which of the following molecules would cause the following experimental results: bromine test - orange potassium permanganate test - purple sodium fusion test - precipitant formed $\ce{ch3-ch2-br}$ $\ce{ch3ch=chch2ch2ch2ch3}$ $\ce{ch3chchch2ch2ch2ch3}$ (note: the original images third options structure may have a typo, here based on ocr and common structure correction attempt)

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Accepted Answer

To solve this, we analyze each test and the compounds:

### 1. Bromine Test (Orange Disappearance)
Bromine water (orange) decolorizes in the presence of alkenes (due to addition across the $ \text{C}=\text{C} $ double bond). Alkanes and alkyl bromides do not react with bromine water.

### 2. Potassium Permanganate Test (Purple Disappearance)
Acidified $ \text{KMnO}_4 $ (purple) is reduced (decolorizes) by alkenes (oxidation of $ \text{C}=\text{C} $). Alkanes and alkyl bromides do not react with $ \text{KMnO}_4 $.

### 3. Sodium Fusion Test (Precipitate Formation)
Sodium fusion test detects halogens (like Br) in organic compounds. The test forms a precipitate (e.g., $ \text{AgBr} $ for bromine).

### Analyzing Each Compound:
- **$ \boldsymbol{\text{CH}_3-\text{CH}_2-\text{Br}} $ (Alkyl Bromide):**
  - Reacts with sodium fusion (forms precipitate, as it contains Br).
  - Does **not** react with bromine water (no $ \text{C}=\text{C} $) or $ \text{KMnO}_4 $ (no $ \text{C}=\text{C} $ to oxidize). Thus, it fails the bromine and $ \text{KMnO}_4 $ tests.

- **$ \boldsymbol{\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3} $ (Alkene):**
  - Reacts with bromine water (decolorizes orange, addition across $ \text{C}=\text{C} $).
  - Reacts with $ \text{KMnO}_4 $ (decolorizes purple, oxidation of $ \text{C}=\text{C} $).
  - Does **not** contain Br (so no precipitate in sodium fusion test). Thus, it fails the sodium fusion test.

- **$ \boldsymbol{\text{CH}_3\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3} $ (Alkane, assuming typo; likely a saturated hydrocarbon):**
  - Does not react with bromine water (no $ \text{C}=\text{C} $) or $ \text{KMnO}_4 $ (no $ \text{C}=\text{C} $).
  - Does not contain Br (no precipitate in sodium fusion test). Fails all three tests.

Wait, there’s a mistake in the analysis. The second compound ($ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $) is an alkene, so it reacts with bromine and $ \text{KMnO}_4 $ (passes those tests) but does not contain Br (fails sodium fusion). The first compound ($ \text{CH}_3\text{CH}_2\text{Br} $) contains Br (passes sodium fusion) but fails bromine and $ \text{KMnO}_4 $.

But the question requires a compound that causes *all three* experimental results:
- Bromine test: orange disappears (so reacts with bromine).
- $ \text{KMnO}_4 $ test: purple disappears (so reacts with $ \text{KMnO}_4 $).
- Sodium fusion: precipitate (so contains Br).

Wait, no—maybe the second compound has a typo, or I misread. Wait, the second compound is $ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $? No, the structure is $ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $? Wait, no, the middle structure is $ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $ (alkene with $ \text{C}=\text{C} $) and the first is alkyl bromide, third is alkane.

But the only compound with $ \text{C}=\text{C} $ (to react with bromine and $ \text{KMnO}_4 $) is the middle one ($ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $), but it lacks Br. The first has Br but no $ \text{C}=\text{C} $.

Wait, maybe the middle compound has a Br? No, the structure shows $ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $ (no Br). The first has Br.

Wait, perhaps the question’s “sodium fusion test - precipitate” is for bromine, so the compound must have Br *and* $ \text{C}=\text{C} $. But none of the given compounds have both. This suggests a possible typo, but among the options, the **alkene** ($ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $) is the only one with $ \text{C}=\text{C} $ (to react with bromine and $ \text{KMnO}_4 $), even though it lacks Br. The sodium fusion test result might be a red herring, or there’s a structural error.

Assuming the sodium fusion test is a distractor, or the compound has Br and $ \text{C}=\text{C} $ (but the given structures don’t show that), the most logical answer is the **alkene** ($ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $) because it reacts with bromine (orange disappears) and $ \text{KMnO}_4 $ (purple disappears), matching the first two tests. The sodium fusion test might be an error, or the compound is intended to have Br (but the structure doesn’t show it).

# Brief Explanations:
The compound must react with bromine (decolorize orange), react with $ \text{KMnO}_4 $ (decolorize purple), and show a precipitate in sodium fusion (indicating Br). The alkene ($ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $) reacts with bromine (addition across $ \text{C}=\text{C} $) and $ \text{KMnO}_4 $ (oxidation of $ \text{C}=\text{C} $), matching the first two tests. Though it lacks Br (suggesting a possible structural typo), it is the only compound with $ \text{C}=\text{C} $ to trigger the bromine and $ \text{KMnO}_4 $ results.

# Answer:
The middle compound: $ \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 $ (the alkene with the double bond)

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QUESTION IMAGE

Question

Explanation:

1. Bromine Test (Orange Disappearance)

2. Potassium Permanganate Test (Purple Disappearance)

3. Sodium Fusion Test (Precipitate Formation)

Analyzing Each Compound:

Answer: