Q.The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is ____.

Concept & Intuition

The key is to work backwards from the S$_N$1 hydrolysis condition. S$_N$1 reactions favor tertiary or secondary carbocations. The compound Y is an alkyl bromide formed by adding HBr to the alkene X. If Y hydrolyzes via S$_N$1, then Y must be a tertiary (or possibly secondary) bromide. That tells us the alkene X must be one that, upon HBr addition, gives a tertiary (or stable secondary) carbocation intermediate. Among the four C$_4$H$_8$ isomers, only 2-methylpropene (isobutylene) yields a tertiary carbocation. Once we identify X, ozonolysis tells us the carbonyl fragments.

Step-by-step reasoning

1. Identify possible alkene isomers of C$_4$H$_8$

   The four structural isomers are:

   • But-1-ene: CH$_2$=CH–CH$_2$–CH$_3$
   • But-2-ene: CH$_3$–CH=CH–CH$_3$ (cis/trans)
   • 2-Methylpropene: (CH$_3$)$_2$C=CH$_2$
   • Cyclobutane (not an alkene, ignore) and methylcyclopropane (also not an alkene here).

2. Determine which alkene gives a tertiary bromide with HBr

   Addition of HBr follows Markovnikov’s rule: the H adds to the less substituted carbon, Br to the more substituted carbon.

   • But-1-ene → secondary carbocation → 2-bromobutane (secondary).
   • But-2-ene → secondary carbocation → 2-bromobutane (secondary).
   • 2-Methylpropene → tertiary carbocation → 2-bromo-2-methylpropane (tertiary). Only the tertiary bromide undergoes S$_N$1 hydrolysis readily. So X must be 2-methylpropene.

3. Confirm S$_N$1 hydrolysis of Y

   Y = (CH$_3$)$_3$C–Br. In water or aqueous conditions, it forms a tertiary carbocation, then reacts with water to give (CH$_3$)$_3$C–OH (tert-butyl alcohol). This is a classic S$_N$1 reaction.

4. Ozonolysis of X (2-methylpropene)

   Ozonolysis cleaves the C=C double bond, replacing each carbon with a carbonyl group.

   For (CH$_3$)$_2$C=CH$_2$: …

The central compound is 2-methylpropene, (CH3)2C=CH2.

Formation of X: 2-methylpropene + HBr adds by Markovnikov's rule; H goes to the terminal CH2 and Br to the more substituted carbon, giving 2-bromo-2-methylpropane (tert-butyl bromide). This is a tertiary halide, which ionises easily to a stable 3-degree carbocation, so it undergoes nucleophilic substitution by the SN1 mechanism. …

The key to this question is knowing the chemical structures of common artificial sweeteners. Many students memorize names but forget the functional groups or elements present. Let’s break it down.

1. Identify the sweetener that contains chlorine.

   Sucralose is made by chlorinating sucrose — three hydroxyl groups are replaced by chlorine atoms. Its chemical name is 1,6-dichloro-1,6-dideoxy-β-D-fructofuranosyl-4-chloro-4-deoxy-α-D-galactopyranoside. The presence of chlorine is a defining feature. No other common sweetener has chlorine as a structural element.

2. Identify the sweetener that contains sulphur.

   Alitame is a dipeptide sweetener (L-aspartic acid and D-alanine) with a terminal amide that includes a sulphur-containing group — specifically, a tetramethylthietanyl moiety. Sulphur is present in its structure. Among the options, only Alitame fits this description.

3. Check the other sweeteners for clarity.

   • Aspartame: a methyl ester of a dipeptide (aspartic acid + phenylalanine). Contains no chlorine or sulphur. …

The key concept here is drug classification by mechanism of action. An antihistamine works by blocking histamine receptors (usually H₁ receptors) to reduce allergic reactions like sneezing, itching, and runny nose. The question tests whether you can distinguish between drugs used for allergies, pain, depression, or sleep — each belongs to a different pharmacological family.

Let’s examine each option:

1. Heroin (A) — This is an opioid, a powerful analgesic and recreational drug. It acts on mu-opioid receptors in the brain, not on histamine receptors. It has no antihistamine effect; in fact, opioids can sometimes release histamine, causing itching. So this is not an antihistamine.

2. Dimetapp (B) — This is a brand name for a combination cold and allergy medication. Its active ingredients typically include an antihistamine (like brompheniramine or diphenhydramine) and a decongestant. It is explicitly used to relieve allergy symptoms. Therefore, this is an antihistamine.

