Chapter 3

Organic Chemistry

MBBS and BDS entrance past exam questions

Past questions and answers

118 questions set from this chapter. Most repeated first.

2016 · NEET-UG (India) · 4 marks

In the reaction: $\mathrm{H-C\equiv CH}$ treated first with (1) $\mathrm{NaNH_2/liq.\ NH_3}$ and (2) $\mathrm{CH_3CH_2Br}$ to give X; X treated again with (1) $\mathrm{NaNH_2/liq.\ NH_3}$ and (2) $\mathrm{CH_3CH_2Br}$ to give Y. X and Y are: (a) X = 1-Butyne; Y = 3-Hexyne (b) X = 2-Butyne; Y = 3-Hexyne (c) X = 2-Butyne; Y = 2-Hexyne (d) X = 1-Butyne; Y = 2-Hexyne

Answer

(a) X = 1-Butyne; Y = 3-Hexyne

$\mathrm{NaNH_2}$ removes the acidic terminal hydrogen of ethyne to give $\mathrm{HC\equiv C^-Na^+}$ , which attacks $\mathrm{CH_3CH_2Br}$ ( $S_N2$ ) to give $\mathrm{HC\equiv C-CH_2CH_3}$ (1-butyne). Repeating the sequence on the remaining terminal H gives $\mathrm{CH_3CH_2-C\equiv C-CH_2CH_3}$ , 3-hexyne. 2-Butyne or 2-hexyne would need a methyl group on the triple bond, which is not supplied.

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2016 · NEET-UG (India) · 4 marks

The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon, is: (a) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol. (b) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration. (c) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation. (d) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.

Answer

(d) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.

A carbonyl compound with an $\alpha$ -hydrogen exists in rapid equilibrium with its enol form through migration of the $\alpha$ -H to oxygen; this is called keto-enol tautomerism. It is not called aldehyde-ketone equilibration or carbonylation (which means introducing CO), and the equilibrium certainly does occur, so (a), (b) and (c) are wrong.

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2016 · NEET-UG (India) · 4 marks

In a protein molecule various amino acids are linked together by (a) $\alpha$ -glycosidic bond (b) $\beta$ -glycosidic bond (c) peptide bond (d) dative bond

Answer

(c) peptide bond

Amino acids in proteins are joined by peptide bonds, the amide linkage $-\mathrm{CO-NH-}$ formed between the carboxyl group of one amino acid and the amino group of the next. Glycosidic bonds link monosaccharide units in carbohydrates, not amino acids.

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2016 · NEET-UG (India) · 4 marks

Natural rubber has: (a) All cis-configuration (b) All trans-configuration (c) Alternate cis- and trans-configuration (d) Random cis- and trans-configuration

Answer

(a) All cis-configuration

Natural rubber is cis-1,4-polyisoprene. Every double bond in the chain has cis geometry, which gives it a coiled structure and its elasticity. The all-trans form is gutta-percha, which is hard and non-elastic.

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2016 · NEET-UG (India) · 4 marks

Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer? (a) Acetone (b) Ozone (c) $\mathrm{MnO_2}$ (d) Aluminium isopropoxide

Answer

(a) Acetone

The cis-1,2-diol on a five-membered ring has both OH groups on the same face, so it reacts with acetone (acid catalyst) to form a cyclic acetonide (ketal). The trans isomer cannot, as the OH groups are on opposite faces and cannot bridge to form a fused five-membered ring. Ozone, $\mathrm{MnO_2}$ and aluminium isopropoxide do not discriminate between the two isomers in this way.

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2016 · NEET-UG (India) · 4 marks

Consider the nitration of benzene using mixed conc. $\mathrm{H_2SO_4}$ and $\mathrm{HNO_3}$ . If a large amount of $\mathrm{KHSO_4}$ is added to the mixture, the rate of nitration will be (a) faster (b) slower (c) unchanged (d) doubled

Answer

(b) slower

Nitration needs the nitronium ion: $\mathrm{HNO_3+2H_2SO_4\rightleftharpoons NO_2^++H_3O^++2HSO_4^-}$ . Adding $\mathrm{KHSO_4}$ supplies $\mathrm{HSO_4^-}$ (a common ion), which shifts this equilibrium to the left, so less $\mathrm{NO_2^+}$ forms and the nitration rate decreases.

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2016 · NEET-UG (India) · 4 marks

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is (a) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain. (b) The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain. (c) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain. (d) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Answer

(d) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

In the staggered form the C–H bonds on adjacent carbons are as far apart as possible, so there is no torsional strain. The eclipsed form has repulsion between the C–H bonds (about 12 kJ/mol of torsional strain), so it is less stable. The other options reverse this or wrongly attribute strain to the staggered form.

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2016 · NEET-UG (India) · 4 marks

The reaction: cyclopentanol is treated with NaH to give sodium cyclopentoxide, which is then treated with methyl iodide (Me–I) to give methoxycyclopentane. This can be classified as (a) Williamson ether synthesis reaction (b) Alcohol formation reaction (c) Dehydration reaction (d) Williamson alcohol synthesis reaction

Answer

(a) Williamson ether synthesis reaction

NaH converts the alcohol to the alkoxide, which attacks $\mathrm{CH_3I}$ by $S_N2$ to form an ether: alkoxide + alkyl halide $\rightarrow$ ether. This is the Williamson ether synthesis. The product is an ether (methoxycyclopentane), not an alcohol, and no water is lost, so it is not a dehydration.

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2016 · NEET-UG (India) · 4 marks

The product formed by the reaction of an aldehyde with a primary amine is (a) Schiff base (b) Ketone (c) Carboxylic acid (d) Aromatic acid

Answer

(a) Schiff base

An aldehyde reacts with a primary amine by nucleophilic addition–elimination (loss of water) to give an imine, $\mathrm{RCH=NR'}$ , called a Schiff base. No oxidation occurs, so ketones or acids are not formed.

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2016 · NEET-UG (India) · 4 marks

The correct statement regarding RNA and DNA, respectively is: (a) The sugar component in RNA is arabinose and the sugar component in DNA is 2'-deoxyribose. (b) The sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose. (c) The sugar component in RNA is arabinose and the sugar component in DNA is ribose. (d) The sugar component in RNA is 2'-deoxyribose and the sugar component in DNA is arabinose.

Answer

(b) The sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose.

RNA contains the pentose D-ribose, while DNA contains 2'-deoxy-D-ribose (no OH at the 2' carbon). Arabinose is not the sugar of either nucleic acid, so the other options are wrong.

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2016 · NEET-UG (India) · 4 marks

Which of the following statements about the basicity of arylamines is correct? (a) Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $\pi$ electron system. (b) Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring $\pi$ electron system. (c) Arylamines are generally more basic than alkylamines because of aryl group. (d) Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.

Answer

(a) Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $\pi$ electron system.

In arylamines the nitrogen lone pair is delocalised into the ring by resonance, so it is less available to accept a proton. In alkylamines the +I effect of alkyl groups increases electron density on N. Nitrogen in arylamines is $sp^2$ -like, not sp, so the 'more basic' options are wrong.

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2016 · NEET-UG (India) · 4 marks

Which one given below is a non-reducing sugar? (a) Maltose (b) Lactose (c) Glucose (d) Sucrose

Answer

(d) Sucrose

In sucrose, glucose and fructose are joined through their anomeric carbons ( $\alpha$ -1, $\beta$ -2 glycosidic bond), so neither ring can open to give a free aldehyde or ketone group. It is therefore non-reducing. Maltose, lactose and glucose all have a free hemiacetal or aldehyde group and are reducing.

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2016 · NEET-UG (India) · 4 marks

The pair of electrons in the given carbanion, $\mathrm{CH_3C\equiv C^{-}}$ , is present in which of the following orbitals? (a) 2p (b) $sp^3$ (c) $sp^2$ (d) sp

Answer

(d) sp

The carbanion $\mathrm{CH_3C\equiv C^-}$ is formed by loss of a proton from a terminal alkyne carbon, which is sp hybridised. The lone pair therefore stays in an sp orbital (50% s character), which holds it close to the nucleus and makes terminal alkynes fairly acidic.

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2016 · NEET-UG (India) · 4 marks

Which of the following can be used as the halide component for Friedel-Crafts reaction? (a) Chlorobenzene (b) Bromobenzene (c) Chloroethene (d) Isopropyl chloride

Answer

(d) Isopropyl chloride

Friedel-Crafts alkylation needs an alkyl halide that forms a carbocation or polarised complex with $\mathrm{AlCl_3}$ . Isopropyl chloride gives a stable secondary carbocation. Chlorobenzene, bromobenzene and chloroethene are aryl/vinyl halides whose C-X bond has partial double-bond character and does not ionise.

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2016 · NEET-UG (India) · 4 marks

In pyrrole (a five-membered ring with the nitrogen atom, carrying a hydrogen, at position 1 and the carbon atoms numbered 2, 3, 4, 5 around the ring), the electron density is maximum on (a) 2 and 3 (b) 3 and 4 (c) 2 and 4 (d) 2 and 5

Answer

(d) 2 and 5

Nitrogen's lone pair is delocalised into the ring, and the resonance structures put negative charge on C-2 and C-5 (the $\alpha$ positions), so these have the highest electron density. That is why electrophilic substitution in pyrrole occurs mainly at C-2.

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2016 · NEET-UG (India) · 4 marks

Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct elimination reaction only? (a) cyclopropane (b) $\mathrm{CH_3CH_2CH_2OH}$ (c) $\mathrm{H_2C=C=O}$ (d) $\mathrm{CH_3CH_2CH_2Br}$

Answer

(c) $\mathrm{H_2C=C=O}$

Cyclopropane adds HBr to give 1-bromopropane, which eliminates to propene. Propan-1-ol (dehydration) and 1-bromopropane (dehydrohalogenation) give propene directly. Ketene has only two carbons, so it cannot give a three-carbon product.

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2016 · NEET-UG (India) · 4 marks

The central dogma of molecular genetics states that the genetic information flows from (a) Amino acids $\rightarrow$ Proteins $\rightarrow$ DNA (b) DNA $\rightarrow$ Carbohydrates $\rightarrow$ Proteins (c) DNA $\rightarrow$ RNA $\rightarrow$ Proteins (d) DNA $\rightarrow$ RNA $\rightarrow$ Carbohydrates

Answer

(c) DNA $\rightarrow$ RNA $\rightarrow$ Proteins

The central dogma: DNA is transcribed to mRNA, which is translated into protein. Carbohydrates are not part of the information flow.

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2016 · NEET-UG (India) · 4 marks

Consider the reaction $\mathrm{CH_3CH_2CH_2Br+NaCN\rightarrow CH_3CH_2CH_2CN+NaBr}$ . This reaction will be the fastest in (a) ethanol (b) methanol (c) $N,N'$ -dimethylformamide (DMF) (d) water

Answer

(c) $N,N'$ -dimethylformamide (DMF)

This is an $S_N2$ reaction of a primary halide with $\mathrm{CN^-}$ . Polar aprotic solvents like DMF solvate the cation but leave the anion "naked" and highly nucleophilic. Ethanol, methanol and water hydrogen-bond to $\mathrm{CN^-}$ and lower its reactivity.

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2016 · NEET-UG (India) · 4 marks

The compound that will react most readily with gaseous bromine has the formula (a) $\mathrm{C_3H_6}$ (b) $\mathrm{C_2H_2}$ (c) $\mathrm{C_4H_{10}}$ (d) $\mathrm{C_2H_4}$

Answer

(a) $\mathrm{C_3H_6}$

Bromine reacts with alkenes by electrophilic addition. Propene is more reactive than ethene because the methyl group stabilises the bromonium/carbocation intermediate. Alkynes are less reactive than alkenes, and the alkane $\mathrm{C_4H_{10}}$ needs light for substitution.

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2015 · NEET-UG (India) · 4 marks

Which one of the following esters gets hydrolysed most easily under alkaline conditions? (a) 4-chlorophenyl acetate (chlorine para to the $\mathrm{OCOCH_3}$ group) (b) 4-nitrophenyl acetate (nitro group para to the $\mathrm{OCOCH_3}$ group) (c) 4-methoxyphenyl acetate (methoxy group para to the $\mathrm{OCOCH_3}$ group) (d) phenyl acetate

Answer

(b) 4-nitrophenyl acetate

Alkaline hydrolysis is attacked by $\mathrm{OH^-}$ on the carbonyl carbon and the leaving group is the phenoxide. The strongly electron-withdrawing $\mathrm{-NO_2}$ makes the carbonyl more electrophilic and stabilises 4-nitrophenoxide best. $\mathrm{OCH_3}$ is donating and slows hydrolysis.

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2015 · NEET-UG (India) · 4 marks

Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional groups? (a) $-\mathrm{CHO}$ (b) $-\mathrm{CH_2Cl}$ (c) $-\mathrm{COOH}$ (d) $-\mathrm{CHCl_2}$

Answer

(a) $-\mathrm{CHO}$

This is the Reimer-Tiemann reaction: phenol with $\mathrm{CHCl_3}$ and base generates dichlorocarbene, which attacks the ring (ortho) to give $\mathrm{-CHCl_2}$ , hydrolysed under the basic conditions to $-\mathrm{CHO}$ (salicylaldehyde).

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2015 · NEET-UG (India) · 4 marks

Which of the following reaction(s) can be used for the preparation of alkyl halides? (I) $\mathrm{CH_3CH_2OH+HCl}$ with anhydrous $\mathrm{ZnCl_2}$ (II) $\mathrm{CH_3CH_2OH+HCl}$ (III) $\mathrm{(CH_3)_3COH+HCl}$ (IV) $\mathrm{(CH_3)_2CHOH+HCl}$ with anhydrous $\mathrm{ZnCl_2}$ (a) (III) and (IV) only (b) (I), (III) and (IV) only (c) (I) and (II) only (d) (IV) only

Answer

(b) (I), (III) and (IV) only

The Lucas-type route works for 2 and 3 alcohols with HCl, and $\mathrm{ZnCl_2}$ allows primary ethanol to react on heating. Tertiary alcohol reacts with HCl alone. Ethanol with HCl alone (II) does not react appreciably.

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2015 · NEET-UG (India) · 4 marks

In an $S_N1$ reaction on chiral centres, there is: (a) 100% inversion (b) 100% racemization (c) inversion more than retention leading to partial racemization (d) 100% retention

Answer

(c) inversion more than retention leading to partial racemization

$S_N1$ goes through a planar carbocation that the nucleophile can attack from either face, giving mostly racemisation. In practice the leaving group partly shields one face, so inversion slightly exceeds retention (partial racemisation). Hence "100% racemization" is idealised and not the key answer.

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2015 · NEET-UG (India) · 4 marks

Which of the following statements is not correct for a nucleophile? (a) Nucleophiles are not electron seeking (b) Nucleophile is a Lewis acid (c) Ammonia is a nucleophile (d) Nucleophiles attack low $e^-$ density sites

Answer

(b) Nucleophile is a Lewis acid

A nucleophile donates an electron pair, so it is a Lewis base, not a Lewis acid. Statements (a), (c) and (d) are correct: nucleophiles are nucleus-seeking, $\mathrm{NH_3}$ is a nucleophile, and they attack electron-poor sites.

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2015 · NEET-UG (India) · 4 marks

Two possible stereo-structures of $\mathrm{CH_3CHOH{\cdot}COOH}$ , which are optically active, are called: (a) Mesomers (b) Diastereomers (c) Atropisomers (d) Enantiomers

Answer

(d) Enantiomers

Lactic acid $\mathrm{CH_3CH(OH)COOH}$ has one chiral carbon, so it has two stereoisomers that are non-superimposable mirror images, i.e. enantiomers. Mesomers need an internal plane of symmetry, and diastereomers need two or more stereocentres, so neither applies.

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2015 · NEET-UG (India) · 4 marks

Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is: (a) sodium hydrogen sulphite (b) a Grignard reagent (c) hydrazine in presence of feebly acidic solution (d) hydrocyanic acid

Answer

(c) hydrazine in presence of feebly acidic solution

Hydrazine ( $\mathrm{NH_2NH_2}$ ) adds to $\mathrm{C=O}$ to give a carbinolamine, which loses water to form a hydrazone $\mathrm{C=N-NH_2}$ : addition followed by elimination. $\mathrm{NaHSO_3}$ , Grignard reagents and HCN only add across $\mathrm{C=O}$ with no loss of water.

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2015 · NEET-UG (India) · 4 marks

Method by which aniline cannot be prepared is: (a) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution. (b) hydrolysis of phenylisocyanide with acidic solution. (c) degradation of benzamide with bromine in alkaline solution. (d) reduction of nitrobenzene with $\mathrm{H_2/Pd}$ in ethanol.

Answer

(a) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Gabriel phthalimide synthesis needs an alkyl halide (an $\mathrm{S_N2}$ attack). Chlorobenzene is an aryl halide whose C-Cl bond has partial double-bond character, so it does not react and no aniline forms. Hofmann bromamide degradation of benzamide, acid hydrolysis of phenyl isocyanide, and catalytic reduction of nitrobenzene all give aniline.

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2015 · NEET-UG (India) · 4 marks

2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid? (a) $\mathrm{(CH_3)_2CH-CH_2-CH=CH_2}$ (b) $\mathrm{(CH_3)_2CH-CH(CH_3)-CH=CH_2}$ (c) $\mathrm{(CH_3)_3C-CH=CH_2}$ (d) $\mathrm{(CH_3)_2C=CH-CH_2-CH_3}$

Answer

(b) $\mathrm{(CH_3)_2CH-CH(CH_3)-CH=CH_2}$

Protonating the terminal double bond gives a secondary carbocation next to a tertiary C-H. A 1,2-hydride shift gives the tertiary cation $\mathrm{(CH_3)_2C^+-CH(CH_3)_2}$ . Loss of H $^+$ gives the most substituted alkene, $\mathrm{(CH_3)_2C=C(CH_3)_2}$ , 2,3-dimethyl-2-butene. The other options give different carbon skeletons.

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2015 · NEET-UG (India) · 4 marks

The oxidation of benzene by $\mathrm{V_2O_5}$ in the presence of air produces: (a) benzaldehyde (b) benzoic anhydride (c) maleic anhydride (d) benzoic acid

Answer

(c) maleic anhydride

Benzene is oxidised by air over $\mathrm{V_2O_5}$ at about 773 K with ring cleavage, giving maleic anhydride (cis-butenedioic anhydride). Benzaldehyde and benzoic acid retain the ring and are formed from toluene, not by this reaction.

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2015 · NEET-UG (India) · 4 marks

Caprolactam is used for the manufacture of: (a) Nylon-6,6 (b) Nylon-6 (c) Teflon (d) Terylene

Answer

(b) Nylon-6

Caprolactam is the cyclic amide monomer; its ring-opening polymerisation on heating with water gives Nylon-6. Nylon-6,6 comes from adipic acid and hexamethylenediamine, Teflon from $\mathrm{CF_2=CF_2}$ and terylene from terephthalic acid and ethylene glycol.

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2015 · NEET-UG (India) · 4 marks

The following reaction: aniline reacts with benzoyl chloride in the presence of NaOH to give N-phenylbenzamide (benzanilide). It is known by the name: (a) Schotten-Baumann reaction (b) Friedel-Craft's reaction (c) Perkin's reaction (d) Acetylation reaction

Answer

(a) Schotten-Baumann reaction

Acylation of an amine (or alcohol) with an acid chloride in the presence of aqueous NaOH, which removes the HCl formed, is the Schotten-Baumann reaction. Friedel-Crafts acylates the ring using $\mathrm{AlCl_3}$ , and Perkin's reaction condenses an aromatic aldehyde with an anhydride. Option (d) is a general term, not the named reaction.

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2014 · NEET-UG (India) · 4 marks

Which of the following compounds will undergo racemisation when solution of KOH hydrolyses? (i) benzyl chloride, $\mathrm{C_6H_5CH_2Cl}$ (ii) $\mathrm{CH_3CH_2CH_2Cl}$ (iii) $\mathrm{(CH_3)_2CH-CH_2Cl}$ (iv) 2-chlorobutane, $\mathrm{CH_3-CH(Cl)-C_2H_5}$ (chiral carbon bearing H, $\mathrm{CH_3}$ , $\mathrm{C_2H_5}$ and Cl) (a) (ii) and (iv) (b) (iii) and (iv) (c) (i) and (iv) (d) (i) and (ii)

Answer

(c) (i) and (iv)

Racemisation in hydrolysis goes through a planar carbocation ( $\mathrm{S_N1}$ ) attacked from both faces. 2-Chlorobutane, (iv), has a chiral secondary carbon and gives the racemic alcohol; the benzyl chloride, (i), reacts through the resonance-stabilised benzyl cation. Primary alkyl chlorides (ii) and (iii) react by $\mathrm{S_N2}$ and cannot form a stable carbocation. Note that benzyl chloride itself is not chiral; (c) is the option given in the official key.

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2014 · NEET-UG (India) · 4 marks

Which of the following will be most stable diazonium salt $\mathrm{RN_2^+X^-}$ ? (a) $\mathrm{C_6H_5N_2^+X^-}$ (b) $\mathrm{CH_3CH_2N_2^+X^-}$ (c) $\mathrm{C_6H_5CH_2N_2^+X^-}$ (d) $\mathrm{CH_3N_2^+X^-}$

Answer

(a) $\mathrm{C_6H_5N_2^+X^-}$

Only the arenediazonium ion is relatively stable (stable at 273-278 K), because the positive charge on the $\mathrm{N_2^+}$ group is delocalised into the benzene ring by resonance. Alkyldiazonium salts have no such resonance and decompose rapidly, even at low temperature, to give $\mathrm{N_2}$ .

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2014 · NEET-UG (India) · 4 marks

Which of the following organic compounds polymerizes to form the polyester Dacron? (a) Benzoic acid and ethanol (b) Terephthalic acid and ethylene glycol (c) Benzoic acid and para $\mathrm{HO-(C_6H_4)-OH}$ (d) Propylene and para $\mathrm{HO-(C_6H_4)-OH}$

Answer

(b) Terephthalic acid and ethylene glycol

Dacron (terylene) is a polyester formed by condensation of terephthalic acid, $p$ - $\mathrm{HOOC-C_6H_4-COOH}$ , with ethylene glycol, $\mathrm{HOCH_2CH_2OH}$ . The other pairs have a monofunctional benzoic acid or lack a diol and a diacid, so they cannot build a polyester chain.

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2014 · NEET-UG (India) · 4 marks

Which of the following will not be soluble in sodium hydrogen carbonate? (a) Benzoic acid (b) $o$ -Nitrophenol (c) Benzenesulphonic acid (d) 2,4,6-trinitrophenol

Answer

(b) $o$ -Nitrophenol

$\mathrm{NaHCO_3}$ dissolves only acids stronger than carbonic acid ( $\mathrm{p}K_a\approx6.4$ ) by releasing $\mathrm{CO_2}$ . Benzoic acid, benzenesulphonic acid and picric acid are all stronger acids. $o$ -Nitrophenol ( $\mathrm{p}K_a\approx7.2$ ) is a weaker acid than carbonic acid, so it does not dissolve.

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2014 · NEET-UG (India) · 4 marks

Which one of the following is an example of a thermosetting polymer? (a) poly(vinyl chloride), $\mathrm{-(CH_2-CHCl)_n-}$ (b) nylon-6,6, $\mathrm{-[NH-(CH_2)_6-NH-CO-(CH_2)_4-CO]_n-}$ (c) novolac-type phenol-formaldehyde polymer (phenol rings joined by $\mathrm{-CH_2-}$ bridges) (d) neoprene, $\mathrm{-[CH_2-C(Cl)=CH-CH_2]_n-}$

Answer

(c) novolac-type phenol-formaldehyde polymer

Thermosetting polymers are cross-linked, infusible and cannot be remoulded; the phenol-formaldehyde resin (novolac, which cures to Bakelite with $\mathrm{-CH_2-}$ cross-links) is the example. PVC and nylon-6,6 are thermoplastics, and neoprene is an elastomer.

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2014 · NEET-UG (India) · 4 marks

Which one is most reactive towards nucleophilic addition reaction? (a) acetophenone, $\mathrm{C_6H_5COCH_3}$ (b) 4-methylbenzaldehyde (c) 4-nitrobenzaldehyde (d) benzaldehyde

Answer

(c) 4-nitrobenzaldehyde

Nucleophilic addition is faster when the carbonyl carbon is more electrophilic. The $-\mathrm{NO_2}$ group withdraws electrons by $-I$ and $-R$ , making 4-nitrobenzaldehyde the most reactive. The $-\mathrm{CH_3}$ group donates electrons, and the ketone acetophenone has two electron-releasing groups and more steric hindrance, so they are less reactive.

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2014 · NEET-UG (India) · 4 marks

Which of the following organic compounds has same hybridization as its combustion product ( $\mathrm{CO_2}$ )? (a) Ethyne (b) Ethene (c) Ethanol (d) Ethane

Answer

(a) Ethyne

$\mathrm{CO_2}$ (O=C=O) has an sp-hybridised carbon. In ethyne, $\mathrm{HC\equiv CH}$ , both carbons are sp as well. Ethene carbons are $sp^2$ , while ethane and ethanol carbons are $sp^3$ .

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2014 · NEET-UG (India) · 4 marks

Among the following sets of reactants which one produces anisole? (a) $\mathrm{C_6H_5OH;\ NaOH;\ CH_3I}$ (b) $\mathrm{C_6H_5OH;}$ neutral $\mathrm{FeCl_3}$ (c) $\mathrm{C_6H_5-CH_3;\ CH_3COCl;\ AlCl_3}$ (d) $\mathrm{CH_3CHO;\ RMgX}$

Answer

(a) $\mathrm{C_6H_5OH;\ NaOH;\ CH_3I}$

NaOH converts phenol to sodium phenoxide, which attacks $\mathrm{CH_3I}$ in a Williamson ether synthesis to give anisole ( $\mathrm{C_6H_5OCH_3}$ ). Neutral $\mathrm{FeCl_3}$ only gives a colour test with phenol, Friedel-Crafts acylation of toluene gives a ketone, and $\mathrm{CH_3CHO}$ with RMgX gives an alcohol.

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2013 · NEET-UG (India) · 4 marks

Reaction by which benzaldehyde cannot be prepared (a) toluene $+\ \mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ followed by $\mathrm{H_3O^+}$ (b) benzoyl chloride $+\ \mathrm{H_2}$ in presence of $\mathrm{Pd-BaSO_4}$ (c) benzene $+\ \mathrm{CO+HCl}$ in presence of anhydrous $\mathrm{AlCl_3}$ (d) benzoic acid $+\ \mathrm{Zn/Hg}$ and conc. HCl

Answer

(d) benzoic acid $+\ \mathrm{Zn/Hg}$ and conc. HCl

Zn/Hg with conc. HCl is the Clemmensen reduction, which reduces carbonyl groups to $\mathrm{CH_2}$ ; it cannot reduce a carboxylic acid to an aldehyde. The others do give benzaldehyde: (a) Etard reaction, (b) Rosenmund reduction, (c) Gattermann-Koch reaction.

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2013 · NEET-UG (India) · 4 marks

Which of these is not a monomer for a high molecular mass silicone polymer? (a) $\mathrm{MeSiCl_3}$ (b) $\mathrm{Me_2SiCl_2}$ (c) $\mathrm{Me_3SiCl}$ (d) $\mathrm{PhSiCl_3}$

Answer

(c) $\mathrm{Me_3SiCl}$

A silicone polymer needs monomers that can link on at least two sides. $\mathrm{Me_3SiCl}$ has only one Si-Cl bond, so it is a chain terminator (end-blocker) and cannot build a high molecular mass polymer. $\mathrm{Me_2SiCl_2}$ gives chains, while $\mathrm{MeSiCl_3}$ and $\mathrm{PhSiCl_3}$ give cross-linked polymers.

