Q.Primary alkyl halide $C_4H_9Br$

Concept & Intuition

For a molecular formula C₃H₉N, the nitrogen can be primary (‑NH₂), secondary (‑NH‑), tertiary (‑N‑), or quaternary (‑N⁺‑ with a counterion, but here we treat neutral amines). The carbon skeleton can be a straight chain (propyl) or branched (isopropyl). Each arrangement of the nitrogen along the chain and its degree of substitution gives a distinct structural isomer. We systematically list all possibilities, being careful not to double-count.

Step-by-step reasoning

1. Identify possible carbon skeletons

   With three carbons, only two skeletons exist:

   • Straight chain: C–C–C (propyl)
   • Branched: C–C(C) (isopropyl, i.e., a central carbon with two methyl groups)

2. Place nitrogen as a primary amine (–NH₂)

   • On the straight chain:
     • 1‑aminopropane: CH₃–CH₂–CH₂–NH₂
     • 2‑aminopropane: CH₃–CH(NH₂)–CH₃
   • On the branched skeleton:
     • The only distinct primary amine is 2‑aminopropane again (same as above). So no new isomer. → 2 primary amines (1‑aminopropane and 2‑aminopropane).

3. Place nitrogen as a secondary amine (–NH–)

   The nitrogen is inserted between two carbon groups.

   • Straight chain possibilities:
     • N‑methyl‑ethylamine: CH₃–NH–CH₂–CH₃ (ethyl group + methyl group on N)
     • N‑ethyl‑methylamine is the same compound.
   • Branched skeleton:
     • N‑methyl‑isopropylamine: (CH₃)₂CH–NH–CH₃
   • Also consider N‑propylamine? That would be primary. So only these two. → 2 secondary amines (N‑methylethylamine and N‑methylisopropylamine).

4. Place nitrogen as a tertiary amine (–N– with three carbon groups)

   • All three carbons must be attached to nitrogen.
   • The only possibility is trimethylamine: (CH₃)₃N
   • No other arrangement (e.g., ethyldimethylamine would need 4 carbons). → 1 tertiary amine.

5. Check for quaternary ammonium (salt) structures

   The formula C₃H₉N is neutral; a quaternary ammonium would require a counterion (e.g., Cl⁻) and would have formula C₃H₁₀N⁺, so not counted here.

   → 0 quaternary isomers.

6. Total count

   Primary: 2

   Secondary: 2

   Tertiary: 1

   Total = 5? Wait — we must check for duplicates.

   • 2‑aminopropane (primary) and N‑methylisopropylamine (secondary) are different.
   • However, note that N‑methylethylamine and N‑methylisopropylamine are distinct. So total distinct structural isomers = 2 + 2 + 1 = 5. But the options given are 2, 3, 4, 5. The correct answer is 4? Let’s re-examine carefully.

• (1) CH₃CH₂CH₂NH₂ (1‑aminopropane)
• (2) CH₃CH(NH₂)CH₃ (2‑aminopropane)
• (3) CH₃CH₂NHCH₃ (N‑methylethylamine)
• (4) (CH₃)₂CHNHCH₃ (N‑methylisopropylamine)
• (5) (CH₃)₃N (trimethylamine) …

Step 1 — List possible skeletons

For C$_5$H$_{12}$, the carbon skeleton can be arranged as:

1. A straight, unbranched chain of 5 carbons: n-pentane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$)
2. A 4-carbon main chain with one methyl branch: 2-methylbutane / isopentane
3. A 3-carbon main chain with two methyl branches on the central carbon: 2,2-dimethylpropane / neopentane

Step 2 — Confirm no further distinct skeletons exist …

1. Identify alcohol A from the formula + the word "tertiary"

$\mathrm{C_4H_{10}O}$ has four alcohol isomers:

A tertiary alcohol is one whose carbinol carbon (the C bearing the $-\mathrm{OH}$) is attached to three other carbons. Only $\mathrm{(CH_3)_3C{-}OH}$ qualifies.

2. Recognise the reaction

$\mathrm{20\%\ H_3PO_4}$ at 358 K is the standard acid-catalysed dehydration condition, and the product $\mathrm{C_4H_8}$ has one degree of unsaturation more than $\mathrm{C_4H_{10}O}$ minus water — exactly $\mathrm{C_4H_{10}O - H_2O = C_4H_8}$. So B is an alkene.

Note how mild the conditions are: tertiary alcohols dehydrate most easily ($3^\circ > 2^\circ > 1^\circ$) because the E1 mechanism goes through a stable tertiary carbocation. That is why only 20% acid and a modest 358 K are needed — a primary alcohol would need ~95% $\mathrm{H_2SO_4}$ at 440 K.

3. Mechanism (E1)

1. Protonation of the $-\mathrm{OH}$ to make a good leaving group:

2. Loss of water ⇒ the stable tertiary carbocation:

3. Loss of a β-hydrogen from one of the three (equivalent) methyl groups:

4. The product is unique …

Step 1 — The molecular formula.

