A) \[1\text{ }mol\]of \[HCOONa,\]\[1\text{ }mol\] of \[PhC{{H}_{2}}OH\] and \[1\text{ }mol\]of \[PhCOONa\]
B) \[1\text{ }mol\]of \[HCOONa,\]\[1\text{ }mol\] of \[PhC{{H}_{2}}OH\]and \[0.5\text{ }mol\]of \[~PhCOONa\]
C) \[1\text{ }mol\]of \[HCOONa,\]\[1.5\text{ }mol\] of \[PhC{{H}_{2}}OH,\]and \[0.5\text{ }mol\]of \[PhCOONa\]
D) \[1\text{ }mol\]of \[HCOONa,\]\[2\text{ }mol\] of \[PhC{{H}_{2}}OH,\]and \[2\text{ }mol\]of \[PhCOONa\]
Correct Answer: C
Solution :
[c] (i) Crossed Cannizzaro reaction: \[\underset{1\,mol}{\mathop{HCHO}}\,+\underset{1\,mol}{\mathop{PhCHO}}\,\xrightarrow{\overset{\bigcirc -}{\mathop{O}}\,H}\underset{1\,mol}{\mathop{HCOONa}}\,+\underset{1\,\,mol}{\mathop{PhC{{H}_{2}}OH}}\,\](ii) Cannizzaro reaction: (1 mole of \[PhCHO\]left) \[\underset{\frac{1}{2}mol}{\mathop{PhCHO}}\,+\underset{\frac{1}{2}mol}{\mathop{PhCHO}}\,\xrightarrow{\overset{\bigcirc -}{\mathop{O}}\,H}\underset{\frac{1}{2}\,mol}{\mathop{PhCOONa}}\,+\underset{\frac{1}{2}\,\,mol}{\mathop{PhC{{H}_{2}}OH}}\,\] Total moles: \[=\underset{1\,\,mol}{\mathop{HCOONa}}\,+\underset{0.5\,mol}{\mathop{PhCOONa}}\,+\underset{1.5\,\,mol}{\mathop{PhC{{H}_{2}}OH}}\,\]
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