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2013 · NEET-UG (India) · 4 marks

Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI? (a) $\mathrm{CH_3CH_2CH_2CH_2-O-CH_3}$ (b) $\mathrm{CH_3CH_2CH(CH_3)-O-CH_3}$ (c) $\mathrm{(CH_3)_3C-O-CH_3}$ (d) $\mathrm{(CH_3)_2CH-CH_2-O-CH_3}$

Answer

(c) $\mathrm{(CH_3)_3C-O-CH_3}$

Normally HI cleaves a mixed ether at the smaller alkyl group, giving methyl iodide and the other alcohol. But with a tertiary alkyl group the cleavage goes by an $S_N1$ path, forming the stable $\mathrm{(CH_3)_3C^+}$ and so tert-butyl iodide plus methanol. In (a), (b) and (d) the cleavage is $S_N2$ at the methyl carbon, giving $\mathrm{CH_3I}$ and the larger alcohol.

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2013 · NEET-UG (India) · 4 marks

Nylon is an example of (a) Polyester (b) Polysaccharide (c) Polyamide (d) Polythene

Answer

(c) Polyamide

Nylon (nylon-6,6 or nylon-6) contains repeating amide ( $-\mathrm{CONH}-$ ) linkages, so it is a polyamide. Polyesters have ester links, polysaccharides are sugar polymers and polythene is a hydrocarbon polymer.

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2013 · NEET-UG (India) · 4 marks

The structure of isobutyl group in an organic compound is (a) $\mathrm{(CH_3)_2CH-CH_2-}$ (b) $\mathrm{CH_3-CH(-)-CH_2-CH_3}$ (attachment at the second carbon) (c) $\mathrm{CH_3-CH_2-CH_2-CH_2-}$ (d) $\mathrm{(CH_3)_3C-}$

Answer

(a) $\mathrm{(CH_3)_2CH-CH_2-}$

The isobutyl group is a four-carbon branched chain attached through its terminal $\mathrm{CH_2}$ : $\mathrm{(CH_3)_2CHCH_2-}$ . Option (b) is sec-butyl, (c) is n-butyl and (d) is tert-butyl.

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2013 · NEET-UG (India) · 4 marks

Nitrobenzene on reaction with conc. $\mathrm{HNO_3/H_2SO_4}$ at 80-100 °C forms which one of the following products? (a) 1,2-Dinitrobenzene (b) 1,3-Dinitrobenzene (c) 1,4-Dinitrobenzene (d) 1,2,4-Trinitrobenzene

Answer

(b) 1,3-Dinitrobenzene

The $-\mathrm{NO_2}$ group is strongly deactivating and meta-directing, because ortho and para attack put a positive charge on the carbon bearing the electron-withdrawing group. Forcing conditions (conc. $\mathrm{HNO_3/H_2SO_4}$ , 80-100 °C) are needed for the second nitration, and the new $\mathrm{NO_2}$ enters the meta position. The ortho and para dinitro products (a, c) are minor, and a trinitro product (d) needs even harsher conditions.

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2013 · NEET-UG (India) · 4 marks

Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? (a) $-\mathrm{C\equiv N}$ (b) $-\mathrm{SO_3H}$ (c) $-\mathrm{COOH}$ (d) $-\mathrm{NO_2}$

Answer

(d) $-\mathrm{NO_2}$

Deactivation depends on how strongly the group withdraws electrons by $-I$ and $-M$ effects. Nitro has a full positive charge on nitrogen and both strong effects, so it deactivates the ring most. $-\mathrm{SO_3H}$ , $-\mathrm{C\equiv N}$ and $-\mathrm{COOH}$ are also deactivating but weaker than $-\mathrm{NO_2}$ .

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2013 · NEET-UG (India) · 4 marks

Which is the monomer of Neoprene in the following? (a) $\mathrm{CH_2=CH-C\equiv CH}$ (b) $\mathrm{CH_2=C(CH_3)-CH=CH_2}$ (c) $\mathrm{CH_2=C(Cl)-CH=CH_2}$ (d) $\mathrm{CH_2=CH-CH=CH_2}$

Answer

(c) $\mathrm{CH_2=C(Cl)-CH=CH_2}$

Neoprene is a polymer of chloroprene (2-chlorobuta-1,3-diene), formed by free-radical addition polymerisation. Option (b) is isoprene, the monomer of natural rubber, and option (d) is 1,3-butadiene, used for Buna rubbers.

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2013 · NEET-UG (India) · 4 marks

The radical $\mathrm{C_6H_5CH_2^{\bullet}}$ (benzyl radical) is aromatic because it has (a) 6 p-orbitals and 6 unpaired electrons (b) 7 p-orbitals and 6 unpaired electrons (c) 7 p-orbitals and 7 unpaired electrons (d) 6 p-orbitals and 7 unpaired electrons

Answer

(c) 7 p-orbitals and 7 unpaired electrons

In the benzyl radical the $\mathrm{CH_2}$ carbon is $sp^2$ hybridised, so its p-orbital overlaps with the six p-orbitals of the ring. This gives 7 p-orbitals in a conjugated planar system, each carrying one electron, which makes it aromatic-like and stabilised by resonance. Strictly only one electron is unpaired in the ground state, but among the options (c) fits the 7 p-orbital, 7 electron count, as in NTA answer key.

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2013 · NEET-UG (India) · 4 marks

The order of stability of the following tautomeric compounds is: I: $\mathrm{CH_2=C(OH)-CH_2-C(=O)-CH_3}$ (enol form) II: $\mathrm{CH_3-C(=O)-CH_2-C(=O)-CH_3}$ (diketo form) III: $\mathrm{CH_3-C(OH)=CH-C(=O)-CH_3}$ (enol form) (a) I > II > III (b) III > II > I (c) II > I > III (d) II > III > I

Answer

(b) III > II > I

Enol III (the enol of acetylacetone) is conjugated with the C=O and is stabilised by intramolecular hydrogen bonding in a six-membered chelate ring, so it is the most stable. The diketo form II is next. Enol I has no conjugation or H-bonding with the carbonyl (the C=C is isolated from C=O by a $\mathrm{CH_2}$ ), so it is the least stable.

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2013 · NEET-UG (India) · 4 marks

Which of the following compounds will not undergo Friedel-Crafts reaction easily? (a) Cumene (b) Xylene (c) Nitrobenzene (d) Toluene

Answer

(c) Nitrobenzene

The $-\mathrm{NO_2}$ group strongly deactivates the ring by $-I$ and $-M$ effects. Friedel-Crafts alkylation and acylation need an electron-rich ring, and the $\mathrm{AlCl_3}$ catalyst also complexes with nitro oxygen. Cumene, xylene and toluene have activating alkyl groups, so they react easily.

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2017 · NEET-UG (India) · 4 marks

With respect to the conformers of ethane, which of the following statements is true?
(a) Bond angle remains same but bond length changes (b) Bond angle changes but bond length remains same (c) Both bond angle and bond length change (d) Both bond angles and bond length remains same

Answer

(d) Both bond angles and bond length remains same

Conformers of ethane differ only by rotation about the C-C single bond (staggered, eclipsed, gauche forms). Rotation changes the dihedral (torsional) angle, not the bond angles or bond lengths, so both stay the same. Only the torsional strain, and hence the energy, changes.

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2017 · NEET-UG (India) · 4 marks

Consider the reactions: compound X ( $\mathrm{C_2H_6O}$ ) on heating with Cu at 573 K gives A; A gives a silver mirror with $\mathrm{[Ag(NH_3)_2]^+}$ ; A on heating with $\mathrm{{}^-OH}$ gives Y; A reacts with $\mathrm{NH_2-NH-CO-NH_2}$ to give Z. Identify A, X, Y and Z.
(a) A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-hydrazine (b) A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide (c) A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone (d) A-Ethanol, X-Acetaldehyde, Y-Butanone, Z-Hydrazone

Answer

(c) A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone

$\mathrm{C_2H_6O}$ that gives an aldehyde on Cu/573 K (dehydrogenation) is ethanol, X, giving ethanal, A. Ethanal gives the Tollens (silver mirror) test. With dilute $\mathrm{OH^-}$ it undergoes aldol condensation to give but-2-enal, Y. With semicarbazide it forms the semicarbazone, Z. Methoxymethane is an ether and cannot give these reactions.

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2017 · NEET-UG (India) · 4 marks

Which one is the correct order of acidity?
(a) $\mathrm{CH_2=CH_2} > \mathrm{CH_3-CH=CH_2} > \mathrm{CH_3-C\equiv CH} > \mathrm{HC\equiv CH}$ (b) $\mathrm{HC\equiv CH} > \mathrm{CH_3-C\equiv CH} > \mathrm{CH_2=CH_2} > \mathrm{CH_3-CH_3}$ (c) $\mathrm{HC\equiv CH} > \mathrm{CH_2=CH_2} > \mathrm{CH_3-C\equiv CH} > \mathrm{CH_3-CH_3}$ (d) $\mathrm{CH_3-CH_3} > \mathrm{CH_2=CH_2} > \mathrm{CH_3-C\equiv CH} > \mathrm{HC\equiv CH}$

Answer

(b) $\mathrm{HC\equiv CH} > \mathrm{CH_3-C\equiv CH} > \mathrm{CH_2=CH_2} > \mathrm{CH_3-CH_3}$

Acidity rises with the s-character of the C-H carbon: $sp$ (50%) > $sp^2$ (33%) > $sp^3$ (25%), because a more s-rich carbon stabilises the conjugate carbanion better. Ethyne and propyne are both $sp$ , but the electron-donating $-\mathrm{CH_3}$ in propyne destabilises the carbanion, so ethyne is more acidic. Option (c) wrongly puts ethene above propyne.

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2017 · NEET-UG (India) · 4 marks

The correct statement regarding electrophile is
(a) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile (b) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile (c) Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile (d) Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile

Answer

(d) Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile

An electrophile is electron-loving: it is electron deficient and accepts an electron pair from a nucleophile. It may be positively charged (e.g. $\mathrm{NO_2^+}$ , $\mathrm{H^+}$ ) or neutral (e.g. $\mathrm{AlCl_3}$ , $\mathrm{BF_3}$ , carbenes). Options (a) and (b) are wrong because electrophiles are not negatively charged, and (c) is wrong because "generally neutral" omits the charged ones.

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2017 · NEET-UG (India) · 4 marks

Which of the following statements is not correct?
(a) Insulin maintains sugar level in the blood of a human body (b) Ovalbumin is a simple food reserve in egg-white (c) Blood proteins thrombin and fibrinogen are involved in blood clotting (d) Denaturation makes the proteins more active

Answer

(d) Denaturation makes the proteins more active

This statement is not correct. Denaturation (by heat, pH change etc.) destroys the secondary and tertiary structure, so the protein loses its biological activity. The other statements are true: insulin controls blood sugar, ovalbumin is a simple storage protein in egg white, and thrombin and fibrinogen take part in blood clotting.

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2017 · NEET-UG (India) · 4 marks

An example of a sigma bonded organometallic compound is:
(a) Ruthenocene (b) Grignard's reagent (c) Ferrocene (d) Cobaltocene

Answer

(b) Grignard's reagent

In Grignard's reagent, $\mathrm{R-Mg-X}$ , the carbon is bound to magnesium through a single $\sigma$ bond. Ferrocene, ruthenocene and cobaltocene are sandwich compounds where the metal is bonded to the $\pi$ cloud of cyclopentadienyl rings, so they are $\pi$ -bonded organometallics.

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2017 · NEET-UG (India) · 4 marks

The heating of phenyl-methyl ethers with HI produces
(a) Ethyl chlorides (b) Iodobenzene (c) Phenol (d) Benzene

Answer

(c) Phenol

In phenyl methyl ether (anisole) the $\mathrm{O-C_6H_5}$ bond has partial double-bond character because of resonance, so it does not break. $\mathrm{I^-}$ attacks the methyl carbon, giving phenol and $\mathrm{CH_3I}$ . Hence iodobenzene and benzene are not formed.

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2017 · NEET-UG (India) · 4 marks

Which of the following reactions is appropriate for converting acetamide to methanamine?
(a) Carbylamine reaction (b) Hoffmann hypobromamide reaction (c) Stephens reaction (d) Gabriels phthalimide synthesis

Answer

(b) Hoffmann hypobromamide reaction

$\mathrm{CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$ . The Hofmann bromamide degradation converts an amide into a primary amine with one carbon fewer (acetamide to methanamine). The carbylamine reaction makes isocyanides from primary amines, and Gabriel synthesis makes primary amines from alkyl halides, so neither fits.

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2018 · NEET-UG (India) · 4 marks

Regarding cross-linked or network polymers, which of the following statements is incorrect?
(a) They contain covalent bonds between various linear polymer chains. (b) Examples are bakelite and melamine. (c) They are formed from bi- and tri-functional monomers. (d) They contain strong covalent bonds in their polymer chains.

Answer

(d) They contain strong covalent bonds in their polymer chains.

This is the statement marked incorrect. In cross-linked (network) polymers such as bakelite and melamine, the distinguishing feature is covalent cross-links between separate linear chains, formed from bi- and tri-functional monomers, which statements (a), (b) and (c) correctly describe. Statement (d) is not the defining property and is the one the key marks as incorrect, since every polymer chain has covalent bonds along its backbone.

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2018 · NEET-UG (India) · 4 marks

Nitration of aniline in strong acidic medium also gives m-nitroaniline because
(a) Inspite of substituents nitro group always goes to only m-position. (b) In absence of substituents nitro group always goes to m-position. (c) In electrophilic substitution reactions amino group is meta directive. (d) In acidic (strong) medium aniline is present as anilinium ion.

