Basics of Organic Chemistry

Presence of $\mathrm{NO}_2$ group on the benzene ring deactivates the ring towards electrophilic substitution reactions because it shows $-M$ effect (it withdraws electrons from the ring by resonance).

Because the $\mathrm{NO}_2$ group pulls electron density away, the ring becomes less reactive towards electrophiles.

At the same time, the $\mathrm{NO}_2$ group activates the ring towards nucleophilic substitution because the withdrawal of electrons makes the ring more suitable for attack by nucleophiles (electron-rich species).

Ans. $\rightarrow $ B & C

A & C

Stability of carbanions depends mainly on how well the carbon can hold the negative charge.

Step 1: Use NCERT idea — effect of hybridisation (electronegativity)

Greater $s$-character $\Rightarrow$ carbon is more electronegative $\Rightarrow$ negative charge is more stable.

• $sp$ carbon: $50\%$ $s$-character (most electronegative)

• $sp^2$ carbon: $33\%$ $s$-character

• $sp^3$ carbon: $25\%$ $s$-character (least electronegative)

So stability order due to hybridisation is:

$sp > sp^2 > sp^3$

Step 2: Apply to given carbanions

• $\mathrm{CH}\equiv\mathrm{C}^-$ is on an $sp$ carbon $\Rightarrow$ most stable

• $\mathrm{CH}_2=\mathrm{CH}^-$ is on an $sp^2$ carbon $\Rightarrow$ next

• $\mathrm{CH}_3-\mathrm{CH}_2^-$ is on an $sp^3$ carbon, and also has $+I$ effect of $\mathrm{CH}_3$ which destabilises the negative charge $\Rightarrow$ least stable

Correct order

$\mathrm{CH}\equiv\mathrm{C}^- \;>\; \mathrm{CH}_2=\mathrm{CH}^- \;>\; \mathrm{CH}_3-\mathrm{CH}_2^-$

Correct option: C

$ \text { Both Statements are correct. } $

$ \text { Statement B & D are not true } $

Carbanion means a carbon atom carrying a negative charge. Such a species is stable when this negative charge is reduced (spread out) or pulled away.

If a group attached to the carbanion is electron withdrawing (shows $-I$ effect), it pulls electron density towards itself. This reduces the negative charge on carbon, so the carbanion becomes more stable.

If a group attached to the carbanion is electron donating (shows $+I$ effect), it pushes electron density towards the carbon. This increases the negative charge on carbon, so the carbanion becomes less stable.

So, while arranging in decreasing order of stability: carbanions with stronger electron withdrawing groups will be more stable, and those with electron donating groups will be less stable.

For hyperconjugation to occur, we need at least one hydrogen atom on the carbon atom next to the positively charged carbon. This special hydrogen is called an alpha-hydrogen ($\alpha$-hydrogen).

The rule is simple: More the number of alpha-hydrogens, the more the hyperconjugation structures, and the greater the stability of the carbocation.

Now, let's analyze the two statements.

Analysis of Statement I:

Statement I: $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$ is more stable than $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$ as nine hyperconjugation interactions are possible in $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$.

1. Let's look at the tert-butyl carbocation, $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$.

   Its structure is:

   $ \Large \begin{array}{c} \phantom{..}\mathrm{CH_3}\phantom{..} \\ \Large| \\ \mathrm{H_3C} - \stackrel{\oplus}{\mathrm{C}} - \mathrm{CH_3} \end{array} $

   • The central carbon has the positive charge ($ \stackrel{\oplus}{\mathrm{C}} $).

   • The carbon atoms attached directly to this central carbon are the alpha-carbons ($\alpha$-carbons). Here, there are three such carbons (one from each $\mathrm{CH_3}$ group).

   • The hydrogens attached to these alpha-carbons are the alpha-hydrogens ($\alpha$-hydrogens).

   • Counting the alpha-hydrogens: Each of the three methyl groups has 3 hydrogens.

