Q.The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________.

Concept: Allylic Halide Classification — the question is about classifying the position of a bromine atom relative to a C=C double bond. The compound is $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$.

Step 1: Identify the carbon bearing the –Br. It is the carbon attached to two methyl groups and the alkene chain: $\mathrm{C(Br)(CH_3)_2}$.

Step 2: Check the relationship of this carbon to the double bond. The double bond is between the second and third carbons: $\mathrm{CH_3CH{=}CH-}$. The bromine-bearing carbon is directly attached to the $\mathrm{CH}$ of the double bond (the allylic position). …

Why this is about “allyl” vs “vinyl” vs “aryl”

In organic chemistry, the classification of a halide (or any substituent) depends on the hybridisation and bonding of the carbon it’s attached to, and its relationship to a double bond or aromatic ring.

• Vinyl halide: halogen directly on a $sp^2$ carbon of a C=C bond.
• Allyl halide: halogen on a carbon adjacent to a C=C bond (i.e., one $sp^3$ carbon away from the double bond).
• Aryl halide: halogen directly on a carbon of an aromatic ring.
• Secondary/primary/tertiary: refers to the number of carbon atoms attached to the halogen-bearing carbon (ignoring the double bond’s influence).

The given compound is $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$. Let’s decode its structure step by step.

1. Draw the full structure

The formula $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ means:

• Start with a three-carbon chain: $\mathrm{CH_3-CH=CH-}$
• Then a carbon that has a bromine and two methyl groups: $\mathrm{-C(Br)(CH_3)_2}$

So the carbon skeleton is:

with a Br attached to the fourth carbon (the one with two methyls).

Numbering from the left:

1. $\mathrm{CH_3-}$ (C1)
2. $\mathrm{=CH-}$ (C2, $sp^2$)
3. $\mathrm{-CH=}$ (C3, $sp^2$)
4. $\mathrm{-C(Br)(CH_3)_2}$ (C4, $sp^3$)

The double bond is between C2 and C3.

2. Locate the bromine

The bromine is on C4. Now ask: what is the relationship of C4 to the double bond?

• C4 is not one of the $sp^2$ carbons of the double bond (those are C2 and C3).
• C4 is directly attached to C3, which is an $sp^2$ carbon of the double bond.

That is the defining feature of an allylic position: the halogen is on a carbon adjacent to a C=C bond.

3. Eliminate the other options

• Vinyl: would require Br directly on C2 or C3 (the $sp^2$ carbons). Not the case.
• Aryl: would require an aromatic ring. There is no benzene ring here. …

Concept: Classification of Alkyl Halides Based on the Carbon–Halogen Bond

The type of halide (allyl, vinyl, aryl, etc.) depends on which carbon the halogen is attached to, and what that carbon is bonded to.

Method: Identify the Halogen-Bearing Carbon and Its Neighbourhood

Step 1 -- Identify the structure

The given compound is $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$: a but-2-ene backbone (C1=CH3, C2=CH, C3=CH, double bond between C2-C3) with a fourth carbon (C4) attached to C3, bearing Br and two methyl groups.

Step 2 -- Identify the halogen-bearing carbon and count its neighbours

C4 (the Br-bearing carbon) is bonded to:

1. C3 (the alkene carbon)
2. a methyl group
3. a second methyl group
4. Br

That's three carbon neighbours and one Br -- C4 is a tertiary carbon, with zero hydrogens.

Step 3 -- Check the relationship to the double bond

C4 itself is not part of the C=C double bond (the double bond is between C2 and C3) -- it is one bond away, directly attached to C3, one of the alkene carbons.

Step 4 -- Apply the classification rules

• Vinyl halide: halogen directly ON an sp² carbon of the C=C bond -- not the case here (Br is on C4, not C2 or C3).
• Allylic halide: halogen on an sp³ carbon directly ADJACENT to a C=C bond -- this matches C4 exactly. …

Here is the breakdown of the common mistakes students make on this classification problem, along with the correct reasoning.

The Correct Answer

The correct classification is (i) Allyl.

Why it is Allyl (The Concept)

To classify a halogen (or any substituent), you must look at the carbon atom to which it is directly attached.

1. Identify the Halogen-bearing Carbon: In the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$, the bromine is attached to the carbon that also has two methyl groups ($\mathrm{(CH_3)_2}$) and the alkene $\mathrm{CH}$ carbon — four bonds in all (Br, two $\mathrm{CH_3}$, one C), so it carries no hydrogen.
2. Identify the Adjacent Carbon: Look at the carbon atom next to the one bearing the Br. That adjacent carbon is part of a double bond ($\mathrm{CH{=}CH}$).
3. Definition of Allyl: An allyl group is defined as $\mathrm{CH_2{=}CH{-}CH_2{-}X}$. The key is that the halogen (X) is on a carbon that is adjacent to a carbon-carbon double bond ($\mathrm{C=C{-}C{-}X}$). This is exactly the case here.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Allyl" with "Vinyl"

• The Error: Students see the double bond ($\mathrm{CH{=}CH}$) and immediately classify the Br as Vinyl.
• Why it's Wrong: A Vinyl halide is $\mathrm{CH_2{=}CH{-}X}$. Here, the halogen is attached directly to one of the doubly-bonded carbons. In our compound, the Br is not on the double bond; it is one carbon away.
• How to Avoid: Draw the structure. Ask: "Is the halogen directly on the $\mathrm{C{=}C}$ bond?"
  • Yes → Vinyl (or Aryl if it's a benzene ring).
  • No, but it's next to it → Allyl.

Mistake 2: Misidentifying the "Secondary" Carbon

• The Error: Students see the carbon with Br is attached to two other carbons (the $\mathrm{CH}$ from the chain and two $\mathrm{CH_3}$ groups) and classify it as Secondary (2°) .
• Why it's Wrong: The Br-bearing carbon is bonded to: (1) the $\mathrm{CH}$ of the double bond, (2) a $\mathrm{CH_3}$, (3) another $\mathrm{CH_3}$, and (4) Br -- three carbon neighbours, making it a tertiary carbon (not secondary). But the question asks for the classification of the position (Allyl, Aryl, Vinyl), not the degree (primary, secondary, tertiary) -- a different axis of classification entirely. …