Answer

(d) In acidic (strong) medium aniline is present as anilinium ion.

In strongly acidic nitrating mixture the $-\mathrm{NH_2}$ group is protonated to $-\mathrm{NH_3^+}$ (anilinium ion). This group is strongly electron-withdrawing and meta-directing, so a substantial amount of m-nitroaniline forms along with the para product. Option (c) is wrong because the free amino group is ortho/para directing, not meta directing.

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2018 · NEET-UG (India) · 4 marks

The difference between amylose and amylopectin is
(a) Amylopectin have $\alpha$ -1,4 linkage and $\beta$ -1,6 linkage (b) Amylopectin have $\alpha$ -1,4 linkage and $\alpha$ -1,6 linkage (c) Amylose have $\alpha$ -1,4 linkage and $\beta$ -1,6 linkage (d) Amylose is made up of glucose and galactose

Answer

(b) Amylopectin have $\alpha$ -1,4 linkage and $\alpha$ -1,6 linkage

Amylose is an unbranched chain of $\alpha$ -D-glucose units joined by $\alpha$ -1,4 glycosidic links. Amylopectin is branched: the main chain has $\alpha$ -1,4 links and the branch points are $\alpha$ -1,6 links. Options (a) and (c) wrongly state a $\beta$ -1,6 linkage, and (d) is wrong because both are made only of glucose.

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2018 · NEET-UG (India) · 4 marks

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. $\mathrm{H_2SO_4}$ . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
(a) 1.4 (b) 2.8 (c) 3.0 (d) 4.4

Answer

(b) 2.8

Formic acid: $\mathrm{HCOOH \xrightarrow{conc.\,H_2SO_4} CO + H_2O}$ . Moles $= 2.3/46 = 0.05$ , giving 0.05 mol CO. Oxalic acid: $\mathrm{(COOH)_2 \rightarrow CO + CO_2 + H_2O}$ . Moles $= 4.5/90 = 0.05$ , giving 0.05 mol CO and 0.05 mol $\mathrm{CO_2}$ . KOH absorbs $\mathrm{CO_2}$ , leaving $0.05+0.05 = 0.1$ mol CO, mass $= 0.1\times 28 = 2.8$ g.

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2018 · NEET-UG (India) · 4 marks

In the Reimer-Tiemann reaction, phenol reacts with $\mathrm{CHCl_3}$ and NaOH to give sodium salicylaldehyde (sodium 2-formylphenoxide). The electrophile involved is
(a) Dichloromethyl cation $\mathrm{{}^+CHCl_2}$ (b) Dichloromethyl anion $\mathrm{{}^-CHCl_2}$ (c) Formyl cation $\mathrm{{}^+CHO}$ (d) Dichlorocarbene $\mathrm{:CCl_2}$

Answer

(d) Dichlorocarbene $\mathrm{:CCl_2}$

In the Reimer-Tiemann reaction, $\mathrm{CHCl_3}$ is deprotonated by NaOH to $\mathrm{{}^-CCl_3}$ , which loses $\mathrm{Cl^-}$ to give the neutral, electron-deficient dichlorocarbene $\mathrm{:CCl_2}$ . This is the electrophile that attacks the ortho position of the phenoxide ring; hydrolysis then gives the aldehyde group. The carbanion in (b) is a nucleophile, not the attacking species.

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2018 · NEET-UG (India) · 4 marks

Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
(a) Formation of intramolecular H-bonding (b) More extensive association of carboxylic acid via van der Waals force of attraction (c) Formation of carboxylate ion (d) Formation of intermolecular H-bonding

Answer

(d) Formation of intermolecular H-bonding

Carboxylic acids form hydrogen-bonded dimers through two intermolecular O-H...O bonds, which are stronger and more extensive than the H-bonding in alcohols. Aldehydes and ketones have no O-H and so no such H-bonding. Option (a) is wrong because the bonding is between molecules, and (b) is wrong because van der Waals forces are far weaker than H-bonds.

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2018 · NEET-UG (India) · 4 marks

Which of the following molecules represents the order of hybridisation $sp^2, sp^2, sp, sp$ from left to right atoms?
(a) $\mathrm{HC\equiv C-C\equiv CH}$ (b) $\mathrm{CH_2=CH-CH=CH_2}$ (c) $\mathrm{CH_2=CH-C\equiv CH}$ (d) $\mathrm{CH_3-CH=CH-CH_3}$

Answer

(c) $\mathrm{CH_2=CH-C\equiv CH}$

Count the hybridisation from left to right: the first carbon in $\mathrm{CH_2=}$ has a double bond and is $sp^2$ , the second ( $\mathrm{=CH-}$ ) is $sp^2$ , the third (in $\mathrm{-C\equiv}$ ) is $sp$ and the last ( $\mathrm{\equiv CH}$ ) is $sp$ . This matches $sp^2, sp^2, sp, sp$ . Butadiene gives all $sp^2$ and diacetylene gives all $sp$ .

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2018 · NEET-UG (India) · 4 marks

Which of the following is correct with respect to $-I$ effect of the substituents? (R = alkyl)
(a) $-\mathrm{NH_2} < -\mathrm{OR} < -\mathrm{F}$ (b) $-\mathrm{NH_2} > -\mathrm{OR} > -\mathrm{F}$ (c) $-\mathrm{NR_2} < -\mathrm{OR} < -\mathrm{F}$ (d) $-\mathrm{NR_2} > -\mathrm{OR} > -\mathrm{F}$

Answer

(c) $-\mathrm{NR_2} < -\mathrm{OR} < -\mathrm{F}$

The $-I$ effect increases with the electronegativity of the atom attached to the carbon chain: $\mathrm{N} < \mathrm{O} < \mathrm{F}$ . So the order of electron-withdrawing inductive effect is $-\mathrm{NR_2} < -\mathrm{OR} < -\mathrm{F}$ . Option (d) reverses this order, and (a)/(b) use $-\mathrm{NH_2}$ , which is not the choice matching the key.

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2018 · NEET-UG (India) · 4 marks

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
(a) $\mathrm{HC\equiv CH}$ (b) $\mathrm{CH_3-CH_3}$ (c) $\mathrm{CH_2=CH_2}$ (d) $\mathrm{CH_4}$

Answer

(d) $\mathrm{CH_4}$

Alkanes undergo substitution with bromine. $\mathrm{CH_4}\rightarrow \mathrm{CH_3Br}$ , and Wurtz coupling gives $\mathrm{CH_3CH_3}$ , a gas with two carbons (fewer than four). Ethane would give $\mathrm{C_2H_5Br}$ and then butane (four carbons), which does not fit. Ethene and ethyne react with bromine by addition, not substitution.

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2018 · NEET-UG (India) · 4 marks

The compound A on treatment with Na gives B, and with $\mathrm{PCl_5}$ gives C. B and C react together to give diethyl ether. A, B and C are in the order
(a) $\mathrm{C_2H_5OH}$ , $\mathrm{C_2H_6}$ , $\mathrm{C_2H_5Cl}$ (b) $\mathrm{C_2H_5Cl}$ , $\mathrm{C_2H_6}$ , $\mathrm{C_2H_5OH}$ (c) $\mathrm{C_2H_5OH}$ , $\mathrm{C_2H_5Cl}$ , $\mathrm{C_2H_5ONa}$ (d) $\mathrm{C_2H_5OH}$ , $\mathrm{C_2H_5ONa}$ , $\mathrm{C_2H_5Cl}$

Answer

(d) $\mathrm{C_2H_5OH}$ , $\mathrm{C_2H_5ONa}$ , $\mathrm{C_2H_5Cl}$

Ethanol (A) reacts with Na to give sodium ethoxide (B) and with $\mathrm{PCl_5}$ to give ethyl chloride (C). Williamson synthesis, $\mathrm{C_2H_5ONa+C_2H_5Cl\rightarrow C_2H_5OC_2H_5+NaCl}$ , then gives diethyl ether. Option (a) is wrong because Na and ethanol give the alkoxide, not ethane.

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2018 · NEET-UG (India) · 4 marks

Which of the following compounds can form a zwitterion?
(a) Aniline (b) Benzoic acid (c) Acetanilide (d) Glycine

Answer

(d) Glycine

A zwitterion needs both an acidic group ( $-\mathrm{COOH}$ ) and a basic group ( $-\mathrm{NH_2}$ ) in the same molecule, so the proton transfers internally to give $\mathrm{{}^+H_3N-CH_2-COO^-}$ . Glycine, an amino acid, does this. Aniline has only a basic group, benzoic acid only an acidic group, and acetanilide has neither.

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2018 · NEET-UG (India) · 4 marks

The compound $\mathrm{C_7H_8}$ undergoes the following reactions: $\mathrm{C_7H_8} \xrightarrow{3Cl_2/h\nu} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C$ . The product 'C' is
(a) m-bromotoluene (b) 3-bromo-2,4,6-trichlorotoluene (c) o-bromotoluene (d) p-bromotoluene

Answer

(a) m-bromotoluene

$\mathrm{C_7H_8}$ is toluene. Free-radical chlorination gives benzotrichloride ( $\mathrm{C_6H_5CCl_3}$ ). The $-\mathrm{CCl_3}$ group is electron-withdrawing and meta-directing, so $\mathrm{Br_2/Fe}$ gives m-bromobenzotrichloride. Zn/HCl then reduces $-\mathrm{CCl_3}$ to $-\mathrm{CH_3}$ , giving m-bromotoluene.

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2019 · NEET-UG (India) · 4 marks

The most suitable reagent for the conversion of 2-butyne ( $\mathrm{CH_3-C\equiv C-CH_3}$ ) into cis-2-butene is:
(a) $\mathrm{H_2}$ , Pd/C, quinoline (b) Zn/HCl (c) $\mathrm{Hg^{2+}/H^+}$ , $\mathrm{H_2O}$ (d) Na/liquid $\mathrm{NH_3}$

Answer

(a) $\mathrm{H_2}$ , Pd/C, quinoline

Lindlar-type partial hydrogenation (poisoned Pd catalyst) adds $\mathrm{H_2}$ syn to the alkyne, stopping at the cis-alkene. Na/liquid $\mathrm{NH_3}$ gives the trans-alkene by anti addition, $\mathrm{Hg^{2+}/H^+}$ hydrates the alkyne to a ketone, and Zn/HCl is not used for this.

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2019 · NEET-UG (India) · 4 marks

The non-essential amino acid among the following is:
(a) Leucine (b) Alanine (c) Lysine (d) Valine

Answer

(b) Alanine

Leucine, lysine and valine are essential amino acids because the body cannot synthesise them and they must come from the diet. Alanine can be synthesised in the body, so it is non-essential.

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2019 · NEET-UG (India) · 4 marks

The number of sigma ( $\sigma$ ) and pi ( $\pi$ ) bonds in pent-2-en-4-yne is
(a) 8 $\sigma$ bonds and 5 $\pi$ bonds (b) 11 $\sigma$ bonds and 2 $\pi$ bonds (c) 13 $\sigma$ bonds and no $\pi$ bonds (d) 10 $\sigma$ bonds and 3 $\pi$ bonds

Answer

(d) 10 $\sigma$ bonds and 3 $\pi$ bonds

Pent-2-en-4-yne is $\mathrm{CH_3-CH=CH-C\equiv CH}$ . It has 4 C-C $\sigma$ bonds and 6 C-H $\sigma$ bonds, giving 10 $\sigma$ bonds. The double bond adds 1 $\pi$ bond and the triple bond adds 2, giving 3 $\pi$ bonds.

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2019 · NEET-UG (India) · 4 marks

The biodegradable polymer is:
(a) Nylon-2-nylon 6 (b) Nylon-6 (c) Buna-S (d) Nylon-6,6

Answer

(a) Nylon-2-nylon 6

Nylon-2-nylon 6 is a copolymer of glycine and amino caproic acid. Its amide linkages resemble peptide bonds, so microbes can break it down. Nylon-6, nylon-6,6 and Buna-S are not biodegradable.

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2019 · NEET-UG (India) · 4 marks

The correct order of the basic strength of methyl substituted amines in aqueous solution is:
(a) $\mathrm{(CH_3)_3N} > \mathrm{CH_3NH_2} > \mathrm{(CH_3)_2NH}$ (b) $\mathrm{(CH_3)_3N} > \mathrm{(CH_3)_2NH} > \mathrm{CH_3NH_2}$ (c) $\mathrm{CH_3NH_2} > \mathrm{(CH_3)_2NH} > \mathrm{(CH_3)_3N}$ (d) $\mathrm{(CH_3)_2NH} > \mathrm{CH_3NH_2} > \mathrm{(CH_3)_3N}$

Answer

(d) $\mathrm{(CH_3)_2NH} > \mathrm{CH_3NH_2} > \mathrm{(CH_3)_3N}$

In aqueous solution, basic strength depends on the +I effect of the alkyl groups, solvation of the ammonium ion by H-bonding, and steric hindrance. The ion from the tertiary amine has only one N-H and is poorly solvated, so $\mathrm{(CH_3)_3N}$ is the weakest. The secondary amine is the strongest, giving the order $2^\circ>1^\circ>3^\circ$ for methylamines.

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2020 · NEET-UG (India) · 4 marks

Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as:
(a) Cross Cannizzaro's reaction (b) Cross Aldol condensation (c) Aldol condensation (d) Cannizzaro's reaction

Answer

(b) Cross Aldol condensation

Benzaldehyde has no $\alpha$ -hydrogen, while acetophenone does. In dilute NaOH, the acetophenone enolate attacks benzaldehyde, and dehydration gives chalcone. Two different carbonyl compounds are involved, so it is a cross aldol condensation. Cannizzaro needs strong base and two aldehydes without $\alpha$ -H.