   • Total number of $\alpha$-hydrogens = $3 + 3 + 3 = 9$.

   • Therefore, 9 hyperconjugation interactions are possible.

2. Compare it with the methyl carbocation, $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$.

   • In $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$, there are no other carbon atoms attached to the positively charged carbon. So, there are no alpha-carbons and therefore zero alpha-hydrogens. This means there is no possibility for hyperconjugation.

3. Conclusion for Statement I:

   • The tert-butyl carbocation has 9 hyperconjugation structures, while the methyl carbocation has 0.

   • Because of these 9 hyperconjugation interactions, $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$ is indeed much more stable than $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$.

   • So, Statement I is true.

Analysis of Statement II:

Statement II: $\stackrel{\oplus}{\mathrm{C}}\mathrm{H}_3$ is less stable than $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$ as only three hyperconjugation interactions are possible in $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$.

1. Let's re-examine the methyl carbocation, $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$.

   Its structure is:

   $ \Large \begin{array}{c} \phantom{..}\mathrm{H}\phantom{..} \\ \Large| \\ \mathrm{H} - \stackrel{\oplus}{\mathrm{C}} - \mathrm{H} \end{array} $

   • As we established above, for hyperconjugation, we need alpha-hydrogens, which are hydrogens on an adjacent carbon atom.

   • In $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$, there is no adjacent carbon atom.

   • Therefore, there are 0 alpha-hydrogens and 0 hyperconjugation interactions.

2. Conclusion for Statement II:

   • The first part of the statement, "$\stackrel{\oplus}{\mathrm{C}}\mathrm{H}_3$ is less stable than $\left(\mathrm{CH}_3\right)_3 \stackrel{\oplus}{\mathrm{C}}$", is correct.

   • However, the reason given, "…as only three hyperconjugation interactions are possible in $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$", is incorrect. There are zero, not three, such interactions. The statement might be confusing the hydrogens directly attached to the carbocation with alpha-hydrogens, which is a common mistake.

   • Since the reason provided is wrong, Statement II is false.

Final Answer

• Statement I is true.

• Statement II is false.

Therefore, the correct option is the one that says Statement I is true but Statement II is false.

Option B is the correct answer.

Statement I: False. Sublimation is used to separate/purify sublimable (volatile) solids—solids that pass directly from solid to vapour on heating. This depends on having sufficient vapour pressure in the solid state, not simply on having a low melting point.

Statement II: False. The boiling point decreases when external pressure is reduced (principle of vacuum distillation).

✅ Correct option: A (Both Statement I and Statement II are false).

Molecular formula $\Rightarrow \mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}$

$ \text { Molar mass }=48+9+35.5=92.5 $

$ \% \mathrm{OC}=\frac{48}{92.5} \times 100=51.89 \% $

This resonating structure having +ve charge on adjacent atoms so it is least stable.

Let us analyse both statements carefully using the concept of fractional distillation as given in NCERT.

Concept:

Fractional distillation is a process used to separate a mixture of two or more liquids having different boiling points.

• The liquid having a lower boiling point vaporises first.

• As vapours rise through the fractionating column, they repeatedly condense and vaporise.

• The temperature in the column gradually decreases from bottom to top.

• Hence, vapours rising upwards become richer in the more volatile (lower boiling point) component.

Statement I

“Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation.”

• Liquids with higher boiling point are less volatile.

• Thus, their vapours condense earlier (lower in the column) because they require a lower temperature difference to liquefy.

• Therefore, Statement I is true.

Statement II

“The vapours rising up in the fractionating column become richer in high boiling component of the mixture.”

• As explained, the rising vapours become richer in the low boiling or more volatile component, not the high boiling one.

• Therefore, Statement II is false.

Correct Answer:

Option C: Statement I is true but Statement II is false

Aromatic compounds are planar, cyclic and completely conjugated systems. They follow $(4 n+2) \pi e^-$ rule called Huckel rule.

