NEET PYQs

Hint

(a) Alkanes, alkenes and alkynes follow the following trend in their acidic behaviour :

This is because sp-hybridised carbon is more electronegative than \(s p^{2}\)-hybridised carbon which is further more electronegative than \(s p^{3}\)-hybridised carbon. Hence, in ethyne proton can be released more easily than ethene and ethane.
Among alkynes the order of acidity is

This is due to \(+I\) effect of \(-\mathrm{CH}_{3}\) group.

Hint

(c) In case of unsymmetrical alkynes addition of \(\mathrm{H}_2 \mathrm{O}\) occurs in accordance with Markownikoff’s rule.

Hint

(c) There is no change in bond angles and bond lengths in the conformations of ethane.

Hint

(d) Friedel-Crafts reaction :

Chlorobenzene, bromobenzene and chloroethene are not suitable halide components as \(\mathrm{C}-X\) bond acquires some double bond character due to resonance of lone pair of electrons with \(\pi\) bond.

Hint

(a) Biphenyl is coplanar as all C-atoms are \(s p^{2}\) hybridised.

Hint

(d) Pyrrole has maximum electron density on 2 and 5 . It generally reacts with electrophiles at the \(\mathrm{C}-2\) or \(\mathrm{C}-5\) due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbocation that is a hybrid of structures (I) and (II). Attack at position 2 or 5 yields a carbocation that is a hybrid not only of structures (III) and (IV) (analogous to I and II) but also of structure (V). The extra stabilization conferred by (V) makes this ion the more stable one.

Also, attack at position 2 or 5 is faster because the developing positive charge is accommodated by three atoms of the ring instead of by only two.

Hint

(c)

Hint

(c)

Hint

(a) Bromine gas reacts with alkene to give allylic substituted product by free radical mechanism. And we know that reaction of free radical depends upon stability of free radical, which here will be stabilised by the conjugation with \(\pi\) bond.

Hint

Magnitude of torsional strain depends upon the angle of rotation about \(\mathrm{C}-\mathrm{C}\) bond. Staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. So, the staggered conformation of ethane is more stable than the eclipsed conformation.

Hint

(d) Mechanism of nitration is :
\(
\mathrm{HNO}_{3}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NO}_{2}^{+}+2 \mathrm{HSO}_{4}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}
\)
If a large amount of \(\mathrm{KHSO}_{4}\) is added then conc. of \(\mathrm{HSO}_{4}^{-}\)ions increases and the reaction will be shifted in backward direction hence, the rate of nitration will be slower.

Hint

(b) \(
\mathrm{CH}_3-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}^{-}
\)

Thus, pair of electrons is present in \(s p\)-hybridised orbital.

Hint

(c)

Hint

(c)

Hint

(a)

Hint

Hint

(d) Enthalpy of hydrogenation is inversely proportional to the stability of alkenes.
Stability of alkenes : I \(>\) II \(>\) III
Enthalpy of hydrogenation : I \( \lt \) II \( \lt \) III

Hint

(c) \(-\mathrm{CH}_{3}\) group is o,p-directing. Because of crowding, no substitution occurs at the carbon atom between the two \(-\mathrm{CH}_{3}\) groups in \(m\)-Xylene, even though two \(-\mathrm{CH}_{3}\) groups activate that position.

Hint

(a)

Hint

(b)

Hint

(a) Nitrobenzene is strongly deactivated, hence will not undergo Friedel-Craft’s reaction.

Hint

(d) The molecules which do not satisfy Huckel rule or \((4 n+2) \pi\)-electron rule are said to be non-aromatic. The compound (d) has total \(4 \pi e^{-}\). It does not follow \((4 n+2)\) rule. So it is non-aromatic compound. All other compounds \((\mathrm{a}, \mathrm{b}, \mathrm{c})\) are planar and have \(6 \pi e^{-}\), so they are aromatic

Hint

Acetaldehyde does not give Victor Meyer test.