$\mathrm{C_7H_8O}$, degree of unsaturation $=\dfrac{2(7)+2-8}{2}=4$ — a benzene ring plus no other unsaturation. The candidates are the three cresols ($\mathrm{CH_3{-}C_6H_4{-}OH}$), benzyl alcohol ($\mathrm{C_6H_5CH_2OH}$) and anisole.

Step 2 — Test 1: dissolves in NaOH ⇒ it is acidic ⇒ phenolic $\mathrm{-OH}$.

Phenols ($\mathrm{p}K_a\approx10$) are acidic enough to react with NaOH, because the phenoxide ion is resonance-stabilised over the ring:

Alcohols are not: benzyl alcohol ($\mathrm{p}K_a\approx16$, no resonance stabilisation of its alkoxide) is insoluble in NaOH. ⇒ (C) benzyl alcohol is eliminated.

Step 3 — Test 2: violet colour with $\mathrm{FeCl_3}$ ⇒ phenol confirmed.

Phenols form coloured iron(III)–phenoxide complexes of the type $[\mathrm{Fe(OAr)_6}]^{3-}$. This is the classic confirmatory test for a phenolic $\mathrm{-OH}$, and it again rules out benzyl alcohol and anisole. So the compound is one of the cresols.

Step 4 — Test 3: bromination gives a TRIbromo derivative — this is what selects the isomer.

Both $\mathrm{-OH}$ (strongly) and $\mathrm{-CH_3}$ (weakly) are activating, o/p-directing groups. Ask, for each isomer, how many ring positions are activated by both groups.

• m-Cresol ($\mathrm{-OH}$ at C-1, $\mathrm{-CH_3}$ at C-3):
  • ortho/para to $\mathrm{-OH}$ (C-1) ⇒ C-2, C-4, C-6.
  • ortho/para to $\mathrm{-CH_3}$ (C-3) ⇒ C-2, C-4, C-6. The two groups reinforce each other at exactly three free positions — 2, 4 and 6 — all of which are vacant. Bromination therefore substitutes cleanly at all three: $\mathrm{C_7H_8O} + 3\,\mathrm{Br_2} \longrightarrow \underset{\text{2,4,6-tribromo-3-methylphenol}}{\mathrm{C_7H_5OBr_3}} + 3\,\mathrm{HBr}$ …

The key here is to recognise that alcoholic ammonia ($\text{alc. NH}_3$) acts as a nucleophile in an $\text{S}_\text{N}2$ displacement. Benzyl chloride ($\mathrm{C_6H_5CH_2Cl}$) has a benzylic carbon that is highly reactive toward nucleophilic substitution because the developing positive charge in the transition state is stabilised by resonance with the benzene ring. Ammonia, being a good nucleophile, attacks this carbon, displacing chloride and forming a primary amine.

1. First step — formation of A:

   $\mathrm{C_6H_5CH_2Cl} + 2\mathrm{NH_3} \rightarrow \mathrm{C_6H_5CH_2NH_2} + \mathrm{NH_4Cl}$

   The product A is benzylamine (phenylmethanamine). Two equivalents of ammonia are needed: one acts as the nucleophile, the other picks up the liberated HCl.

2. Second step — exhaustive methylation:

   Benzylamine (A) is now treated with excess methyl chloride ($2\mathrm{CH_3Cl}$). This is a classic Hofmann alkylation: the amine nitrogen, being nucleophilic, attacks methyl chloride repeatedly.

   • First methylation: $\mathrm{C_6H_5CH_2NH_2} + \mathrm{CH_3Cl} \rightarrow \mathrm{C_6H_5CH_2NH(CH_3)}^+ \mathrm{Cl}^-$ (a secondary ammonium salt).
   • In the presence of excess methyl chloride and the basic conditions provided by the excess ammonia (or by the amine itself), the free base is regenerated and undergoes a second methylation: $\mathrm{C_6H_5CH_2NH(CH_3)} + \mathrm{CH_3Cl} \rightarrow \mathrm{C_6H_5CH_2N(CH_3)_2}^+ \mathrm{Cl}^-$ The product after two methylations is the tertiary amine — N,N-dimethylbenzylamine — as its hydrochloride salt. The question likely intends the neutral amine, which is N,N-dimethyl phenyl methanamine (IUPAC: N-benzyl-N-methylmethanamine).

3. Identifying the correct option: …

The key insight here is that you cannot simply swap the OH group from one carbon to another in one step. You need to break and reform bonds in a controlled way. The strategy is to first create a double bond (alkene) from the starting alcohol, then add HBr across that double bond in the anti-Markovnikov fashion so that the bromine ends up on the terminal carbon, and finally replace the bromine with an OH group.

Let's walk through each step.

1. Dehydration of propan-2-ol to propene Propan-2-ol is a secondary alcohol. When heated with a strong acid like concentrated $H_2SO_4$ (or passed over alumina at high temperature), it undergoes dehydration (elimination of water) to form propene.

