JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 56

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Chapter-wise Objective Questions: Organic Chemistry Some Basic Principles and Techniques

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Multiple Choice Questions

Question 1: In nucleophilic aliphatic substitution the leaving power of a group increases as its basicity decreases. The group with the maximum leaving power among -OSO2CF3 (triflate), -OSO2C4F9 (nonaflate), tosylate and Cl- is
(a) Cl-
(b) triflate, -OSO2CF3
(c) tosylate
(d) nonaflate, -OSO2C4F9
Show Answer & Explanation

Answer: (d) nonaflate, -OSO2C4F9

Explanation:
1. Strongly electron withdrawing perfluoroalkyl groups stabilise the sulphonate anion.
2. The longer perfluorobutyl group stabilises the negative charge a little more than CF3, so nonaflate is the weakest base of the four and the best leaving group.

Teacher's Note:
1. Compare the conjugate acids.
2. Cl- is far less stable than the sulphonates.

Question 2: Which of the following is/are ambident nucleophile(s)? (A) NO2- (B) CN- (C) NaHSO3 (HSO3-) (D) Cl-
(a) A, B and C only
(b) B and C only
(c) A, B, C and D only
(d) A and B only
Show Answer & Explanation

Answer: (a) A, B and C only

Explanation:
1. An ambident nucleophile has two different nucleophilic sites.
2. NO2- (N or O), CN- (C or N) and HSO3- (S or O) are ambident; Cl- has only one atom.

Teacher's Note:
1. Look for more than one donor atom.
2. Cl- is monoatomic.

Question 3: Which of the following is an electrophilic reagent?
(a) OH-
(b) None of these
(c) NO2+
(d) H2O
Show Answer & Explanation

Answer: (c) NO2+

Explanation:
1. An electrophile is electron deficient and accepts an electron pair.
2. The nitronium ion NO2+ carries a positive charge and has a vacant orbital on nitrogen, so it is an electrophile.
3. H2O and OH- have lone pairs and are nucleophiles.

Teacher's Note:
1. Positive or electron deficient species seek electrons.
2. Species with lone pairs act as nucleophiles.

Question 4: The decreasing order of ease of alkaline hydrolysis of ethyl esters of (I) benzoic acid, (II) 4-chlorobenzoic acid, (III) 4-nitrobenzoic acid and (IV) 4-methoxybenzoic acid is
(a) II > III > I > IV
(b) IV > II > III > I
(c) III > II > IV > I
(d) III > II > I > IV
Show Answer & Explanation

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

Explanation:
1. Alkaline hydrolysis proceeds by attack of OH- on the carbonyl carbon; electron withdrawing groups make it more electrophilic and stabilise the tetrahedral intermediate.
2. NO2 (strong -M, -I) > Cl (-I) > H > OCH3 (+M).

Teacher's Note:
1. Withdrawing groups speed up hydrolysis.
2. Donors slow it.

Question 5: The decreasing order of the ratio of HCN addition to (a) HCHO, (b) CH3COCH3, (c) PhCOCH3 and (d) PhCOPh is
(a) d > b > c > a
(b) d > c > b > a
(c) c > d > b > a
(d) a > b > c > d
Show Answer & Explanation

Answer: (d) a > b > c > d

Explanation:
1. Equilibrium and rate of cyanohydrin formation fall with increasing hindrance and decreasing electrophilicity of the carbonyl group.
2. HCHO > acetone > acetophenone > benzophenone.

Teacher's Note:
1. Aldehyde > aliphatic ketone > aromatic ketone.
2. Two phenyl groups hinder most.

Question 6: The treatment of CH3MgX with CH3-C≡C-H produces
(a) CH3-C≡C-CH3
(b) CH4
(c) CH3CH=CHCH3
(d) CH3-CH=CH2
Show Answer & Explanation

Answer: (b) CH4

Explanation:
1. The terminal alkyne C-H is acidic (sp carbon).
2. CH3MgX is protonated to methane, giving CH3C≡CMgX.

Teacher's Note:
1. Grignard reagents deprotonate terminal alkynes.
2. Methane escapes as gas.

Question 7: In R-CO-Z + Nu- → R-CO-Nu + Z-, the rate of the reaction is fastest when Z is
(a) OC2H5
(b) NH2
(c) OCOCH3
(d) Cl
Show Answer & Explanation

Answer: (d) Cl

Explanation:
1. Rate depends on the leaving group ability of Z-.
2. Cl- is the weakest base, so it leaves fastest.

Teacher's Note:
1. Order: Cl > OCOR > OR > NH2.
2. Compare basicity of Z-.