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2020 · NEET-UG (India) · 4 marks

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?
(a) $-R$ effect of $\mathrm{-CH_3}$ group (b) Hyperconjugation (c) $-I$ effect of $\mathrm{-CH_3}$ groups (d) $+R$ effect of $\mathrm{-CH_3}$ groups

Answer

(b) Hyperconjugation

A tertiary carbocation has nine $\alpha$ C-H bonds that can overlap with the empty p orbital, while a secondary one has six. More hyperconjugation structures give greater stability. The +I effect of methyl also helps, but the methyl group does not show a -I, +R or -R effect as the other options claim.

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2020 · NEET-UG (India) · 4 marks

Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give:
(a) Tert. butyl alcohol (b) Isobutyl alcohol (c) Isopropyl alcohol (d) Sec. butyl alcohol

Answer

(a) Tert. butyl alcohol

$\mathrm{CH_3MgCl}$ adds to the carbonyl carbon of acetone, $\mathrm{(CH_3)_2C=O}$ , to give $\mathrm{(CH_3)_3C-OMgCl}$ . Hydrolysis then gives $\mathrm{(CH_3)_3C-OH}$ . A ketone gives a tertiary alcohol, while formaldehyde gives a primary and other aldehydes a secondary alcohol.

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2020 · NEET-UG (India) · 4 marks

Which of the following is a basic amino acid?
(a) Tyrosine (b) Lysine (c) Serine (d) Alanine

Answer

(b) Lysine

Lysine has a second amino group in its side chain, $\mathrm{(CH_2)_4NH_2}$ , giving more $\mathrm{-NH_2}$ than $\mathrm{-COOH}$ groups, so it is basic. Tyrosine, serine and alanine are neutral amino acids.

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2020 · NEET-UG (India) · 4 marks

Elimination reaction of 2-bromopentane to form pent-2-ene is:
(a) $\beta$ -Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction
(a) (b), (c), (d) (b) (a), (b), (d) (c) (a), (b), (c) (d) (a), (c), (d)

Answer

(c) (a), (b), (c)

Removal of H from the $\beta$ -carbon and Br from the $\alpha$ -carbon is $\beta$ -elimination, which is dehydrohalogenation (loss of HBr). The more substituted alkene, pent-2-ene, is the major product, as Zaitsev rule predicts. It is not dehydration, because dehydration removes water from an alcohol, not HBr from an alkyl halide.

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2020 · NEET-UG (India) · 4 marks

Which of the following alkanes cannot be made in good yield by Wurtz reaction?
(a) n-Heptane (b) n-Butane (c) n-Hexane (d) 2,3-Dimethylbutane

Answer

(a) n-Heptane

Wurtz coupling gives good yield only when two identical alkyl halides couple (symmetrical alkane $R-R$ ) from primary halides: n-butane (from $\mathrm{C_2H_5X}$ ), n-hexane (from $\mathrm{C_3H_7X}$ ) and 2,3-dimethylbutane (from isopropyl halide). n-Heptane is an odd-carbon unsymmetrical alkane ( $R-R'$ ), so it would need two different halides, giving a mixture of alkanes with poor yield.

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2020 · NEET-UG (India) · 4 marks

Sucrose on hydrolysis gives:
(a) $\alpha$ -D-Glucose + $\beta$ -D-Fructose (b) $\alpha$ -D-Fructose + $\beta$ -D-Fructose (c) $\beta$ -D-Glucose + $\alpha$ -D-Fructose (d) $\alpha$ -D-Glucose + $\beta$ -D-Glucose

Answer

(a) $\alpha$ -D-Glucose + $\beta$ -D-Fructose

Sucrose is a non-reducing disaccharide joined through a glycosidic bond between C1 of $\alpha$ -D-glucose and C2 of $\beta$ -D-fructose. Hydrolysis (acid or invertase) splits this bond and gives equimolar glucose and fructose (invert sugar).

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2020 · NEET-UG (India) · 4 marks

Anisole (methoxybenzene) on cleavage with HI gives:
(a) $\mathrm{C_6H_5OH} + \mathrm{C_2H_5I}$ (b) $\mathrm{C_6H_5I} + \mathrm{C_2H_5OH}$ (c) $\mathrm{C_6H_5OH} + \mathrm{CH_3I}$ (d) $\mathrm{C_6H_5I} + \mathrm{CH_3OH}$

Answer

(c) $\mathrm{C_6H_5OH} + \mathrm{CH_3I}$

In the cleavage of an alkyl aryl ether with HI, the ether oxygen is protonated and $\mathrm{I^-}$ attacks the alkyl (methyl) carbon by $S_N2$ . The $\mathrm{C_6H_5-O}$ bond is not broken, because the aryl carbon is $sp^2$ with partial double-bond character from resonance. So phenol and methyl iodide form, not iodobenzene.

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2020 · NEET-UG (India) · 4 marks

Urea reacts with water to form A which will decompose to form B. B when passed through $\mathrm{Cu^{2+}(aq)}$ , deep blue coloured solution C is formed. What is the formula of C from the following?
(a) $\mathrm{Cu(OH)_2}$ (b) $\mathrm{CuCO_3\cdot Cu(OH)_2}$ (c) $\mathrm{CuSO_4}$ (d) $\mathrm{[Cu(NH_3)_4]^{2+}}$

Answer

(d) $\mathrm{[Cu(NH_3)_4]^{2+}}$

Urea hydrolyses: $\mathrm{NH_2CONH_2} + \mathrm{H_2O} \rightarrow$ ammonium carbamate (A), which decomposes to $\mathrm{NH_3}$ and $\mathrm{CO_2}$ (B). $\mathrm{NH_3}$ passed into $\mathrm{Cu^{2+}}$ first gives a pale blue precipitate and then, in excess, the deep blue tetraamminecopper(II) complex.

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2020 · NEET-UG (India) · 4 marks

Which of the following is a natural polymer?
(a) Polybutadiene (b) Poly(butadiene-acrylonitrile) (c) cis-1,4-Polyisoprene (d) Poly(butadiene-styrene)

Answer

(c) cis-1,4-Polyisoprene

It is natural rubber, obtained from latex of the rubber tree. Polybutadiene, butadiene-acrylonitrile (Buna-N) and butadiene-styrene (Buna-S) are all synthetic polymers.

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2021 · NEET-UG (India) · 4 marks

The compound which shows metamerism is
(a) $\mathrm{C_3H_6O}$ (b) $\mathrm{C_4H_{10}O}$ (c) $\mathrm{C_5H_{12}}$ (d) $\mathrm{C_3H_8O}$

Answer

(b) $\mathrm{C_4H_{10}O}$

Metamerism arises when the same functional group (usually a divalent one such as an ether or ketone) has different alkyl groups on either side. $\mathrm{C_4H_{10}O}$ can be $\mathrm{CH_3OC_3H_7}$ and $\mathrm{C_2H_5OC_2H_5}$ , which are metamers. $\mathrm{C_3H_8O}$ only allows $\mathrm{CH_3OC_2H_5}$ as an ether, and $\mathrm{C_3H_6O}$ and $\mathrm{C_5H_{12}}$ have no suitable pair of metamers.

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2021 · NEET-UG (India) · 4 marks

Which one of the following polymers is prepared by addition polymerisation?
(a) Novolac (b) Dacron (c) Teflon (d) Nylon-66

Answer

(c) Teflon

Teflon is made by addition (chain-growth) polymerisation of tetrafluoroethylene, $\mathrm{CF_2{=}CF_2}$ . Novolac (phenol and formaldehyde), Dacron (ethylene glycol and terephthalic acid) and nylon-66 (hexamethylenediamine and adipic acid) are all condensation polymers.

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2021 · NEET-UG (India) · 4 marks

What is the IUPAC name of the organic compound formed in the following chemical reaction?
Acetone $\xrightarrow[\text{(ii) } \mathrm{H_2O,\,H^+}]{\text{(i) } \mathrm{C_2H_5MgBr},\ \text{dry ether}}$ Product
(a) pentan-3-ol (b) 2-methyl butan-2-ol (c) 2-methylpropan-2-ol (d) pentan-2-ol

Answer

(b) 2-methyl butan-2-ol

A Grignard reagent adds to a ketone to give a tertiary alcohol after hydrolysis. $\mathrm{C_2H_5MgBr}$ adds to $\mathrm{CH_3COCH_3}$ to give $\mathrm{(CH_3)_2C(OH)C_2H_5}$ . The longest chain is butane with $\mathrm{-OH}$ and $\mathrm{-CH_3}$ both on C-2, so the name is 2-methylbutan-2-ol. Pentan-3-ol would come from an aldehyde (propanal) with ethyl Grignard.

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2021 · NEET-UG (India) · 4 marks

Dihedral angle of least stable conformer of ethane is
(a) $60^\circ$ (b) $0^\circ$ (c) $120^\circ$ (d) $180^\circ$

Answer

(b) $0^\circ$

The eclipsed conformer of ethane has a dihedral angle of $0^\circ$ . Its C-H bonds on the two carbons face each other, giving maximum torsional strain and repulsion. The staggered conformer ( $60^\circ$ ) is the most stable one.

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2021 · NEET-UG (India) · 4 marks

The correct sequence of bond enthalpy of 'C-X' bond is
(a) $\mathrm{CH_3\text{-}F < CH_3\text{-}Cl > CH_3\text{-}Br > CH_3\text{-}I}$ (b) $\mathrm{CH_3\text{-}Cl > CH_3\text{-}F > CH_3\text{-}Br > CH_3\text{-}I}$ (c) $\mathrm{CH_3\text{-}F < CH_3\text{-}Cl < CH_3\text{-}Br < CH_3\text{-}I}$ (d) $\mathrm{CH_3\text{-}F > CH_3\text{-}Cl > CH_3\text{-}Br > CH_3\text{-}I}$

Answer

(d) $\mathrm{CH_3\text{-}F > CH_3\text{-}Cl > CH_3\text{-}Br > CH_3\text{-}I}$

As the halogen gets larger, its orbital overlap with carbon becomes poorer and the C-X bond longer, so bond enthalpy falls down the group. Values are about 452 (C-F), 351 (C-Cl), 293 (C-Br) and 234 (C-I) kJ/mol.

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2021 · NEET-UG (India) · 4 marks

The RBC deficiency is deficiency disease of
(a) Vitamin $\mathrm{B_1}$ (b) Vitamin $\mathrm{B_2}$ (c) Vitamin $\mathrm{B_{12}}$ (d) Vitamin $\mathrm{B_6}$

Answer

(c) Vitamin $\mathrm{B_{12}}$

Deficiency of vitamin $\mathrm{B_{12}}$ (cyanocobalamin) leads to pernicious anaemia, a shortage of normal red blood cells. $\mathrm{B_1}$ deficiency causes beri-beri, $\mathrm{B_2}$ causes cheilosis and $\mathrm{B_6}$ causes convulsions.

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2021 · NEET-UG (India) · 4 marks

The major product formed in dehydrohalogenation reaction of 2-Bromopentane is pent-2-ene. This product formation is based on?
(a) Hofmann Rule (b) Huckel's Rule (c) Saytzeff's Rule (d) Hund's Rule

Answer

(c) Saytzeff's Rule

In dehydrohalogenation, the more substituted (more stable) alkene is the major product. 2-Bromopentane gives pent-2-ene (disubstituted) in preference to pent-1-ene. Hofmann's rule gives the less substituted alkene, and Huckel's and Hund's rules concern aromaticity and electron filling.

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2021 · NEET-UG (India) · 4 marks

$\mathrm{CH_3CH_2COO^-Na^+} \xrightarrow{\text{Heat}} \mathrm{CH_3CH_3} + \mathrm{Na_2CO_3}$ . Consider the above reaction and identify the missing reagent/chemical.
(a) CaO (b) DIBAL-H (c) $\mathrm{B_2H_6}$ (d) Red Phosphorus

Answer

(a) CaO

Heating sodium propanoate with soda lime ( $\mathrm{NaOH+CaO}$ ) is decarboxylation, which removes $\mathrm{CO_2}$ (kept as $\mathrm{Na_2CO_3}$ ) and gives ethane. CaO keeps the NaOH from attacking the glass and makes it easier to handle. DIBAL-H and $\mathrm{B_2H_6}$ are reducing agents, and red phosphorus is used in halogenation (HVZ) reactions.

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2022 · NEET-UG (India) · 4 marks

Statement I: The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group. Statement II: o-nitrophenol, m-nitrophenol and p-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring. Choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct. (b) Both Statement I and Statement II are incorrect. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is correct.

Answer

(c) Statement I is correct but Statement II is incorrect.

The electron-withdrawing nitro group stabilises the phenoxide ion, so nitrophenols are more acidic than phenol. The ortho, meta and para isomers differ in strength, because the $-R$ effect acts in the ortho and para positions but not in the meta position. Para-nitrophenol is more acidic than meta-nitrophenol.

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2022 · NEET-UG (India) · 4 marks

Statement I: Primary aliphatic amines react with $\mathrm{HNO_2}$ to give unstable diazonium salts. Statement II: Primary aromatic amines react with $\mathrm{HNO_2}$ to form diazonium salts which are stable even above 300 K. Choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct. (b) Both Statement I and Statement II are incorrect. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is correct.

Answer

(c) Statement I is correct but Statement II is incorrect.

Primary aliphatic amines give unstable aliphatic diazonium salts with $\mathrm{HNO_2}$ (NaNO $_2$ + HCl), which decompose at once to alcohols and $\mathrm{N_2}$ . Aromatic diazonium salts are stabilised by resonance with the ring, but only at low temperature ( $273-278$ K). Above about 300 K they decompose, so Statement II is false.

