NEET PYQs

Hint

(c) The o- and p-nitrophenols are separated by steam distillation since o-isomer is steam volatile due to intramolecular H-bonding while p-isomer is not steam volatile due to association of molecules by intermolecular H-bonding. 

Hint

(d)

Hint

(c) On the other hand, nucleophile is the opposite of the electrophile, it will donate the electrons, which are not used up in bonding or which are non-bonding electrons in valence shell and it will have extra electron which will be used up to donate to the species which do not have electron.

The difference between the electrophile and nucleophile is that the electrophile is Lewis acid nucleophile is Lewis base, electrophile is positively charged and nucleophile is negatively charged. Electrophile accept electrons as it is electron deficient, nucleophile donates electrons and it is electron rich. Electrophile will accept electron pairs and form covalent bonds; nucleophiles will donate electron pain to form covalent bonds. They can be neutral or have positive charge, hydronium ion is the example of the electrophile. The example of a nucleophile is chloride ion.Some other examples of electrophiles include \(\mathrm{NO}_2^{+}, \mathrm{FeCl}_3, \mathrm{Br}^{+}\), etc.

Hence, we conclude that the correct statement regarding electrophile is that electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

Hint

(a) \(\alpha\)-Hydrogen at bridge carbon never participate in tautomerism. Thus, only (III) exhibits tautomerism.

Hint

 (d) o-Substituted biphenyls are optically active as both the rings are not in one plane and their mirror images are non-superimposable.

Hint

(d)

\(
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{KOH} \rightarrow \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}
\)

Saturated compound is converted into unsaturated compound by removal of group of atoms hence, it is an elimination reaction.

\(-\mathrm{Br}\) group is replaced by \(-\mathrm{OH}\) group hence, it is a substitution reaction.

Addition of \(\mathrm{Br}_2\) converts an unsaturated compound into a saturated compound hence, it is an addition reaction.

Hint

(d) Nucleophiles are electron rich species and hence, they are lewis bases.

Hint

(a)

Hint

Hint

(d) There are four double bonds. Hence, no. of \(\pi\)-electrons \(=2 \times 4=8\)

Hint

(d)

\(
\mathrm{XeO}_4 \text { contain equal number of } \sigma \text { and } \pi \text {-bonds. Number of } \sigma \text { and } \pi \text {-bonds present in species are as follows: }
\)

Hint

Hint

(a) Nucleophile will attack a stable carbocation \(\left(\mathrm{S}_{\mathrm{N}} 1\right.\) reaction).

Hint

(a) Mass of organic compound \(=0.25 \mathrm{~g}\) Experimental values, At STP,
\(V_{1}=40 \mathrm{~mL}\)
\(V_{2}=\) ?
\(T_{1}=300 \mathrm{~K}\)
\(T_{2}=273 \mathrm{~K}\)
\(P_{1}=725-25=700 \mathrm{~mm}\)
\(P_{2}=760 \mathrm{~mm}\)
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
\(V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{700 \times 40 \times 273}{300 \times 760}=33.52 \mathrm{~mL}\)
\(22400 \mathrm{~mL}\) of \(\mathrm{N_2}\) at STP weighs \(=28 \mathrm{~g}\)
\(\therefore \quad 33.52 \mathrm{~mL}\) of \(\mathrm{N}_{2}\) at STP weighs \(=\frac{28 \times 33.52}{22400}\)
\(=0.0419 \mathrm{~g}\)
\(\%\) of \(\mathrm{N}=\frac{\text { Mass of nitrogen at } \mathrm{STP}}{\text { Mass of organic compound taken }} \times 100\)
\(=\frac{0.0419}{0.25} \times 100=16.76 \%\)

Hint

(a) Hyperconjugation occurs through the \(\mathrm{H}\) – atoms present on the carbon atom next to the double bond i.e alpha hydrogen atoms. There is no alpha \(-\mathrm{H}\) in the structure \(\mathrm{I}\) and II.
So, hyperconjugation occurs in structure III only i.e.;

Hint

• (c) Sigma \((\sigma)\) bonds are the strongest form of covalent chemical bond. These are made up of head-on
  overlapping between atomic orbitals. Sigma bonding is more clearly described for diatomic molecules using the symmetry group language and tools. Through this systematic solution, the bond becomes symmetrical with respect to the orientation of the bond axis.
• Sigma bonds are the strongest type of covalent bonding due to the close overlap of orbitals, and the electrons in such bonding are often referred to as sigma electrons.
  Usually, a single bond is a sigma bond, whereas a multiple bond is a sigma bond with pi or other bonds. A double bond contains a sigma and a pi bond, while a triple bond has a sigma and two pi bonds.
  After counting the bonds very carefully, we come to this conclusion-
  The above structure has 18 sigma \((\sigma)\) bonds and 2 pi \((\pi)\) bonds.
  Hence it is clear that option \(\)\mathrm{C}\(\) is the correct option.
  Note: Pi bonds are covalent chemical bonds where two orbital lobes on one atom overlap two orbital lobes on another atom and this overlap happens laterally. Both of such atomic orbitals have zero electron density on a common nodal plane, going through the two bonded nodes. The same plane is also a nodal plane for the pi bond molecular orbital. Pi Bonds that develop in double and triple bonds, but in most cases do not shape in single bonds.