Question 8: The order of reactivity of phenylmagnesium bromide with (I) acetone, (II) acetaldehyde and (III) benzophenone is
(a) II > I > III
(b) all react at the same rate
(c) II > III > I
(d) I > III > II
Show Answer & Explanation

Answer: (a) II > I > III

Explanation:
1. Aldehydes are more reactive than ketones.
2. Acetone is more reactive than benzophenone (steric hindrance and conjugation with the two rings).

Teacher's Note:
1. Fewer and smaller substituents mean faster addition.
2. Aromatic ketones are the least reactive.

Question 9: 3-(Aminomethyl)benzamide is treated with acetic anhydride in CH2Cl2. The major product is
(a) the ammonium acetate salt of the free diamine
(b) 3-(acetamidomethyl)benzamide + CH3COOH
(c) the diacetylated compound + H2O
(d) the N-acetylated amide with the free amine, + CH3COOH
Show Answer & Explanation

Answer: (b) 3-(acetamidomethyl)benzamide + CH3COOH

Explanation:
1. The benzylic NH2 is much more nucleophilic than the amide nitrogen (lone pair delocalised into C=O).
2. Acylation therefore occurs at the amine: ArCH2NHCOCH3 + CH3COOH, leaving the amide unchanged.

Teacher's Note:
1. More basic nitrogen reacts first.
2. Amide N is non-nucleophilic.

Question 10: A compound P is treated with PhMgBr followed by H2O to give CH3CH(OH)Ph as a racemic (d,l) mixture. P can be
(a) CH3COOH
(b) HCOOCH3
(c) CH3CHO
(d) CH3COCl
Show Answer & Explanation

Answer: (c) CH3CHO

Explanation:
1. CH3CHO + PhMgBr gives CH3CH(OMgBr)Ph, then CH3CH(OH)Ph, a secondary alcohol with one stereocentre, formed as a racemic mixture (attack from either face).
2. Acid derivatives would give tertiary alcohols or fail (CH3COOH just loses a proton).

Teacher's Note:
1. The product is a secondary alcohol, so P must be an aldehyde.
2. Planar C=O is attacked from both faces.

Question 11: The reaction (CH3)2C=CH2 + H2O (H+) → (CH3)3COH is an example of
(a) an elimination reaction
(b) a substitution reaction
(c) an addition reaction
(d) an acid-base reaction
Show Answer & Explanation

Answer: (c) an addition reaction

Explanation:
1. The pi bond is broken and H and OH add across the double bond.
2. This is electrophilic addition (hydration) following Markovnikov's rule.

Teacher's Note:
1. The product is saturated.
2. Two atoms or groups add to a pi bond.

Question 12: A table gives three columns for reactions: Column-1 reactions, Column-2 reaction type, Column-3 lab test for reactants. (I) CH3CH(OH)CH2CH2COOH with H+ ; (II) 2-naphthol with PhMgBr ; (III) cyclohexyl methyl ketone with NaBD4 / H2O ; (IV) CH3CH2NH2 with CH3MgBr. Types: (i) acid base reaction, (ii) nucleophilic addition, (iii) nucleophilic substitution, (iv) Fischer esterification. Tests: (P) 2,4-DNP test, (Q) carbylamine test, (R) Lucas test, (S) neutral FeCl3 test. The only correct combination that gives two different stereoisomeric products is
(a) (III) (iii) (P)
(b) (II) (i) (S)
(c) (I) (iv) (R)
(d) (IV) (iii) (Q)
Show Answer & Explanation

Answer: (c) (I) (iv) (R)

Explanation:
1. (I) 4-hydroxypentanoic acid (racemic) cyclises by intramolecular Fischer esterification (iv) to a gamma-lactone with one stereocentre, formed as two enantiomers; the alcohol is secondary so it responds to the Lucas test (R).
2. Other options either have the wrong type or give no pair of stereoisomers.

Teacher's Note:
1. A racemic mixture contains two stereoisomers.
2. Secondary alcohols react with Lucas reagent in a few minutes.

Question 13: A sweet smelling compound reacts with LiAlH4 to give only ethanol. Its molecular weight is
(a) 46
(b) 60
(c) 88
(d) 74
Show Answer & Explanation

Answer: (c) 88

Explanation:
1. An ester gives two alcohols on reduction; to give only ethanol both parts must be ethyl-derived: ethyl ethanoate CH3COOC2H5.
2. Molar mass = 4(12) + 8(1) + 2(16) = 88.

Teacher's Note:
1. Reduction splits an ester into acyl-derived and alkyl-derived alcohols.
2. Both are ethanol here.

Question 14: In the reaction C6H5CH2-X + NaCN → C6H5CH2CN, the rate is highest when X is
(a) -OH
(b) -NH2
(c) -S(=O)2OCH3 (attached through sulphur)
(d) -OSO2CH3 (mesylate, attached through oxygen)
Show Answer & Explanation

Answer: (d) -OSO2CH3 (mesylate, attached through oxygen)

Explanation:
1. The rate depends on the leaving group ability of X.
2. Mesylate CH3SO3- is a very weak base (conjugate acid of a strong sulphonic acid), so it is an excellent leaving group, whereas OH- and NH2- are strong bases and poor leaving groups.