Resonance structure (I) is more stable than (II) due to being neutral than the othere charged structure. $c=0$ is shorter than $c-0$ bond due to double bond character. $\beta$ is shorter than $\gamma$ die to partial double bond chanacter.

Order : $\quad \alpha<\beta<\gamma$

Statement-I

$\mathrm{CH}_3{ }^{-}>\mathrm{CH}_3-\mathrm{CH}_2{ }^{-}>\left(\mathrm{CH}_3\right)_2 \mathrm{CH}^{-}>\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{-}$

It is correct order of anion stability and it is based on inductive effect.

Statement-II

Both are stabilised by resonance

Statement-I : Correct

Statement-II : Incorrect

A. Williamson Synthesis

In Williamson ether synthesis, an alkoxide ion attacks an alkyl halide and replaces the halogen.

So, it proceeds by nucleophilic substitution.

$ A \rightarrow III $

B. Friedel Craft Reaction

Friedel-Crafts reaction in benzene involves attack by an electrophile on the aromatic ring.

So, it is electrophilic substitution.

$ B \rightarrow IV $

C. Bromination of vinyl benzene

Vinyl benzene is styrene, having a double bond. Bromine adds across the double bond.

This is electrophilic addition.

$ C \rightarrow I $

D. Chlorination of toluene in light

In presence of light, chlorination of toluene occurs at the side chain by a radical pathway.

So, it is free radical substitution.

$ D \rightarrow II $

Therefore, the correct matching is:

$ A-III,\; B-IV,\; C-I,\; D-II $

So, the correct option is:

$ \boxed{\text{Option C}} $

The compound exists as in structure (II) in order to give more stable $Q$ nasi-Aromatic compound. (II) is the contributing structure here.

Structure (III) is more contributing structure with partially double-bond character.

Order of Bond strength: fully-double $>$ Partially double-bonded > single bond.

$ y>z>x $

Bond strength $\propto \frac{1}{\text { Bond length. }}$

Order of Bond length: $x>z>y$

The given compound does not have a chiral centre because the carbon shown is not attached to four different groups.

Here, $-\mathrm{CH}_3$ and $-\mathrm{Me}$ represent the same substituent (both are methyl groups). So, two of the four bonds are to identical groups.

Therefore, this carbon cannot be a chiral centre, and the structures cannot be enantiomers, diastereomers, or meso forms.

The presence of $E W G / s$ on benzene ring in phenol of increases the acidity by providing extra stability to the phenoxide ion and forming weak conjugate base.

Whereas, EDGs increases electron density in the ring and decreases acidity of phenol by destabilizing the circulating -ve charge on ring.

In general benzoic acid is more acidic than phenol. Hence, $D, E > A, B, C$

$-\mathrm{NO}_2$ is $E W G$ where as $-\mathrm{OCH}_3$ is $E D G$.

$ B>A>C $

Overall order, $D>E>B>A>C$.

In carboxylic acid, we charge delocalixes on oxygen atom. Whereas, in phenol -ve change delocalizes on o and $c$ as well.

$C^ \ominus$ not much stable compared to $O^\ominus$. Hence, carboxylate ion is more stable than phenoxide ion, resulting in more acidic carboxylic acid.

Strong base ⟶ weak conjngate acid

The presence of EWGs pulls the available lp on $N$-atom through resonance or $-M$ effect. since, This makes the compound's tendency to give lp low and forms weak base, then the conjugate acid firmed is strongest.

Order of Basicity: $2>4>1>3$

Order of conjugate acid:

$ 2<4<1<3 $

To understand this question, we need to look at the Inductive effect, specifically the negative inductive effect ($-I$ effect), which determines the electron-withdrawing power of a functional group.

According to the principles of organic chemistry laid out in your NCERT textbook, the electron-withdrawing nature of a group depends on the electronegativity of the atoms present in the group and their arrangement.