Hint

(d)

is most stable due to hyperconjugation

Hint

(a) \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NH}_{3} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)
\(10 \mathrm{~mL}\) of \(1 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}=10 \mathrm{~mmol}\)

\(\quad\left[\because M \times V_{(\mathrm{mL})}=\mathrm{mmol}\right]\)
\(\mathrm{NH}_{3}\) consumed \(=20 \mathrm{~mmol}\)
Acid used for the absorption of ammonia
\(=20-10 \mathrm{~mmol}\)
\(=10 \mathrm{~mL}\) of \(2 \mathrm{~N}\) (or \(1 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}\)
\(\%\) of \(\mathrm{N}=\frac{1.4 \times N \times V}{W}=\frac{1.4 \times 2 \times 10}{0.75}=37.33 \%\)

Hint

(c)

Hint

(c)

Hint

(d)

Hint

(b) \(-\mathrm{NO}_{2}\) is most deactivating due to \(-I\) and \(-M\) effect.

Deactivating power :
\(
-\mathrm{NO}_2>-\mathrm{C} \equiv \mathrm{N}>-\mathrm{SO}_3 \mathrm{H}>\mathrm{COOH}
\)

Hint

(a) Greater the number of electron donating alkyl groups ( \(+I\) effect), greater is the stability of carbocations.
\(+I\) effect is in the order:

Hence the order of stability of carbocations is \(5<4<3<1<2\)

Hint

(a)

An atom is considered as a neutral species but when it loses or gains any electrons, then it forms ions. If the charge on the species is positive, then it is known as cation and if the charge on the species is negative, then it is known as anion.

So, we can define carbonium as a species in which the carbon atom possesses a positive charge which is also known as carbocation.

Hence, the benzyl carbonium is the species with a carbocation attached to a benzene ring and its structure is given as:
So, we can see that the given structure has carbocation (carbonium) and we know that carbocation has a triangular planar structure and have the hybridisation of \(s p^2\).

Hint

(b)  

\(
3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+2 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \rightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+6 \mathrm{Na}_2 \mathrm{SO}_4
\)
\(\text { Sodium ferrocyanide } \quad \quad \quad \text {Ferric ferrocyanide (Prussian blue) }\)

Hint

(b) More the number of hyperconjugative structures, the greater is the stability.

Stability of radical follows similar order as carbocation follows. (methyl \(<1^{\circ}<2^{\circ}<3^{\circ}\) ).
Moreover, stability of free radical \(\propto\) Number of \(\mathrm{a}-\mathrm{H}\)

Hint

(c) As the \(+I\) effect increases reactivity towards electrophilic reactions increases and as \(-I\) or \(-M\) effect increases reactivity towards electrophilic reactions decreases. Thus, the order is

Hint

(a) \(
\mathrm{Br}-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{1}{\mathrm{C}} \mathrm{H}_2
\)
\(
\text { 3-Bromopropene }
\)

Hint

(b) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
In case of \(s p^{3}\) hybridised carbon, bond angle is \(109^{\circ} 28^{\prime}\); \(s p^{2}\) hybridised carbon, bond angle is \(120^{\circ}\) and \(s p\) hybridised carbon, bond angle is \(180^{\circ}\)

So, only \(\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_3 \text { is linear. }\)

Hint

(b) \(+R\) effect of \(-\mathrm{OH}\) group is greater than that of \(-\mathrm{OCH}_{3}\) group

Hint

Hint

(b) Given \(V_{1}=55 \mathrm{~mL}, V_{2}=\) ?
\(P_{1}=715-15=700 \mathrm{~mm}, P_{2}=760 \mathrm{~mm}\)
\(T_{1}=300 \mathrm{~K}, T_{2}=273 \mathrm{~K}\)
General gas equation, \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
Volume of nitrogen at \(\mathrm{STP}\)
\(
V_{2}=\frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}=\frac{700 \times 55 \times 273}{760 \times 300}=46.099 \mathrm{~mL}
\)
\(\%\) of nitrogen \(=\frac{V_{2}}{8 W}\), were \(W=\) the mass of organic compound.
\(\%\) of \(\mathrm{N}=\frac{46.099}{8 \times 0.35}=16.46\)

Hint

 (c) Nucleophilic substitution reaction involves the displacement of a nucleophile by another.