Hint

Hint

(a) Maleic acid shows geometrical isomerism and not optical isomerism.

Hint

(a) Terminal alkynes (1-butyne) react with \(\mathrm{NaNH}_{2}\) to form sodium acetylide and evolve hydrogen but 2-butyne do not.

Hint

(b) The anti-conformation is the most stable conformation of \(n\)-butane. In this, the bulky methyl groups are as far apart as possible thereby keeping steric repulsion at a minimum.

Hint

(b) Cracking : The process of cracking converts higher alkanes into smaller alkanes and alkenes. This process can be used for production of natural gas.

Hint

(c) The reaction of \(\mathrm{Cl}_{2}\), in presence of \(\mathrm{FeCl}_{3}\), with benzene yields a ring substitution product.

Hint

Hint

(d) the conformation d is most stable due to Intermolecular H-bonding.

Hint

Hint

(d) Cis-trans isomerism is exhibited by compounds having \(\mathrm{C}=\mathrm{C}, \mathrm{C}=\mathrm{N}\) and \(\mathrm{N}=\mathrm{N}\) groups, due to restricted rotation around the double bond. Among the given options only 2-butene qualifies to exhibit geometrical isomerism.

Hint

(d)

Hint

Hint

The state of hybridisation of carbon in 1,3 and 5 position are \(s p, s p^3, s p^2\).

Hint

According to Markownikoff’s rule, during hydrohalogenation to unsymmetrical alkene, the negative part of the addendum adds to less hydrogenated (i.e. more substituted) carbon atom.

Hint

Hint

(a) The relative rates of hydrogenation decrease with the increase of steric hindrance. In order of stability of alkene, most stable the alkene slowly it gives the product.

Hence alkene which will react faster with \(\mathrm{H}_2\) is that which is most unstable.

Hint

(b) On ozonolysis, higher alkynes form diketones which are further oxidised to dicarboxylic acid.

Hint

(c)

\(
\mathrm{C}_6 \mathrm{H}_5 \mathrm{H}+\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 \xrightarrow[95^{\circ}, \text { pressure }]{\mathrm{AlCl}_3, \mathrm{HCl}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3
\)

Hint

(d) The formation of \(n\)-propyl bromide in presence of peroxide can be explained as follows: Step 1 : Peroxide undergoes fission to give free radicals.
\(
R-\mathrm{O}-\mathrm{O}-{R} \longrightarrow 2 R-\dot{\mathrm{O}}
\)

Step 2 : HBr combines with free radical to form bromine free radical.
\(
R-\dot{\mathrm{O}}+\mathrm{HBr} \longrightarrow R-\mathrm{OH}+\dot{\mathrm{Br}}
\)

Step \(3: \dot{\mathrm{Br}}\) attacks the double bond of the alkene to form a more stable free radical.

Step 4 : More stable free radical attacks on \(\mathrm{HBr}\).
\(
\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{HCH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{Br}}
\)
\(
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad n \text {-propyl bromide }
\)
\(
\text { Step } 5: \dot{\mathrm{Br}}+\dot{\mathrm{Br}} \longrightarrow \mathrm{Br}_2
\)

Hint

Hint

Hint

(d)

\(-\mathrm{NH}_2\) group is electron donating hence increases electron density on ring. Benzene is also electron rich due to delocalisation of electrons. \(-\mathrm{NO}_2\) group is electron withdrawing hence, decreases electron density on ring. Thus, correct order for electrophilic substitution is:

I \(>\) II \(>\) III.

Hint

(b)

The product is benzoic acid.

Hint

(b)

\(
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2 \xrightarrow{\mathrm{NaNH}_2} \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH}
\)

Hint

(b)  (b) Alcohols may be dehydrated to the corresponding olefins. The order of ease of dehydration is
\(
3^{\circ} \text { alcohol }>2^{\circ} \text { alcohol }>1^{\circ} \text { alcohol. }
\)
\(
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{Al}_2 \mathrm{O}_3, 620 \mathrm{~K}} \mathrm{CH}_2=\mathrm{CH}_2
\)

Hint

(b) Phenol exist as resonance hybrid of the following structures:

Thus, due to resonance the oxygen atom of the – \(\mathrm{OH}\) group acquires a positive charge and hence attracts electron pair of the \(\mathrm{O}-\mathrm{H}\) bond leading to the release of hydrogen atom as proton.

