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.

(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.

Carbocation intermediate is stabilised by $+\mathrm{I},+\mathrm{M} \&$ hyperconjugation effect. Since in option 2 carbocation is in conjugation with stronger +M group $-\mathrm{OCH}_3$ hence it will be most stable.

q is aromatic r is nonaromatic p is antiaromatic

Due to +M effect of -OMe at para position

To determine the number of different stereoisomers for a molecule, use the formula:

$ \text{Number of stereoisomers} = 2^n $

where $ n $ is the number of chiral centers in the molecule.

For the given molecule, there are 2 chiral centers, so:

$ n = 2 $

Therefore, the number of possible stereoisomers is:

$ 2^2 = 4 $

This means there are 4 different stereoisomers possible for the molecule in question.

(A) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$. GI is not possible

(B) Trans isomer has identical atoms/groups on the opposite side of double bond.

There are three stereo center

So No of stereoisomer $=2^3=8$

Firstly higher the number of bonds or say complete octet higher will be stability, then the negative charge situated over more electronegative element more will be stability. Hence, overall order of stability is I > II > III.

: Due to strength of I effect of substituents

Hence the correct order of basic strength

Higher the electron withdrawing strength of substituents higher will be acidic strength of phenols and lesser will be $\mathrm{pK}_{\mathrm{a}}$

The methods used for the purification of organic compounds are fundamentally based on both the nature of the compound and the presence of the impurity. This is because the selection of a suitable purification technique requires understanding not only what the organic compound is (i.e., its physical and chemical properties) but also the type of impurities that need to be removed.

For example, if an organic compound is a solid with a high melting point and the impurity is a low melting solid or a liquid, techniques like sublimation or crystallization could be used, which rely on the differences in physical properties between the compound and the impurity. On the other hand, if the compound needs to be purified from substances with similar boiling points, then a more refined method such as fractional distillation might be employed, relying on slight differences in boiling points. In cases where the compound is mixed with substances that have similar physical properties but different chemical reactivities, chemical methods such as extraction or chromatography could be used, which exploit these chemical differences.

Therefore, the correct answer is:

Option C: nature of compound and presence of impurity.

$\begin{aligned} & \mathrm{HCOOH}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}> \\ & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} \end{aligned}$