(1) is electrophilic addition reaction, water molecule get added to the alkene.
(2) is nucleophilic addition reaction across carbonyl group where the nucleophile attacks the carbonyl carbon.
(3) is nucleophilic substitution reaction where \(\mathrm{Br}\) is substituted by \(\mathrm{NH}_2\) group.
(4) is nucleophilic addition reaction of \(\mathrm{HCN}\) across carbonyl group

Hint

(a) In case of Lassaigne’s test of halogens, it is necessary to remove sodium cyanide and sodium sulphide from the sodium extract if nitrogen and sulphur are present. This is done by boiling the sodium extract with conc. \(\mathrm{HNO}_{3}\).
\(\mathrm{NaCN}+\mathrm{HNO}_{3} \rightarrow \mathrm{NaNO}_{3}+\mathrm{HCN}_{2} \uparrow\) \(\mathrm{Na}_{2} \mathrm{~S}+2 \mathrm{HNO}_{3} \rightarrow 2 \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{~S} \uparrow\)

Hint

(a) Electron withdrawing groups like \(-\mathrm{NO}_{2}\) facilitates nucleophilic substitution reaction in chlorobenzene.
Explanation: The electron-withdrawing or electron-donating groups influence the nucleophilic substitution reaction. The substitution reaction is enhanced by the electron-withdrawing group, whereas the nucleophilic attack is decreased by the electron-donating group. The nitro group \(\left(-\mathrm{NO}_2\right)\) in p-nitrochlorobenzene is an electron-withdrawing group. When in the para position, the electron density is withdrawn to the nitro group, resulting in extended conjugation. The electron density on the ring is reduced as a result of this. As a result, the bond between the chlorine and the carbon atom (from the ring) weakens. When a more reactive nucleophile, such as \(\mathrm{OH}^{-}\), attacks the \(\mathrm{p}\)-nitrochlorobenzene, the nucleophilic substitution reaction is easily accomplished. When the aromatic ring is attached to the electron-withdrawing group, electron density shifts towards the electron withdrawing group, making the aromatic ring electrophilic and hence easy to attack by nucleophiles. When an aromatic ring is attached to an electron-donating or releasing group, electron density shifts towards the aromatic ring, making it nucleophilic and making it difficult for nucleophiles to approach it. The ring can easily undergo the electrophilic substitution reaction in such instances. Methoxy is an electron donating group. It contributes its electron density to the aromatic ring when it is attached to it. As a result, the electron density surrounding the ring’s (existing) nucleophilic groups increases. The existing nucleophile is difficult to knock off because of the high electron density. In the presence of an electron-donating group, nucleophilic substitution reactions are difficult to perform. The nucleophilic substitution process is difficult to perform on p-methylchlorobenzene. The decreasing order of the ease towards the nucleophilic substitution reaction is as shown below:

The p-chloronitrobenzene can easily undergo the nucleophilic substitution reaction.

Hint

(a)

Hint

(c)

1. Electron deficient species are known as electrophiles.
2. \(\mathrm{H}_{3}\stackrel{\oplus}{\mathrm{O}}\) has a lone pair of electrons over oxygen for donation, so it is not an electron-deficient species, hence not an electrophile

The explanation for the incorrect option:
(A) \(\stackrel{\oplus}{\mathrm{Cl}}\) has one vacant \(\mathrm{p}\)-orbital and can accept an electron to attain a stable gas configuration, hence it is an electrophile.
(B) \(\mathrm{BH}_3\) has one vacant \(\mathrm{p}\)-orbital and can accept an electron, hence it is an electrophile.
(D) \(\stackrel{\oplus}{\mathrm{N}} \mathrm{O}_2\) has one vacant \(\mathrm{p}\)-orbital and can accept an electron, hence it is an electrophile.
hence, Option(c) \(\mathrm{H}_{3}\stackrel{\oplus}{\mathrm{O}}\) is not an electrophile

Hint

(b)

\(
\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}=\stackrel{3}{\mathrm{C}} \mathrm{H} \stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H}\)
\(\text { Pent-3-en-1-yne }\)

Fact: If a molecule contains both carbon-carbon double or triple bonds, the two are treated as per in seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chain, then lowest number is given to the \(\mathrm{C}=\mathrm{C}\) double bond

Hint

(c) 

\(
\stackrel{4}{\mathrm{C}} \mathrm{H} \equiv \stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{1}{\mathrm{C}} \mathrm{H}_2
\)
\(
\text { 1-Butene-3-yne }
\)

Since the sum of numbers starting from either side of the carbon chain turns out to be the same, so lowest number is given to the \(\mathrm{C}=\mathrm{C}\) end.

Hint

(b)

\(
\mathrm{H}-\underset{s p}{\mathrm{C}} \equiv \underset{s p}{\mathrm{C}}-\mathrm{H}>\underset{s p^2}{\mathrm{CH}_2}=\underset{s p^2}{\mathrm{CH}_2}>\underset{s p^3}{\mathrm{CH}_3} \mathrm{CH}_3
\)
\(
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text { (Acidic character) }
\)
Conjugate base of the given acid :
\(
\stackrel{\ominus}{\mathrm{C}} \equiv \mathrm{C}-\mathrm{H}<\stackrel{\ominus}{\mathrm{C}} \mathrm{H}=\mathrm{CH}_2<\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2 \mathrm{CH}_3
\)
\(
\quad \quad \quad \quad \quad \quad \quad \quad \text { (Basic character) }
\)
Conjugate base of stronger acid is weaker and vice versa.

