General Organic Chemistry

Condensed form of $n$ butane

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \text {. } $

Bond line form of $n$ butane

Complete formulae of $n$ butane

Distillation under reduced pressure (vacuum distillation) is generally used for the purification of high-boiling organic liquids that decompose at or below their normal boiling points.

In this method, the external pressure is reduced so that the liquid boils at a lower temperature than its normal boiling point. This prevents thermal decomposition and allows safe purification of heat-sensitive liquids such as glycerol, aniline, and nitrobenzene.

An aromatic species must be cyclic, planar, have a continuous ring of $p$-orbitals, and contain ( $4 n+2$ ) $\pi$-electrons (Huckel's rule). Let's apply these rules to each option

(c) Cycloheptatrienyl cation : It is cyclic, planar, and has $6 \pi$-electrons. This follows Huckel's rule aromatic.

(b) Cyclopentadienly anion: It is cyclic, planar, and has $6 \pi$-electrons. This follows Huckel's rule aromatic.

(a) Cyclohexadienyl cation: It is not aromatic. It is anti-aromatic as it has $4 n$ electrons, which violates the Huckel's rule for aromaticity.

(d) Naphthalene : It is cyclic, planar, and has $10 \pi$-electrons. This follows Huckel's rule aromatic.

Therefore, the cyclohexadienyl cation is the only species that is not aromatic.

$\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Ethyl carbocation) :

This is a primary carbocation. It's stabilised by the $+I$ (inductive) effect of the methyl group and some hyperconjugation from the three alpha-hydrogens. The positive carbon is $s p^2$-hybridised.

(b) $\mathrm{CH}_2=\mathrm{C} \mathrm{H}$ (Vinyl carbocation) : In this carbocation, the positive charge is on an $s p$ hybridised carbon. An $s p$ hybridised carbon is more electronegative than an $s p^2$ or $s p^3$ hybridised carbon, meaning it's less capable of accommodating a positive charge. The double bond itself cannot provide effective resonance stabilisation to this directly attached positive charge, and in fact, the proximity of the electron-withdrawing $s p$ hybridised carbon makes it highly unstable. This is due to the electron-withdrawing nature of the double bond.

(c) $\mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Allyl carbocation) : This carbocation is highly stabilised by resonance. The positive charge is delocalised over two carbon atoms due to conjugation with the double bond.

(d) $\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Benzyl carbocation) This carbocation is also highly stabilised by resonance. The positive charge is delocalised into

the benzene ring due to conjugation.

• Comparing these, the vinyl carbocation $\left(\mathrm{CH}_2=\stackrel{+}{\mathrm{C}} \mathrm{H}\right)$ is the least stable.

Smaller the $\mathrm{p} K_a$ value, stronger is the acid. Among the given compounds

$ \text { lowest } \mathrm{p} K_a \text { value. } $

Hyperconjugation involves delocalisation of electrons from CH $\sigma$-bond adjacent to a $\pi$ system or carbocation.

Hence, option (a) shows hyperconjugation.

$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$ molecules shows hyperconjugation as electron from CH bond of methyl group interact with $\pi$ bond of alkene.

The correct increasing order of acidic strength of given compounds is I $<$ II $<$ III $<$ IV.

As acidity increases with electron withdrawing group.

$ -\mathrm{OCH}_3<-\mathrm{CH}_3<-\mathrm{CN}<-\mathrm{NO}_2 $

A carbocation is stable when it thes resonance, hyperconjugation, I effect.

In option (c), Benzylic carbocation (resonance with phenyl ring) and methyl group (hyperconjugation) lead to more stability.

Hydrogen bonding occurs when a molecule has a hydrogen atom bonded to a highly electronegative atom like N, O, or F — and there is a lone pair on another N, O, or F atom that can participate in bonding.

1. Ethanol (CH₃CH₂OH)

• Has an –OH group

✅ Can form hydrogen bonds (O–H···O)

2. Acetic acid (CH₃COOH)

• Has both C=O and –OH groups

✅ Forms strong hydrogen bonds (dimers, intramolecular and intermolecular H-bonding possible)

3. Ethylamine (CH₃CH₂NH₂)

• Has an –NH₂ group

✅ Can form hydrogen bonds (N–H···N or N–H···O with other molecules)

4. Trimethylamine ((CH₃)₃N)

• Has nitrogen with a lone pair but no N–H bond

❌ Cannot form H-bonds as donor, though it can accept them — but “having H‑bonding” typically means being capable of intermolecular H-bonding as both donor and acceptor. Since it cannot donate an H, we generally do not count it as hydrogen‑bonding on its own.