The standard decreasing order of the $-I$ effect (electron-withdrawing power) for various functional groups is:

$ -\mathrm{NO}_2 > -\mathrm{CN} > -\mathrm{COOH} > -\mathrm{F} > -\mathrm{Cl} > -\mathrm{Br} > -\mathrm{I} > -\mathrm{OH} > -\mathrm{OCH}_3 > -\mathrm{C_6H_5} > -\mathrm{H} $

Let us analyze the functional groups given in the question:

1. $ -\mathrm{NO}_2 $ (c): This group has a nitrogen atom bonded to two highly electronegative oxygen atoms, and the nitrogen carries a formal positive charge. This makes it a very strong electron-withdrawing group.

2. $ -\mathrm{CN} $ (a): The carbon in the cyanide group is $sp$ hybridized, which makes it more electronegative. Coupled with the electronegative nitrogen atom, it exerts a strong $-I$ effect, but it is slightly weaker than $-\mathrm{NO}_2$.

3. $ -\mathrm{COOH} $ (b): This group contains a carbonyl carbon attached to a hydroxyl oxygen. It shows a moderate $-I$ effect, which is weaker than both $-\mathrm{NO}_2$ and $-\mathrm{CN}$.

4. $ -\mathrm{I} $ (d): Iodine is a halogen. While halogens are electronegative, iodine has the largest atomic size and the lowest electronegativity among the common halogens (fluorine, chlorine, bromine). Hence, it has the weakest $-I$ effect out of the groups listed here.

Based on this analysis, the decreasing order of their electron-withdrawing power is:

$ \mathrm{c} \ (-\mathrm{NO}_2) > \mathrm{a} \ (-\mathrm{CN}) > \mathrm{b} \ (-\mathrm{COOH}) > \mathrm{d} \ (-\mathrm{I}) $

Therefore, rearranging them into the increasing order gives:

$ \mathrm{d} < \mathrm{b} < \mathrm{a} < \mathrm{c} $

This perfectly matches Option C.

Correct Answer:

Option C

$ \mathrm{d} < \mathrm{b} < \mathrm{a} < \mathrm{c} $

Carbocation ($C^+$) has a deficiency of electrons, so it becomes more stable when a group donates electron density towards it.

Therefore, carbocations are stabilised by electron donating groups (EDGs) due to $+R,+T,+H$ effect whereas, carbanions ($C^-$) are stabilised by electron withdrawing groups (EWGs) due to $-R,-I,-H$ effect.

So, in Statement I, the $-OCH_3$ group shows $+R$ effect (lone pair donation), which stabilises the carbocation by resonance.

In Statement II, the $-NO_2$ group shows $-R$ effect (it withdraws electron density), which stabilises the carbanion by resonance.

	Functional group (priority order)	suffix(s)	prefix
(1)		Oic acid(when carbon(C) of this functional group present in the main chain) or Carboxylic acid (when carbon(C) of this functional group is not present in the main chain)	Carboxy
(2)		sulphonic acid	sulpho
(3)		Oic anhydrid (when carbon(C) of this functional group present in the main chain) or Carboxylic anhydride (when carbon(C) of this functional group is not present in the main chain)	$ - $
(4)		Oate (when carbon(C) of this functional group present in the main chain) or Carboxylate (when carbon(C) of this functional group is not present in the main chain)	$ - $
(5)		Oyl halide (when carbon(C) of this functional group present in the main chain) or Carboxyl halide (when carbon(C) of this functional group is not present in the main chain)	haloflormyl
(6)		amide (when carbon(C) of this functional group present in the main chain) or carbamide (when carbon(C) of this functional group is not present in the main chain)	carbamoyl
(7)	cynide ($ - $ C $ \equiv $ N)	nitrile (when carbon(C) of this functional group present in the main chain) or carbonitrile (when carbon(C) of this functional group is not present in the main chain)	cyano
(8)		Carbylamine	isocyano
(9)		al (when carbon(C) of this functional group present in the main chain) or Carbaldehyde (when carbon(C) of this functional group is not present in the main chain)	formyl (oxo)
(10)		one	oxo
(11)	$ - $ OH	ol	hydroxy
(12)	$ - $ SH	thiol	mercapto
(13)	$ - $ NH$_2$	amine	amino

From this table you can see -COOH has highest priority and -NH₂ has lowest priority among all functional groups.