Hint

(a) Groups like, \(-\mathrm{Cl}\) and \(-\mathrm{NO}_{2}\) shows \(-I\) effect. \(-I\) groups attached to the benzene ring decrease the electron density and hence less prone to electrophilic attack. \(-\mathrm{OH}\) not only shows \(-I\) effect but also \(+M\) effect which predominates the \(-I\) character and electron density is increased in the benzene ring which facilitates electrophilic attack

Hint

(c)  Higher the no. of electron releasing group lower will be stability of carbanion, and vice versa. So the order of stability of carbanions is

Hint

Hint

 (a)  In case of different nucleophiles, but present in the same group in the periodic table, then larger is the atomic mass, higher is the nucleophilicity. Hence the decreasing order of nucleophilicity of the halide ions is

\(
\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{F}^{-} .
\)

Hint

(c) Electrophiles are electron loving chemical species. They attack at the highest electron-density site of the substrate. Electron donating substances \((+I\) effect) increases the electron density of the molecule. \(+I\) effect decreases in the order \(-\mathrm{OH}>-\mathrm{CH}_{3}>-\mathrm{H}>-\mathrm{Cl}\).
Hence order of decreasing reactivity towards electrophile is
\(
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}>\mathrm{C}_{6} \mathrm{H}_{6}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}
\)

Hint

(c) General molecular formula for alkanols is \(\mathrm{C}_{n} \mathrm{H}_{2 n+2} \mathrm{O}\) or \(\left[\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{OH}\right]\)

Hint

Hint

Hint

\((\mathrm{d}) \mathrm{S}_{\mathrm{N}}1\) reaction is favoured by heavy (bulky) groups an the carbon atom attached to halogens and nature of carbonium ion in substrate is

Betuyl > Allyl > Tertiary > Secondary > Primary > Methyl halides

Hint

Lowest priority atom is always away from the viewer. Priority is seen on the basis of atomic no. and if atomic no. are same then on the basis of atomic mass. See in the order of higher to lower priority.
If clockwise then it is \(R\).
If anticlockwise then it is \(S\). Full name of the molecule is
(R) 1-bromo-1-chloroethane

Hint

\(3^{\circ} \mathrm{C}\) is more stable due to the stabilization of the charge by three methyl groups (or inductive effect). It can also be explained on the basis of hyperconjugation.

Hint

(b) Optical isomerism and geometrical isomerism are the two parts of stereoisomerism

Hint

(b) Sublimation method is used for those organic substances which pass directly from solid to vapour state on heating and vice-versa on cooling. e.g. benzoic acid, naphthalene, camphor, anthracene, etc. Naphthalene is volatile and benzoic acid is non-volatile due to the formation of the dimer.

Hint

(a) \(\mathrm{ROH}+\mathrm{PCl}_{5} \rightarrow R \mathrm{Cl}+\mathrm{POCl}_{3}+\mathrm{HCl}\)
\(R \mathrm{COOH}+\mathrm{PCl}_{5} \rightarrow R \mathrm{COCl}+\mathrm{POCl}_{3}+\mathrm{HCl}\)

Hint

(b)  Only four structural isomers are possible for diphenyl methane.

Hint

Hint

(b) Carboxylic acid is much stronger than water and alcohol. Since the carboxylate ion after the removal of proton is stabilised by resonating structures. The – \(\mathrm{OH}\) in alcohols is almost neutral. Acetylene is also weakest acid.

Hint

Therefore, empirical formula is \(\mathrm{CH}_4 \mathrm{~N}\). Right option is c.

Hint

(b)

\(
\stackrel{1}{\mathrm{C}} \mathrm{H}_2=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\stackrel{5}{\mathrm{C}} \equiv \stackrel{6}{\mathrm{C}} \mathrm{H}
\)

The double bond gets priority over triple bond. Therefore correct IUPAC name is 1-hexene-5-yne.

Hint

(c) Geometrical isomers are those isomers which possess the same molecular and structural formula but differ in the arrangement of atoms or groups in space due to hindered rotation around the double bonded atoms.

Hint

(c) Tollen’s reagent is a solution of ammoniacal silver nitrate and used for the detection of \(-\mathrm{CHO}\) group. Aldehydes reduce Tollen’s reagent and itself get oxidised to give \(\mathrm{Ag}^{+}\)ions to \(\mathrm{Ag}\) powder which forms the silver coloured mirror in the test tube. So this test is also known as silver mirror test. \(R-\mathrm{CHO}+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \rightarrow R-\mathrm{COO}^{-}+\mathrm{Ag}\)
\(\text { (Powder) }\)

Hint

Hint

(b) Steam distillation is essentially Co-distillation with water and is carried out when a solid or liquid is insoluble in water and is volatile with steam but the impurities are non-volatile.

