The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________. (i) Allyl (ii) Aryl (iii) Vinyl (iv) Secondary

> [!TLDR] > The key is to identify the carbon bearing the bromine and check its immediate neighbours. The bromine is attached to a carbon that is one bond away from a C=C double bond, making it an **allylic** halide. The correct option is **(i)**. ## 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: $$ \mathrm{CH_3-CH=CH-C(CH_3)_2} $$ 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

The position of –Br in the compound mathrmCH₃CH=CHC(Br)(CH₃)₂ can be classified as ______. (i) Allyl (ii) Aryl

NCERT Exemplar · Q12

\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}

(i) Allyl

(ii) Aryl

(iii) Vinyl

(iv) Secondary

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (acidified potassium dichromate): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones. (Not to be confused with Jones reagent, which is specifically CrO₃ dissolved in dilute aqueous H₂SO₄, often used in acetone — a related but distinct oxidant with the same general 1°→acid / 2°→ketone outcome.) …

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

Primary alcohol ( $R-CH_2OH$ ): Has two hydrogens on the carbon attached to –OH.
Secondary alcohol ( $R_2CHOH$ ): Has one hydrogen on that carbon.
Tertiary alcohol ( $R_3COH$ ): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ( $R-CH_2OH$ ) is oxidized, the first product is an aldehyde ( $R-CHO$ ).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$ ) removes two hydrogens:

One from the –OH group
One from the carbon atom

The carbon–oxygen bond becomes a double bond ( $C=O$ ), forming the aldehyde.

R-CH_2OH \xrightarrow{[O]} R-CHO + H_2O

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$ , heat), the aldehyde is further oxidized to a carboxylic acid ( $R-COOH$ ).

R-CHO \xrightarrow{[O]} R-COOH

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ( $R_2CHOH$ ) has only one hydrogen on the carbon with –OH. Oxidation removes:

One hydrogen from –OH
One hydrogen from the carbon

This forms a ketone ( $R_2C=O$ ).

R_2CHOH \xrightarrow{[O]} R_2C=O + H_2O

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ( $R_3COH$ ) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$ , heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table

Alcohol Type	Hydrogens on C–OH	Product	Why?
Primary ( $1^\circ$ )	2	Aldehyde → Carboxylic acid	Two hydrogens available; aldehyde still has one more

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The key is to identify the carbon bearing the bromine and check its immediate neighbours. The bromine is attached to a carbon that is one bond away from a C=C double bond, making it an allylic halide. The correct option is (i).

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:

\mathrm{CH_3-CH=CH-C(CH_3)_2}

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

Numbering from the left:

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

Tip

A quick way: if the carbon with the halogen is one bond away from a C=C, it’s allylic. If it’s on the C=C itself, it’s vinylic. If it’s on an aromatic ring, it’s aryl.

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. …

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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.

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.
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 ( $\mathrm{CH{=}CH}$ ).
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. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

GUJCET 2025Set 031 markMCQ
$Q. Identify R', R'' and R''' for the following reaction. [FIGURE: a ketone R'R''C=O reacting via$
(i) R'''MgX
(ii) H2O to give 2-methylbutane-2-ol.] (A) $R'=C_{2}H_{5}, R''=C_{2}H_{5}, R'''=CH_{3}$ (B) $R'=CH_{3}, R''=C_{2}H_{5}, R'''=CH_{3}$ (C) $R'=C_{2}H_{5}, R''=CH_{3}, R'''=C_{2}H_{5}$ (D) $R'=CH_{3}, R''=CH_{3}, R'''=CH_{3}$
›Reveal solutionSolution
Ketone $R'R''C{=}O$ + $R'''MgX \rightarrow R'R''R'''C\!-\!OH$ ; the product's three alkyls are CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

Concept — Grignard synthesis of tertiary alcohols. A ketone gives a tertiary alcohol whose carbinol carbon bears the ketone's two groups plus the Grignard's group.

Steps.

2-methylbutan-2-ol: $\mathrm{CH_3\!-\!C(OH)(CH_3)\!-\!CH_2CH_3}$ . The C–OH carbon carries CH $_3$ , CH $_3$ , C $_2$ H $_5$ .

