Which of the following reaction(s) produce th | vedclass

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

Question

(B) The Saytzeff product is the more substituted,and therefore more stable,alkene formed during an elimination reaction.$A$: $CH_3CH_2CH(F)CH_3 \xrigh

Vedclass · Accepted Answer

$CH_3CH_2CH(Cl)CH_3 \xrightarrow{alc. KOH, \Delta}$

Vedclass · Answer

(B) The Saytzeff product is the more substituted,and therefore more stable,alkene formed during an elimination reaction.$A$: $CH_3CH_2CH(F)CH_3 \xrightarrow{NH_2^-, \Delta}$ This reaction involves a poor leaving group $(F^-)$ and a strong base,which often leads to the less substituted alkene (Hofmann product) as the major product due to the high transition state energy for the more substituted alkene.$B$: $CH_3CH_2CH(Cl)CH_3 \xrightarrow{alc. KOH, \Delta}$ This is a standard $E2$ elimination with a small base $(OH^-)$. The major product is the more substituted alkene,$CH_3CH=CHCH_3$ (but$-2-$ene),which is the Saytzeff product.$C$: $CH_3CH(CH_3)CH(Cl)CH_3 \xrightarrow{Me_3CO^-K^+, \Delta}$ This reaction uses a bulky base $(Me_3CO^-)$. Bulky bases prefer to abstract the most accessible proton,leading to the less substituted alkene (Hofmann product) as the major product.$D$: $CH_3CH_2CH(NMe_3^+)CH_3 \xrightarrow{OR^-, \Delta}$ This is a Hofmann elimination. The bulky leaving group $(NMe_3^+)$ forces the base to abstract the most accessible proton,resulting in the less substituted alkene (Hofmann product) as the major product.Therefore,only reaction $B$ produces the Saytzeff product as the major product.

Which of the following reaction$(s)$ produce the Saytzeff product as a major product?

Similar Questions

Which of the following compounds will give a racemic mixture by $S_N1$ reaction?

Which information below regarding this reaction is applicable?
$CH_3-O-CH_2-Cl \xrightarrow{aq. OH^-, \Delta} CH_3-O-CH_2-OH$

Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

The product formed on the reaction of ethyl alcohol with bleaching powder is:

When $tert$-butyl bromide is heated with silver fluoride,the major product obtained is

Vedclass Test Series

Exam Paper Generator

Online Exam Module