Hint

(a) The electronegativity of \(\mathrm{F}, \mathrm{O}\), and \(\mathrm{N}\) follows the order: \(\mathrm{N} \lt \mathrm{O} \lt \mathrm{F}\)

Therefore the negative inductive effect of \(-\mathrm{N} R_{2}\), \(-\mathrm{OR}\) and \(-\mathrm{F}\) follows the order:
\(-\mathrm{N} R_{2} \lt -\mathrm{OR} \lt -\mathrm{F}\)

Hint

(a) \(\mathrm{S}_{\mathrm{N}} 2\) reaction are bimolecular reactions where rate of reaction depends on the concentration of both substrate and nucleophile. When \(\mathrm{OH}^{-}\) attacks the substrate from the opposite side of the leaving group i.e., \(\mathrm{Br}^{-}\)a transition state results, to which both \(\mathrm{OH}\) and \(\mathrm{Br}\) are partially bonded to carbon atom.

Hint

(b) It is a special type of functional isomerism, in which both the isomers are represented by one and the same substance and are always present in equilibrium.

Explanation: The two structures formed in the process of tautomerism are known as tautomers and these types of isomer compounds usually differ only in the number of electrons and protons. They also exist in dynamic equilibrium. When a reaction occurs between these compounds there is only transfer of protons. Tautomerism is also known by some other name called desmotropism. Tautomerism is generally the shifting of a hydrogen from one place to another by exchanging the covalent bond. There are many types of tautomerism like keto-enol, lactam-lactim etc. There is a very simple method of tautomerism which extracts the acidic hydrogen and undergoes through the process of resonance and during this process it adds the hydrogen to another place. So the basic condition for compounds to show tautomerism is that it must have acidic hydrogen and that will undergo resonance too. Out of all the given option \(\mathrm{RCH}_2 \mathrm{NO}_2\) is the only case which have acidic hydrogen and go through the resonance process while \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\) also have acidic hydrogen but it does not give resonance structures.

Hint

(d) In the IR spectroscopy, each functional group appears at a certain peak (in \(\mathrm{cm}^{-1}\) ). So, cyclohexanone can be identified by carbonyl peak.

Hint

(a) There are 7 isomers in \(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\). Out of these, 4 are alcohols and 3 are ethers.

Hint

Hint

(c) Tetraethyl lead \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{4} \mathrm{~Pb}\), is used as an antiknocking agent in gasoline used for running automobiles

Hint

(d)

Hint

(a)

As – \(\mathrm{COOH}\) group is highest priority group, it is numbered one. So, the IUPAC name is 3-amino-5-heptenoic acid.

Hint

(b) Sigma bond is stronger than \(\pi\)-bond because of better overlap. All single bonds are \(\sigma\) bonds and all multiple bonds contain one \(\sigma\) and other \(\pi\) bonds.

Hint

(d) Angle increases progressively \(s p^3\left(109^{\circ} 28^{\prime}\right), s p^2\left(120^{\circ}\right), s p\left(180^{\circ}\right)\)

Hint

(c) Urea \((46.6 \% \mathrm{~N}) . \%\) of \(\mathrm{N}\) in other compounds are: \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}=21.2 \% ; \mathrm{CaCN}_{2}=35.0 \%\) and \(\mathrm{NH}_{4} \mathrm{NO}_{3}=35.0 \%\)

Hint

Restricted rotation is due to sideways overlap of two \(p\)-orbitals.

Hint

(d) Carboxylic acids dissolve in \(\mathrm{NaHCO}_{3}\) but phenols do not.

Hint

Hint

(b)

Hint

(d) Isomers must have same molecular formula but different structural formula.

Hint

(c) \(3^{\circ}>2^{\circ}>1^{\circ}\) more the delocalisation of positive charge, more is its stability.

Hint

(b) Sodium cyanide \((\mathrm{Na}+\mathrm{C}+\mathrm{N} \rightarrow \mathrm{NaCN})\)

Hint

(d) Shortest \(\mathrm{C}-\mathrm{C}\) distance (1.20 Å) is in acetylene.

Hint

(a) \(s p^{3}\) orbital has \(1 / 4(25 \%) s\)-character

Hint

Hint

(a) Kjeldahl method is an analytical chemistry method that helps in the quantitative measurement and determination of nitrogen present in organic substances as well as inorganic compounds. The method was first developed by Johan Kjeldahl in the year 1883. This process plays an important role in the method of analyzing proteins. This method was developed to determine the nitrogen contents in organic and inorganic substances  (like foodstuffs, fertilizers, wastewater, soil, feed, grain, and other substances). This method is also used for estimating the protein content in food.

Hint

Hence it is homocyclic (as the ring system is made of one type of atoms, i.e., carbon) but not aromatic. Option (a) is the correct answer.

Hint

(c)

1, 2-dichloroethene exhibits cis-trans (geometrical) isomerism.

