JEE Chemistry Haloalkanes and Haloarenes MCQs Set 04

Answer: (c) \(\ce{C2H5OH + NaCl ->}\)

Explanation:
1. Alcohol with HCl and ZnCl2 (Lucas reagent) gives an alkyl chloride, HBr adds to propene to give 2-bromopropane, and the silver salt with bromine gives an alkyl bromide (Borodin-Hunsdiecker reaction).
2. Sodium chloride cannot replace the OH group of an alcohol, because OH- is a very poor leaving group and there is no acid to protonate it.

Teacher's Note:
1. An alcohol needs HX, PX3 or SOCl2 to be converted into an alkyl halide.
2. A neutral chloride salt will not do it.

Answer: (b) \(\ce{CHCl3}\)

Explanation:
1. Bleaching powder supplies chlorine and calcium hydroxide.
2. Chlorine trichlorinates the methyl group of acetone to give trichloroacetone, which is then cleaved by the base to give chloroform and calcium acetate.

Teacher's Note:
1. This is the laboratory and industrial route to chloroform.
2. Ethanol also gives chloroform by the same route.

Answer: (c) \(\ce{Cl3C-CH(OH)2}\)

Explanation:
1. Two hydroxyl groups on one carbon normally lose water to form a carbonyl compound.
2. In chloral hydrate the strong electron-withdrawing effect of the \(\ce{CCl3}\) group and intramolecular hydrogen bonding stabilise the gem-diol, so it can be isolated.

Teacher's Note:
1. Chloral hydrate is the stable gem-diol.
2. Its stability comes from the three chlorine atoms.

Answer: (a) b > c > d

Explanation:
1. The rate of an \(\ce{S_N1}\) reaction depends on the stability of the carbocation formed.
2. 7-Chlorocycloheptatriene gives the aromatic tropylium ion (six pi electrons), which is exceptionally stable.
3. 1-Methylcyclohexyl is a tertiary cation, while neopentyl chloride is primary and cannot ionise easily. Order: b > c > d.

Teacher's Note:
1. Aromatic carbocations are extra stable.
2. Tertiary beats primary in SN1.

Answer: (d) silver nitrate solution

Explanation:
1. The C - I bond is much weaker than the C - Cl bond, so iodoform ionises on heating with silver nitrate and gives a yellow precipitate of AgI.
2. Chloroform, with its strong C - Cl bonds, does not give a precipitate.
3. The other reagents do not separate them in a clear way.

Teacher's Note:
1. Weaker C - X bond means easier reaction with silver ions.
2. AgI is yellow, AgCl is white.

Answer: (d) 2,4,6-trinitrochlorobenzene

Explanation:
1. Hydrolysis of aryl halides proceeds by nucleophilic aromatic substitution, which is helped by electron-withdrawing groups at the ortho and para positions.
2. Each nitro group stabilises the intermediate carbanion by resonance.
3. So the compound with three nitro groups (all ortho and para) is most reactive.

Teacher's Note:
1. More ortho and para nitro groups mean faster SNAr.
2. Picryl chloride hydrolyses even with warm water.

Answer: (a) D.D.T.

Explanation:
1. D.D.T. (dichlorodiphenyltrichloroethane) is a well-known insecticide.
2. Chloretone is a hypnotic and preservative, and chloroform is an anaesthetic and solvent.

Teacher's Note:
1. DDT is formed from chloral and chlorobenzene.
2. It is now banned in many places because it persists in the environment.

Answer: (d) tert-butyl bromide with \(\ce{CH3ONa}\)

Explanation:
1. A tertiary halide with a strong base and no good nucleophile pathway follows \(\ce{E2}\) elimination, giving 2-methylpropene.
2. The reactions in (a), (b) and (c) are substitution reactions of methyl or primary halides.

Teacher's Note:
1. Tertiary halide plus strong base gives an alkene.
2. Methyl and primary halides prefer substitution.

Answer: (b) \(\ce{R-H}\)

Explanation:
1. A Grignard reagent behaves as a strong base and takes a proton from any compound with an acidic N - H or O - H bond.
2. The alkyl group of the Grignard reagent therefore leaves as the alkane \(\ce{R-H}\), and the amide \(\ce{RNH-MgX}\) is formed.

Teacher's Note:
1. Grignard reagents are destroyed by amines, alcohols and water.
2. The R group becomes R-H.