5. Salicylic acid (o-hydroxybenzoic acid)

• Has both –OH and –COOH groups; strong intramolecular hydrogen bond and intermolecular as well.

✅ H-bonding present

6. Ethanal (CH₃CHO)

• Has a C=O group only, no –OH or –NH–

❌ Cannot form hydrogen bonds with itself (can accept H bonds but not donate).

✅ Substances with H‑bonding:

1. Ethanol

2. Acetic acid

3. Ethylamine

4. Salicylic acid

Total = 4

✅ Correct Answer: Option A (4)

Toluene would be a suitable example of $X$ and nitrobenzene for $Y$.

In toluene, the methyl group CH bond can participate in hyperconjugation with the pi system of benzene ring.

Nitrobenzene exhibit resonance due to the interaction of the lone pair on the oxygen atom of the nitrogroup with the pi electrons of the benzene ring.

Salicylaldehyde has - OH and - CHO ortho to each other, enabling intramolecular H-bonding.

$m$-hydroxy and benzoic acid form intermolecular H-bonds.

The stability of carbocation decreases with increase in $S$-character on carbon. So, the correct order will be I $>$ III $>$ IV $>$ II $>$ V.

The molar mass of the compound ' $ Z $ ' i.e aspirin is $ 180 \mathrm{~g} / \mathrm{mol} $. It contain 9 carbon atoms

$ \begin{aligned} C \% & =\frac{9 \times 12}{180} \times 100 \\\\ & =60 \% \end{aligned} $

Less is the acidic strength more will be the $\mathrm{p} K_a$ value. So, the correct order increasing $\mathrm{p} K_a$ is

$ \text { IV }<\text { II }<\mathrm{I}<\text { III. } $

Considering electronic effects for resonance/mesomeric ( $\mathrm{R} / \mathrm{M}$ ) which essentially actives or deactivates an organic molecules towards a reagent. The activating groups are:

and the deactivating groups are:

$ \text { The correct order of acidic strength is } $

Acidic character $\propto$ stability of conjugate base Stability of anion $\propto-M \propto-I$ effect and

$ \frac{1}{+I \text { effect }} \propto \frac{1}{\text { hyperconjugation }} $

A deactivating group attached to a benzene ring reduces the electron density within the ring, thereby slowing down and complicating electrophilic aromatic substitution reactions compared to reactions with benzene alone.

In the given list, the following are deactivating groups:

$-\mathrm{Cl}$

$-\mathrm{SO}_3\mathrm{H}$

$-\mathrm{COOCH}_3$

These groups withdraw electron density from the benzene ring, making it less reactive to electrophiles.

The electron withdrawing group increases the basicity while electron donating group decreases the basicity. So , the correct order is

$ B>A>C $

Nucleophiles are electron-rich species that donate an electron pair. In the given list:

$\mathrm{CH}_3 \mathrm{NH}_2$ (methylamine): This compound contains a nitrogen atom with a lone pair of electrons, making it an effective nucleophile.

$\mathrm{CH}_3 \mathrm{SH}$ (methanethiol): The sulfur atom in this molecule also has a lone pair of electrons, allowing it to act as a nucleophile.

On the other hand, electrophiles are electron-deficient species that can accept an electron pair. In the list:

$\mathrm{CH}_3 \mathrm{CHO}$ (acetaldehyde): This compound is an electrophile because the carbonyl carbon is electron-deficient and can accept electron pairs.

Lastly:

$\mathrm{C}_2 \mathrm{H}_4$ (ethylene): This acts as a substrate and is not typically considered a nucleophile.

Therefore, the number of nucleophiles in the list is two: $\mathrm{CH}_3 \mathrm{NH}_2$ and $\mathrm{CH}_3 \mathrm{SH}$.

Tropolone is an example of non-benzenoid aromatic compound.

Tropolone is an organic compound that is pale yellow in colour. It is a seven-membered cyclic ring.

The following three compound shows negative resonance $(-R)$ effect when attached to conjugate system. $A \rightarrow$ formyl, $D \rightarrow$ cyano, $E \rightarrow$ nitro. The negative resonance effect occurs when the groups withdraw the electrons from other molecules by delocalisation process.