So $\{R',R'',R'''\}=\{CH_3, CH_3, C_2H_5\}$ . …
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GUJCET 2025Set 031 markMCQ
$KMnO_{4}/H_{2}SO_{4}$
›Reveal solutionSolution
[!TLDR]

Oxidising a primary alcohol to an aldehyde requires the mild, selective reagent PCC ( $C_5H_5NH^+CrO_3Cl^-$ ).

Concept

Primary alcohols are oxidised to aldehydes and can be further oxidised to carboxylic acids by strong oxidants. To stop cleanly at the aldehyde, a mild oxidant such as PCC (in anhydrous dichloromethane) is used.

Solution

The substrate (hydroxymethyl)cyclohexane has a $-CH_2OH$ group that must become $-CHO$ (cyclohexanecarbaldehyde) — a controlled oxidation to the aldehyde.

(A) $KMnO_4/H_2SO_4$ : strong oxidant, over-oxidises to the carboxylic acid.

(B) $O_3/H_2O$ –Zn: ozonolysis, cleaves C=C double bonds — not applicable to an alcohol. …
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GUJCET 2023Set 091 markMCQ
$Q. Which of the following alcohol undergo dehydration reaction with Cu (Copper) metal at 573 K temperature? (A) Secondary and Tertiary (B) Primary & Secondary (C) Primary and Tertiary (D) Only Tertiary$
›Reveal solutionSolution
With Cu at 573 K: 1° and 2° alcohols dehydrogenate (→ aldehyde/ketone), while only tertiary alcohols dehydrate (→ alkene).

Concept. Passing alcohol vapour over heated copper at 573 K:

Primary → aldehyde (dehydrogenation, loss of $H_2$ )

Secondary → ketone (dehydrogenation)

Tertiary → has no α-H on the carbinol carbon to lose as $H_2$ , so it instead loses water and dehydrates to an alkene. …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. R'-X --Na/ether--> 2,3-dimethylbutane. Identify R'.$
(a) (CH3)2CH-
(b) (C2H5)2CH-
(c) (CH3CH2)3C-
(d) (CH3)3C-
›Reveal solutionSolution

2,3-dimethylbutane is symmetric, made by Wurtz coupling of two isopropyl (2-propyl) groups.

Wurtz reaction: 2 R'-X + 2 Na --dry ether--> R'-R' + 2 NaX. It couples two alkyl groups to give a symmetrical alkane.

…

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GUJCET 2022Set 171 markMCQ
$-CH_2-CO-OCH_3$
›Reveal solutionSolution

$NaBH_4$ is a mild reducing agent — it reduces aldehydes/ketones to alcohols but does NOT reduce esters.

Concept: Sodium borohydride selectively reduces the cyclohexanone carbonyl ( $C=O$ ) to a secondary alcohol ( $CH-OH$ ), converting the ring ketone into a ring alcohol. The methyl ester group $-CH_2-CO-OCH_3$ is unreactive toward $NaBH_4$ and is retained unchanged. Among the options, only the choic …
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GUJCET 2021Set 151 markMCQ
$CH_3-CH_2-CH_2-Mg-X$
›Reveal solutionSolution

Grignard + methanal → primary alcohol $R-CH_2OH$ ; work backwards to find R.

Concept: $R-MgX + HCHO \rightarrow R-CH_2-OMgX \xrightarrow{H_2O} R-CH_2OH$ . Methanal always adds one carbon and gives a primary alcohol. …
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GUJCET 2021Set 151 markMCQ
$O_3/H_2O$
›Reveal solutionSolution

PCC cleanly oxidises 1° alcohol → aldehyde and leaves the double bond intact.

Concept: Pyridinium chlorochromate (PCC) is a mild oxidant that stops at the aldehyde stage and does not attack C=C.

$CH_2=CH-CH_2OH \xrightarrow{PCC} CH_2=CH-CHO \ (\text{propenal})$ …
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GUJCET 2020Set 071 markMCQ
$-CH_2-C(=O)-OCH_3$
›Reveal solutionSolution

NaBH₄ reduces the ketone (→ cyclohexanol) and does not touch the ester. …

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GUJCET 2020Set 071 markMCQ
$CrO_3$
›Reveal solutionSolution
[!TLDR] Secondary alcohol → ketone needs a mild oxidant; PCC cleanly gives cyclohexanone.