This step creates the carbon-carbon double bond that we will later use to attach the bromine at the correct position.

2. Addition of HBr in the presence of peroxide (anti-Markovnikov addition) Normally, HBr adds to an unsymmetrical alkene following Markovnikov's rule — the hydrogen attaches to the carbon with more hydrogens, and bromine goes to the more substituted carbon. For propene, that would give 2-bromopropane, which would take us back to a secondary alkyl halide. However, in the presence of organic peroxides (like benzoyl peroxide), the addition follows a free radical mechanism that reverses the regioselectivity. The bromine radical attacks the less substituted carbon (the terminal carbon), so the product is 1-bromopropane.

This is the critical step that moves the halogen to the terminal position.

3. Heating with aqueous KOH (hydrolysis) The final step is a nucleophilic substitution. 1-bromopropane is a primary alkyl halide, so it undergoes $S_N2$ reaction with the hydroxide ion from aqueous KOH. The OH group replaces the bromine atom, giving propan-1-ol.

Aqueous KOH is used here because it provides free hydroxide ions; alcoholic KOH would favour elimination (forming propene again), which is not what we want.

Now, let's check the options against this sequence: …

The starting material is $HOH_2C-CH=CH-CH_2-CHO$ — it carries three reducible/oxidisable handles: a primary alcohol (left end), a C=C double bond (middle) and an aldehyde (right end). Each arrow attacks a different one, so this is a pure chemoselectivity question.

Step 1 — Reaction A: $HOH_2C{-}CH{=}CH{-}CH_2{-}\mathbf{CHO} \;\to\; HOH_2C{-}CH{=}CH{-}CH_2{-}\mathbf{CH_2OH}$

What changed: the aldehyde has become a primary alcohol. What did NOT change: the C=C is still drawn — it survives.

So A must be a reducing agent that attacks C=O but leaves C=C alone. That is exactly sodium borohydride, NaBH$_4$: the hydride $H^-$ adds to the electron-poor (electrophilic) carbonyl carbon, but an isolated alkene is electron-rich and is not attacked by a nucleophilic hydride.

Could A be H$_2$/Pd? No — Pd would hydrogenate the C=C as well, and the product still shows the double bond. ⇒ A = NaBH$_4$.

That single observation already eliminates options (A) and (D), both of which begin with H$_2$/Pd.

Step 2 — Reaction B: $\mathbf{HOH_2C}{-}CH{=}CH{-}CH_2{-}CHO \;\to\; \mathbf{OHC}{-}CH{=}CH{-}CH_2{-}CHO$

What changed: the primary alcohol has been oxidised to an aldehyde (giving the dialdehyde). What did NOT change: the C=C survives, and the new –CHO has not been over-oxidised to –COOH.

This demands a mild, selective oxidant that stops at the aldehyde: PCC (pyridinium chlorochromate). Being anhydrous (in CH$_2$Cl$_2$), it gives no gem-diol intermediate, so the reaction cannot proceed to the carboxylic acid. …

Step 1 — The Wurtz reaction.

With two different halides $R{-}X$ and $R'{-}X$ you get the cross-coupled alkane $R{-}R'$ (along with $R{-}R$ and $R'{-}R'$ as by-products). To design the synthesis, you disconnect the target alkane at one C–C bond and put a halogen on each fragment.

Step 2 — Write and number the target.

2-Methylpentane:

It has 6 carbons in total (5 in the main chain + 1 methyl branch).

Step 3 — Disconnect at C2–C3.

Breaking the bond between C-2 and C-3 gives two 3-carbon fragments:

• Left fragment: $\mathrm{CH_3-CH(CH_3)-}$ $=$ isopropyl group, $(\mathrm{CH_3})_2\mathrm{CH}-$ (C-1, C-2 and the branch methyl).
• Right fragment: $\mathrm{-CH_2CH_2CH_3}$ $=$ n-propyl group (C-3, C-4, C-5).

So the halides are $(\mathrm{CH_3})_2\mathrm{CHCl}$ and $\mathrm{CH_3CH_2CH_2Cl}$, and

Step 4 — Rule out the other pairs (check the product each would give). …

Step 1 — Why a double bond is essential.

A C=C consists of a $\sigma$ bond plus a $\pi$ bond formed by sideways overlap of $p$-orbitals. Rotating about the axis would break that $\pi$ overlap, which costs far too much energy at ordinary temperature. This restricted rotation locks the substituents in place, so two spatially distinct arrangements can exist and be isolated. In a single-bonded (saturated) compound, free rotation instantly interconverts such arrangements — they are mere conformers, not isomers.

Step 2 — The second condition.

Writing the alkene as

geometrical isomerism requires $a \neq b$ and $c \neq d$. If either carbon carries two identical groups, flipping them gives the same molecule.

Step 3 — Test each option. …