The acidity of compound depends upon the strength of electron withdrawing group present in it. The order of $+I$ effect (electron withdrawing effect) is

$ \mathrm{NO}_2>\mathrm{CN}>\mathrm{F} $

Hence, the order of acidity is

The delocalisation of $\sigma$ electrons of $\mathrm{C}-\mathrm{H}$ bond of an alkyl group linked directly to an atom of unsaturated system or to an atom having unshared $p$-orbital (as in benzene) is observed in hyperconjugation effect.

The uncharged resonating structure is more stable than charged ones. So, (II) is more stable than I and III. Out of I and III, I is more stable as negative charge is acquired by more electronegative atom i.e., oxygen while in (III) positive charge is more oxygen atom.

So, correct order of stability of resonance structures are : II > I > III.

Hyperconjugation is not possible in compounds in which $\alpha$-hydrogen is not present.

Thuscompound does notshow hyperconjugation.

In electrophilic aromatic substitution (EAS) reactions, the nitro group ($ \text{NO}_2 $) is known to be a strong electron-withdrawing group due to its resonance and inductive effects. This property significantly influences the electron distribution in the aromatic ring, thereby affecting the position where incoming electrophiles are most likely to attack.

Let's analyze the options provided in this context:

• Option A: Decreases electron density at ortho-and para-positions
• Option B: Decreases electron density at meta-position
• Option C: Increases electron density at meta-position
• Option D: Increases electron density at ortho-and para-positions

The nitro group, being electron-withdrawing, pulls electron density away from the aromatic ring through two main mechanisms:

• Inductive Effect: The nitro group withdraws electron density through the sigma bonds due to its high electronegativity.
• Resonance Effect: The nitro group can participate in resonance, pulling electron density away from the ring, particularly affecting the ortho and para positions relative to the nitro group.

Due to the electron-withdrawing nature of the nitro group, electron density is significantly reduced at the ortho and para positions more than the meta positions. This makes the ortho and para positions less reactive towards electrophiles since electrophiles are attracted to electron-rich areas.

As a result, the meta position remains relatively electron-rich in comparison to the ortho and para positions. Thus, the electrophilic substitution is more likely to occur at the meta position.

Based on this understanding, the correct answer is:

Option A: Decreases electron density at ortho-and para-positions

If acid is weak, its conjugate base (nucleophile) is strong and vice-versa.

I. is conjugate base of

II. CH$_3$O$^-$ is a conjugate base of CH$_3$OH

III. CN$^-$ is a conjugate base of HCN

IV. is a conjugate base of

Order of acidic nature is IV > I > III > II.

So, nucleophilicity order is II > III > I > IV.

I, III and V contain both $s p$ and $s p^2$ hybridised carbons.

Acidic strength depends on the stability of conjugate base formed. More stable the conjugate base, stronger is the acid. The order of stability of conjugate base formed will be

$ \begin{aligned} \mathrm{H}-\mathrm{C} \equiv & \mathrm{C}-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_2=\mathrm{CH}-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}^{\ominus} \end{aligned} $

(Increase in \% character increases stability of conjugate base)

∴ Correct order of acidic strength will be-

$ \begin{aligned} \mathrm{H}-\mathrm{C} \equiv & \mathrm{C}-\mathrm{COOH} \\ & >\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \\ & >\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \\ & >\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} . \end{aligned} $

For hyperconjugation, compound/ molecule should possess $\alpha$-hydrogen atoms.

A. $\mathrm{H}_3 \mathrm{C}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2-3 \alpha \mathrm{H}$-atoms

D. $-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$-zero $\alpha \mathrm{H}$-atom.

Therefore, $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$ lacks hyperconjugative stability.

Radicals on carbon atom are stabilised when they are in more substituted position that means tertiary radical is more stable than secondary and secondary is more stable than primary radical.

Therefore, $[B]$ is least stable followed by $[C,[A]$ and $[D]$ are most stable ones.

Out of $[A]$ and $[D],[A]$ is more stable as it has more $\alpha$-hydrogen atom than $[D]$, which lead to more number of hyperconjugated structures.

Correct order of stability is $[B]<[C]<[D]<[A]$

Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a $\sigma$-bond (usually $\mathrm{C}-\mathrm{H}$ ) with an adjacent empty or partially filled p-orbital or a $\pi$-orbital to give an extended molecular orbital that increases the stability of the

system. Consider the molecule propene, where $\pi$ bond is in conjugation with $3 \sigma \mathrm{C}-\mathrm{H}$ bond of adjacent methyl group. So, higher the number of adjacent $\sigma \mathrm{C}-\mathrm{H}$ bond to $\pi$-bond higher will be the hyperconjugation and higher will be the stability.