Concept

Secondary alcohols are oxidised to ketones. PCC (pyridinium chlorochromate) is a mild, selective, non-aqueous oxidant that converts secondary alcohols to ketones without further oxidation. Ozonolysis reagents and DIBAL-H are not alcohol-oxidation reagents.

Solution

(A) Anhydrous CrO $_3$ : a strong Cr(VI) oxidant, generally used with acid; not the selective mild reagent intended here. …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. Substance A, on reaction with Cu at 573 K, gives Isobutylene. Which is the structural formula of substance A in this reaction?$
(a) CH3-CH(OH)-CH2-CH3
(b) CH3-CH2-CH2-CH2-OH
(c) CH3-CH(CH3)-CH2-OH
(d) CH3-C(CH3)(CH3)-OH (i.e. (CH3)3C-OH)
›Reveal solutionSolution
Passed over hot copper at 573 K, a TERTIARY alcohol cannot dehydrogenate (it has no H on the carbinol carbon to lose alongside the O-H), so it instead undergoes dehydration to an alkene; a primary or secondary alcohol would dehydrogenate to an aldehyde or ketone instead.

Alcohols passed over copper catalyst at 573 K behave differently by class:

Primary alcohols dehydrogenate to aldehydes (R-CH2-OH -> R-CHO + H2).

Secondary alcohols dehydrogenate to ketones. …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Give the IUPAC name of the product obtained when Phenol is oxidized by chromic acid. (Na2Cr2O7 + Conc. H2SO4).$
(a) Cyclohexa-2,5-diene-1,4-dione
(b) Cyclohexa-1,4-dione
(c) Cyclohexanone
(d) Cyclohexa-1,4-diene-2,5-dione
›Reveal solutionSolution

Phenol + Na2Cr2O7/conc. H2SO4 -> benzoquinone = cyclohexa-2,5-diene-1,4-dione.

Phenol on oxidation with chromic acid (from Na2Cr2O7 + conc. H2SO4) is converted to para-benzoquinone. Its IUPAC name is cyclohexa-2,5-diene-1,4-dione: a six-membered ring with C= …

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Identify Pyridinium chlorochromate from the following.$
(a) pyridine ring, N+ - H . CrO3Cl^-
(b) pyridine ring, N+ - H . CrO2Cl^-
(c) pyridine ring, N+ - CrO3Cl^-
(d) pyridine ring, N+ - H2 . CrO3Cl^-
›Reveal solutionSolution
PCC = pyridinium (C5H5N-H+) chlorochromate (CrO3Cl-), i.e. option (a).

Pyridinium chlorochromate (PCC) is a mild oxidising reagent (C5H5NH+ ClCrO3-) used to oxidise primary alcohols to aldehydes (without over-oxidation to acids). It consists of:

the pyridinium cation: pyridine protonated at nitrogen, so N carries a positive charge and an N-H bond, and …
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Q.The position of –Br in the compound $\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}$ can be classified as ____________.

Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

1. The Core Idea: Loss of Hydrogen

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

Step 2: Carboxylic acid formation

3. Why Secondary Alcohols Give Ketones (and Stop)

4. Why Tertiary Alcohols Do NOT Oxidize

5. The "Why" in One Table

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

1. Draw the full structure

2. Locate the bromine

3. Eliminate the other options

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

Method: Identify the Halogen-Bearing Carbon and Its Neighbourhood

The Correct Answer

Why it is Allyl (The Concept)

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Allyl" with "Vinyl"

Mistake 2: Misidentifying the "Secondary" Carbon

Concept

Solution

Concept

Solution

Q.The position of –Br in the compound CH3CH=CHC(Br)(CH3)2\mathrm{CH_3CH{=}CHC(Br)(CH_3)_2}CH3​CH=CHC(Br)(CH3​)2​ can be classified as ____________.

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

1. Draw the full structure

2. Locate the bromine

3. Eliminate the other options

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

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