Alcohol, Phenol and Ether MCQ Questions & Answers in Organic Chemistry | Chemistry
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41.
A compound of the formula $${C_4}{H_{10}}O$$ reacts with sodium and undergoes oxidation to give a carbonyl compound which does not reduce Tollen’s reagent, the original compound is
A
Diethyl ether
B
$$n$$ - Butyl alcohol
C
Isobutyl alcohol
D
$$sec$$ - Butyl alcohol
Answer :
$$sec$$ - Butyl alcohol
Since the compound $$\left( {{C_4}{H_{10}}O} \right)$$ react with sodium, it must be alcohol ( option B, C, or D ). As it is oxidised to carbonyl compound which does not reduce Tollen’s reagent, the carbonyl compound should be a ketone and thus $${{C_4}{H_{10}}O}$$ should be a secondary alcohol, i.e. $$sec$$ - butyl alcohol; other two given alcohols are $${1^ \circ }.$$
42.
In the following reaction,
A
B
C
D
Answer :
43.
Identity $$Z$$ in the sequence of reactions, \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow{\frac{HBr}{{{H}_{2}}{{O}_{2}}}}\] \[Y\xrightarrow{{{C}_{2}}{{H}_{5}}ONa}Z\]
A
$$C{H_3} - {\left( {C{H_2}} \right)_3} - O - C{H_2}C{H_3}$$
B
$${\left( {C{H_3}} \right)_2}C{H_2} - O - C{H_2}C{H_3}$$
C
$$C{H_3}{\left( {C{H_2}} \right)_4} - O - C{H_3}$$
D
$$C{H_3}C{H_2} - CH\left( {C{H_3}} \right) - O - C{H_2}C{H_3}$$