Teacher's Note:
1. A good leaving group is a weak base.
2. Sulphonates are among the best leaving groups.

Question 15: On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is
(a) CH3Cl + C2H5COONa
(b) CH3COONa + C2H5OH
(c) CH3COOC2H5 + NaCl
(d) CH3COCl + C2H5OH + NaOH
Show Answer & Explanation

Answer: (c) CH3COOC2H5 + NaCl

Explanation:
1. Cl- is a very weak nucleophile and a better leaving group than C2H5O-, so substitution would be unfavourable.
2. No reaction occurs; the solution contains ethyl acetate and NaCl.

Teacher's Note:
1. Substitution proceeds towards the weaker base.
2. The incoming Cl- is a weaker base than EtO-.

Question 16: A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was
(a) CH3COOH
(b) HCHO
(c) CH3COCH3
(d) CH3OH
Show Answer & Explanation

Answer: (a) CH3COOH

Explanation:
1. A fruity smell indicates an ester.
2. Ethanol + a carboxylic acid with H2SO4 catalyst forms an ester (Fischer esterification), so the liquid is CH3COOH, giving ethyl acetate.

Teacher's Note:
1. Esters smell fruity.
2. Acid + alcohol + H+ gives ester.

Question 17: Which one of the following has the maximum nucleophilicity?
(a) CH3O-
(b) (CH3)3CO-
(c) NH2-
(d) CH3-
Show Answer & Explanation

Answer: (d) CH3-

Explanation:
1. In a period nucleophilicity falls from left to right with increasing electronegativity of the donor atom.
2. The carbanion CH3- (carbon donor) is the least electronegative and so the strongest nucleophile; (CH3)3CO- is also hindered.

Teacher's Note:
1. Compare C, N and O as donor atoms.
2. Order: CH3- > NH2- > OH- > F-.

Question 18: 2-Methylbutanal (one pure enantiomer, with a stereocentre at C-2) reacts with KCN followed by H+. The products obtained are
(a) diastereomers
(b) enantiomers
(c) a single optically pure product
(d) a meso compound
Show Answer & Explanation

Answer: (a) diastereomers

Explanation:
1. The new cyanohydrin carbon is a new stereocentre formed with either configuration.
2. The existing stereocentre keeps its configuration, so the pair of products are diastereomers (not mirror images).

Teacher's Note:
1. Pair of products with one fixed centre and one new centre.
2. Diastereomers have different physical properties.

Question 19: delta-Valerolactone (tetrahydropyran-2-one) is treated with excess PhMgBr and then H2O. The product is
(a) 5-phenylpentan-1-ol
(b) 2,2-diphenyltetrahydropyran
(c) 1,5-diphenoxypentane
(d) 1,1-diphenylpentane-1,5-diol
Show Answer & Explanation

Answer: (d) 1,1-diphenylpentane-1,5-diol

Explanation:
1. The Grignard reagent opens the lactone ring to give a ketone (with an alkoxide at the other end).
2. The ketone reacts with a second PhMgBr to give a tertiary alkoxide.
3. After hydrolysis: HOCH2CH2CH2CH2C(OH)Ph2, 1,1-diphenylpentane-1,5-diol.

Teacher's Note:
1. Lactones behave as cyclic esters.
2. Two equivalents add to the carbonyl carbon.

Question 20: The major product obtained when an amino alcohol of the type Ph-C(OH)(R)-CH(CH3)-NH2 (benzylic OH and a neighbouring primary amine) is treated with one equivalent of (CH3CO)2O/pyridine at room temperature is the
(a) O-acetyl compound (ester, free NH2)
(b) alkene with NHCOCH3
(c) ring acylated ketone with free OH and NH2
(d) N-acetyl compound (amide, free OH)
Show Answer & Explanation

Answer: (d) N-acetyl compound (amide, free OH)

Explanation:
1. The amine nitrogen is much more nucleophilic than the (tertiary, hindered) alcohol oxygen.
2. Acylation occurs on nitrogen to give the amide, the OH remains free.

Teacher's Note:
1. N is a better nucleophile than O.
2. Tertiary alcohols are hindered.

Question 21: Which of the following is NOT a nucleophile?
(a) (CH3)2NH
(b) C2H5OH
(c) H2O
(d) AlCl3
Show Answer & Explanation

Answer: (d) AlCl3

Explanation:
1. AlCl3 has an incomplete octet on aluminium and accepts an electron pair: a Lewis acid, hence an electrophile.
2. The other three have lone pairs on N or O.

