Alcohol, Phenol and Ether MCQ Questions & Answers in Organic Chemistry | Chemistry

Order of reactivity of different alcohols towards esterification is $${1^ \circ }$$ alcohol > $${2^ \circ }$$ alcohol > $${3^ \circ }$$ alcohol due to increased steric hindrance in $${2^ \circ }$$ and $${3^ \circ }$$ alcohols.

In case of unsymmetrical ether, the alkyl halide is always formed from smaller alkyl group. This happens, because $${I^ - }\,ion$$   being larger in size approaches smaller alkyl group to avoid steric hindrance.

When sodium phenoxide $$\left( {{C_6}{H_5}{O^ - }{N^ + }a} \right)$$    is heated with ethyl iodide $$\left( {{C_2}{H_5}I} \right)$$   it form ethyl phenyl ether which is also called phenetole. This reaction is called Williamson's synthesis

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NOTE : This reaction is an example of Williamson's synthesis.
$${{\text{C}}_2}{H_5}{O^ - }$$  will abstract proton from phenol converting the latter into phenoxide ion. This would then make nucleophilic attack on the methylene carbon of alkyl iodide forming $${C_6}{H_5}O{C_2}{H_5}.$$   But if $${C_2}{H_5}{O^ - }$$  is in excess, $${C_6}{H_5}O{C_2}{H_5}$$   will be formed. $${C_2}{H_5}{O^ - }$$  is a better nucleophile than $${C_2}{H_5}{O^ - }$$   (phenoxide) ion because in the former the negative charge is localised over oxygen, while in the latter it is delocalised over the whole molecular framework. So, it is $${C_2}{H_5}{O^ - }$$  ion that would make nucleophilic attack at ethyl iodide to give diethyl ether (Williamson’s synthesis).

Phenol has active (acidic) hydrogen so it reacts with $$C{H_3}MgI$$   to give $$C{H_4},$$  and not anisole $${C_6}{H_5}OH + C{H_3}MgI \to $$      $$C{H_4} + {C_6}{H_5}OMgI$$

$$ROH + Na \to RONa + \frac{1}{2}{H_2} \uparrow $$
We have to get molecular mass of alcohol corresponding to half mole of $${H_2}$$  only.
$$\frac{{1120}}{{11200}} = \frac{{7.4}}{M} \Rightarrow M = 74$$
$${C_n}{H_{2n + 1}}OH = 74 \to {C_n}{H_{2n + 1}}$$       $$ = 74 - 17 = 57$$
$$ \Rightarrow {C_n}{H_{2n}} = 57 - 1 = 56\,\,\,\,{\text{i}}{\text{.e}}{\text{.}}$$       $$12n + 2n = 14n = 56$$
$$ \Rightarrow n = \frac{{56}}{{14}} = 4$$
Thus, molecular formula of $$(A)$$  is $${C_4}{H_9}OH.$$   As $$(A)$$  gives Lucas test within $$5\,\min .,$$  thus $${2^ \circ }$$ alcohol corresponding to molecular formula $${C_4}{H_9}OH$$  is $$C{H_3}CH\left( {OH} \right)C{H_2}C{H_3}$$     ( butan-2-ol ).

$$-OH$$  group in phenol can release electrons to the ring better than $$ - C{H_3}$$  group in toluene. $$Cl$$  atom has electron withdrawing effect which inhibits electrophilic attack.