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\( 0.44 \mathrm{~g} \) of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at
S.T.P., \( 112 \mathrm{~cm}^{3} \) of methane. With PCC the same alcohol forms a carbonyl compound that
answers silver mirror test. The monohydric alcohol is
Options:
S.T.P., \( 112 \mathrm{~cm}^{3} \) of methane. With PCC the same alcohol forms a carbonyl compound that
answers silver mirror test. The monohydric alcohol is
- A \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH} \)
- B \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH} \)
- C
- D
Solution:
1602 Upvotes
Verified Answer
The correct answer is:
\( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH} \)
At STP $112 \mathrm{~cm}^{3}$ of methane liberated, $22400 \mathrm{~cm}^{3}=1 \mathrm{~mol}$ at ST
$1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}$
$112 \mathrm{~cm}^{3}=\frac{1}{22400} \times 112 \mathrm{~mol}$
$=0.005$ mol of methane liberated
1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces $1 \mathrm{~mol}$ of methane, according to
the reaction,
Monohydric alcohol+ $\mathrm{CH}_{3} \mathrm{M} \mathrm{gl}-\overline{\text { Either }}$
$\Rightarrow 0.005$ mol of monohydric alcohol produces $0.005$ mol of methane
Now, molar mass of monohydric alcohol
$=\frac{\text { Given mass of monohydric alcohol }}{\mathrm{Number} \text { of moles of monohydric alcohol }}$
$=\frac{0.44}{0.005}=88 \mathrm{~g}$
Monohydric alcohol from the given option having $88 \mathrm{~g}$ molar mass is $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{M} \mathrm{gI}-\mathrm{Either}$
When react with $\mathrm{PCC}$, aldehyde is formed which give positive silver mirror test.
( $\mathrm{CH}_{3}$ ) $\left._{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}-\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OM} \mathrm{gI}+\mathrm{CH}_{4}$
( $\mathrm{CH}_{3}$ ) $_{3} \mathrm{C}-\mathrm{CHO}$
$1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}$
$112 \mathrm{~cm}^{3}=\frac{1}{22400} \times 112 \mathrm{~mol}$
$=0.005$ mol of methane liberated
1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces $1 \mathrm{~mol}$ of methane, according to
the reaction,
Monohydric alcohol+ $\mathrm{CH}_{3} \mathrm{M} \mathrm{gl}-\overline{\text { Either }}$
$\Rightarrow 0.005$ mol of monohydric alcohol produces $0.005$ mol of methane
Now, molar mass of monohydric alcohol
$=\frac{\text { Given mass of monohydric alcohol }}{\mathrm{Number} \text { of moles of monohydric alcohol }}$
$=\frac{0.44}{0.005}=88 \mathrm{~g}$
Monohydric alcohol from the given option having $88 \mathrm{~g}$ molar mass is $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{M} \mathrm{gI}-\mathrm{Either}$
When react with $\mathrm{PCC}$, aldehyde is formed which give positive silver mirror test.
( $\mathrm{CH}_{3}$ ) $\left._{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}-\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OM} \mathrm{gI}+\mathrm{CH}_{4}$
( $\mathrm{CH}_{3}$ ) $_{3} \mathrm{C}-\mathrm{CHO}$
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