Q.Identify A, B, C, D, E, R and $R^1$ in the following: Bromocyclohexane, shown below, $+ \ Mg \xrightarrow{\text{dry ether}} A \xrightarrow{H_2O} B$

Concept: Structural Isomerism — the same molecular formula can give different products depending on the structure of the alkyl halide.

Step 1: First sequence

Bromocyclohexane with Mg in dry ether gives the Grignard reagent A: cyclohexylmagnesium bromide.

Hydrolysis of A yields B: cyclohexane ($C_6H_{12}$).

Step 2: Second sequence

The product after $D_2O$ quench is $CH_3CHDCH_3$, which is propane with one deuterium at the middle carbon. This means the Grignard reagent C must be isopropylmagnesium bromide.

Hence $R\text{-}Br$ is isopropyl bromide ($CH_3CHBrCH_3$), so R is isopropyl ($CH_3CHCH_3$).

Step 3: Third sequence

The Wurtz reaction with Na/ether gives 2,2,3,3-tetramethylbutane. This is the coupling product of two tert-butyl radicals. …

Let’s unpack the logic. The question gives you three separate reaction sequences, each built around the chemistry of Grignard reagents and the Wurtz reaction. The trick is to identify the unknown organic groups (R, R¹) and the intermediates (A through E) by reasoning backwards from the known products.

Why this approach works: Grignard reagents (R–Mg–X) are formed by reacting an alkyl/aryl halide with magnesium metal in dry ether. They act as strong nucleophiles and bases. When you quench them with water (H₂O) or heavy water (D₂O), you replace the MgX group with H or D, giving the corresponding alkane. The Wurtz reaction (2R–X + 2Na → R–R + 2NaX) couples two alkyl halides, but only works well for symmetrical, primary alkyl halides. By matching the products to these known reactions, we can deduce each unknown.

1. First sequence: Bromocyclohexane → A → B

   Bromocyclohexane (C₆H₁₁Br) reacts with Mg in dry ether. This is the classic Grignard formation: the bromine is replaced by MgBr, giving A = cyclohexylmagnesium bromide (C₆H₁₁–Mg–Br).

   When A is treated with water (H₂O), the Grignard reagent is protonated: the MgBr group is replaced by H. The product is B = cyclohexane (C₆H₁₂).

   $\ce{C6H11Br + Mg ->[dry ether] C6H11MgBr ->[H2O] C6H12 + Mg(OH)Br}$

2. Second sequence: R–Br → C → CH₃CHDCH₃

   Here, an unknown alkyl bromide R–Br forms a Grignard reagent C (R–Mg–Br). This is then quenched with D₂O (heavy water). The product is CH₃CHDCH₃ — that’s propane with one deuterium on the middle carbon.

   The product tells us the structure of R. If the Grignard reagent R–Mg–Br is quenched with D₂O, the D ends up on the carbon that was bonded to Mg. So the product is R–D. Here, R–D = CH₃CHDCH₃, which means R must be the isopropyl group: CH₃CHCH₃ (with the free bond on the middle carbon). Therefore, R = isopropyl (CH₃)₂CH–, and C = isopropylmagnesium bromide, (CH₃)₂CH–Mg–Br.

   Tip

   The deuterium label is a tracer: it lands exactly where the MgBr was. So the product’s structure directly reveals the original alkyl group R.

3. Third sequence: R¹–X → (CH₃)₃C–C(CH₃)₃ and R¹–X → D → E

   This sequence has two branches from the same starting material R¹–X. …

Method: Reaction Tracking with Reagent Logic (Grignard & Wurtz Reactions)

This method uses the reagent-specific behaviour of alkyl halides to deduce unknown structures step-by-step.

Step 1 – Identify A and B (Grignard formation & hydrolysis)

Reaction 1:

• A is the Grignard reagent: cyclohexylmagnesium bromide

• B is the product after hydrolysis: cyclohexane

Key idea: Grignard reagents react with water to give the corresponding alkane.

Step 2 – Identify R (from the second reaction)

Reaction 2:

• C is the Grignard reagent: $R\text{MgBr}$
• D$_2$O replaces the MgBr group with deuterium (D).
• The product is 2-deuteriopropane: $\text{CH}_3\text{CHDCH}_3$

This means the original alkyl group $R$ must be isopropyl:

So:

• $R\text{-}Br$ = 2-bromopropane (isopropyl bromide)
• C = isopropylmagnesium bromide

Step 3 – Identify R¹ (from Wurtz reaction)

Reaction 3 (first part):

• The product is 2,2,3,3-tetramethylbutane (a symmetrical alkane).
• In the Wurtz reaction, two identical alkyl halides couple:

• Since $R^1\text{-}R^1 = (\text{CH}_3)_3\text{C-C}(\text{CH}_3)_3$, each $R^1$ must be:

So $R^1\text{-}X$ = tert-butyl halide (e.g., $(\text{CH}_3)_3\text{CCl}$ or $(\text{CH}_3)_3\text{CBr}$).

Step 4 – Identify D and E (Grignard from R¹)

Reaction 3 (second part): …

Here are the common mistakes students make in this exact problem and how to avoid each.

1. Mistake: Thinking A is a Grignard reagent with a different carbon skeleton

Why it happens:

Students see Bromocyclohexane + Mg and assume the Mg inserts between Br and the ring, but then they forget the product is still a cyclohexyl group.

How to avoid:

Remember: Grignard reagents preserve the carbon skeleton of the alkyl/aryl halide.

• C₆H₁₁Br (bromocyclohexane) + Mg → C₆H₁₁MgBr (cyclohexylmagnesium bromide).
• A = Cyclohexylmagnesium bromide (not a ring-opened or rearranged product).

2. Mistake: Writing B as cyclohexene instead of cyclohexane

Why it happens:

Students think A + H₂O always gives the alkane directly.

How to avoid:

Grignard reagent + water gives the corresponding hydrocarbon via protonation:

RMgX + H₂O → R–H + Mg(OH)X

• Here, R = cyclohexyl, so B = cyclohexane ($C_6H_{12}$).
• B is not cyclohexene — no elimination occurs.

3. Mistake: Misidentifying R from the deuterium product

Why it happens:

Students see CH₃CHDCH₃ and think R must be isopropyl, but they forget the Grignard step.

How to avoid:

Work backwards:

• C + D₂O → CH₃CHDCH₃ means C must be isopropylmagnesium bromide ((CH₃)₂CHMgBr).
• Therefore, R–Br = (CH₃)₂CHBr (2-bromopropane).
• R = isopropyl group ((CH₃)₂CH–).

4. Mistake: Confusing the Wurtz reaction with Grignard formation for R¹

Why it happens:

The same R¹–X is used in two different reactions — students mix up the products.

How to avoid:

• Wurtz reaction (Na/ether): 2R¹–X + 2Na → R¹–R¹ + 2NaX Given product: (CH₃)₃C–C(CH₃)₃ (2,2,3,3-tetramethylbutane). So R¹ = tert-butyl group (CH₃)₃C–.
• Grignard reaction (Mg then H₂O): R¹–X + Mg → R¹MgX → R¹–H So E = isobutane (CH₃)₃CH.

5. Mistake: Writing D as the Grignard reagent but forgetting the solvent condition

Why it happens:

Students write D = (CH₃)₃CMgX but don’t specify X or the ether condition.

How to avoid:

Always note:

• D = tert-butylmagnesium halide (e.g., (CH₃)₃CMgCl or (CH₃)₃CMgBr).
• The reaction requires dry ether — water destroys Grignard reagents.

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