geometrical and optical isomerism are stereoisomerism.

Answer: (d) 37.33

Explanation:
1. 10 mL of 1 M H2SO4 = 10 mmol, which neutralises 20 mmol NH3.
2. Nitrogen = 20 mmol x 14 = 280 mg = 0.28 g.
3. Percentage = (0.28/0.75) x 100 = 37.33.

Teacher's Note:
1. Each mole of H2SO4 neutralises two moles of NH3.
2. 1 mol NH3 corresponds to 1 mol of N.

Answer: (d) decompose \(\ce{Na2S}\) and \(\ce{NaCN}\) formed

Explanation:
1. Cyanide and sulphide would also give precipitates (AgCN, Ag2S) with silver nitrate and interfere.
2. Boiling with HNO3 destroys them as HCN and H2S, which escape.

Teacher's Note:
1. Always boil with HNO3 before adding AgNO3.
2. If N and S are both present, extra care is needed.

Answer: (b) \(\ce{Cl-}\)

Explanation:
1. Cl- has lone pairs and is a nucleophile.
2. BH3 is electron deficient, H3O+ can donate a proton, and NO2+ is a strong electrophile.

Teacher's Note:
1. Electrophiles are electron deficient.
2. Anions with lone pairs are nucleophiles.

Answer: (b) \(\ce{Cl- < Br- < I-}\)

Explanation:
1. In protic solvents, smaller ions are strongly solvated by hydrogen bonding, which reduces their nucleophilicity.
2. Larger, more polarisable ions are better nucleophiles, so Cl- < Br- < I-.

Teacher's Note:
1. The order is reversed in aprotic solvents.
2. Iodide is an excellent nucleophile in water or alcohol.

Answer: (b) \(\ce{RCOOH > HOH > ROH > HC#CH}\)

Explanation:
1. pKa values: RCOOH about 5, H2O 15.7, ROH 16-18 and HC≡CH 25.
2. The alkyl group in ROH increases the electron density on oxygen, so alcohols are weaker acids than water.
3. So the order is RCOOH > HOH > ROH > HC≡CH.

Teacher's Note:
1. Alkyl groups reduce acidity (+I effect).
2. Terminal alkynes are weaker acids than water.

Answer: (d) Tollens' reagent is used in the detection of unsaturation

Explanation:
1. Unsaturation is detected using Baeyer's reagent (cold dilute alkaline KMnO4) or bromine water.
2. Tollens' reagent detects aldehydes (reducing sugars).

Teacher's Note:
1. Tollens' and Fehling's are for reducing groups.
2. Bisulphite forms addition products with aldehydes and ketones.

Answer: (d) sodium bicarbonate

Explanation:
1. Carboxylic acids are strong enough to react with NaHCO3, releasing CO2 and forming a water-soluble salt.
2. Phenols are weaker acids than carbonic acid and do not react with bicarbonate, while NaOH would dissolve both.

Teacher's Note:
1. Use the bicarbonate test to distinguish carboxylic acids from phenols.
2. pKa of phenol is 10, higher than that of carbonic acid (6.4).

Answer: (b) 7

Explanation:
1. Structural ethers: methyl butyl, methyl sec-butyl, methyl isobutyl, methyl tert-butyl, ethyl propyl and ethyl isopropyl = 6.
2. Methyl sec-butyl ether has a chiral carbon, adding one more stereoisomer, so the total is 7.

Teacher's Note:
1. Count enantiomers separately when stereoisomers are included.
2. Systematically vary the alkyl groups on the two sides of oxygen.

Answer: (d) 2

Explanation:
1. The alcohols are butan-1-ol, butan-2-ol (R and S), 2-methylpropan-1-ol and 2-methylpropan-2-ol.
2. Only butan-2-ol has a chiral carbon, so its two enantiomers are the optically active forms: 2.

Teacher's Note:
1. A chiral centre needs four different groups.
2. Count each enantiomer separately.

Answer: (c) 6

Explanation:
1. Alkenes of C5H10: pent-1-ene, cis- and trans-pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene.
2. Counting cis and trans separately, the total is 6.

Teacher's Note:
1. Geometrical isomers count as separate isomers here.
2. Only pent-2-ene shows cis-trans isomerism.

Answer: (c) 4

Explanation:
1. For n dissimilar chiral centres there are 2^n optically active forms.
2. With n = 2, the number is 4 (two enantiomeric pairs).

Teacher's Note:
1. Dissimilar centres give no meso form.
2. Number of stereoisomers = 2^n.

Answer: (c) 7

Explanation:
1. The smallest chiral alkane is 3-methylhexane, \(\ce{CH3CH2CH(CH3)CH2CH2CH3}\), whose C3 has H, CH3, C2H5 and C3H7 groups.
2. It has 7 carbon atoms.

Teacher's Note:
1. Smaller alkanes cannot have four different groups on one carbon.
2. 2-Methylbutane and 3-methylpentane are not chiral.