The adjacent $\sigma \mathrm{C}-\mathrm{H}$ bond to $\pi$ bond increases in the order of IV $>$ III $>$ II $>$ I.

Therefore, stability follows the same order.

Electron rich groups act as ortho and para directing groups as they increases electron density at these two positions.

Out of the given groups, following are electron rich and deficient,

Electron rich $-\mathrm{OH},-\mathrm{OCH}_3,-\mathrm{NHCOCH}_3$

Electron deficient $-\mathrm{CN},-\mathrm{CO}_2 \mathrm{H},-\mathrm{CHO}$

Hence, $-\mathrm{OH},-\mathrm{OCH}_3$ and $-\mathrm{NHCOCH}_3$ acts as ortho and para-directing groups towards electrophilic substitution reactions.

is not an aromatic compound as it has only 4$\pi$ electrons. So, it follows 4n $\pi e^-$ rule, which makes it anti-aromatic compound.

$\mathrm{p} K_b \propto \frac{1}{\text { Basicity }}$

$A$ and $B$ are more basic than $C$ and $D$ due to $+I$-effect of methyl group $A$ which is more basic than $B$. This is because of high/more steric hindrance in $B$ due to three methyl groups. Out of $C$ and $D, C$ is more basic than $D$ because of more steric hindrance in $B$. So, order of basicity is

Thus, decreasing pK_b order of given compounds is

i.e. D > C > B > A.

Methoxybenzene in the presence of anhydrous aluminium chloride (anhyd. AlCl$_3$) reacts with ethanoyl chloride to form 2-methoxy acetophenone and 4-methoxy acetophenone which is further converted to 4-ethyl phenol.

The most reactive towards electrophilic reagent is $o$-cresol.

The phenolic $(-\mathrm{OH})$ group increases the electron density on benzene ring through resonance. With increase in electron density, the reactivity towards electrophilic reagent increases. The decreasing order of activating influence of substituents towards electrophilic reagent is

$\begin{aligned} & \mathrm{NR}_2>-\mathrm{NH} R>-\mathrm{NH}_2>-\mathrm{OH}>-\mathrm{OR}> \\ & -\mathrm{NHCOR}>-\mathrm{ph}>-R \end{aligned}$

Hyperconjugation is a stabilising interaction between one or more $\sigma$-bonds and one or more $\pi$-bonds.

It helps in explaining stability order of free radicals, carbocations, alkanes but cannot explain stability of carbanion due to presence of vacant orbitals.

According to Markownikov rule, in addition reaction of unsymmetrical alkenes, electron rich component of reagent add to carbon atom with fewer hydrogen atoms bonded to it.

The complete shifting of $\pi$-electron pair of a multiple bond to one of the bonded atoms in presence of the attacking reagents is called electromeric effect.

Positive electromeric effect (+E)

Negative electromeric effect [$-$E]

In positive $+$E effect, $\pi$-electrons of the multiple bonds are transfered to that atom to which the reagent gets attached.

In negative $-$E effect, $\pi$-electrons of the multiple bonds are transferred to that atom to which the attacking reagent does not get attached.

Bond dissociation energy or bond energy (BE) of $\mathrm{C}-\mathrm{H}$ bond of a hydrocarbon $(R \mathrm{H})$ can be predicted from stability of free radical ( $R^{\bullet}$ ) formed by homolytic bond fission of $\mathrm{C}-\mathrm{H}$ bond of the hydrocarbon.

$ \text { So, }(\mathrm{BE})_{\mathrm{C}-\mathrm{H}} \propto \frac{1}{\text { Stability of free radical }\left(R^{\bullet}\right)} $

$ \begin{aligned} &\text { Stability order of free radicals }\\ &(C)>(D)>(B)>(A) \end{aligned} $

$ \text { In the reaction, } $

The hybridisation of carbon- 2 in $(P)$ is $s p^3$ and in $(Q)$ in $s p^2$, because of 4 sigma bonds in ( $P$ ) and one pi $(\pi)$ bond in $(Q)$.

In option (c), the hydrogen of $-$CH group is

surrounded by three groups as:

Here, O of CO attracts, the electron density towards itself. Thus, (c) has most acidic hydrogen among given compounds.

Or

The anion formed after removal of

acidic proton in is stablised by $-R$-effect of three $\mathrm{C}=\mathrm{O}$ groups.

Thus, (c) has most acidic hydrogen among given compounds.