Select the incorrect option:

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

Question

(A) In the chlorination of isobutane $(CH_3-CH(CH_3)-CH_3)$,the relative reactivity of $1^\circ$ and $3^\circ$ hydrogens is $1 : 5$. Number of $1^\cir

Vedclass · Accepted Answer

$CH_3-CH(CH_3)-CH_3 \xrightarrow[h
u]{Cl_2} CH_3-CCl(CH_3)-CH_3$ (Major)

Vedclass · Answer

(A) In the chlorination of isobutane $(CH_3-CH(CH_3)-CH_3)$,the relative reactivity of $1^\circ$ and $3^\circ$ hydrogens is $1 : 5$. Number of $1^\circ$ $H$ atoms = $9$. Number of $3^\circ$ $H$ atoms = $1$. Relative yield of $1^\circ$ product = $9 	imes 1 = 9$. Relative yield of $3^\circ$ product = $1 	imes 5 = 5$. Since $9 > 5$,the primary chloride $(CH_3-CH(CH_3)-CH_2Cl)$ is the major product,not the tertiary chloride. Therefore,option $(A)$ is incorrect. $(B)$ Allylic bromination with $NBS$ is correct. $(C)$ Anti-Markovnikov addition of $HBr$ in the presence of peroxide is correct. $(D)$ Peroxide effect is only applicable to $HBr$,not $HCl$. Thus,$HCl$ follows Markovnikov's rule,making this option also incorrect. However,$(A)$ is the most standard textbook example of an incorrect major product prediction based on reactivity ratios.

Similar Questions

What are $X, Y$ and $Z$ in the following reactions?
$(CH_3)_3 C-O^- Na^+ + CH_3 CH_2 Br \longrightarrow X + NaBr$
$(CH_3)_3 C-Br + CH_3 CH_2 O^- Na^+ \longrightarrow Y + Z$

Ethanolic $KOH$ gives

Considering the above reactions,the compound $A$ and compound $B$ respectively are:

Explain why:
$(i)$ the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
$(ii)$ alkyl halides,though polar,are immiscible with water?
$(iii)$ Grignard reagents should be prepared under anhydrous conditions?

Which of the following is most reactive towards alcoholic $KOH$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module