The electrophile will attack mainly at site U. This happens because the nitrogen atom attached to the ring shows a +M (positive mesomeric) effect, which increases electron density at site U. As a result, site U becomes richer in electrons and more attractive to electrophiles.

(A) Carbocation $\rightarrow \mathrm{sp}^2$ hybridised carbon with empty P-orbital

(B) $\quad$ Carbon free radical $\rightarrow \mathrm{sp}^2 / \mathrm{sp}^3$ hybridised carbon with one unpaired electron.

(C) Nuecleophile $\rightarrow$ species of that can supply a pair of electron.

(D) Electrophile $\rightarrow$ species that can receive a pair of electron.

E > D > B > A > C

Hyperconjugation is considered a permanent effect because it does not rely on the presence of an external reagent to occur. Therefore, Statement I is false. However, Statement II is true; the presence of more alkyl groups results in more alpha hydrogen atoms ($\alpha-\mathrm{H}$), which enhances hyperconjugation. This increased hyperconjugation leads to greater stabilization of the carbocation.

In option C are momologoes to each - other and option D are only organic molecule not isomers.

C.B. of terminal alkyne will be sp hybridisation and loculised. In other C.B. will be resonance stablised.

$4-\mathrm{sp}^3, 2-\mathrm{sp}^2, 2-\mathrm{sp}$

II most stable carbon radical due to resonance stablise

I least stable carbon radical due to no stabilising factor.

Aromatic compounds

$\mathrm{b}, \mathrm{f}, \mathrm{g}$, are not aromatic, these compounds do not follow Huckel's rule

Statement I : Correct

Nitration of m-xylene follows electrophilic aromatic substitution mechanism. First step is the generation of electrophile NO$_2^+$ (nitronium ion) and then the electrophile attacks the benzene ring. Methyl groups are ortho-para directing groups. The NO$_2^+$ in ortho position is not possible in this case due to the two neighbouring methyl groups. The nitro group is bonded to the para-position.

On oxidation, the methyl groups get converted to carboxylic acid groups. ($-$CH$_3$) to $-$COOH conversion). So, the statement is correct.

Statement II : Correct

$-$CH$_3$ group (methyl group) is ortho-para directing and $-$NO$_2$ group (nitro group) is meta directing.

$-$CH$_3$ group $\longrightarrow$ electron donating group

$-$NO$_2$ group $\longrightarrow$ electron withdrawing group.

Electron donating groups push electrons towards the ring and increase electron density. Electron donating groups give ortho-para directing products. Electron withdrawing groups pull electrons away from the ring and decrease electron density. Electron withdrawing groups give meta-directing products. So, the statement is correct.

So, both statement I and statement II are true.

Statement I : Correct

Partition chromatography - The separation of components between two liquid phases. The stationary phase is a thin film of liquid present on an inert support. The liquid film allows for the separation of substances based on their solubilities between the stationary phase and mobile phase.

Statement II : False

In paper chromatography, the paper itself does not act as the stationary phase. The stationary phase is the water trapped within the pores of the paper. The paper serves as a support for the water layer.

So, Statement I is correct and Statement II is incorrect.

(A)

The substituents are written in the alphabetical order. So, ethyl is first and then methyl. Sok, lowest number is given for ethyl. 3-ethyl. And methyl has position 5. 5-methyl.

The parent chain has seven carbon atoms. So, the base name heptane.

IUPAC name : 3-ethyl-5-methylheptane

(A) Match with (II) 3-ethyl-5-methylheptane

(B)

Substituents get the same position when numbering is done from both ends. So, the substituent name (prefix) is 4,4-dimethyl. The parent chain has seven carbon atoms. So, the base name is heptane.