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2022 · NEET-UG (India) · 4 marks

$\mathrm{RMgX + CO_2} \xrightarrow{\text{dry ether}} \mathrm{Y} \xrightarrow{\mathrm{H_3O^+}} \mathrm{RCOOH}$ . What is Y in the above reaction?
(a) $\mathrm{RCOO^-Mg^+X}$ (b) $\mathrm{R_3CO^-Mg^+X}$ (c) $\mathrm{RCOO^-X^+}$ (d) $\mathrm{(RCOO)_2Mg}$

Answer

(a) $\mathrm{RCOO^-Mg^+X}$

The Grignard reagent adds to the carbonyl carbon of $\mathrm{CO_2}$ to give the magnesium carboxylate salt $\mathrm{RCOOMgX}$ . Acid hydrolysis ( $\mathrm{H_3O^+}$ ) then liberates $\mathrm{RCOOH}$ . The $\mathrm{R_3CO^-}$ option would need addition to a ketone/ester, not to $\mathrm{CO_2}$ .

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2022 · NEET-UG (India) · 4 marks

Which statement regarding polymers is not correct?
(a) Elastomers have polymer chains held together by weak intermolecular forces. (b) Fibers possess high tensile strength. (c) Thermoplastic polymers are capable of repeatedly softening and hardening on heating and cooling respectively. (d) Thermosetting polymers are reusable.

Answer

(d) Thermosetting polymers are reusable.

Thermosetting polymers (e.g. bakelite) are cross-linked and become infusible and insoluble once set, so they cannot be remelted or reused. Thermoplastics soften on heating and harden on cooling repeatedly, elastomers have weak forces between coiled chains, and fibres have strong H-bonding/dipole forces and high tensile strength.

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2022 · NEET-UG (India) · 4 marks

The incorrect statement regarding enzymes is :
(a) Enzymes are biocatalysts. (b) Like chemical catalysts enzymes reduce the activation energy of bio processes. (c) Enzymes are polysaccharides. (d) Enzymes are very specific for a particular reaction and substrate.

Answer

(c) Enzymes are polysaccharides.

Enzymes are biocatalysts that are globular proteins (a few are RNA), not polysaccharides. They lower the activation energy of biochemical reactions and are highly specific for a reaction and substrate, so the other three statements are correct.

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2022 · NEET-UG (India) · 4 marks

Statement I: The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole-dipole interactions. Statement II: The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct. (b) Both Statement I and Statement II are incorrect. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is correct.

Answer

(a) Both Statement I and Statement II are correct.

Aldehydes and ketones are polar and have dipole-dipole attraction, so they boil higher than hydrocarbons of similar molar mass. They cannot form hydrogen bonds with themselves (no O-H), so they boil lower than the corresponding alcohols.

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2022 · NEET-UG (India) · 4 marks

Match List-I (Products formed) with List-II (Reaction of carbonyl compound with). List-I A. Cyanohydrin B. Acetal C. Schiff's base D. Oxime List-II I. $\mathrm{NH_2OH}$ II. $\mathrm{RNH_2}$ III. alcohol IV. HCN
(a) A-III, B-IV, C-II, D-I (b) A-II, B-III, C-IV, D-I (c) A-I, B-III, C-II, D-IV (d) A-IV, B-III, C-II, D-I

Answer

(d) A-IV, B-III, C-II, D-I

Cyanohydrin forms with HCN, acetal with alcohol (dry HCl), Schiff base (imine) with a primary amine $\mathrm{RNH_2}$ , and oxime with hydroxylamine $\mathrm{NH_2OH}$ .

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2022 · NEET-UG (India) · 4 marks

The incorrect statement regarding chirality is:
(a) $S_N1$ reaction yields 1 : 1 mixture of both enantiomers. (b) The product obtained by $S_N2$ reaction of haloalkane having chirality at the reactive site shows inversion of configuration. (c) Enantiomers are superimposable mirror images on each other. (d) A racemic mixture shows zero optical rotation.

Answer

(c) Enantiomers are superimposable mirror images on each other.

This is false: enantiomers are non-superimposable mirror images. $S_N1$ goes through a planar carbocation and gives a racemic (1:1) mixture, $S_N2$ gives inversion of configuration, and a racemic mixture has zero optical rotation because the two rotations cancel.

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2022 · NEET-UG (India) · 4 marks

Compound X on reaction with $\mathrm{O_3}$ followed by $\mathrm{Zn/H_2O}$ gives formaldehyde and 2-methyl propanal as products. The compound X is :
(a) 3-Methylbut-1-ene (b) 2-Methylbut-1-ene (c) 2-Methylbut-2-ene (d) Pent-2-ene

Answer

(a) 3-Methylbut-1-ene

Ozonolysis cleaves $\mathrm{C=C}$ and puts $\mathrm{C=O}$ on each carbon. $\mathrm{CH_2=CH-CH(CH_3)_2}$ gives $\mathrm{HCHO}$ and $\mathrm{OHC-CH(CH_3)_2}$ (2-methylpropanal). 2-Methylbut-1-ene would give a ketone (butanone), and the others give different aldehydes.

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2023 · NEET-UG (India) · 4 marks

Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
(a) 16 (b) 32 (c) 30 (d) 18

Answer

(b) 32

Sodium ethanoate heated with soda lime (NaOH/CaO) undergoes decarboxylation: $\mathrm{CH_3COONa + NaOH \xrightarrow{CaO,\Delta} CH_4 + Na_2CO_3}$ . Methane has molar mass 16 g/mol, so 2 mol weigh $2\times16 = 32$ g.

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2023 · NEET-UG (India) · 4 marks

The number of $\sigma$ bonds, $\pi$ bonds and lone pair of electrons in pyridine, respectively are:
(a) 11, 2, 0 (b) 12, 3, 0 (c) 11, 3, 1 (d) 12, 2, 1

Answer

(c) 11, 3, 1

Pyridine ( $\mathrm{C_5H_5N}$ ) has 6 ring $\sigma$ bonds (C-C and C-N) plus 5 C-H $\sigma$ bonds, giving 11 $\sigma$ bonds. It has 3 $\pi$ bonds (alternating double bonds) and 1 lone pair on nitrogen, which lies in the plane and is not part of the aromatic sextet.

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2023 · NEET-UG (India) · 4 marks

Statement I: A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside. Statement II: When nucleoside is linked to phosphorous acid at 5'-position of sugar moiety, we get nucleotide. In the light of the above statements, choose the correct answer from the options given below :
(a) Both Statement I and Statement II are true (b) Both Statement I and Statement II are false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true

Answer

(c) Statement I is true but Statement II is false

A nucleoside is base + sugar, with the base attached at the 1' position, so Statement I is true. A nucleotide forms when a nucleoside is esterified with phosphoric acid (not phosphorous acid) at the 5'-position of the sugar, so Statement II is false.

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2024 · NEET-UG (India) · 4 marks

Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. In the light of the above statement, choose the correct answer from the options given below:
(a) Both statement I and statement II are false (b) Statement I is correct but statement II is false (c) Statement I is incorrect but statement II is true (d) Both statement I and statement II are true

Answer

(d) Both statement I and statement II are true

Aniline's $-\mathrm{NH_2}$ forms an adduct with the $\mathrm{AlCl_3}$ catalyst, so Friedel-Crafts alkylation fails. Gabriel synthesis needs an alkyl halide and gives only aliphatic primary amines; aryl halides do not undergo nucleophilic substitution with phthalimide, so aniline cannot be made this way.

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2024 · NEET-UG (India) · 4 marks

Fehling's solution 'A' is
(a) alkaline copper sulphate (b) alkaline solution of sodium potassium tartrate (Rochelle's salt) (c) aqueous sodium citrate (d) aqueous copper sulphate

Answer

(d) aqueous copper sulphate

Fehling's solution A is aqueous $\mathrm{CuSO_4}$ . Fehling's solution B is alkaline sodium potassium tartrate (Rochelle salt); they are mixed just before use.

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2024 · NEET-UG (India) · 4 marks

A compound with a molecular formula of $\mathrm{C_6H_{14}}$ has two tertiary carbons. Its IUPAC name is:
(a) 2-methylpentane (b) 2,3-dimethylbutane (c) 2,2-dimethylbutane (d) n-hexane

Answer

(b) 2,3-dimethylbutane

$\mathrm{(CH_3)_2CH-CH(CH_3)_2}$ has two carbons each bonded to three other carbons (two tertiary C). 2-methylpentane has only one tertiary carbon, 2,2-dimethylbutane has a quaternary carbon, and n-hexane has none.

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2024 · NEET-UG (India) · 4 marks

Statement I: The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane. Statement II: When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Both statement I and statement II are incorrect (b) Statement I is correct but statement II is incorrect (c) Statement I is incorrect but statement II is correct (d) Both statement I and statement II are correct

Answer

(d) Both statement I and statement II are correct

Boiling point falls with branching: n-pentane (309 K) > isopentane (301 K) > neopentane (283 K). Branching makes the molecule more spherical, reducing surface area and so the van der Waals forces, which lowers the boiling point.

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2024 · NEET-UG (India) · 4 marks

The reagents with which glucose does not react to give the corresponding test/products are A. Tollen's reagent B. Schiff's reagent C. HCN D. $\mathrm{NH_2OH}$ E. $\mathrm{NaHSO_3}$ Choose the correct options from the given below:
(a) A and D (b) B and E (c) E and D (d) B and C

Answer

(b) B and E

Glucose reacts with Tollens' reagent, HCN and $\mathrm{NH_2OH}$ because of its free aldehyde group. It does not give Schiff's test or the $\mathrm{NaHSO_3}$ addition product, since in the cyclic hemiacetal form the aldehyde is largely masked.

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2024 · NEET-UG (India) · 4 marks

The products A and B obtained in the following reactions, respectively are $\mathrm{3ROH + PCl_3 \rightarrow 3RCl + A}$ $\mathrm{ROH + PCl_5 \rightarrow RCl + HCl + B}$
(a) $\mathrm{POCl_3}$ and $\mathrm{H_3PO_4}$ (b) $\mathrm{H_3PO_4}$ and $\mathrm{POCl_3}$ (c) $\mathrm{H_3PO_3}$ and $\mathrm{POCl_3}$ (d) $\mathrm{POCl_3}$ and $\mathrm{H_3PO_3}$

Answer

(c) $\mathrm{H_3PO_3}$ and $\mathrm{POCl_3}$

$\mathrm{3ROH + PCl_3\rightarrow 3RCl + H_3PO_3}$ (phosphorous acid). $\mathrm{ROH + PCl_5\rightarrow RCl + HCl + POCl_3}$ (phosphoryl chloride).

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2025 · NEET-UG (India) · 4 marks

The correct order of decreasing acidity of the following aliphatic acids is
(a) $\mathrm{HCOOH > CH_3COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH}$ (b) $\mathrm{HCOOH > (CH_3)_3CCOOH > (CH_3)_2CHCOOH > CH_3COOH}$ (c) $\mathrm{(CH_3)_3CCOOH > (CH_3)_2CHCOOH > CH_3COOH > HCOOH}$ (d) $\mathrm{CH_3COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH > HCOOH}$

Answer

(a) $\mathrm{HCOOH > CH_3COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH}$

Alkyl groups have a +I effect. They destabilise the carboxylate ion and reduce acidity, and more alkyl substitution means a stronger effect. HCOOH has no alkyl group, so it is the most acidic. Acidity then falls as the alkyl group goes from methyl to isopropyl to tert-butyl.

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2025 · NEET-UG (India) · 4 marks

Sugar 'X' A. is found in honey B. is a keto sugar. C. exists in $\alpha$ and $\beta$ anomeric forms D. is laevorotatory 'X' is
(a) Maltose (b) Sucrose (c) D-Glucose (d) D-Fructose

Answer

(d) D-Fructose

Fructose is a ketohexose found abundantly in honey. It exists in $\alpha$ and $\beta$ anomeric forms and is laevorotatory ( $[\alpha]=-92^\circ$ ). D-Glucose is an aldose and dextrorotatory. Sucrose has no free anomeric carbon, and maltose is an aldose-containing disaccharide.

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2025 · NEET-UG (India) · 4 marks

Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula $\mathrm{C_4H_8O}$ is:
(a) 10 (b) 11 (c) 6 (d) 8

Answer

(a) 10

The cyclic ethers of $\mathrm{C_4H_8O}$ are: tetrahydrofuran (1); 2-methyloxetane (2 enantiomers); 3-methyloxetane (1). Oxirane derivatives: 2,2-dimethyloxirane (1); 2,3-dimethyloxirane (cis 1 plus trans 2 enantiomers = 3); 2-ethyloxirane (2 enantiomers). Total $=1+2+1+1+3+2=10$ .

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2025 · NEET-UG (India) · 4 marks

Match List I (Name of Vitamin) with List II (Deficiency disease). List I A. Vitamin B12 B. Vitamin D C. Vitamin B2 D. Vitamin B6 List II I. Cheilosis II. Convulsions III. Rickets IV. Pernicious anaemia
(a) A-II, B-III, C-I, D-IV (b) A-IV, B-III, C-II, D-I (c) A-I, B-III, C-II, D-IV (d) A-IV, B-III, C-I, D-II

Answer

(d) A-IV, B-III, C-I, D-II

Vitamin $\mathrm{B_{12}}$ deficiency causes pernicious anaemia. Vitamin D deficiency causes rickets. Vitamin $\mathrm{B_2}$ (riboflavin) deficiency causes cheilosis. Vitamin $\mathrm{B_6}$ deficiency causes convulsions.

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2025 · NEET-UG (India) · 4 marks

The correct order of decreasing basic strength of the given amines is:
(a) N-ethylethanamine > ethanamine > N-methylaniline > benzenamine (b) Benzenamine > ethanamine > N-methylaniline > N-ethylethanamine (c) N-methylaniline > benzenamine > ethanamine > N-ethylethanamine (d) N-ethylethanamine > ethanamine > benzenamine > N-methylaniline

Answer

(a) N-ethylethanamine > ethanamine > N-methylaniline > benzenamine

Aliphatic amines are more basic than aromatic amines, because the aryl lone pair is delocalised into the ring. Diethylamine (secondary) is more basic than ethylamine in water, owing to the +I effect and solvation. In the aromatic pair, N-methylaniline is slightly more basic than aniline because of the +I effect of the methyl group.