Once the phenoxide ion is formed it stabilises itself by resonance which is more stable than the parent phenol as there is no charge separation.

Effect of substituent \(\rightarrow\) Presence of electron withdrawing groups \(\left(-\mathrm{NO}_2,-X,-\mathrm{CN}\right)\) increase the acidity of phenols while the presence of electron releasing groups \(\left(-\mathrm{NH}_2,-\mathrm{CH}_3\right)\) decrease the acidity of phenols. This explains the following order of acidity.

\(
p \text {-nitrophenol }>\text { phenol }>p \text {-cresol. }
\)

Hint

(a) Due to \(-I\) effect of \(\mathrm{F}\) atom, \(\mathrm{CF}_3\) in benzene ring deactivates the ring and does not favour electrophilic substitution. While \(-\mathrm{CH}_3\) and \(-\mathrm{OCH}_3\) are ‘ \(+I\) group’ which favours electrophilic substitution in the benzene ring at ‘ortho’ and ‘para’ positions. The \(+I\) effect of \(-\mathrm{OCH}_3\) is more than \(-\mathrm{CH}_3\), therefore the correct order for electrophilic substitution is

Hint

(a) In Friedel-Crafts reaction toluene is obtained by the action of \(\mathrm{CH}_3 \mathrm{Cl}\) on benzene in presence of \(\mathrm{AlCl}_3\).

Hint

(b) Propene adds to diborane \(\left(\mathrm{B}_2 \mathrm{H}_6\right)\) giving an addition product. The addition compound on oxidation gives 1-propanol. Here addition of water takes place according to anti-Markownikoff’s rule.

Hint

(c)

This is most stable as the repulsion between two methyl groups is least.

Hint

(a) Due to restricted rotation about double bond, 2-butene shows geometrical isomerism.

Hint

(c) The staggered form of ethane has the following structure and the dihedral angle is \(60^{\circ}\), which means ‘ \(\mathrm{H}\) ‘ atoms are at an angle of \(60^{\circ}\) to each other.

Hint

(d)

Hint

(a) In Friedel-Crafts reaction, an alkyl group is introduced into the benzene ring in presence of a Lewis acid \(\left(\mathrm{AlCl}_3\right)\) catalyst. The reaction is 

Hint

(a) \(-\mathrm{OH},-\mathrm{Cl}\) and \(-\mathrm{CH}_3\) groups in benzene are ortho-para directing groups and activate the ring towards electrophilic substitution reaction. But among these \(-\mathrm{OH}\) group is strongly activating while \(-\mathrm{CH}_3\) is weakly activating and \(-\mathrm{Cl}\) is deactivating. Thus, phenol will be most easily attacked by an electrophile.

Hint

(a) Arenes undergo nucleophilic substitution reaction and are resistant to addition reactions, due to delocalisation of \(\pi\)-electrons. These are also stabilized by resonance

Hint

Hint

 (c)  When an aqueous solution of sodium or potassium salt of carboxylic acid is electrolysed, hydrocarbon is evolved at anode.

Hint

(d)

\(
\text { Therefore, } M=\mathrm{CH}_2 \mathrm{Cl}-\mathrm{CH}_2 \mathrm{OH} \text { and } R=\text { aq. } \mathrm{NaHCO}_3
\)

Hint

(b) In alkyne, two carbon atoms constituting the triple bond are \(s p\)-hybridised. Carbon undergoes \(s p\)-hybridisation to form two \(s p\)-hybrid orbitals. The two \(2 p\)-orbitals remain unhybridised. Hybrid orbitals form one sigma bond while two \(\pi\)-bonds are formed by unhybridised orbitals.

