An alcohol having molecular weight 74 gives w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The molecular weight $74$ corresponds to the formula $C_4H_{10}O$ $(12 	imes 4 + 10 	imes 1 + 16 = 74)$.Lucas reagent giving white turbidity in

Vedclass · Accepted Answer

$CH_3-CH(OCH_3)-CH_2-CH_3$

Vedclass · Answer

(C) The molecular weight $74$ corresponds to the formula $C_4H_{10}O$ $(12 	imes 4 + 10 	imes 1 + 16 = 74)$.Lucas reagent giving white turbidity in $5-10$ minutes indicates that the alcohol is a secondary $(2^\circ)$ alcohol.The only secondary alcohol with this formula is butan-$2$-ol $(CH_3-CH(OH)-CH_2-CH_3)$.Reaction with $Na$:$CH_3-CH(OH)-CH_2-CH_3 + Na ightarrow CH_3-CH(ONa)-CH_2-CH_3$ $(Y)$ $+ 1/2 H_2$.Reaction with $CH_3-Br$ (Williamson synthesis):$CH_3-CH(ONa)-CH_2-CH_3 + CH_3-Br ightarrow CH_3-CH(OCH_3)-CH_2-CH_3$ $(Z)$ $+ NaBr$.Thus,$Z$ is $2$-methoxybutane.

An alcohol having molecular weight $74$ gives white turbidity with Lucas reagent within $5-10$ minutes. It reacts with sodium metal to give $Y$. $Y$ reacts with $CH_3-Br$ to give $Z$. $Z$ is:

Similar Questions

Identify $Z$ in the following sequence of reactions:
$CH_{3}-CH_{2}-CH_{2}-OH$ $\xrightarrow{PCl_{3}} X$ $\xrightarrow[\Delta]{alco.KOH} Y$ $\xrightarrow[H-OH/heat]{conc.H_{2}SO_{4}} Z$

Identify the secondary alcohol from the following:

In the following sequence of reactions,
$CH_3CH_2CH_2Br$ $\xrightarrow{KOH(alc.)}$ $(A)$ $\xrightarrow{HBr}$ $(B)$ $\xrightarrow{KOH(aq.)}$ $(C)$
The product $(C)$ is

Absolute alcohol ($100\%$ alcohol) is obtained by the distillation of rectified spirit over which of the following?

Which of the following alcohols is most easily dehydrated with concentrated ${H_2SO_4}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module