3. Nardil (C) — This is a brand name for phenelzine, a monoamine oxidase inhibitor (MAOI) used as an antidepressant. It affects neurotransmitter levels (serotonin, norepinephrine), not histamine. So it is not an antihistamine. …

The question asks which metallic fluoride can convert an alkyl chloride into an alkyl fluoride. This is a classic example of a halogen-exchange reaction, specifically the Finkelstein reaction (or a variant). The challenge is that fluoride ion ($\mathrm{F^-}$) is very small, highly electronegative, and strongly solvated in polar protic solvents, making it a poor nucleophile in such environments. Therefore, we need a fluoride source that is soluble in organic solvents and provides a "naked" or weakly solvated fluoride ion.

Let’s evaluate each option:

1. $\mathrm{NaF}$ and $\mathrm{LiF}$ — These are ionic salts with high lattice energies. They are insoluble in most organic solvents (like acetone, ether, or chloroform) and dissolve only in water or highly polar solvents. In water, the fluoride ion is heavily hydrated and loses its nucleophilicity. Thus, they are ineffective for converting alkyl chlorides to alkyl fluorides in organic media.

2. $\mathrm{BF_3}$ — This is a Lewis acid, not a source of fluoride ions. It actually accepts fluoride ions to form $\mathrm{BF_4^-}$. It would not donate fluoride to an alkyl chloride; instead, it might catalyze other reactions (like Friedel-Crafts) but not a simple substitution.

3. $\mathrm{Hg_2F_2}$ (mercurous fluoride) — This is a covalent compound, soluble in organic solvents. It can undergo a metathesis reaction with alkyl chlorides:

   $$$$

The key to this question is understanding what controls the rate of an $\mathrm{S_N^1}$ reaction. Unlike $\mathrm{S_N^2}$, where the nucleophile attacks the carbon bearing the leaving group in a single step, $\mathrm{S_N^1}$ proceeds in two steps. First, the leaving group departs on its own, forming a carbocation. Then the nucleophile attacks that carbocation. The slow, rate-determining step is the first step — the formation of the carbocation.

That means the rate depends entirely on how easily the C–Br bond breaks, which in turn depends on how stable the resulting carbocation is. A more stable carbocation forms faster. So the problem reduces to ranking the stability of the carbocations that would form from each alkyl bromide.

Let’s look at each compound.

1. Compound I: $\mathrm{CH_3CH_2CH_2CH_2Br}$ (1-bromobutane)

   This is a primary alkyl halide. If the Br leaves, the carbocation formed is a primary carbocation — $\mathrm{CH_3CH_2CH_2CH_2^+}$. Primary carbocations are very unstable because there is only one alkyl group donating electron density to the positive carbon. They are so unstable that $\mathrm{S_N^1}$ reactions on primary halides are practically never observed under normal conditions. So this will be the slowest.

2. Compound III: $(\mathrm{CH_3})_2\mathrm{CHCH_2Br}$ (1-bromo-2-methylpropane, isobutyl bromide)

   This looks like a primary halide at first glance — the Br is on a primary carbon. But look at the carbocation that forms: $\mathrm{(CH_3)_2CHCH_2^+}$. That is still a primary carbocation. However, there is a subtlety: this primary carbocation is adjacent to a tertiary carbon (the isopropyl group). It can undergo a 1,2-hydride shift to form a much more stable tertiary carbocation: $(\mathrm{CH_3})_3\mathrm{C}^+$. In an $\mathrm{S_N^1}$ reaction, such rearrangements are common and fast. So although the initial carbocation is primary, the actual intermediate that reacts is the tertiary one. This makes compound III react faster than a simple primary halide like I, but still slower than a halide that directly gives a secondary or tertiary carbocation because the rearrangement step adds a small energy barrier.

3. Compound II: $\mathrm{H_3CCH_2CHBrCH_3}$ (2-bromobutane) …

The key concept here is that acid-mediated dehydration of alcohols proceeds via formation of a carbocation intermediate. The alcohol is protonated by the acid, water leaves, and a carbocation is formed. The rate-determining step is the formation of this carbocation. So, the more stable the carbocation that can be formed, the faster the reaction.

We need to compare the carbocation stability for each compound. The compounds are alcohols, and we must consider the structure of the carbocation that would result after loss of water. The stability order for carbocations is: tertiary > secondary > primary > methyl. Also, resonance stabilisation (allylic or benzylic) can make a carbocation even more stable than a simple tertiary one.

Let’s examine each compound:

1. Compound I: This is a primary alcohol. The carbon bearing the –OH group is attached to only one other carbon. After protonation and loss of water, it would form a primary carbocation. Primary carbocations are very unstable and difficult to form. This compound will have the slowest rate of dehydration.

2. Compound II: This is a secondary alcohol. The –OH carbon is attached to two other carbons. Loss of water gives a secondary carbocation. Secondary carbocations are more stable than primary but less stable than tertiary. So, this will be faster than I but slower than a tertiary alcohol.

3. Compound III: This is a tertiary alcohol. The –OH carbon is attached to three other carbons. Loss of water gives a tertiary carbocation. Tertiary carbocations are the most stable among simple alkyl carbocations due to hyperconjugation and inductive effects. This will be very fast.