Teacher's Note:
1. Nucleophiles are Lewis bases.
2. Look for a lone pair.

Question 22: The reaction RCOOH + PCl5 → RCOCl + POCl3 + HCl is an example of
(a) an addition reaction
(b) an elimination reaction
(c) an acid-base reaction
(d) a substitution reaction
Show Answer & Explanation

Answer: (d) a substitution reaction

Explanation:
1. The OH of the acid is replaced by Cl.
2. Replacement of one group by another is substitution (nucleophilic acyl substitution).

Teacher's Note:
1. No double bond is formed or lost.
2. A group is swapped, not added or removed.

Question 23: Phenylmagnesium bromide reacts with tert-butyl alcohol to give
(a) Ph-C(CH3)3
(b) Ph-OH
(c) Ph-O-C(CH3)3
(d) Ph-H
Show Answer & Explanation

Answer: (d) Ph-H

Explanation:
1. The Grignard reagent is a strong base and the alcohol has an acidic O-H.
2. PhMgBr + (CH3)3COH gives benzene + (CH3)3COMgBr.

Teacher's Note:
1. Alcohols protonate Grignard reagents.
2. No C-C bond forms.

Question 24: The correct decreasing reactivity order(s) towards hydrolysis under identical conditions is/are (A) CH3COCl > CH3CONH2 (B) CH3COCl > (CH3CO)2O (C) CH3COOCH3 > CH3COCl (D) (CH3CO)2O > CH3CONH2
(a) A and B only
(b) A, B, C and D only
(c) A, B and D only
(d) B and D only
Show Answer & Explanation

Answer: (c) A, B and D only

Explanation:
1. Reactivity: acyl chloride > anhydride > ester > amide.
2. A, B and D follow this order; C reverses it.

Teacher's Note:
1. Leaving group ability decides.
2. The acyl chloride is always the most reactive.

Question 25: HCN adds with the fastest rate to
(a) acetophenone
(b) benzophenone
(c) acetone
(d) ethanal
Show Answer & Explanation

Answer: (d) ethanal

Explanation:
1. Aldehydes are more reactive than ketones towards nucleophilic addition.
2. Ethanal is the only aldehyde and has the least hindrance.

Teacher's Note:
1. Order of reactivity: HCHO > RCHO > R2CO > ArCOR > Ar2CO.
2. Bulky groups hinder the approach of CN-.

Question 26: Sodium ethoxide reacted with ethanoyl chloride. The compound produced is
(a) ethyl ethanoate
(b) diethyl ether
(c) 2-butanone
(d) ethyl chloride
Show Answer & Explanation

Answer: (a) ethyl ethanoate

Explanation:
1. EtO- attacks the acyl carbon of CH3COCl and displaces Cl- (nucleophilic acyl substitution).
2. The product is CH3COOC2H5, ethyl ethanoate.

Teacher's Note:
1. Cl is the leaving group.
2. Alkoxide + acid chloride gives ester.

Question 27: The increasing order of reactivity with LiAlH4 of (A) C2H5CONH2, (B) C2H5COOCH3, (C) C2H5COCl and (D) (C2H5CO)2O is
(a) B < A < C < D
(b) A < B < D < C
(c) A < B < C < D
(d) B < A < D < C
Show Answer & Explanation

Answer: (b) A < B < D < C

Explanation:
1. Reactivity follows leaving group ability: NH2- (poorest) < CH3O- < RCOO- < Cl-.
2. Hence amide < ester < anhydride < acid chloride.

Teacher's Note:
1. Same order as nucleophilic acyl substitution.
2. Acid chloride is the most reactive.

Question 28: 2-Butanone is reduced (i) with NaBD4 in H2O to give A, (ii) with NaBD4 in D2O to give B, and (iii) with NaBH4 in D2O to give C. A, B and C are respectively
(a) CH3CD(OH)CH2CH3, CH3CH(OH)CH2CH3, CH3CH(OD)CH2CH3
(b) CH3CD(OH)CH2CH3, CH3CD(OD)CH2CH3, CH3CH(OD)CH2CH3
(c) CH3CD(OH)CH2CH3 in all three cases
(d) CH3CH(OH)CH2CH3 in all three cases
Show Answer & Explanation

Answer: (b) CH3CD(OH)CH2CH3, CH3CD(OD)CH2CH3, CH3CH(OD)CH2CH3

Explanation:
1. The hydride/deuteride from the borohydride bonds to carbon; the solvent supplies the proton/deuteron for oxygen.
2. A: D on C, H on O. B: D on C, D on O. C: H on C, D on O.

Teacher's Note:
1. Reagent labels carbon; solvent labels oxygen.
2. BD4- delivers D, BH4- delivers H.

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