Hint

Hint

(d) All the properties mentioned in the question suggest that it is a benzene molecule. Since in benzene all carbons are \(s p^{2}\)-hybridized, therefore, \(\mathrm{C}-\mathrm{C}-\mathrm{C}\) angle is \(120^{\circ}\)

Hint

(d) All the three ( \(\mathrm{N}, \mathrm{S}\), halogens)

Hint

(b) 5-chain isomers are obtained from alkane \(\mathrm{C}_6 \mathrm{H}_{14} \text {. }\)

Hint

(a) Tetrachloroethene being an alkene has \(s p^{2}\) -hybridized \(\mathrm{C}\)-atoms and hence the angle \(\mathrm{Cl}-\mathrm{C}-\mathrm{Cl}\) is \(120^{\circ}\) while in tetrachloromethane, carbon is \(s p^{3}\) hybridized, therefore the angle \(\mathrm{Cl}-\mathrm{C}-\mathrm{Cl}\) is \(109^{\circ} 28^{\prime}\)

Hint

(c) Stability of carbocation \(\propto\) No. of \(\alpha-\mathrm{H}\) Among the given carbocations,
\(
\stackrel{\alpha-\mathrm{H}}{\mathrm{C}} \mathrm{H}_3-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}-\stackrel{\alpha-\mathrm{H}}{\mathrm{C}} \mathrm{H}_2-\mathrm{CH}_2-\mathrm{CH}_3
\) is the most stable carbocation \((5 \alpha-\mathrm{H})\)

Hint

(a) The rate of electrophilic substitution reaction is directly proportional to nucleophilicity of Benzene ring. The group which increase the electron density in benzene ring i.e. electron donating group increase the nucleophilicity of Benzene ring which results in an increase in the rate of reaction towards electrophilic substitution \(\mathrm{Rn}^{\mathrm{n}}\). In series of activating group \(\mathrm{OH}\) comes first then \(\mathrm{OCH}_{3}\), hence, phenol is most reactive towards electrophilic substitution reaction.

Hint

(b)

1. Basicity of amines is the tendency to donate the lone pair of electrons of \(\mathrm{N}\) (nitrogen) in a chemical reaction, more easily it can donate more will be the basicity.
2. The presence of electron-withdrawing and conjugation of lone pair decreases the basicity.
3. The presence of an electron-donating group increases the basicity.
4. Here, the lone pair of \(\mathrm{N}\) is engaged in conjugation with two carbonyl groups and the lone pair is not available for donation conjugation.
5. So, it will be the least basic.
In this case choice  b is the right answer.

Hint

(d)

Hint

(d) Rate of Dehydrohalogenation : II < III < I

Hint

(b) To address these statements, let’s analyze each one :
Statement I: “In an organic compound, when inductive and electromeric effects operate in opposite directions, the inductive effect predominates.”
The inductive effect refers to the transmission of charge through a chain of atoms in a molecule by electrostatic induction. It’s a permanent effect.
The electromeric effect is a temporary effect, occurring when a reagent approaches a double bond or a polar molecule and causes electrons to shift entirely from one atom to another.
Generally, in organic chemistry, the inductive effect is considered to be a weaker effect compared to the electromeric effect. When both effects are present and oppose each other, the electromeric effect, being a more dominant electronic effect, usually prevails.
Statement II : “Hyperconjugation is observed in o-xylene.”
Hyperconjugation is a phenomenon where electrons in a sigma bond (usually C-H or CC) are delocalized into an adjacent empty or partially filled p-orbital or pi-orbital, stabilizing the system.
o-Xylene is an aromatic compound with two methyl groups attached to the benzene ring in ortho positions. However, hyperconjugation typically occurs in alkenes or carbocations where there is an adjacent p-orbital. In o-xylene, the aromatic system is already stabilized by resonance, and the hyperconjugation of the methyl groups does not significantly contribute to this stability.
Based on these explanations :
Statement I is false because the electromeric effect is generally stronger than the inductive effect.
Statement II is true as hyperconjugation can be observed in o-xylene, although its impact on the molecule’s stability is minimal compared to the aromatic resonance.
Thus, the correct option is :
Option b : Statement-I is false but Statement-II is true.

Hint

(b)

Hint

n case nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed, it give blood red colour and no prussian blue since there are no free cyanide lons
\(
\begin{array}{l}
\mathrm{Na}+\mathrm{C}+\mathrm{N}+\mathrm{S} \rightarrow \mathrm{NaSCN} \\
\mathrm{Fe}^{+3}+\mathrm{SCN}^{\ominus} \longrightarrow[\mathrm{Fe}(\mathrm{SCN})]^{2+} \text { Blood red }
\end{array}
\)

Hint

\(
\left.\begin{array}{l}
\mathrm{H}_2 \mathrm{O} \\
\mathrm{OH}^{-} \\
\mathrm{NH}_3
\end{array}\right\}
\)
can not act as lewis acid because they does not contain vacant orbital
\(\mathrm{BF}_3 \rightarrow\) Contains vacant orbital on central atom (Boron).

Hint

Huckle’s rule \(=(4 n+2) \pi\) electrons Comp (i), (ii), (v), (vii) obey Huckle’s rule

Hint

(d) \(\mathrm{Cl}\) has pronounced \(-\mathrm{I}\) effect than \(+\mathrm{R}\) effect due to large size difference between carbon and chlorine, also due to high electronegativity of chlorine.