4. Compound IV: This is also a secondary alcohol, but with a twist. The –OH carbon is attached to a carbon that is part of a benzene ring (benzylic position). Loss of water here gives a secondary carbocation that is benzylic. A benzylic carbocation is stabilised by resonance with the aromatic ring — the positive charge can be delocalised into the ring. This makes it even more stable than a simple tertiary carbocation. In fact, a benzylic carbocation is often comparable in stability to a tertiary carbocation, and in many cases more stable. So, compound IV will form a carbocation that is more stable than that from compound III.

So, the order of carbocation stability (and hence rate of dehydration) is: IV (benzylic secondary) > III (tertiary) > II (simple secondary) > I (primary).

Thus, the correct order is IV > III > II > I. But wait — look at the options. Option (B) says IV > III > I > II, which is wrong because II (secondary) is faster than I (primary). Option (D) says III > IV > II > I, which puts III before IV, which is incorrect. Option (A) says II > III > IV > I, which is completely reversed. None of these match IV > III > II > I exactly.

Let’s re-check the options carefully:

• (A) II > III > IV > I
• (B) IV > III > I > II
• (C) III > II > IV > I
• (D) III > IV > II > I …

Concept and Intuition

The Lucas test is a classic method in organic chemistry used to distinguish between primary, secondary, and tertiary alcohols. The test relies on the difference in reactivity of these alcohols with a specific reagent, known as Lucas reagent.

Lucas reagent is a solution of concentrated hydrochloric acid ($\text{HCl}$) and anhydrous zinc chloride ($\text{ZnCl}_2$). The reaction involves the substitution of the hydroxyl group ($-\text{OH}$) of the alcohol with a chloride ion ($-\text{Cl}$) to form an alkyl chloride ($\text{R-Cl}$).

The key to the test is that alkyl chlorides are generally insoluble in the aqueous Lucas reagent, causing the solution to turn cloudy or turbid. The rate at which this turbidity appears indicates the type of alcohol:

• Tertiary alcohols react immediately, producing immediate turbidity.
• Secondary alcohols react within 5-10 minutes, producing turbidity after a short delay.
• Primary alcohols do not react significantly at room temperature, and no turbidity is observed.

This difference in reactivity is due to the mechanism of the reaction, which is primarily an $\text{S}_{\text{N}}1$ (Substitution Nucleophilic Unimolecular) pathway, especially for tertiary and secondary alcohols. The rate-determining step in an $\text{S}_{\text{N}}1$ reaction is the formation of a carbocation intermediate.

The stability of carbocations follows the order: tertiary ($3^\circ$) > secondary ($2^\circ$) > primary ($1^\circ$). Since tertiary carbocations are the most stable, they form most readily, leading to the fastest reaction rate for tertiary alcohols. Anhydrous $\text{ZnCl}_2$ acts as a Lewis acid, coordinating with the oxygen of the alcohol and making the $-\text{OH}$ group a better leaving group ($\text{H}_2\text{O}$).

Step-by-step Evaluation

1. Evaluate Assertion (A): Tertiary alcohols produce turbidity immediately with Lucas reagent.
   • As explained above, tertiary alcohols form the most stable carbocations, leading to a very fast $\text{S}_{\text{N}}1$ reaction with Lucas reagent. …

The $\mathrm{S_N^1}$ reaction proceeds through a two-step mechanism: first, the leaving group (bromide) departs, generating a carbocation intermediate; then the nucleophile attacks this carbocation. The rate-determining step is the formation of the carbocation, so anything that stabilises the carbocation speeds up the reaction. More stable carbocation → faster $\mathrm{S_N^1}$ reaction.

The question gives four benzylic bromides. Each can lose $\mathrm{Br^-}$ to form a carbocation where the positive charge is on the carbon that was attached to bromine. The stability of that carbocation depends on how many aryl (phenyl) groups and alkyl groups are attached to the positively charged carbon — because these groups can delocalise or donate electron density to stabilise the charge.

Let’s examine each option.

1. Option (A): $\mathrm{C_6H_5CH_2Br}$

   Loss of $\mathrm{Br^-}$ gives $\mathrm{C_6H_5CH_2^+}$, a primary benzylic carbocation. The positive charge is stabilised by resonance with the phenyl ring, but there is only one phenyl group and no alkyl substituent on the carbocation carbon. This is the least stabilised among the four.

2. Option (B): $\mathrm{C_6H_5CH(CH_3)Br}$

   The carbocation formed is $\mathrm{C_6H_5CH^+(CH_3)}$, a secondary benzylic carbocation. Here, the positive carbon has one phenyl group (resonance stabilisation) and one methyl group (inductive + hyperconjugative stabilisation). This is more stable than (A).