Answer: (a) 1,2-dibromocyclohexane with Zn dust on heating gives cyclohexa-1,3-diene

Explanation:
1. Zinc removes two vicinal bromine atoms, so 1,2-dibromocyclohexane gives cyclohexene, not a diene.
2. The other reactions are correct: Finkelstein reaction in dry acetone, dehalogenation of a vicinal dibromide to but-2-ene, and double dehydrohalogenation of a gem-dihalide with strong base giving propyne after protonation.

Teacher's Note:
1. Zn with a vicinal dihalide removes both halogens and forms one double bond.
2. Excess amide base converts a gem-dihalide into a terminal alkyne.

Answer: (c) \(\ce{C6H5CH2Cl}\)

Explanation:
1. The faster the carbocation forms, the faster the \(\ce{S_N1}\) reaction.
2. The benzyl cation is stabilised by resonance with the ring, which is better than the secondary isopropyl cation or the primary ethyl cation.
3. Chlorobenzene does not undergo SN1 at all because the C - Cl bond has partial double bond character and the phenyl cation is unstable.

Teacher's Note:
1. Benzylic halides are very reactive in SN1.
2. Aryl halides do not ionise.

Answer: (b) cyclohexanone

Explanation:
1. The methyl ketone group gives the iodoform reaction: A is \(\ce{CHI3}\) and B is the sodium salt of 2-oxocyclohexanecarboxylic acid.
2. Acidification gives the beta-keto acid C.
3. A beta-keto acid loses \(\ce{CO2}\) on heating, so D is cyclohexanone.

Teacher's Note:
1. Beta-keto acids decarboxylate on warming.
2. The haloform reaction converts the acetyl group into a carboxylate.

Answer: (c) \(\ce{-NO2}\)

Explanation:
1. This is nucleophilic aromatic substitution, which is favoured by electron-withdrawing groups at the ortho and para positions.
2. The nitro group stabilises the negative charge of the intermediate by resonance.
3. Donors like methoxy and methyl slow the reaction.

Teacher's Note:
1. SNAr needs a strong -M group at ortho or para.
2. Donor groups reduce the rate.

Answer: (a) 2,2-bis(4-chlorophenyl)-1,1,1-trichloroethane (D.D.T.)

Explanation:
1. Chloral acts as an electrophile in an electrophilic substitution on two molecules of chlorobenzene, with sulphuric acid as the dehydrating catalyst.
2. Each ring is attacked at the para position and water is lost.
3. The product is \(\ce{(ClC6H4)2CH-CCl3}\), D.D.T.

Teacher's Note:
1. DDT: two chlorophenyl groups on a CH attached to CCl3.
2. Chlorine directs the attack to the para position.

Answer: (b) \(\ce{C6H5CH3}\)

Explanation:
1. Magnesium converts benzyl bromide into the Grignard reagent \(\ce{C6H5CH2MgBr}\).
2. On treatment with aqueous acid the carbon-magnesium bond is protonated, giving toluene.

Teacher's Note:
1. Grignard reagent plus water gives the hydrocarbon.
2. The alcohol would need an aldehyde or oxygen.

Answer: (b) (a), (b) and (c)

Explanation:
1. Ethanol (oxidised to ethanal), propanone (a methyl ketone) and propan-2-ol (oxidised to propanone) all contain or form the \(\ce{CH3CO-}\) group.
2. Methanol cannot, because it has no methyl group attached to carbinol carbon.

Teacher's Note:
1. Only ethanol among primary alcohols gives the test.
2. Methanol never does.

Answer: (c) 1-(4-chlorophenyl)ethanone

Explanation:
1. Acetic anhydride with anhydrous \(\ce{AlCl3}\) acylates the ring (Friedel-Crafts acylation).
2. Chlorine is ortho/para directing but deactivating, and the bulky acylium ion prefers the para position over the hindered ortho position.
3. So the major product is 4-chloroacetophenone.

Teacher's Note:
1. Halogens direct ortho and para.
2. Para is usually major in acylation.

Answer: (b) III > II > I

Explanation:
1. Among isomeric alkyl halides, the boiling point falls with increased branching, because the molecule becomes more compact and van der Waals forces fall.
2. The straight chain n-butyl bromide boils highest (about 101 C), then sec-butyl (91 C), then tert-butyl (73 C). Order: III > II > I.

Teacher's Note:
1. More branching means a lower boiling point.
2. Compare surface areas.

Answer: (a) \(\ce{CH3Cl}\)

Explanation:
1. Dipole moment depends on both the charge separation and the bond length.
2. Although C - F is the most polar bond, it is very short, and the dipole moment of CH3Cl (1.87 D) is slightly greater than CH3F (1.85 D).
3. CH3Br (1.81 D) and CH3I (1.62 D) are lower.

Teacher's Note:
1. The order is CH3Cl > CH3F > CH3Br > CH3I.
2. Dipole equals charge times distance.

Answer: (b) salicylic acid

Explanation:
1. With carbon tetrachloride the Reimer-Tiemann reaction forms a trichloromethyl group at the ortho position.
2. Alkaline hydrolysis converts \(\ce{-CCl3}\) into \(\ce{-COO-}\), and acidification gives salicylic acid.

Teacher's Note:
1. CHCl3 gives -CHO but CCl4 gives -COOH.
2. The attacking species is dichlorocarbene for CHCl3.

Answer: (d) 2-methylpropane and \(\ce{CH3CH2OMgBr}\)

Explanation:
1. The Grignard reagent takes the acidic hydrogen of ethanol: \(\ce{(CH3)2CHCH2MgBr + C2H5OH -> (CH3)2CHCH3 + C2H5OMgBr}\).
2. The alkane formed is 2-methylpropane (isobutane).

Teacher's Note:
1. Alcohol protonates the carbanion of the Grignard reagent.
2. The R group becomes R-H.

Answer: (b) \(\ce{CH3-CH(OH)-CH2Cl}\)

Explanation:
1. Aqueous NaOH gives propan-1-ol (X), and heating with alumina dehydrates it to propene (Y).
2. Chlorine water adds \(\ce{Cl-OH}\) across the double bond.
3. The chloronium ion is opened by water at the more substituted carbon, so the product is 1-chloropropan-2-ol, \(\ce{CH3CH(OH)CH2Cl}\).

Teacher's Note:
1. Cl2 with water gives a halohydrin.
2. OH adds to the more substituted carbon.

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

Explanation:
1. Aromatic nucleophilic substitution is accelerated by electron-withdrawing groups, especially by -M groups at the ortho and para positions.
2. The para nitro compound (d) is strongly activated by resonance.
3. The meta nitro groups act only by their -I effect, so the dinitro compound (c) is next, then the mononitro (b), and chlorobenzene (a) is the least reactive. Order: d > c > b > a.

Teacher's Note:
1. Resonance activation from a para nitro group beats the inductive effect of meta groups.
2. More meta nitro groups help a little.

Answer: (d) triphenylmethane

Explanation:
1. Chloroform acts as an alkylating agent in the Friedel-Crafts reaction: each of its three C - Cl bonds can be replaced by a phenyl group in the presence of \(\ce{AlCl3}\).
2. The product is triphenylmethane, \(\ce{(C6H5)3CH}\).

Teacher's Note:
1. Polyhalides give multiple Friedel-Crafts substitutions.
2. CCl4 gives triphenylchloromethane.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Alkyl halides are polar but cannot form hydrogen bonds with water, since the halogen is a very poor hydrogen bond acceptor.
2. The energy released in a few weak dipole-dipole interactions does not compensate for breaking the strong hydrogen bonds of water, so they are insoluble.
3. The reason explains the assertion.

Teacher's Note:
1. Dissolving in water needs hydrogen bonding.
2. Polarity alone is not enough.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Isobutanal, \(\ce{(CH3)2CH-CHO}\), does not contain the \(\ce{CH3CO-}\) or \(\ce{CH3CH(OH)-}\) group, so it does not give the iodoform test. The assertion is true.
2. But it does have an alpha-hydrogen (on the CH carbon), so the reason is false.

Teacher's Note:
1. The iodoform test needs a methyl group on the carbonyl, not just an alpha hydrogen.
2. Check the reason statement separately.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. The proton adds to the terminal carbon to give the resonance-stabilised benzylic carbocation, so bromide adds to give 1-bromo-1-phenylethane (Markovnikov addition). The assertion is true.
2. The benzyl radical is indeed more stable than an alkyl radical, but the reaction goes through a carbocation, not a radical, so this statement does not explain it.

Teacher's Note:
1. The ionic mechanism involves the benzylic carbocation.
2. Radical stability is a different matter.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Chlorine withdraws electrons by its -I effect more strongly than it donates by +R, so the ring is deactivated and chlorobenzene reacts more slowly than benzene. The assertion is true.
2. The resonance effect of chlorine (donation of its lone pair) in fact stabilises the carbocation intermediate, so the reason is false.

Teacher's Note:
1. Chlorine is deactivating through -I but directs ortho and para through +R.
2. The two effects act in opposite directions.