Hint

(d) Lesser the steric hinderance on halide carbon, more will be the reactivity of alkyl halide towards \(\mathrm{S}_{\mathrm{N}} 2\) reaction.
So correct order towards \(\mathrm{S}_{\mathrm{N}} 2\) reactivity is :

\(
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Cl}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_3>\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}
\)
\(\quad \quad \quad 1^{\circ} \quad \quad \quad \quad  \quad \quad \quad \quad \quad \quad 1^{\circ} \quad \quad \quad \quad \quad \quad \quad \quad  2^{\circ} \quad \quad \quad \quad \quad \quad \quad \quad 3^{\circ}\)

Hint

(b)

\(-\mathrm{COOCH}_3\) has higher priority than \(-\mathrm{C}=\mathrm{C}-\) and \(-\mathrm{CHO}\) in IUPAC nomenclature.

\(
\begin{array}{l}
C_1=s p^2\\
\mathrm{C}_2=\mathrm{sp}^3
\end{array}
\)

Hint

Due to cross-conjugation and \(3 \alpha-\mathrm{H}\) [Hyperconjugation], (1) is most stable.

\(\longrightarrow\) Cross conjugation and \(3 \alpha-\mathrm{H}\) [Hyperconjugation]

Hint

(a) Planar, cyclic, conjugated species containing \((4 n+2) \pi\) electrons will be aromatic in nature ( \({n}\) is an integer)

Hint

(c) The stereoisomers related to each other as non-superimposable mirror image are called enantiomers.

Hint

(2) Kjeldahl method is not applicable to compounds containing nitrogen in nitro group, azo groups and nitrogen present in the ring (e.g., pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Hint

(d) 1-Bromo-5-chloro-4-methylhexan-3-ol

Hint

(a) Metamerism compound which have same molecular formula but different number of carbon atoms on either sides of functional group are known as metamers and this phenomenon is known as metamerism.

Compounds with formula \(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\) has ether functional group in which following arrangements are possible. So, it shows metamerism. For example
\(
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3 \text {, }
\)

and \(\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3\) are metamers as structure of alkyl chains are different around the functional group.

Hint

(a) Ethane has two conformers (i) Eclipsed (ii) Staggered

Eclipsed conformer is least stable while staggered conformer is most stable. In eclipsed conformer the dihedral angle is \(0^{\circ}\).

Hint

(a) Paper chromatography is a type of partition chromatography in which a special quality paper known as chromatography paper is used.

Hint

(c) More the number of \(\alpha-\mathrm{H}\) atoms, more will be the hyper conjugation effect hence more will be the stability of carbocation.

Hint

(d) In \(\mathrm{Ph}-\mathrm{OH}\), the lone pair of electrons on \(\mathrm{O}\) atom causes delocalization which makes it difficult to protonate due to less availability of electrons.
Due to involvement of lone pair of electron in reasons in phenol, it will have positive charge (Partial ), Hence incoming proton will not be able to attack easily.

Hint

(a, b)

– I effect increases on increasing electronegativity of atom. So, correct order of – I effect is
\(
-\mathrm{NH}_2<-\mathrm{OR}<-\mathrm{F}
\)

Also, \(-\mathrm{NR}_2<-\mathrm{OR}<-\mathrm{F}\)

Hint

Hint

\(-\mathrm{NO}_2\) group is an electron withdrawing group and exhibit – \(I\) effect. This effect increases with decrease in distance of positive charge present on \(\mathrm{C}\)-atom and hence lesser is the stability of carbocation.
In option (a), the positive charge is at maximum distance to \(\mathrm{NO}_2\) group, so-\(I\) effect due to \(\mathrm{NO}_2\) group will be minimum and stability will be maximum.

Hint

(d) It is Reimer-Tiemann reaction. The electrophile formed is: \(\left(: \mathrm{CCl}_2\right)\) (Dichlorocarbene) according to the following mechanism:

Hint

(a)

\(
\text { 4-ethyl-1-fluoro-2-nitrobenzene }
\)

Hint

(d) Steam distillation is a separation technique used for purifying organic compounds that are immiscible with water and have high boiling points. In this method, the organic compound co-distills with water. The key concept here is that the total vapor pressure of the mixture is the sum of the vapor pressures of the organic compound and water. When the total vapor pressure equals atmospheric pressure, the mixture will boil.

For a compound that has a boiling point of \(250^{\circ} C\) and is immiscible with water, during steam distillation, the boiling point of the mixture will be influenced by both the vapor pressure of water and the organic compound. Since water has a boiling point of \(100^{\circ} C\), steam distillation allows the mixture to boil at a temperature lower than the boiling point of the organic compound when in pure form. This is because the partial pressures of both the water and the organic compound add up to reach the atmospheric pressure more quickly than either would alone.

Practically, the mixture will boil at a temperature close to the boiling point of water, which is \(100^{\circ} C\). Hence, the correct answer is:

Option D close to but below \(100^{\circ} C\)

Hint

(d)

\(
\begin{array}{ll}
\text { Lake test } & – Al ^{3+} \\
\text { Nessler’s reagent } & – NH _4^{+} \\
\text {Potassium sulphocyanide } & – Fe ^{3+} \\
\text { Brown ring test } & – NO _3^{-}
\end{array}
\)

Hint

(c)

Methyl group attached to a positively charged carbon atom stabilizes the carbocation due to hyperconjugation and +I effect.