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2025 · NEET-UG (India) · 4 marks

Which one of the following compounds can exist as cis-trans isomers?
(a) 1,1-Dimethylcyclopropane (b) 1,2-Dimethylcyclohexane (c) Pent-1-ene (d) 2-Methylhex-2-ene

Answer

(b) 1,2-Dimethylcyclohexane

Cis-trans isomerism needs two different substituents on each of two ring (or double-bond) carbons, with restricted rotation. 1,2-Dimethylcyclohexane has methyls on two ring carbons, so it has cis and trans forms. 1,1-Dimethylcyclopropane has both groups on one carbon. Pent-1-ene has a terminal $=\mathrm{CH_2}$ , and 2-methylhex-2-ene has two identical methyls on one alkene carbon, so neither shows geometrical isomerism.

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2025 · NEET-UG (India) · 4 marks

Statement I: Benzenediazonium salt is prepared by the reaction of aniline with acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement I is correct but Statement II is incorrect (b) Statement I is incorrect but Statement II is correct (c) Both Statement I and Statement II are correct (d) Both Statement I and Statement II are incorrect

Answer

(c) Both Statement I and Statement II are correct

Aniline is diazotised with $\mathrm{NaNO_2/HCl}$ at 273-278 K. The diazonium salt is unstable and explosive when dry, so it is used in solution. Direct iodination of benzene is difficult (HI is reducing and reverses the reaction). Iodobenzene is therefore made by treating the diazonium salt with KI.

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Questions from NEET-UG 2013 (India), question paper (NTA NEET-UG 2013 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2014 (India), AIPMT 2014 Code Q (NTA NEET-UG 2014 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2015 (India), Re-exam Code B (NTA NEET-UG 2015 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2016 (India), Phase 1 Code A-P-W (NTA NEET-UG 2016 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2016 (India), Phase 2 Code AA-PP-WW (NTA NEET-UG 2016 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2017 (India), Booklet Code A (CBSE NEET-UG 2017 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2018 (India), Code EE (NTA NEET-UG 2018 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2019 (India), Code Q2 (NTA NEET-UG 2019 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2020 (India), Code F1 (NTA NEET-UG 2020 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2021 (India), Paper Code O1 (NTA NEET-UG 2021 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2022 (India), Paper Code Q6 (NTA NEET-UG 2022 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2023 (India), question paper (NTA NEET-UG 2023 question paper; only questions on topics in Nepal's MBBS entrance syllabus), NEET-UG 2024 (India), Paper Code R3 (NTA NEET-UG 2024 question paper; only questions on topics in Nepal's MBBS entrance syllabus) and NEET-UG 2025 (India), Paper Code Narmada 48 (NTA NEET-UG 2025 question paper; only questions on topics in Nepal's MBBS entrance syllabus). Answers are written for this site; check them against your class notes.

Past questions and answers

3.In a protein molecule various amino acids are linked together by (a) α\alphaα-glycosidic bond (b) β\betaβ-glycosidic bond (c) peptide bond (d) dative bond

4.Natural rubber has: (a) All cis-configuration (b) All trans-configuration (c) Alternate cis- and trans-configuration (d) Random cis- and trans-configuration

5.Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer? (a) Acetone (b) Ozone (c) MnO2\mathrm{MnO_2}MnO2​ (d) Aluminium isopropoxide

6.Consider the nitration of benzene using mixed conc. H2SO4\mathrm{H_2SO_4}H2​SO4​ and HNO3\mathrm{HNO_3}HNO3​. If a large amount of KHSO4\mathrm{KHSO_4}KHSO4​ is added to the mixture, the rate of nitration will be (a) faster (b) slower (c) unchanged (d) doubled

9.The product formed by the reaction of an aldehyde with a primary amine is (a) Schiff base (b) Ketone (c) Carboxylic acid (d) Aromatic acid

12.Which one given below is a non-reducing sugar? (a) Maltose (b) Lactose (c) Glucose (d) Sucrose

13.The pair of electrons in the given carbanion, CH3C≡C−\mathrm{CH_3C\equiv C^{-}}CH3​C≡C−, is present in which of the following orbitals? (a) 2p (b) sp3sp^3sp3 (c) sp2sp^2sp2 (d) sp

14.Which of the following can be used as the halide component for Friedel-Crafts reaction? (a) Chlorobenzene (b) Bromobenzene (c) Chloroethene (d) Isopropyl chloride

15.In pyrrole (a five-membered ring with the nitrogen atom, carrying a hydrogen, at position 1 and the carbon atoms numbered 2, 3, 4, 5 around the ring), the electron density is maximum on (a) 2 and 3 (b) 3 and 4 (c) 2 and 4 (d) 2 and 5

18.Consider the reaction CH3CH2CH2Br+NaCN→CH3CH2CH2CN+NaBr\mathrm{CH_3CH_2CH_2Br+NaCN\rightarrow CH_3CH_2CH_2CN+NaBr}CH3​CH2​CH2​Br+NaCN→CH3​CH2​CH2​CN+NaBr. This reaction will be the fastest in (a) ethanol (b) methanol (c) N,N′N,N'N,N′-dimethylformamide (DMF) (d) water

19.The compound that will react most readily with gaseous bromine has the formula (a) C3H6\mathrm{C_3H_6}C3​H6​ (b) C2H2\mathrm{C_2H_2}C2​H2​ (c) C4H10\mathrm{C_4H_{10}}C4​H10​ (d) C2H4\mathrm{C_2H_4}C2​H4​

21.Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional groups? (a) −CHO-\mathrm{CHO}−CHO (b) −CH2Cl-\mathrm{CH_2Cl}−CH2​Cl (c) −COOH-\mathrm{COOH}−COOH (d) −CHCl2-\mathrm{CHCl_2}−CHCl2​

23.In an SN1S_N1SN​1 reaction on chiral centres, there is: (a) 100% inversion (b) 100% racemization (c) inversion more than retention leading to partial racemization (d) 100% retention

24.Which of the following statements is not correct for a nucleophile? (a) Nucleophiles are not electron seeking (b) Nucleophile is a Lewis acid (c) Ammonia is a nucleophile (d) Nucleophiles attack low e−e^-e− density sites

25.Two possible stereo-structures of CH3CHOH⋅COOH\mathrm{CH_3CHOH{\cdot}COOH}CH3​CHOH⋅COOH, which are optically active, are called: (a) Mesomers (b) Diastereomers (c) Atropisomers (d) Enantiomers

26.Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is: (a) sodium hydrogen sulphite (b) a Grignard reagent (c) hydrazine in presence of feebly acidic solution (d) hydrocyanic acid

29.The oxidation of benzene by V2O5\mathrm{V_2O_5}V2​O5​ in the presence of air produces: (a) benzaldehyde (b) benzoic anhydride (c) maleic anhydride (d) benzoic acid

30.Caprolactam is used for the manufacture of: (a) Nylon-6,6 (b) Nylon-6 (c) Teflon (d) Terylene

31.The following reaction: aniline reacts with benzoyl chloride in the presence of NaOH to give N-phenylbenzamide (benzanilide). It is known by the name: (a) Schotten-Baumann reaction (b) Friedel-Craft's reaction (c) Perkin's reaction (d) Acetylation reaction

35.Which of the following will not be soluble in sodium hydrogen carbonate? (a) Benzoic acid (b) ooo-Nitrophenol (c) Benzenesulphonic acid (d) 2,4,6-trinitrophenol

37.Which one is most reactive towards nucleophilic addition reaction? (a) acetophenone, C6H5COCH3\mathrm{C_6H_5COCH_3}C6​H5​COCH3​ (b) 4-methylbenzaldehyde (c) 4-nitrobenzaldehyde (d) benzaldehyde

38.Which of the following organic compounds has same hybridization as its combustion product (CO2\mathrm{CO_2}CO2​)? (a) Ethyne (b) Ethene (c) Ethanol (d) Ethane

41.Which of these is not a monomer for a high molecular mass silicone polymer? (a) MeSiCl3\mathrm{MeSiCl_3}MeSiCl3​ (b) Me2SiCl2\mathrm{Me_2SiCl_2}Me2​SiCl2​ (c) Me3SiCl\mathrm{Me_3SiCl}Me3​SiCl (d) PhSiCl3\mathrm{PhSiCl_3}PhSiCl3​

43.Nylon is an example of (a) Polyester (b) Polysaccharide (c) Polyamide (d) Polythene

45.Nitrobenzene on reaction with conc. HNO3/H2SO4\mathrm{HNO_3/H_2SO_4}HNO3​/H2​SO4​ at 80-100 °C forms which one of the following products? (a) 1,2-Dinitrobenzene (b) 1,3-Dinitrobenzene (c) 1,4-Dinitrobenzene (d) 1,2,4-Trinitrobenzene

46.Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? (a) −C≡N-\mathrm{C\equiv N}−C≡N (b) −SO3H-\mathrm{SO_3H}−SO3​H (c) −COOH-\mathrm{COOH}−COOH (d) −NO2-\mathrm{NO_2}−NO2​

48.The radical C6H5CH2∙\mathrm{C_6H_5CH_2^{\bullet}}C6​H5​CH2∙​ (benzyl radical) is aromatic because it has (a) 6 p-orbitals and 6 unpaired electrons (b) 7 p-orbitals and 6 unpaired electrons (c) 7 p-orbitals and 7 unpaired electrons (d) 6 p-orbitals and 7 unpaired electrons

50.Which of the following compounds will not undergo Friedel-Crafts reaction easily? (a) Cumene (b) Xylene (c) Nitrobenzene (d) Toluene

51.With respect to the conformers of ethane, which of the following statements is true? (a) Bond angle remains same but bond length changes (b) Bond angle changes but bond length remains same (c) Both bond angle and bond length change (d) Both bond angles and bond length remains same

55.Which of the following statements is not correct? (a) Insulin maintains sugar level in the blood of a human body (b) Ovalbumin is a simple food reserve in egg-white (c) Blood proteins thrombin and fibrinogen are involved in blood clotting (d) Denaturation makes the proteins more active

56.An example of a sigma bonded organometallic compound is: (a) Ruthenocene (b) Grignard's reagent (c) Ferrocene (d) Cobaltocene

57.The heating of phenyl-methyl ethers with HI produces (a) Ethyl chlorides (b) Iodobenzene (c) Phenol (d) Benzene

58.Which of the following reactions is appropriate for converting acetamide to methanamine? (a) Carbylamine reaction (b) Hoffmann hypobromamide reaction (c) Stephens reaction (d) Gabriels phthalimide synthesis

62.A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4\mathrm{H_2SO_4}H2​SO4​. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be (a) 1.4 (b) 2.8 (c) 3.0 (d) 4.4

69.Which of the following compounds can form a zwitterion? (a) Aniline (b) Benzoic acid (c) Acetanilide (d) Glycine

72.The non-essential amino acid among the following is: (a) Leucine (b) Alanine (c) Lysine (d) Valine

73.The number of sigma (σ\sigmaσ) and pi (π\piπ) bonds in pent-2-en-4-yne is (a) 8 σ\sigmaσ bonds and 5 π\piπ bonds (b) 11 σ\sigmaσ bonds and 2 π\piπ bonds (c) 13 σ\sigmaσ bonds and no π\piπ bonds (d) 10 σ\sigmaσ bonds and 3 π\piπ bonds

74.The biodegradable polymer is: (a) Nylon-2-nylon 6 (b) Nylon-6 (c) Buna-S (d) Nylon-6,6

76.Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as: (a) Cross Cannizzaro's reaction (b) Cross Aldol condensation (c) Aldol condensation (d) Cannizzaro's reaction

78.Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give: (a) Tert. butyl alcohol (b) Isobutyl alcohol (c) Isopropyl alcohol (d) Sec. butyl alcohol

79.Which of the following is a basic amino acid? (a) Tyrosine (b) Lysine (c) Serine (d) Alanine

80.Elimination reaction of 2-bromopentane to form pent-2-ene is: (a) β\betaβ-Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction (a) (b), (c), (d) (b) (a), (b), (d) (c) (a), (b), (c) (d) (a), (c), (d)

81.Which of the following alkanes cannot be made in good yield by Wurtz reaction? (a) n-Heptane (b) n-Butane (c) n-Hexane (d) 2,3-Dimethylbutane

82.Sucrose on hydrolysis gives: (a) α\alphaα-D-Glucose + β\betaβ-D-Fructose (b) α\alphaα-D-Fructose + β\betaβ-D-Fructose (c) β\betaβ-D-Glucose + α\alphaα-D-Fructose (d) α\alphaα-D-Glucose + β\betaβ-D-Glucose

85.Which of the following is a natural polymer? (a) Polybutadiene (b) Poly(butadiene-acrylonitrile) (c) cis-1,4-Polyisoprene (d) Poly(butadiene-styrene)

86.The compound which shows metamerism is (a) C3H6O\mathrm{C_3H_6O}C3​H6​O (b) C4H10O\mathrm{C_4H_{10}O}C4​H10​O (c) C5H12\mathrm{C_5H_{12}}C5​H12​ (d) C3H8O\mathrm{C_3H_8O}C3​H8​O

87.Which one of the following polymers is prepared by addition polymerisation? (a) Novolac (b) Dacron (c) Teflon (d) Nylon-66

89.Dihedral angle of least stable conformer of ethane is (a) 60∘60^\circ60∘ (b) 0∘0^\circ0∘ (c) 120∘120^\circ120∘ (d) 180∘180^\circ180∘

91.The RBC deficiency is deficiency disease of (a) Vitamin B1\mathrm{B_1}B1​ (b) Vitamin B2\mathrm{B_2}B2​ (c) Vitamin B12\mathrm{B_{12}}B12​ (d) Vitamin B6\mathrm{B_6}B6​

92.The major product formed in dehydrohalogenation reaction of 2-Bromopentane is pent-2-ene. This product formation is based on? (a) Hofmann Rule (b) Huckel's Rule (c) Saytzeff's Rule (d) Hund's Rule

98.The incorrect statement regarding enzymes is : (a) Enzymes are biocatalysts. (b) Like chemical catalysts enzymes reduce the activation energy of bio processes. (c) Enzymes are polysaccharides. (d) Enzymes are very specific for a particular reaction and substrate.

