A compound with the molecular formula {C_4}{H | vedclass

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

Question

(D) $1$. The molecular formula ${C_4}{H_{10}}O$ corresponds to either an alcohol or an ether. $2$. The compound reacts with sodium,which indicates it

Vedclass · Accepted Answer

sec-Butyl alcohol

Vedclass · Answer

(D) $1$. The molecular formula ${C_4}{H_{10}}O$ corresponds to either an alcohol or an ether. $2$. The compound reacts with sodium,which indicates it is an alcohol $(R-OH + Na \rightarrow R-ONa + \frac{1}{2}H_2)$. $3$. Upon oxidation,it yields a carbonyl compound that does not reduce Tollen's reagent. This means the carbonyl compound is a ketone,not an aldehyde. $4$. For an alcohol to yield a ketone upon oxidation,it must be a secondary alcohol. $5$. The secondary alcohol with the formula ${C_4}{H_{10}}O$ is $sec$-butyl alcohol $(CH_3-CH(OH)-CH_2-CH_3)$,which oxidizes to butanone $(CH_3-CO-CH_2-CH_3)$. $6$. Butanone is a ketone and does not give a positive Tollen's test.

$A$ compound with the molecular formula ${C_4}{H_{10}}O$ reacts with sodium and gives a carbonyl compound upon oxidation,which does not reduce Tollen's reagent. What is the original compound?

Similar Questions

When vapours of ethanol are passed over platinised asbestos in excess of air,the compound formed is

Arrange the following alcohols in increasing order of reactivity towards elimination reaction:
$(I)$ $NO_2-CH_2-CH_2-CH_2-CH_2OH$
$(II)$ $CH_3-CH_2-CH(NO_2)-CH_2OH$
$(III)$ $CH_3-CH(NO_2)-CH_2-CH_2OH$
$(IV)$ $CH_3-CH_2-CH_2-CH_2OH$
Select the answer from the codes given below:

The major product of the following reaction is:

When the vapours of tertiary butyl alcohol are passed through heated copper at $573 \ K$,the product formed is

Tertiary butyl alcohol gives tertiary butyl chloride on treatment with

Vedclass Test Series

Exam Paper Generator

Online Exam Module