X is an isomer of C_6H_{14}. It has four prim | vedclass

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

Question

(D) The isomer of $C_6H_{14}$ with four primary carbons and two tertiary carbons is $2,3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$.Let's analyze

Vedclass · Accepted Answer

Wurtz reaction of $2$-bromopropane

Vedclass · Answer

(D) The isomer of $C_6H_{14}$ with four primary carbons and two tertiary carbons is $2,3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$.Let's analyze the options:$A$: Hydrogenation of $2$-methylpent-$2$-ene gives $2$-methylpentane,which has five primary carbons and one tertiary carbon.$B$: Wurtz reaction of $1$-bromopropane $(CH_3CH_2CH_2Br)$ gives $n$-hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$.$C$: Reduction of $2$-chloro-$3$-methylpentane gives $3$-methylpentane,which has three primary carbons and one tertiary carbon.$D$: Wurtz reaction of $2$-bromopropane $(CH_3CH(Br)CH_3)$ with $Na$ in dry ether leads to the coupling of two isopropyl groups to form $2,3$-dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$.Thus,the correct reaction is $D$.

$X$ is an isomer of $C_6H_{14}$. It has four primary carbons and two tertiary carbons. '$X$' can be prepared from which of the following reactions?

Similar Questions

Determine the number of mono-chloro products (structural isomers only) formed by the reaction of isopropylcyclopentane with $Cl_2/\Delta$.

How many isomers of monobromo derivatives are obtained on the bromination of the following compound?

The product obtained by the vapor phase nitration of propane is:

If $CH_4$ is known as methane,then $C_9H_{20}$ is known as

$A$ hydrocarbon $A$ $(V.D. = 36)$ forms only one monochloro substitution product. $A$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module