Hint

(a) The two statements provided deal with molecular structure and boiling points of isomers of pentane. Let’s analyze both statements:

Statement I: This statement lists the boiling point order of the three isomers of pentane as n-pentane > isopentane > neopentane. To verify this statement, let’s consider the molecular structure and boiling points of each isomer:
n-pentane: It is a straight-chain alkane with the formula \(C_5 H_{12}\), and it has the highest boiling point among the isomers because of its larger surface area which allows for greater van der Waals forces (dispersion forces).
isopentane (also known as methylbutane): It has one branch in its carbon chain. The branching reduces the surface area slightly, which slightly weakens the van der Waals forces compared to n-pentane.
neopentane (also known as dimethylpropane): It has a highly branched structure making it almost spherical. This shape minimizes the surface area significantly, thereby reducing the van der Waals forces drastically compared to the other two isomers.

Thus, the boiling point order correctly reflects the influence of molecular structure and intermolecular forces. Hence, Statement I is correct.

Statement II: This statement elaborates on why branching leads to lower boiling points. The assertion is that increased branching gives the molecule a more spherical shape, which then results in a smaller surface area, and thus weaker intermolecular forces (primarily van der Waals forces). Weaker intermolecular forces correspond with a lower energy requirement for the liquid to gas phase transition, thereby lowering the boiling point. This explanation is coherent with concepts in physical chemistry regarding molecular interactions and phase change. Therefore, Statement II is also correct.
In summary, both Statement I and Statement II are correct, reflecting the accurate relationship between molecular structure, intermolecular forces, and physical properties like boiling points. Therefore, the correct answer is:

Option A: Both Statement I and Statement II are correct.

Hint

(d) The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation.

\(
\text { Stability order of carbocations }=(4)>(2)>(1)>(3)
\)

Hint

Hint

(b) The technique described in the question is known as sublimation. Sublimation is a process where a solid turns directly into a gas without passing through the intermediate liquid state. This physical change occurs under specific conditions of temperature and pressure and is characteristic of certain substances.

Option B: Sublimation is the correct answer because it directly describes the process of solid substances transforming into vapour without becoming liquid first. For example, substances like iodine, naphthalene, and dry ice (solid carbon dioxide) sublimate when heated under normal atmospheric conditions. Sublimation is not only a fascinating chemical process but is also exploited in various scientific and industrial applications, including purification of substances, where impurities do not sublimate and can thus be easily separated from the vaporized material.

The other options provided do not describe this process:
Crystallization (Option A) involves the formation of solid crystals from a homogeneous solution. It is typically used to purify solids wherein the impurities remain dissolved in the solvent.
Distillation (Option C) is a process used to separate mixtures based on differences in the conditions required to change the phase (liquid phase) of the components of the mixture. It does not involve direct transition from solid to gas.
Chromatography (Option D) is a method for separating components of a mixture based on differences in their movement through a stationary medium under the influence of a solvent or carrier gas. It does not involve the phase transition of solid to gas directly.

Hint

(d) Mohr’s salt is the ammonium iron(II) sulfate, with the chemical formula \(\left( NH _4\right)_2 Fe \left( SO _4\right)_2 \cdot 6 H _2 O\). It is known for its stability compared to the other iron(II) salts which tend to readily oxidize to iron(III) salts when exposed to the air. To prevent the oxidation and hydrolysis of the \(Fe ^{2+}\) ion during the preparation of Mohr’s salt, an acid is added. This acid serves several purposes: it maintains the acidic environment necessary to prevent hydrolysis, aids in the solubilization of iron(II) sulfate, and minimizes the oxidation of \(Fe ^{2+}\) ions to \(Fe ^{3+}\).

The options provided give four different types of acids, from which one needs to be selected for the preparation of this salt. We can evaluate them based on their appropriateness:

Dilute Hydrochloric Acid: While capable of maintaining a low pH to prevent oxidation, HCl can potentially introduce chloride ions \(\left( Cl ^{-}\right)\), which can form complexes with iron, changing the composition of the solution.
Concentrated Sulphuric Acid: This choice is too strong and can lead to excessive acidity as well as potential safety issues during handling and dilution. It is unnecessary for this application.
Dilute Nitric Acid: Nitric acid is an oxidizing agent and can promote the oxidation of \(Fe ^{2+}\) to \(Fe ^{3+}\), which is undesirable in the preparation of Mohr’s salt. Dilute Sulphuric Acid: This is a very common choice since it helps maintain the stability of the \(Fe ^{2+}\) ion without contributing extraneous ions that could form undesired complexes or products. Moreover, it keeps the solution acidic, helping prevent oxidation and hydrolysis, while not introducing any oxidizing characteristics.

Given these considerations, the best choice for preventing hydrolysis of the \(Fe ^{2+}\) ion during the preparation of Mohr’s salt is Dilute Sulphuric Acid (Option D). This is because it maintains the suitable acidic conditions needed for stabilizing the ferrous ion and does not interfere with the redox stability of the iron by providing an oxidative chemical environment.