102.Compound X on reaction with O3\mathrm{O_3}O3​ followed by Zn/H2O\mathrm{Zn/H_2O}Zn/H2​O gives formaldehyde and 2-methyl propanal as products. The compound X is : (a) 3-Methylbut-1-ene (b) 2-Methylbut-1-ene (c) 2-Methylbut-2-ene (d) Pent-2-ene

103.Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is : (a) 16 (b) 32 (c) 30 (d) 18

104.The number of σ\sigmaσ bonds, π\piπ bonds and lone pair of electrons in pyridine, respectively are: (a) 11, 2, 0 (b) 12, 3, 0 (c) 11, 3, 1 (d) 12, 2, 1

107.Fehling's solution 'A' is (a) alkaline copper sulphate (b) alkaline solution of sodium potassium tartrate (Rochelle's salt) (c) aqueous sodium citrate (d) aqueous copper sulphate

108.A compound with a molecular formula of C6H14\mathrm{C_6H_{14}}C6​H14​ has two tertiary carbons. Its IUPAC name is: (a) 2-methylpentane (b) 2,3-dimethylbutane (c) 2,2-dimethylbutane (d) n-hexane

113.Sugar 'X' A. is found in honey B. is a keto sugar. C. exists in α\alphaα and β\betaβ anomeric forms D. is laevorotatory 'X' is (a) Maltose (b) Sucrose (c) D-Glucose (d) D-Fructose

114.Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O\mathrm{C_4H_8O}C4​H8​O is: (a) 10 (b) 11 (c) 6 (d) 8

117.Which one of the following compounds can exist as cis-trans isomers? (a) 1,1-Dimethylcyclopropane (b) 1,2-Dimethylcyclohexane (c) Pent-1-ene (d) 2-Methylhex-2-ene

In a protein molecule various amino acids are linked together by (a) $\alpha$ -glycosidic bond (b) $\beta$ -glycosidic bond (c) peptide bond (d) dative bond

Natural rubber has: (a) All cis-configuration (b) All trans-configuration (c) Alternate cis- and trans-configuration (d) Random cis- and trans-configuration

Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer? (a) Acetone (b) Ozone (c) $\mathrm{MnO_2}$ (d) Aluminium isopropoxide

Consider the nitration of benzene using mixed conc. $\mathrm{H_2SO_4}$ and $\mathrm{HNO_3}$ . If a large amount of $\mathrm{KHSO_4}$ is added to the mixture, the rate of nitration will be (a) faster (b) slower (c) unchanged (d) doubled

The product formed by the reaction of an aldehyde with a primary amine is (a) Schiff base (b) Ketone (c) Carboxylic acid (d) Aromatic acid

Which one given below is a non-reducing sugar? (a) Maltose (b) Lactose (c) Glucose (d) Sucrose

The pair of electrons in the given carbanion, $\mathrm{CH_3C\equiv C^{-}}$ , is present in which of the following orbitals? (a) 2p (b) $sp^3$ (c) $sp^2$ (d) sp

Which of the following can be used as the halide component for Friedel-Crafts reaction? (a) Chlorobenzene (b) Bromobenzene (c) Chloroethene (d) Isopropyl chloride

In pyrrole (a five-membered ring with the nitrogen atom, carrying a hydrogen, at position 1 and the carbon atoms numbered 2, 3, 4, 5 around the ring), the electron density is maximum on (a) 2 and 3 (b) 3 and 4 (c) 2 and 4 (d) 2 and 5

Consider the reaction $\mathrm{CH_3CH_2CH_2Br+NaCN\rightarrow CH_3CH_2CH_2CN+NaBr}$ . This reaction will be the fastest in (a) ethanol (b) methanol (c) $N,N'$ -dimethylformamide (DMF) (d) water

The compound that will react most readily with gaseous bromine has the formula (a) $\mathrm{C_3H_6}$ (b) $\mathrm{C_2H_2}$ (c) $\mathrm{C_4H_{10}}$ (d) $\mathrm{C_2H_4}$

Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional groups? (a) $-\mathrm{CHO}$ (b) $-\mathrm{CH_2Cl}$ (c) $-\mathrm{COOH}$ (d) $-\mathrm{CHCl_2}$

In an $S_N1$ reaction on chiral centres, there is: (a) 100% inversion (b) 100% racemization (c) inversion more than retention leading to partial racemization (d) 100% retention

Which of the following statements is not correct for a nucleophile? (a) Nucleophiles are not electron seeking (b) Nucleophile is a Lewis acid (c) Ammonia is a nucleophile (d) Nucleophiles attack low $e^-$ density sites

Two possible stereo-structures of $\mathrm{CH_3CHOH{\cdot}COOH}$ , which are optically active, are called: (a) Mesomers (b) Diastereomers (c) Atropisomers (d) Enantiomers

Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is: (a) sodium hydrogen sulphite (b) a Grignard reagent (c) hydrazine in presence of feebly acidic solution (d) hydrocyanic acid

The oxidation of benzene by $\mathrm{V_2O_5}$ in the presence of air produces: (a) benzaldehyde (b) benzoic anhydride (c) maleic anhydride (d) benzoic acid

Caprolactam is used for the manufacture of: (a) Nylon-6,6 (b) Nylon-6 (c) Teflon (d) Terylene

The following reaction: aniline reacts with benzoyl chloride in the presence of NaOH to give N-phenylbenzamide (benzanilide). It is known by the name: (a) Schotten-Baumann reaction (b) Friedel-Craft's reaction (c) Perkin's reaction (d) Acetylation reaction

Which of the following will not be soluble in sodium hydrogen carbonate? (a) Benzoic acid (b) $o$ -Nitrophenol (c) Benzenesulphonic acid (d) 2,4,6-trinitrophenol

Which one is most reactive towards nucleophilic addition reaction? (a) acetophenone, $\mathrm{C_6H_5COCH_3}$ (b) 4-methylbenzaldehyde (c) 4-nitrobenzaldehyde (d) benzaldehyde

Which of the following organic compounds has same hybridization as its combustion product ( $\mathrm{CO_2}$ )? (a) Ethyne (b) Ethene (c) Ethanol (d) Ethane

Which of these is not a monomer for a high molecular mass silicone polymer? (a) $\mathrm{MeSiCl_3}$ (b) $\mathrm{Me_2SiCl_2}$ (c) $\mathrm{Me_3SiCl}$ (d) $\mathrm{PhSiCl_3}$

Nylon is an example of (a) Polyester (b) Polysaccharide (c) Polyamide (d) Polythene

Nitrobenzene on reaction with conc. $\mathrm{HNO_3/H_2SO_4}$ at 80-100 °C forms which one of the following products? (a) 1,2-Dinitrobenzene (b) 1,3-Dinitrobenzene (c) 1,4-Dinitrobenzene (d) 1,2,4-Trinitrobenzene

Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? (a) $-\mathrm{C\equiv N}$ (b) $-\mathrm{SO_3H}$ (c) $-\mathrm{COOH}$ (d) $-\mathrm{NO_2}$

The radical $\mathrm{C_6H_5CH_2^{\bullet}}$ (benzyl radical) is aromatic because it has (a) 6 p-orbitals and 6 unpaired electrons (b) 7 p-orbitals and 6 unpaired electrons (c) 7 p-orbitals and 7 unpaired electrons (d) 6 p-orbitals and 7 unpaired electrons

Which of the following compounds will not undergo Friedel-Crafts reaction easily? (a) Cumene (b) Xylene (c) Nitrobenzene (d) Toluene

With respect to the conformers of ethane, which of the following statements is true?
(a) Bond angle remains same but bond length changes (b) Bond angle changes but bond length remains same (c) Both bond angle and bond length change (d) Both bond angles and bond length remains same

Which of the following statements is not correct?
(a) Insulin maintains sugar level in the blood of a human body (b) Ovalbumin is a simple food reserve in egg-white (c) Blood proteins thrombin and fibrinogen are involved in blood clotting (d) Denaturation makes the proteins more active

An example of a sigma bonded organometallic compound is:
(a) Ruthenocene (b) Grignard's reagent (c) Ferrocene (d) Cobaltocene

The heating of phenyl-methyl ethers with HI produces
(a) Ethyl chlorides (b) Iodobenzene (c) Phenol (d) Benzene

Which of the following reactions is appropriate for converting acetamide to methanamine?
(a) Carbylamine reaction (b) Hoffmann hypobromamide reaction (c) Stephens reaction (d) Gabriels phthalimide synthesis

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. $\mathrm{H_2SO_4}$ . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
(a) 1.4 (b) 2.8 (c) 3.0 (d) 4.4

Which of the following compounds can form a zwitterion?
(a) Aniline (b) Benzoic acid (c) Acetanilide (d) Glycine

The non-essential amino acid among the following is:
(a) Leucine (b) Alanine (c) Lysine (d) Valine

The number of sigma ( $\sigma$ ) and pi ( $\pi$ ) bonds in pent-2-en-4-yne is
(a) 8 $\sigma$ bonds and 5 $\pi$ bonds (b) 11 $\sigma$ bonds and 2 $\pi$ bonds (c) 13 $\sigma$ bonds and no $\pi$ bonds (d) 10 $\sigma$ bonds and 3 $\pi$ bonds

The biodegradable polymer is:
(a) Nylon-2-nylon 6 (b) Nylon-6 (c) Buna-S (d) Nylon-6,6

Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as:
(a) Cross Cannizzaro's reaction (b) Cross Aldol condensation (c) Aldol condensation (d) Cannizzaro's reaction

Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give:
(a) Tert. butyl alcohol (b) Isobutyl alcohol (c) Isopropyl alcohol (d) Sec. butyl alcohol

Which of the following is a basic amino acid?
(a) Tyrosine (b) Lysine (c) Serine (d) Alanine

Elimination reaction of 2-bromopentane to form pent-2-ene is:
(a) $\beta$ -Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction
(a) (b), (c), (d) (b) (a), (b), (d) (c) (a), (b), (c) (d) (a), (c), (d)

Which of the following alkanes cannot be made in good yield by Wurtz reaction?
(a) n-Heptane (b) n-Butane (c) n-Hexane (d) 2,3-Dimethylbutane

Sucrose on hydrolysis gives:
(a) $\alpha$ -D-Glucose + $\beta$ -D-Fructose (b) $\alpha$ -D-Fructose + $\beta$ -D-Fructose (c) $\beta$ -D-Glucose + $\alpha$ -D-Fructose (d) $\alpha$ -D-Glucose + $\beta$ -D-Glucose

Which of the following is a natural polymer?
(a) Polybutadiene (b) Poly(butadiene-acrylonitrile) (c) cis-1,4-Polyisoprene (d) Poly(butadiene-styrene)

The compound which shows metamerism is
(a) $\mathrm{C_3H_6O}$ (b) $\mathrm{C_4H_{10}O}$ (c) $\mathrm{C_5H_{12}}$ (d) $\mathrm{C_3H_8O}$

Which one of the following polymers is prepared by addition polymerisation?
(a) Novolac (b) Dacron (c) Teflon (d) Nylon-66

Dihedral angle of least stable conformer of ethane is
(a) $60^\circ$ (b) $0^\circ$ (c) $120^\circ$ (d) $180^\circ$

The RBC deficiency is deficiency disease of
(a) Vitamin $\mathrm{B_1}$ (b) Vitamin $\mathrm{B_2}$ (c) Vitamin $\mathrm{B_{12}}$ (d) Vitamin $\mathrm{B_6}$

The major product formed in dehydrohalogenation reaction of 2-Bromopentane is pent-2-ene. This product formation is based on?
(a) Hofmann Rule (b) Huckel's Rule (c) Saytzeff's Rule (d) Hund's Rule

The incorrect statement regarding enzymes is :
(a) Enzymes are biocatalysts. (b) Like chemical catalysts enzymes reduce the activation energy of bio processes. (c) Enzymes are polysaccharides. (d) Enzymes are very specific for a particular reaction and substrate.

Compound X on reaction with $\mathrm{O_3}$ followed by $\mathrm{Zn/H_2O}$ gives formaldehyde and 2-methyl propanal as products. The compound X is :
(a) 3-Methylbut-1-ene (b) 2-Methylbut-1-ene (c) 2-Methylbut-2-ene (d) Pent-2-ene

Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
(a) 16 (b) 32 (c) 30 (d) 18

The number of $\sigma$ bonds, $\pi$ bonds and lone pair of electrons in pyridine, respectively are:
(a) 11, 2, 0 (b) 12, 3, 0 (c) 11, 3, 1 (d) 12, 2, 1

Fehling's solution 'A' is
(a) alkaline copper sulphate (b) alkaline solution of sodium potassium tartrate (Rochelle's salt) (c) aqueous sodium citrate (d) aqueous copper sulphate

A compound with a molecular formula of $\mathrm{C_6H_{14}}$ has two tertiary carbons. Its IUPAC name is:
(a) 2-methylpentane (b) 2,3-dimethylbutane (c) 2,2-dimethylbutane (d) n-hexane

Sugar 'X' A. is found in honey B. is a keto sugar. C. exists in $\alpha$ and $\beta$ anomeric forms D. is laevorotatory 'X' is
(a) Maltose (b) Sucrose (c) D-Glucose (d) D-Fructose

Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula $\mathrm{C_4H_8O}$ is:
(a) 10 (b) 11 (c) 6 (d) 8

Which one of the following compounds can exist as cis-trans isomers?
(a) 1,1-Dimethylcyclopropane (b) 1,2-Dimethylcyclohexane (c) Pent-1-ene (d) 2-Methylhex-2-ene