MCQ
IUPAC name of (CH3)2CH−CH2−CH2Br is:
- A1-Bromopentane
- B2-Methyl-4-bromobutane
- C1-Bromo-3-methylbutane
- D2-Methyl-3-bromobutane
✓
Answer
- 1-Bromo-3-methylbutane
Explanation:
IUPAC name of (CH3)2CH−CH2−CH2Br is 1-Bromo-3-methylbutane. The parent chain has 4 carbon atoms with a methyl substituent (CH3) at C−3 and a bromine atom at C−1.
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Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential ( $\mathrm{E}^{\circ}$ ) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}$ ($V$ with respect to normal hydrogen electrode) values.
→$\mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}\ \ \ \mathrm{E}^{\circ}=0.54 $
$\mathrm{Cl}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \ \ \ \mathrm{E}^{\circ}=1.36 $
$\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} \ \ \ \mathrm{E}^{\circ}=1.50 $
$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} \ \ \ \mathrm{E}^{\circ}=0.77 $
$\mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \ \ \ \mathrm{E}^{\circ}=1.23$
$1.$ Among the following, identify the correct statement.
$(A)$ Chloride ion is oxidized by $\mathrm{O}_2$
$(B)$ $\mathrm{Fe}^{2+}$ is oxidized by iodine
$(C)$ Iodide ion is oxidized by chlorine
$(D)$ $\mathrm{Mn}^{2+}$ is oxidized by chlorine
$2.$ While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
$(A)$ $\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$
$(B)$ $\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$
$(C)$ $\mathrm{Fe}^{3+}$ oxidizes $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$(D)$ $\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
$3.$ Sodium fusion extract, obtained from aniline, on treatment with iron $(II)$ sulphate and $\mathrm{H}_2 \mathrm{SO}_4$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of
$(A)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(B)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$ $(D)$ $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
Give the answer question $1,2$ and $3.$
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