Hint

(c) The branching of chain increases the octane number of a fuel. High octane number means better fuel.

Hint

(a) In structure III, withdrawal of electrons by \(-\mathrm{NO}_2\) causes decrease in reaction rate while in structure I, there is electron releasing effect by \(-\mathrm{OCH}_3\) group which accelerates the reaction.

Hint

(b) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) – \(n\)-butane Newman projection for \(n\)-butane is

The staggered conformation has minimum repulsion between the hydrogen atoms attached tetrahedrally to the two carbon atoms. Thus, it is the most stable conformation.

Hint

(c) 

\(
\mathrm{Cl}_2+\mathrm{FeCl}_3 \rightarrow \mathrm{FeCl}_4^{-}+\mathrm{Cl}^{+}
\)

Hint

(d) In this reaction, \(\mathrm{HBr}\) undergoes heterolytic fission as \(\mathrm{HBr} \rightarrow \mathrm{H}^{+}+\mathrm{Br}^{-}\)
\(
\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_3+\mathrm{HBr} \longrightarrow \mathrm{CH}_3-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3 \xrightarrow{\mathrm{Br}^{-}} \mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CH}_3
\)

Hint

(d)

Both methyl group cancel each other. So net zero dipole moment exists.

Hint

(d)

Hint

(c) In the case of but-2-ene \(\left(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3\right)\) both double bonded carbons are identical. Therefore, it does not observe the anti-Markownikoff’s addition of \(\mathrm{HBr}\).

Hint

(c) Nitronium ion \(\left(\mathrm{NO}_2^{+}\right)\)is an electrophile that actually attacks the benzene ring.

Hint

(c) \(\mathrm{KOH}\) in \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\), when reacts with 1,1-dihaloalkanes form alkynes.

Explanation:

This reaction is an example of dehydrohalogenation Hence, alcoholic \(\mathrm{KOH}\) is used as a reagent.

Hint

(c) Reduction of non-terminal alkynes with \(\mathrm{Na}\) in liq. \(\mathrm{NH}_3\) at \(195-200 \mathrm{~K}\) gives trans alkene.

Hint

(a) Alkynes react with strong bases like \(\mathrm{NaNH}_2\) to form sodium acetylide derivative known as acetylides.

Hint

(d) \(3^{\circ}>2^{\circ}>1^{\circ}\). The reactivity of \(\mathrm{H}\)-atom depends upon the stability of free radicals, therefore reactivity of \(\mathrm{H}\)-atom follows the order.
\(
3^{\circ}>2^{\circ}>1^{\circ}
\)

Hint

(d) Benzene shows kekule structures which are resonating structures and these structures are separated by a double headed arrow \((\leftrightarrow)\)

Hint

(c) Due to resonance all the \(\mathrm{C}-\mathrm{C}\) bonds in the benzene possess same nature and the resonating structures are obtained because of the delocalisation of \(\pi\)-electrons

Hint

(a)

The formation of \(\mathrm{C}-\mathrm{H}\) bond in acetylene involves \(s p\)-hybridised carbon atom. Since \(s\)-electrons are closer to the nucleus than \(p\) electrons, the electrons present in a bond having more \(s\)-character will be more closer to the nucleus. In alkynes \(s\) character is \(50 \%\), the electrons constituting this bond are more strongly bonded by the carbon nucleus. Thus, acetylenic C-atom becomes more electronegative in comparison to \(s p^2, s p^3\) and hence the hydrogen atom present on carbon atom \((\equiv \mathrm{C}-\mathrm{H})\) can be easily removed.

Hint

(d) All the three reagents except ammoniacal \(\mathrm{AgNO}_3\) reacts with 1,2 and 4 compounds. The compound 3 possessing the terminal alkyne only reacts with ammoniacal \(\mathrm{AgNO}_3\)and thus can be distinguished from 1,2 and 4 compounds.

Hint

(b)
In Wurtz reaction, an alkyl halide \((R-X)\) give a symmetrical alkane \((R-R)\) of even number of carbon atom with \(100 \%\) yield.
\(
2 R \text { 一 } \times \frac{2 \mathrm{Na} \text { /ether }}{-2 \mathrm{NaX}} \mathrm{R}-\mathrm{R}(100 \%)
\)
An unsymmetrical alkane \(\left(R-R^{\prime}\right)\) with odd number of carbon atom can be obtained with \(33 \%\) yield only, if we take a mixture of two different alkyl halides, \(R-X\) and \(R^{\prime}-X\)

Hint

(b)
Key Idea Alkanes which contain all equivalent hydrogen atoms forms only one monochloro derivative on halogenation.

Hint

(b)

Hint

(a)

1. With increase in number of carbons in alkane, boiling point increases
2. In case of isomeric alkanes, greater is the number of branches, lesser is the boiling point.

Boiling point order :
Heptane \(>\) Hexane \(>2\) – methylbutane \(>\) butane \(>2-\) methylpropan
\((a)>(c)>(e)>(d)>(b)\)

Hint

(b) is the correct choice

Hint

Alkenes can be prepared
\(
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{X}\xrightarrow{\mathrm{KOH} \text { (aq.) }}\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}
\)
\(
\text { alkyl halide } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text { alcohol }
\)

Hint

(c) The correct structure of 2,6-Dimethyl-dec-4-ene is:
The correct numbering in the present case is one that gives lowest possible number to double bond, thus correct numbering to longest chain is given by

The structure can be identified with the help of branches at 2 and 6 positions and a double bond at 4 position.

Hint

(a) During ozonolysis, an alkene is oxidised in such a way that \(\mathrm{C}=\mathrm{C}\) bond is broken and each \(\mathrm{C}\) gets a new bond \(-\mathrm{C}=\mathrm{O}\). Here, methanal can only be formed if the double bond is with a terminal \(\mathrm{C}\) atom so that after reaction, it gets converted into \(\mathrm{H}_2 \mathrm{C}=\mathrm{O}\).

Hint

(b)

Hint

(d) In birch reduction, Alkynes are reduced to cis/trans alkenes with the help of \(\mathrm{Na} /\) liquid \(\mathrm{NH}_3\) and \(\mathrm{H}_2, \mathrm{Pd} / \mathrm{C}\), quinoline. \(\mathrm{Na} /\) liquid \(\mathrm{NH}_3\) is used to form trans alkene and \(\mathrm{H}_2, \mathrm{Pd} / \mathrm{C}\), quinoline is used to form cis-alkene. \(\mathrm{H}_2, \mathrm{Pd} / \mathrm{C}\), quinoline is a hydrogenation catalyst. It is used for the hydrogenation of alkynes to convert them into alkenes by breaking the triple bond into double bonds. \(\mathrm{H}_2, \mathrm{Pd} / \mathrm{C}\), quinoline is called lindlar’s catalyst. It is a poisonous catalyst which is used for the partial hydrogenation of alkynes. The quinolone present in the reagent stops the complete hydrogenation of alkynes and prevents the formation of alkanes. The lindlar’s catalyst has three components. The conversion of but – 2 yne to cis – 2 – butene is given as:

Hint

(c)

Hint

\(
\mathrm{CH}_4 \xrightarrow[h \nu]{\mathrm{Br}_2}\mathrm{CH}_3-\mathrm{Br}\xrightarrow[\text { ether }]{\mathrm{Na}}\mathrm{CH}_3-\mathrm{CH}_3
\)

Hint

Statement I is incorrect but Statement II is correct.

Hint

Generally alkanes resist oxidation but alkane with tertiary H atom(s) can be oxidised to corresponding alcohols by \(KMnO _4\)

\(
\therefore R _1 \Rightarrow CH _3 \quad R _2 \Rightarrow CH _3 \quad R _3 \Rightarrow CH _3
\)