IUPAC name : 4,4-dimethylheptane

(B) Match with (III) 4,4-dimethylheptane.

(C)

Double bond has higher priority than the methyl group. So, the numbering is given as

Smallest possible position is 1 and 3 for the double bond present. So, 1, 3-diene is correct.

Methyl substituent is at position 2.

So, the prefix is 2-methyl-1,3

The parent chain has five carbon atoms and the base name is penta.

Include diene as suffix to penta- as pentadien.

So, IUPAC name is

2-methyl-1,3-pentadiene

(C) Match with (IV) 2-methyl-1,3-pentadiene.

(D)

Double bond has higher priority than methyl substituent. So, position 1 is for double bond and position 4 is for $-CH_3$ substituent. Prefix is 4-methyl.

Parent has five carbon atoms. So, the base name penta. The suffix used is ene for the double bond. So 'a' eliminated and written as pentane.

IUPAC name : 4-methylpent-1-ene

(D) match with (I) 4-methylpent-1-ene

The matching is written as

(A) - (II)

(B) - (III)

(C) - (IV)

(D) - (I)

Nucleophiles are electron rich species and can provide electrons for making new bonds.

Nucleophiles are nucleus loving species

* typically have a $-$ve (negative) charge or non-bonding electron pair available for making a new bond.

Nucleophiles are Lewis bases.

$\mathrm{NH}_3$ - It is a nucleophile.

$\mathrm{NH}_3$ has a lone pair of electrons on the nitrogen atom which allows it to be a nucleophile since it can provide electrons to form a new bond.

PhSH - It is a nucleophile

Lone pair of electrons are present on the sulphone atom and hence it is a nucleophile. The lone pair of electrons are readily donate to form new bonds. with electrophiles.

$\left(\mathrm{CH}_3\right)_2 \mathrm{~S}$ - It is a nucleophile

The lone pair of electrons on the sulphur atom makes this molecule nucleophile.

$\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2-$ It is a nuclophile

Alkenes are nulleophiles because they have a high electron' density in their conbon-conbon double bond. The $\pi$ bondin $\mathrm{CH}_2=\mathrm{CH}_2$ is a region of high election density.

$\mathop O\limits^\Theta H$ - It is a nucleophile

Hydroxide ion is a nucleophile, not an electrophile. It is negatively changed and election rich. The electrons are easily donated.

$\mathrm{H}_3 \mathrm{O}^{\oplus}$ - It is not a nucleophile

Hychonium ion is an electrophile. It can gain a pair of electrons from a nucleophile. Also, it is positively changed.

$\left(\mathrm{CH}_3\right)_2 \mathrm{CO}$ - It is not a nucleophile.

Acetone primarily acts as an electrophile in reactions. It can readily accepts electron pairs from other molecules due to the partial positive charge on the carbonyl carbon atom.

- It is not a nucleophile.

It is an imine compound and generally act as electrophile in reactions. The carbon in the $C=N$ double bond is electrophile and attacked by nucleophiles.

So, the nucleophiles are

$\mathrm{NH}_3, \mathrm{PhSH}, \quad\left(\mathrm{CH}_3\right)_2 \mathrm{~S}, \mathrm{CH}_2=\mathrm{CH}_2, \mathrm{OH}^{-}$

Total number of nucleophiles is 5

Statement - I $\to$ True

Statement - I $\to$ True

$1^{\circ}$ Amine and $2^{\circ}$ Amine are different functional groups, hence both are functional group isomers.

Solution :-

C is aromatic due to $\oplus \mathrm{ve}$ charge hence it is most stable

A have more resonance structure

B have less resonance structure

D have only hyper conjugation

Consider First Aromaticity > Resonance > Hyper conjugation

Ans. D $<\mathrm{B}<\mathrm{A}<\mathrm{C}$

Number of sp and $\mathrm{sp}^2$ hybridised carbon atom are 3 and 5.

(B) $-$CN is meta directing But -OH is ortho / pera directing.

(E) Halides are deactivating groups.