3. Option (C): $\mathrm{C_6H_5CH(C_6H_5)Br}$

   This gives $\mathrm{C_6H_5CH^+(C_6H_5)}$, a secondary benzylic carbocation with two phenyl groups attached. Two phenyl rings can delocalise the positive charge through resonance, making this carbocation significantly more stable than (B). The extra phenyl group adds a second resonance pathway.

4. Option (D): $\mathrm{C_6H_5C(CH_3)(C_6H_5)Br}$ …

Concept and Intuition

Ether cleavage with hot concentrated HI is a classic reaction. The mechanism is typically SN2: the iodide ion attacks the less hindered carbon of the protonated ether, breaking the C–O bond. With one equivalent of HI, only one C–O bond is cleaved, producing one alcohol and one alkyl iodide. Which alcohol forms depends on which alkyl group becomes the iodide (the better leaving group in SN2) and which becomes the alcohol. Since methyl groups are small and unhindered, they are excellent SN2 targets — but if the methyl group is the one attacked, it becomes methyl iodide, not methanol. To get methanol, the methyl group must remain as the alcohol, meaning the other alkyl group must be the one attacked by iodide. So we need an ether where the methyl group is attached to oxygen but the other alkyl group is more hindered or bulky, so that iodide attacks the other carbon, leaving the methyl group as methanol.

Step-by-step reasoning

1. General reaction: For an ether R–O–R' + HI (1 equiv, hot conc.), the mechanism is:

   • Protonation of oxygen.
   • SN2 attack by I⁻ on the less hindered carbon.
   • Products: R–I + R'–OH (or vice versa, depending on which carbon is attacked). The alcohol is the one whose alkyl group was not attacked.

2. Option (A): $\mathrm{CH_3CH_2CH_2OCH_3}$

   • Two alkyl groups: propyl (primary, straight chain) and methyl.
   • The methyl carbon is less hindered than the propyl carbon. I⁻ attacks the methyl carbon → forms $\mathrm{CH_3I}$ and $\mathrm{CH_3CH_2CH_2OH}$ (propanol).
   • Methyl alcohol is NOT produced.

3. Option (B): $\mathrm{CH_3CH_2OCH_2CH_3}$ (diethyl ether)

   • Both groups are ethyl (primary, identical).
   • I⁻ can attack either ethyl carbon; both give ethyl iodide and ethanol.
   • No methyl alcohol produced.

4. Option (C): $\mathrm{(CH_3)_3C-OCH_3}$ (tert-butyl methyl ether)

   • Groups: tert-butyl (highly hindered, tertiary) and methyl.
   • The methyl carbon is much less hindered than the tert-butyl carbon. However, note: SN2 attack on a tertiary carbon is impossible (steric hindrance). So the only feasible attack is on the methyl carbon.
   • Attack on methyl → $\mathrm{CH_3I}$ and $\mathrm{(CH_3)_3COH}$ (tert-butyl alcohol).
   • Wait — that gives tert-butyl alcohol, not methanol. But the question asks which produces methyl alcohol. Let’s re-evaluate: With one equivalent of HI, the reaction stops after one cleavage. If we attack the methyl, we get methyl iodide. But can we attack the tert-butyl? No, because SN2 at a tertiary carbon is extremely slow. So the only product is tert-butyl alcohol + methyl iodide. …

The relevant concept is the cleavage of ethers by hydrogen iodide. Hot concentrated HI is a strong acid and a good nucleophile. The reaction proceeds by protonation of the ether oxygen, followed by nucleophilic attack by iodide ion. The mechanism can be SN1 or SN2 depending on the alkyl groups. With one equivalent of HI, only one C–O bond is broken, and the alcohol produced corresponds to the alkyl group that does not become an alkyl iodide. The key rule: the more stable carbocation (or the less hindered site for SN2) determines which bond breaks. For methyl ethers, the methyl group is small and can undergo SN2, but if the other group is tertiary, it will form a tertiary carbocation (SN1) and then an iodide, leaving the methyl group to form methanol.

Let’s analyze each option step by step.

1. Option (A): $\mathrm{CH_3CH_2CH_2CH_2-O-CH_3}$

   This is a primary methyl ether. Both alkyl groups are primary (methyl and n-butyl). With HI, the mechanism is typically SN2. Iodide attacks the less hindered carbon. Methyl is less hindered than n-butyl, so attack occurs at the methyl carbon, giving methyl iodide and n-butanol. Thus, methanol is not produced; n-butanol is.

2. Option (B): $\mathrm{CH_3CH_2CH(CH_3)-O-CH_3}$

   This is a secondary methyl ether (the larger group is sec-butyl). Again, the methyl group is less hindered, so SN2 attack at methyl is favored, yielding methyl iodide and sec-butanol. No methanol.

3. Option (C): $\mathrm{CH_3C(CH_3)_2-O-CH_3}$ …